Last updated: July 30, 2015

Analysis, Volume I

Terence Tao

Hindustan Book Agency, January 2006. Third edition, 2014

Hardcover, 368 pages.ISBN 81-85931-62-3 (first edition)

This is basically an expanded and cleaned up version of my lecture notes for Math 131A. In the US, it is available through the American Mathematical Society. It is part of a two-volume series; here is my page for Volume II. It is currently in its third edition.

There are no solution guides for this text.

- Sample chapters (contents, natural numbers, set theory, integers and rationals, logic, decimal system, index)

— Errata —

- p. 2, item 3: “can you add” should be “Can you add”.
- p. 9, line 5: “right-hand side” should be “left-hand side”.
- p. 10, first display: should be .
- p. 5, line 6 from bottom: should be . (Actually, for pedagogical reasons, it may be slightly better to use throughout this example instead of .)
- p. 59, Lemma 3.3.12: f should map Z to W, and h should map X to Y. In the proof of this lemma (on page 60): is a function from X to Z, and is a function from Y to W.
- p. 67, last paragraph: should be .
- p. 98: In Exercise 4.2.1, Corollary 2.3.7 should be Corollary 4.1.9. In Exercise 4.2.6, should be rational numbers, not real.
- p. 101: In Definition 4.3.9, after ““, add “; in particular, we define “.
- p. 127: In Exercise 5.3.4: add “(Hint: use Exercise 5.2.2.)”.
- p. 131, line 12 from bottom: “they cannot be than” should be “they cannot be larger than”.
- p. 175, Exercise 6.6.3: In the hint, replace “introduce” by “recursively introduce”, and insert “; ” after “” (two occurrences), with the parenthetical “(omitting the condition when )” inserted after the recursive definition of .
- p. 183: In the proof of Proposition 7.1.8, should be replaced by in every display of the proof in which it appears.
- p. 197, in second line of proof of Proposition 7.3.4: the second sum should be rather than .
- p. 216, Exercise 8.1.9: It needs to be noted that this exercise requires the axiom of choice from Section 8.4.
- p. 220, Lemma 8.2.5: It needs to be noted that this lemma requires the axiom of choice from Section 8.4. Similarly, the case in Proposition 8.2.6 in which X is uncountable requires the axiom of choice also.
- p. 227, Exercise 8.3.2: should be .
- p. 236, last line: “for any good set Y'” should be “for any good set Y’ with non-empty”.
- p. 255, Proposition 9.3.9(b): should be .
- p. 303, Exercise 10.4.3(a): The limit should be in the set rather than .
- p. 336, line 13: replace “we have made no assumption on ” with “the function could have been arbitrary”.
- p. 337, Exercise 11.8.1: Lemma 11.8.1 should be Lemma 11.8.4.
- p. 337, Exercise 11.8.5: In the last display, should be .
- p. 342, Exercise 11.9.1: “the function f is not differentiable” should be “the function is not differentiable.
- p. 383, first display: should be .
- p. 387, fourth display: should be .

— Errata for the second edition (hardback) —

- p. xii, bottom: “solidifed” –> “solidified”.
- p. xiv, top: “to know how to to” –> “to know how to”.
- p. 19. In footnote 2, add: “In the converse direction, if we have , then we may deduce ; this is the axiom of substitution (see Appendix A.7) applied to the operation .”
- p. 24, after Definition2.2.1: “defined for every integer ” should be “defined for every natural number “.
- p. 26, after Proposition 2.2.6: “these notes” should be “this text”.
- p. 28, Proposition 2.2.14: “and Let” should be “and let”.
- p. 30, Lemma 2.3.3: “Natural numbers have no zero divisors” should read “Positive natural numbers have no zero divisors”.
- p. 32, Definition 2.3.11: Add the remark “In particular, we define to equal .”
- p. 37, Example 3.1.10: “(why?)” should be “(why?))”.
- p. 45: “8-m, where n is a…” should be “8-m, where m is a…”. In Exercise 3.1.2, add Axiom 3.1 to the list of permitted axioms. In Exercise 3.1.1: (3.1.4) should be Definition 3.1.4.
- p. 50: In the first line, should be , and should be .
- p. 55, Exercise 3.3.1: and should be and respectively.
- p. 61: In Exercise 3.4.8, Axiom 3.1 should be added to the list of permitted axioms.
- p. 64: In Example 3.5.9, “” should be ““.
- p. 70, 4th line of proof of Lemma 3.6.9: should be . In the 6th line of proof of Proposition 3.6.8: Proposition 3.6.4 should be Lemma 3.6.9. After Lemma 3.6.9, add the following remark: “Strictly speaking, the expression has not yet been defined. For the purposes of this lemma, we temporarily define it to be the unique natural number such that (which exists and is unique by Lemma 2.2.10).”
- p. 81, before Lemma 4.2.3: “product of a rational number” -> “product of two rational numbers”.
- p. 84, before Definition 4.2.6: a space is missing between “Proposition 4.2.4” and “allows”. Before this paragraph, add “In a similar spirit, we define subtraction on the rationals by the formula , just as we did with the integers.”
- p. 86: In Definition 4.3.2, “real numbers” should be “rational numbers”. In definition 4.3.4, “be a rational number” should be added after “Let “.
- p. 88: In Proposition 4.3.10(b), the hypothesis n>0 should be added.
- p. 104, proof of Lemma 5.3.7; after invoking Proposition 4.3.7, add “(extended in the obvious manner to the case)”.
- p. 105, after Proposition 5.3.10: should be .
- p. 108, proof of Lemma 5.3.15: should be . “This shows that ” should read “This shows that “.
- p. 115: In the hint for Exercise 5.4.8, add “or Corollary 5.4.10” after “use Proposition 5.4.9”.
- p. 120: Add an additional exercise, Exercise 5.5.5: “Establish an analogue of Proposition 5.4.14, in which “rational” is replaced by “irrational”.”
- p. 124, Exercise 5.6.3: Add the hypothesis that x is non-zero (since the roots of 0 are not yet defined).
- p. 126, proof of Proposition 6.1.4: Proposition 5.4.14 should be Proposition 5.4.12.
- p. 134: In Definition 6.2.6(c) (and also on the first line of p. 135), should be .
- p. 135, Theorem 6.2.11(b), (c): Replace “Suppose that ” with “Suppose that ” (two occurrences). Exercise 6.2.2: Proposition 6.2.11 should be Theorem 6.2.11.
- p.144: Cor. 6.4.14: line 4: ” .. for all ” should be ” .. for all “
- p.146: proof of Theorem 6.4.18: Replace “from Corollary 6.1.17” here by “from Lemma 5.1.15 (or more precisely, the extension of that lemma to the real numbers, which is proven in exactly the same fashion)”.
- p. 151, Exercise 6.6.5: Replace “the formula , explaining why the set is non-empty” with “the recursive formula , with the convention , explaining why the set is non-empty”.
- p. 164, Definition 7.2.2: should be .
- p. 169, Exercise 7.2.6: Add “How does the proposition change if we assume that does not converge to zero, but instead converges to some other real number ?”. After Corollary 7.3.2: “conditionally divergent” should be “not conditionally convergent”.
- p. 176: “absolutely divergent series” should be “series that is not absolutely convergent”.
- p. 177, Theorem 7.5.1: “conditionally divergent” should be “not conditionally convergent”, and similarly “absolutely divergent” should be “not absolutely convergent”. Similarly for Corollary 7.5.3 on page 179.
- p. 186, Exercise 8.1.1: This exercise requires the axiom of choice, Axiom 8.1. In Exercise 8.1.4. should be .
- p. 192, proof of Theorem 8.2.8: “absolutely divergent” should be “not absolutely convergent” (two occurrences).
- p. 196, Remark 8.3.6: “Paul Cohen (1934-)” should now be “Paul Cohen (1934-2007)”. :-(
- p. 197, Exercise 8.3.2: should be an injection rather than a bijection. In the definition of , should be (two occurrences).
- p. 200, Exercise 8.4.1: should be .
- p. 206, Exercise 8.5.5: “” should be “latex Y := \{y \in Y’: y < x \}$ should be . In Exercise 8.5.20, the additional hypothesis “Assume that does not contain the empty set ” should be added.
- p. 214, Lemma 9.1.21. One needs the additional hypothesis “We assume that .”
- p. 220, Definition 9.3.6: “ is -close to near ” should be “, after restricting to , is -close to near “.
- p. 228, Proposition 9.4.7: change “three items” to “four items”, and add “(d): For every , there exists a such that for all with .
- p. 232, proof of Proposition 9.5.3: after “Proposition 9.4.7”, add “(applied to the restriction of to the subdomain )”.
- p. 252, Proposition 10.1.7: One needs the additional hypothesis . Similarly for Proposition 10.1.10, Theorem 10.1.13, and Proposition 10.3.1.
- p. 253, Definition 10.1.11: “For every ” should be “For every limit point “.
- p. 254, Remark 10.1.14: Leibnitz should be Leibniz (two occurrences).
- p. 256, Exercise 10.1.1: “ is also limit point of ” should be “, and is also a limit point of “.
- p. 257, Definition 10.2.1: should be .
- p. 262: In the proof of Theorem 10.4.2,”” should be ““.
- p. 271, Remark 11.2.2: “constant on ” should be “constant on “.
- p. 290: In Exercise 11.6.5, add “For this exercise, you may use the second Fundamental Theorem of Calculus (Theorem 11.9.4); there is no circularity, because Corollary 11.6.5 is not used in the proof of that theorem.”
- p. 290: In the proof of Proposition 11.7.1, in the third display, should be .
- p. 299: In Exercise 11.9.1, the hint is misleading (it requires the mean value theorem for integrals rather than for derivatives, which is not covered in this text) and should be deleted.

— Errata to the third edition (hardback) —

- General note: all references to “Analysis II” need to be renumbered to account for the new chapter numbering (basically, all chapter numbers need to be lowered by 11.)
- Page 15: In Section 2.1, “Guiseppe Peano” should be “Giuseppe Peano”.
- Page 37: In Example 3.1.10, “so is singleton set” should be “the singleton set”; also, a right parenthesis is missing after (why?).
- Page 54: In Definition 3.3.17, the remark that a function is onto if should be moved to the next section, because the image is not defined until that section.
- Page 62: In Example 3.3.22, “Axioms 2.2, 2.3, 2.4” should be “Lemma 2.2.10”.
- Page 64: The justification that the product set given in Remark 3.5.8 is not quite correct if one is using the definition of an ordered n-tuple as defined in Exercise 3.5.2 (one has to restrict the range of the tuples to be surjective). As the correct version of this remark is part of Exercise 3.5.2, the second sentence of this remark should be replaced with a reference to that exercise.
- Page 67: In Exercise 3.5.12, should be .
- Page 70: In the proof of Lemma 3.6.9, “Now define the function to ” should be “Now define the function ” . In the 4th line of proof of Lemma 3.6.9: should be .
- Page 72: In Exercise 3.6.8, the additional hypothesis that A is non-empty should be added.
- Page 144: Below the proof of Proposition 6.4.12, a right parenthesis should be added after “(provided that and are finite”. Also, “(c) and (d)” should be “(d) and (e)”.
- Page 150: In Example 6.6.3, should be inserted between and .
- Page 157: In Lemma 7.1.4(c), a period is missing at the end of .
- Page 174: In the proof of Proposition 7.4.1, and should be and respectively.
- Page 176: In the proof of Proposition 7.4.3, “-close to ” should be -close to ” in the last paragraph.
- Page 202: In Exercise 8.4.3, “there exists an injection ; in other words…” should be “there exists an injection with the identity map; in particular…”. (This is needed in order to establish the converse part of the question.)
- Page 209: In Exercise 8.5.16, “” should be ““.
- Page ???: Exercise 8.5.18: A right parenthesis is missing after “… which contains “. “Tthus” should be “Thus”.
- Page 217: In Exercise 9.1.15, the hypothesis that is non-empty should be added.
- Page 226: In Example 9.3.21, all sequences here should start from rather than from .
- Page ???: In the last paragraph of Section 11.8, a right parenthesis should be added at the end of the penultimate sentence.

Note that the first edition paperback page numbers differ from the second (or third) edition hardback page numbers, which should be born in mind when applying the second edition errata to the first edition. (The section, theorem and exercise numbering, however, is mostly unchanged.)

Thanks to José Antonio Lara Benítez, Tai-Danae Bradley, Brian, Eduardo Buscicchio, Gonzales Castillo Cristhian, Ck, Lorenzo Dragani, Evangelos Georgiadis, Ti Gong, Christian Gz., Ulrich Groh, Erik Koelink, Matthis Lehmkühler, Percy Li, Ming Li, Manoranjan Majji, Pieter Naaijkens, Vineet Nair, Cristina Pereyra, David Radnell, Tim Reijnders, Pieter Roffelsen, Luke Rogers, Gabriel Salmerón, Marc Schoolderman, Kent Van Vels, Daan Wanrooy, Yandong Xiao, Luqing Ye, and the students of Math 401/501 and Math 402/502 at the University of New Mexico for corrections.

## 338 comments

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9 July, 2008 at 8:09 am

John FranksThe link to Volume II on this page is broken. It is a link to this page.

Instead it should be

https://terrytao.wordpress.com/books/analysis-ii/

10 July, 2008 at 9:33 am

Terence TaoThanks for the correction!

9 September, 2008 at 10:33 am

luke rogersone of my students points out that in ex 4.2.1 on page 98 a better hint is Corol 4.1.9 rather than Corol 2.3.7.

10 September, 2008 at 7:44 pm

Terence TaoDear Luke: Thanks for the correction!

16 January, 2009 at 6:56 am

Edoardo BuscicchioThe two links to Hindustan Book Agency seem broken.

The Hindustan Book Agency home page’s link is:

http://www.hindbook.com/

but in the whole web site there’s no trace about the textbook (neither volume I nor Volume II)

16 January, 2009 at 7:37 am

Edoardo Buscicchiohere’s the link for the textbook’s page:

Volume I

http://www.hindbook.com/trims_pub.php?pubid=37

Volume II

http://www.hindbook.com/trims_pub.php?pubid=38

16 January, 2009 at 8:50 am

Terence TaoDear Eduardo: Thanks for the correction!

17 January, 2009 at 6:40 am

Edoardo BuscicchioHi prof. Tao,

I’d like to propose a correction if I’m not wrong.

Pag 28 – Lemma 2.2.2 – Line 8:

“The base case 0+0=0 follows since we know that 0+m = m for every natural m, and 0 is a natural number.”

This doesn’t seem wrong, but I think it doesn’t follows since the definition (2.2.1) of addition 0+m=m; conversely, it seems that given definition of addition inductly follows since the statement 0+0=0.

In fact, I think that in the definition 2.2.1 (page 27) induction isn’t used only to define the addition, but, implicitly, induction is used also defining base case’s definition 0+m=m: when we fix a natural m and say that this definition will be useful for each natural, we’re using the induction on the base case 0+0=0. Otherwise, how to define the property 0+m=m for each natural?

So, in Lemma 2.2.2, maybe 0+0=0 is (and doesn’t follow) the statement 0+m=m.

Sorry if my reasonging is in a such kind trivial or wrong, I was just trying to avoid misunderstanding for my studies.

2 October, 2009 at 10:34 pm

Keith DowDear Professor Tao,

Do you have as estimate for when the second edition will be available to purchase?

Thank you for your time,

Keith Dow

22 October, 2009 at 6:04 am

anonymousDear Prof. Tao,

I have a question about Axiom 2.5. Let A be a set such that

if a property P(n) is true for every element of B={0,1,2,…}, then P(n)

is true for every element of A, and A and B are distinct sets.

Let C is the union of A and B. Then can’t we see C as

the set of natural numbers?

22 October, 2009 at 7:44 am

Terence TaoWell, if A and B are disjoint sets, note that the property P(n) := “n lies in B” is true for all n in B, but false for all n in A. So Axiom 2.5 fails in this case.

There is however a subtle loophole here: it may be that you are working in a logical system in which “n lies in B” is not considered a valid property of n. In this case it is indeed possible to have number systems that are strictly larger than the standard natural numbers B, in which the Peano axioms still hold. Such systems are known as non-standard natural numbers and are useful in a number of branches of mathematics. (More generally, it is not possible to pin down any number system precisely using an explicit (or more precisely, recursively enumerable) set of axioms, a consequence of Godel’s completeness and incompleteness theorems; one must always admit the possibility of non-standard models).

This is however a somewhat more advanced topic than what is covered in the text. You can see my blog post

https://terrytao.wordpress.com/2007/06/25/ultrafilters-nonstandard-analysis-and-epsilon-management/

for more discussion.

26 December, 2009 at 7:37 pm

August 17, 2009 « Yandong Xiao(Max Hsiao)'s Blog[…] https://terrytao.wordpress.com/books/analysis-i/ […]

25 May, 2010 at 6:38 am

AnonymousDear Prof. Tao,

I do not mean to criticize, but I do believe that it would be useful to add some references at the back of the book.

With best regards

from a reader

18 June, 2010 at 12:07 am

percyliin first edition, page 60 first line should be “since g0h is a function from X to Z” accroding to the correction on the previous page!

[Clarified, thanks. -T.]18 June, 2010 at 8:18 pm

percyliFirst edition page 63 exercises 3.3.1. Shouldn’t f be Y to Z and g be X to Y? Just as the correction of definition on page 61?

[Corrected, thanks – T.]17 July, 2010 at 1:40 pm

Natural Numbers | Dilawar's Notes[…] Analysis I, by Terrence Tao. This entry was posted in Uncategorized. Bookmark the permalink. […]

9 August, 2010 at 11:34 pm

tusharchimedeSome illumination on the following example from Analysis I would be much appreciated:

Example 1.2.13 (Limits and lengths) Consider the right-angled triangle with vertices (0; 0), (1; 0), and (0; 1), and suppose we wanted to compute the length of the hypotenuse of this triangle. Pythagoras’ theorem tells us that this hypotenuse has length root 2, but suppose for some reason that we did not know about Pythagoras’ theorem, and wanted to compute the length using calculus methods. Well, one way to do so is to approximate the hypotenuse by horizontal and vertical edges. Pick a large number N, and approximate the hypotenuse by a “staircase” consisting of N horizontal edges of equal length, alternating with N vertical edges of equal length. Clearly these edges all have length 1/N, so the total length of the staircase is 2N/N = 2. If one takes limits as N goes to infinity, the staircase clearly approaches the hypotenuse, and so in the limit we should get the length of the hypotenuse. However, as N approaches infinity, the limit of 2N/N is 2, not root 2, so we have an incorrect value for the length of the hypotenuse. How did this happen?

22 September, 2010 at 12:18 pm

Ming LiDear Prof. Tao,

Considering the Example 2.1.9

{0, 0.5, 1, 1.5, 2, 2.5, … …}

I think this example is just combination of two notation system of natural numbers, because 0.5, 1.5, 2.5, 3.5 …, are just the same signs as 0, 1, 2, 3, …, to represent natural numbers. we may not say 0.5 is a natural number because it has the same function as zero 0. Also, by axiom 2.1-2.4, we cannot conclude that this example is right because the 0.5 is unsure–I mean, we are trying to build the set N(natural numbers) from the very fundamental zero and incrementing operation, so we have the beginning element zero 0 and the continuing incrementing operation to get the successive numbers, then if 0.5 does not have its prior number, it may be not considered in our notation system:N.

Sorry maybe my expression is rogue and the understanding is wrong.

22 September, 2010 at 2:16 pm

Terence TaoOne needs to distinguish between the

intuitivedefinition of the natural numbers (namely, that of the system one “gets” when starting at 0 and “counting forward”), and the axiomatic definition, which is any system that obeys the stated axioms (in this case, Axioms 2.1-2.4). Ideally, the latter should precisely capture the former, but the above example shows that this is not the case: the system {0, 0.5, 1, …} obeys the axioms 2.1-2.4 but does not conform to our intuitive understanding of what the natural numbers should be. Note that there is nothing in these axioms themselves that prohibit the presence of a number such as 0.5 that behaves functionally like a zero, without being zero itself. (Once one has the induction axiom (Axiom 2.5), though, one can show that every non-zero number must be the successor of some other natural number, thus finally ruling out this system as a possible candidate for the natural numbers.)One can view a set of axioms as analogous to a set of bureaucratic tests (much like the tests one might have to take to, say, obtain a drivers licence) which are interpreted strictly to the letter, without any appeal to the “spirit” or “intent” of these tests. If the putative number system passes all of the tests (e.g. if the system does not make 0 the successor of any natural number, and if the system ensures that distinct numbers have distinct successors), it is considered to be a valid number system as far as this set of axioms is concerned, even if it contravenes the “spirit” of these axioms by having, say, unnecessary elements or unorthodox rules of arithmetic.

Being capable of interpreting axioms and other mathematical statements strictly to the letter is one of the basic requirements of rigorous mathematical thinking. This is not to say that intuition is unimportant – far from it – but the two modes of thought should not be confused with each other.

25 September, 2010 at 9:30 am

Ming LiThank you very much for this courteous help and reply Prof. Tao. Your example is so terrific. The natural numbers exist and we’re trying to determine the corresponding axioms it shall comply with. Cool.

I was not very sure then, and now things are cleared up. I am wrong with saying “Also, by axiom 2.1-2.4, we cannot conclude that this example is right because the 0.5 is unsure” because axioms 2.1- 2.4 really don’t exclude 0.5. What I really wanted to indicate was that I thought this example of {0, 0.5, 1, 1.5, 2,…} may be not very suitable. My previous opinion partly came from axioms 2.1 and 2.2 using the term “natural number”. This expression suggests me that we have the foreknowledge about natural numbers, like the very fundamental zero and incrementing operation you indicate in the book “we will use two fundamental concepts”. I concluded from this to suppose that 0.5 in the example functioning same as zero is kind of nonsense, seems as just combine two notation systems together. Therefore, maybe at the beginning asserting that the following axioms are just for axiomatic defining will be better, because merely introducing and using intuitive fundamental concepts of zero and incrementing operation is kind of confusing–which lead people to assume all the following axioms are based on the most primitive zero and ++, not just to describe and construct the system natural numbers should conform to.

I truly love your style, which opens my eyes and heart, and redefines some key concepts in my brain, more than ever.

I’m totally intoxicated with your explanation and references in the book, like your remarks just indicating mathematics caring only about what properties the objects have, not what the objects are or what they mean, etc.

25 September, 2010 at 1:39 pm

Terence TaoTechnical terminology in mathematics, such as “natural number”, is deliberately chosen to invoke connotations with more intuitive and non-rigorous concepts, but one should still separate one’s intuitive preconceptions of such terms from their formal, rigorous theory. If you find this difficult, I suggest temporarily renaming the terminology, e.g. changing “natural number” to some term with no

a prioriconnotations or meaning, such as “Peano number”, (and perhaps changing symbols such as + to , for similar reasons) until you are able to comfortably manipulate the formal axioms abstractly and separately from one’s intuition. (This is analogous to how, in algebra, one can formally manipulate quantities such as without having to know what value represents.) At that point you can then recombine the two.See also

https://terrytao.wordpress.com/career-advice/there%E2%80%99s-more-to-mathematics-than-rigour-and-proofs/

(particularly the discussion concerning the transition from the “pre-rigorous” to the “rigorous” stage of thinking) for some further discussion of this point.

27 September, 2010 at 3:28 am

Yandong Xiao(Max Hsiao)But you still have to note your understanding on that ” this example is just combination of two notation system of natural numbers, because 0.5, 1.5, 2.5, 3.5 …, are just the same signs as 0, 1, 2, 3, …, to represent natural numbers” is false. As I have shown in the last post, 0.5 is distinct from 0, and in fact, any natural number defined by Axioms 2.1-2.4. Admittedly, it is really sort of difficult to consider natural number system like the first time we confront and deal with it. However, Terry’s argument or discussion is clear and instructive.

8 January, 2011 at 8:17 pm

AnonymousProf. Tao,

There seems to be an “impossible idea” here that I don’t fully understand. By Axiom 2.5, informally, xince P(0) is true, P(0++)=P(1) is true. Then P(1), P(2), P(3), etc. are all true. How do we know this line of reasoning will never let us conclude that P(0.5) is true? How do we know there is no chance for FINALLY counting to 0.5 from 0?

22 February, 2012 at 9:25 am

Luqing YeI am also very gratitude for your very good explaination.I teach myself using this extremely rigourous book,sometimes I usually get confused because I have never experience the axiomatic way of constructing things,I don;t know what principles I should follow when constructing things.Thanks for your reply to these comments,which is very very helpful .

23 September, 2010 at 12:50 am

Yandong Xiao(Max Hsiao)Unfortunately, Axioms 2.1 to 2.4 do not exclude that 0.5 is legal to be a natural number between 0 and 1, thus it is distinct from 0 and 1 but a natural number. So do any other ones such as 1.5, 2.5,…. As a matter of fact, informally, I try to understand them in the following way. Suppose 0.5 is a natural number (Axiom 2.1); if n+0.5 (the symbol “+” here doesn’t represent addition, it is only for the purpose that such a number is different from the number n, and it is between n and n++.) is a natural number, then (n++)+0.5 is a natural number (Axiom 2.2); any (n++)+0.5 will not be the 0.5(Axiom 2.3); if m,n are two natural numbers and (m++)+0.5=(n++)+0.5, then m+0.5=n+0.5 (Axiom 2.4). Further, for any natural number n, n+0.5 is a natural number between n and n++. The operation of increment doesn’t exclude there exists such a number.

25 September, 2010 at 9:32 am

Ming LiThank you so much!!!!

27 September, 2010 at 4:26 am

Yandong Xiao(Max Hsiao)It is my pleasure in any event.

You should also note that the conclusion you have made which says that “{0, 0.5, 1, 1.5, 2, 2.5, … …} is just combination of two notation system of natural numbers, because 0.5, 1.5, 2.5, 3.5 …, are just the same signs as 0, 1, 2, 3, …, to represent natural numbers.” is supported by neither Axioms 2.1-2.4 nor any other persuasive statements.

Your misunderstanding seems sort of odd in my point of view. It sounds unreasonable that we try to build natural number system from 0 and the operation of increment, for we in fact have to define a plenty of symbols to show how natural numbers are distinct from one another, such as 0++ is different from (0++)++. But it is true that we can only use “0”, “++” and “()” to represent any natural number.

Maybe, it is a good effort to clarify my understanding in more information some day later.

30 September, 2010 at 3:27 pm

AnonymousDear Prof. Dr. Tao,

My question is about Russell Paradox.

1) For each set A:

If A is an element of A, then A isn’t an element of X.

2)For each set A:

If A isn’t element of A, then A is an element of X.

Suppose that there is a set X which satisfies both 1 and 2. Then there are two cases:

i)X is an element of X:

If X is an element of X, then by 1, X isn’t an element of X, contradiction.

ii)X isn’t an element of X:

If X isn’t an element of X, then by 2, X is an element of X, contradiction.

Hence there isn’t a set X which satisfies both 1 and 2.

But then where is the paradox? Or, what is the meaning of “paradox” in here?

I hope it isn’t a dumb question.

Thanks.

30 September, 2010 at 5:47 pm

Terence TaoBy itself, the assertion that there is no set X with the properties (1) and (2) listed above is not yet a paradox. However, it becomes paradoxical when combined with the axioms of “naive” set theory, and in particular with the axiom (or more precisely, axiom schema) of universal specification:

This axiom seems harmless enough, given that one should be able to form the set of all integers, the set of all red objects, and so forth. But if one applies this axiom to the property then one obtains a set X that obeys the properties (1) and (2), thus triggering the Russell paradox.

To evade the Russell paradox, one must therefore discard the axiom of universal specification as being too powerful, and replace it with weaker axioms that do not trigger the paradox. The usual solution (as adopted, for instance, in Zermelo-Frankel set theory) is to localise the axiom to sets:

By doing so, Russell’s paradox gets converted into something more useful, namely Cantor’s theorem.

These topics are discussed in Section 3.2 of my Analysis text.

1 October, 2010 at 12:27 am

Yandong Xiao(Max Hsiao)A few days ago, I made a promise writing something in more information to clarify my understanding on the Axioms 2.1-2.4 of the Peano Axioms and the relationship between the Axiom 2.5 of it and the four axioms. In any event, I have fulfilled my promise in this page.

http://yandongxiao.wordpress.com/2010/10/01/an-insightful-perspective-on-natural-numbers/

2 October, 2010 at 5:04 am

BrianI am using the hardback version of the second edition of Analysis I and I have a question regarding one of the definitions. On pg. 86, Definition 4.3.2 (Distance) starts off with “Let x and y be real numbers.” I think maybe it should say “Let x and y be rational numbers.” because we haven’t defined the reals yet. I just wanted to clarify this because I did not see it in the above errata. Thanks!

[Corrected, thanks – T.]2 October, 2010 at 11:04 pm

Yandong Xiao(Max Hsiao)The following exercise seems difficult to find a solution. Is there anyone who may provide some suggestion to solve it?

Exercise 3.4.7. Let X; Y be sets. Define a partial function from X to Y to be any function f : X’—> Y’ whose domain X’ is a subset of X, and whose range Y’ is a subset of Y. Show that the collection of all partial functions from X to Y is itself a set. (Hint: use Exercise 3.4.6, the power set axiom, the replacement axiom, and the union axiom.)

3 October, 2010 at 12:20 pm

Ming LiThere are mainly two points I suppose.

First, for fixed X’ and Y’, the entire possible choices for the partial function f is exactly the set (by power set axiom).

Second, we need to unite all the sets in the kind of advanced level, that is, for every possible choice of X’ and Y’:

(by union axiom and exercise 3.4.6)

Informally, I think this question is interesting because of it’s kind of invisibly related to the understanding of Russell’s paradox. I mean, there is a hierarchy in this problem, the bottom (or primitive) is the partial functions from X’ to Y’ when X’, Y’ are definite, and then the more advanced level is when . Though actually, the two levels are all about the specific partial function.

Prof. Tao’s explanation is terrific and inspiring.

3 October, 2010 at 11:03 pm

Yandong Xiao(Max Hsiao)Thank you for sharing your solution for the exercise. I consider the replacement axiom may be needed, but I just failed to make a clear train of thought two days before.

3 October, 2010 at 2:39 pm

Ming LiDear Prof. Tao,

Thank you!

I’m using the first edition book and just want to clarify for the following points:

1. p. 36 Def. 2.3.11. Are we supposed to indicate that ?

2. p. 190 Def. 7.2.2. In the fourth line, the lower script is n or N? I mean, is it supposed to be or ?

3. p. 194 EP 7.2.13. From the fourth line, “Thus absolute divergence does not imply conditional divergence, even though absolute convergence implies conditional convergence.”, is it kind of typo?

[Thanks for the corrections. Yes, 0^0 is defined to equal 1, and yes, absolute divergence is a weaker property than conditional divergence, since absolute convergence is a stronger property than conditional convergence. (Because of this, it is often best to avoid the terms “absolute divergence” and “conditional divergence” when there is possibility of confusion.) – T.]Maybe I didn’t see them in the above errata and just want to clarify.

5 October, 2010 at 4:01 am

studentDear Prof. Dr. Tao,

Is it possible that on a line, there are still some numbers except real numbers? If not, how can we know that? Although standard analytic definition of a line limits it by reals, may be we can find an another analytic definition of line which includes real number system and another number system together.

8 October, 2010 at 7:52 am

Terence TaoIn the standard modern axiomatisation of Euclidean geometry, due to Hilbert, one can show that lines in Euclidean geometry are in one-to-one correspondence with the real numbers.

There are however other linear structures that are in some sense “larger” than the real line, such as the long line, the extended real line, and the hyperreals. Whether one would consider these mathematical structures as “genuine” lines is more a matter of definition than anything else, though.

6 October, 2010 at 2:35 am

ErichProfessor Tao,

When will the second edition of your Analysis texts be published?

8 October, 2010 at 11:12 am

Terence TaoThe second edition has been out for some time; I’ve updated the links on the page to point to these editions.

6 October, 2010 at 11:05 pm

Yandong Xiao(Max Hsiao)Dear Terry,

In Chapter 3, we still have not defined the operation of substraction, thus I suggest some replacements for your discussion in the Chapter. I only display three key replacements.

(1) Replace “Lemma 3.6.8. Suppose that n>=1,and X has cardinality n. Then X is non-empty, and if x is any element of X,then the set X-{x} (i.e., X with the element x removed) has cardinality n-1 ” with “Lemma 3.6.8.Suppose that n is a natural number, and X has cardinality n++. Then X is non-empty, and if x is any element of X,then the set X-{x} (i.e., X with the element x removed) has cardinality n.”

(2) In the proof for the Lemma 3.6.8, replace “g(y):= f(y)-1 if f(y) >f(x).” with “g(y)+1:= f(y) if f(y) >f(x).”

(3) In the proof for the Proposition 3.6.7, replace”Let X have cardinality n++; and suppose that X also has some other cardinality m ≠ n++.” with “Let X have cardinality n++; and suppose that X also has some other cardinality m++ ≠ n++ (For X is non-empty, we can find a natural number m++ to represent its cardinality).

8 October, 2010 at 9:44 am

StudentI like the construction of the reals from Cauchy sequences, but I think that

the idea of equivalence classess of rationals seems a bit artificial and not very simple/elegant.

What do professional mathematicians think about that ?

Does an alternative exist ?

Thanks !

Your blog is very cool ! Moreover, I always get good inspiration and motivation from here !

8 October, 2010 at 10:02 am

Terence TaoThere are several other constructions of the reals; one popular one proceeds via Dedekind cuts. Another one which I quite like is based on quasimorphisms of the integers, dubbed the “Eudoxus reals“.

The idea of quotienting out by equivalence may seem like overkill for a basic construction such as the real numbers, but this idea is useful for many other constructions (e.g. of direct and inverse limits, metric completions, ultraproducts, tensor products, etc.), so it is worth knowing about for future reference.

8 October, 2010 at 4:22 pm

Yandong Xiao(Max Hsiao)Anyone who considers this example N := {0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, …}a little difficult to understand may see the following explanation.

See also http://yandongxiao.wordpress.com/2010/10/01/an-insightful-perspective-on-natural-numbers/

Consider the Axioms 2.1-2.4 of the Peano Axioms:

Axiom 2.1. 0 is a natural number.

Axiom 2.2. If n is a natural number, then n++ is also a natural number.

Axiom 2.3. 0 is not the successor of any natural number; i.e., we have n++ ≠ 0 for every natural number n.

Axiom 2.4. Different natural numbers must have different successors; i.e., if n, m are natural numbers and n ≠ m, then n++ ≠ m++. Equivalently, if n++ = m++, then we must have n = m.

Starting from “0” to create a series of natural numbers, we find any of such natural numbers can be represented by no more than three basic symbols, that is, “0”, “( )” and “++”. Nevertheless, We should admit that with the above four axioms, there still exist at least two possibilities to create natural numbers.

(i) One would not lead to an incompatibility through defining or assigning a symbol such as “0.5”, “1.1”, “⊙” which is distinct from any natural number denoted by “0”, “( )” and “++”to a natural number. We can test whether the Axioms 2.1-2.4 still hold. For the Axiom 2.1, certainly it doesn’t exclude “0.5”, “1.1” or “⊙” to be a natural number. For the Axiom 2.2, we recursively define 0.5++ is the successor of 0.5, (0.5++)++ is the successor of 0.5++, and so forth. Also, we can replace 0.5 with 1.1 or ⊙ for the same process. For the Axiom 2.3, evidently, we can have n++ ≠ 0 for every natural number n. Though it may not be necessary, we can also have n++ ≠ 0.5, or n++ ≠ 1.1, or n++ ≠ ⊙ for every natural number n. Bringing in the Axiom 2.4, it would not cause any contradiction considering “0.5”, “1.1” or “⊙” and their corresponding successors as natural numbers. First, we actually have defined 0.5 ≠ 0 in the very beginning; Secondly, we can define 0.5++ ≠ 0++, (0.5++)++ ≠ (0++)++, etc. Thus, it still satisfies the requirement of the Axiom 2.4 that different natural numbers must have different successors. Similarly, we can define 1.1++ ≠ 0++, (1.1++)++ ≠ (0++)++, …; we can also define ⊙++ ≠ 0++, (⊙++)++ ≠ (0++)++, … . Hence, no contradicton would happen even containing the Axiom 2.4. Further, we can define 0.5, 1.1 together with ⊙ to be three different natural numbers. Similar to the process shown above, we can define 1.1++ ≠ 0.5++, (1.1++)++ ≠ (0.5++)++, …, and ⊙++ ≠ 0.5++, (⊙++)++ ≠ (0.5++)++, …, and so forth.

(ii) One may define a different operation from increment to extend the extent of our natural number system. For instance, we can define a new operation such as “insert ⊙ forward”, then “⊙⊙”, “⊙⊙⊙”,…are also natural numbers. Understandably, we define ⊙ ≠ ⊙⊙, ⊙⊙ ≠ ⊙⊙⊙, … .Through the same way shown in (i), we can create a new series of natural numbers from such as “⊙⊙” or “⊙⊙⊙”.

However, the Axiom 2.5 of the Peano Axioms excludes the above two possibilities.

Axiom 2.5. Let P(n) be any property pertaining to a natural number n. Suppose that P(0) is true, and suppose that whenever P(n) is true, P(n++) is also true. Then P(n) is true for every natural number n.

11 October, 2010 at 11:08 pm

Yandong Xiao(Max Hsiao)Dear Terry,

In my view, I consider it is necessary to reiterate that “x-y:=x+(-y) for any two rational numbers x and y” before displaying the Definition 4.2.8.

Also speak of some exercise. I failed to find the Exercise 3.4.7 necessary for the proof required by the Exercise 3.5.2, but it is still ok to work it out without the Exercise 3.4.7.

16 October, 2010 at 11:28 pm

Yandong Xiao(Max Hsiao)In Chapter 4, it says “We do not bother defining a notion of ε-close when is zero or negative” but in the proof for lemma 5.3.7 in Chapter 5, we have found “bn is certainly 0-close to bn”.

17 October, 2010 at 8:18 am

Ming LiHey Max,

I think this is kind of clear and understandable, and just like Tao indicated in Remark 4.3.5: “The ε-closeness is not standard in mathematics textbooks. It’s just auxiliary scaffolding to help you construct the convergence or limit… …because if ε is zero then x and y are only ε-close when they are equal”, the 0-close is a property even better than that ε-close for every ε. The Lemma 5.3.7 could be kind of consistent with Def 4.3.4.

17 October, 2010 at 5:09 pm

Yandong Xiao(Max Hsiao)Dear Li Ming,

In fact, I didn’t consider it is a mistake, but I do believe the expression such as 0-close unnecessary. For the purpose of rigorous mathematics, I also have pointed out some other problems, as you may see them from my previous posts in this page.

By the way, the solution you provided to the Exercise 3.4.7 may not properly solve it. I had worked it out some weeks ago. It looks like that the replacement axiom is the key point to the answer. Other tools’ application such as the use of the power set axiom is simple and obvious.

Finally, I would like to say one more thing. I find the Exercise 3.4.7 which contained in the hint for the Exercise 3.5.2 unnecessary for the solution. Have you also done the exercise?

18 February, 2014 at 3:51 am

Anonymoushi, can you please tell me how to apply the replacement axiom to Exercise 3.4.7?

21 October, 2010 at 11:06 pm

Yandong Xiao(Max Hsiao)It should be more directly to use Corollary 5.4.10 instead of the Proposition 5.4.9 for the Exercise 5.4.8.

22 October, 2010 at 8:31 pm

AnonymousProf. Tao,

Before one uses the Peano axioms to define “natural numbers”, does he need to define what’s “=” mean by the equality axioms first?

Besides, I think the emphasis on rigor and the way to deal with it are very nice characteristic in this book. But when I keep asking “why” and ” what” for the concepts in the book, I find that finally I have to go to the issue of logic (mathematical logic at least) and even philosophy. So when you stress “rigor” in doing math, which is the very spirit of this book, do you just mean complying with the law in mathematical logic and trying to building the math knowledge from the “very beginning” (say, in this book, Peano axioms) without lacking any steps?

25 October, 2010 at 10:37 pm

Yandong Xiao(Max Hsiao)It seems not easy solving the proof for lemma 5.6.6 (a). Though it is clear to prove by contradiction. That is, if y_n0, (y+ε)_nx, then there exists a ε’>0, (y-ε’)_n >x. But what should this ε or ε’ be? I find myself lost.

26 October, 2010 at 4:46 pm

Terence TaoFirst prove that and for all sufficiently small .

29 October, 2010 at 11:27 pm

Yandong XiaoI have worked it out. I finally construct those epsilons I seek.

Another question: though I have proven theorem 6.1.19 (a)-(f), (g) and (h) seem unusual, can you also give me more hints?

26 October, 2010 at 11:36 pm

Yandong Xiao(Max Hsiao)Thanks for your help. I regret that I didn’t show my problem in a good manner and thus misled. Actually, I had also tried this way you mentioned above for working the lemma 5.6.6 (a) out. What stopped me is another question which seems difficult for me. For I never learn how to display such as (y-ε) to the power n, I would not show it here.

31 October, 2010 at 10:52 pm

Yandong XiaoTwo questions:

(1) It seems necessary that x=y when lim max(an,bn)=max(liman, limbn)

n->∞ n->∞ n->∞

So does lim min(an,bn)=min(liman, limbn).

n->∞ n->∞ n->∞

(2) In the extended-real-number-system section, one may fail to prove that -∞≠-∞ or +∞≠+∞ though it has claimed for any real number x, x≠-∞ and x≠+∞. Further, it also says -∞≠+∞.

4 November, 2010 at 11:50 pm

Yandong XiaoIt is unnecessary to emphasize that L-and L+ are limited at the beginning of the proof for thereom 6.4.18 since some steps later the inequality

aN -ε<= L- <= L+ <= aN+ε ensures them limited. Moreover, it should be more direct to derive the inequality from aN -ε<= inf (an),n from N to ∞<= sup(an), n from N to ∞ <= aN+ε by combining Exercise 6.4.2 and Proposition 6.4.12 (c).

6 November, 2010 at 12:50 am

Yandong XiaoI’m afraid the proof for Corollary 6.5.1 is not correct. The number k in this case is a given number by random, and the lemma 5.6.6 doesn’t ensure 1/(n^(1/k)) is a decreasing function for the variable n. To prove it is a decreasing function, we have to apply lemma 5.6.9(b) and (d).

On the other hand, I think it may not be so obvious to prove

L^k=lim 1/n. I have proven that inf(1/(n^(1/k))), n from 1 to ∞, is 0.

n->∞

6 November, 2010 at 11:35 am

Terence TaoLemma 5.6.6(d) ensures that for any given k, the function is decreasing, and so the limit exists. From Theorem 6.1.19(b) and induction, one can then prove that for every natural number , and in particular for .

6 November, 2010 at 11:12 pm

Yandong XiaoYes, you’re right. But I still find Theorem 6.1.19(e) not useful for the Exercise 6.5.1. I simply use to lead a contradiction.

To prove lemma 6.5.3, I mainly apply Lemma 5.6.6(e), I failed to follow your hint.

7 November, 2010 at 12:45 pm

Kent Van VelsDear Dr. Tao,

In exercise 8.5.5, to show that turns into a partially ordered set, do we need to assume that is an injection? For example, consider the function defined as . Then we have since . And we also have since but we don’t have .

Thanks,

Kent

[Oops, “” should be “ or ". This has been added to the errata. -T.]1 December, 2010 at 7:40 pm

Kent Van VelsIn exercise 8.5.12 should the partial ordering be defined as if , or if and ?

The change is the strict ordering on the $x$ terms.

Thanks for these books!

[Corrected, thanks – T.]2 December, 2010 at 9:30 am

StudentIn exercise 12.3.4, the proof of Proposition 12.3.4(b). I think i may have constructed a counter example to the “backward” implication. Let and be the standard metric. Let and , a closed set. Then form . Now, is not closed with respect to since we can construct a sequence in that converges to , say. But is irrational and is therefore not included in . Should we add the stipulation that is closed?

2 December, 2010 at 10:54 am

Terence Taois not an element of Y, so it is not a counterexample to being closed in . (It is a counterexample to being

complete, but that is a different concept.)26 December, 2010 at 11:47 am

AnonymousDear Prof. Tao,

Considering the “mathematical statements” in Chapter A.1, what’s the difference between, “If X, then Y” and “Since X, Y” or “Because X, Y”? are they logically the same? Since in many mathematical argument, “because”, “since” are commonly used, can they be categorized into some logical connectives?

In A.2, (p.315 2nd edition, Vol. I), when you say, “If you know that ‘If X is true, then Y is true’, then it is also true that ‘If Y is false, then X is false'” do you mean the more complicated mathematical statement “If X’, then Y'” is true where X’:=”If X is true, then Y is true” and Y’:=”If Y is false, then X is false”?

What’s more, is the sentence “(because if Y is false, then X can’t be true, since that would imply Y is true, a contradiction)” a PROOF for the statement above? Mathematically, does a “contradiction”, as I understand in the book, mean a false compound statement like “X is true AND X is not true”?

The last question, what’s the difference between X and “X is true” where X is a mathematical statement? It seems that they are logically the same. But as long as they are the same, one will have “‘X is true’ is true”, “‘X is true’ is true” is true… How should I understand such phenomenon?

26 December, 2010 at 12:39 pm

Terence TaoStrictly speaking, if one wants to discuss the theory of logical deduction properly, one should take care to distinguish between the “internal” formal theory under discussion (e.g. propositional logic, or first-order logic), and the more informal (and “external”) metatheory used to discuss that formal theory. With such a careful perspective, deductive rules such as “Given that “If X is true, then Y is true”, one can deduce “If Y is false, then X is false”” are part of the external metatheory, rather than the theory itself. Actually, strictly speaking, the use of phrases such as “is true” or “is false” are already part of the metatheory; if one were to adhere to the formal syntax of propositional logic completely, one should be instead saying things like “Given that ““, one can deduce ““.” (As such, statements such as “”X is true” is true” are part of the meta-metatheory, which one should probably not analyse too much unless one really knows what one is doing and is thinking very clearly.)

A classic example of what goes wrong when one tries to identify the metatheory with the theory is Lewis Carroll’s short dialogue “What the Tortoise said to Achilles“.

Connectives such as “Because” and “since” can only be represented in formal propositional logic by the single symbol , but have much richer connotations in the metatheory; see this page of mine on this topic.

This text, though, is an analysis text rather than a logic text, and these sorts of self-referential fine distinctions, while important in logic, are only of secondary importance in analysis, which rarely needs such levels of logical introspection, and as such does not need to distinguish the theory of logical deduction from its metatheory. (In particular, one usually reasons in analysis using the semi-formal language of mathematical English, rather than using the strict formal syntax of first-order logic.) As I stated at the beginning of that appendix, these rules of logic are not to be memorised or interpreted too formally; they should simply

make sense.26 December, 2010 at 3:29 pm

AnonymousDear Prof. Tao,

Thank you for your clarification.

In your preface to the first edition, you mentioned that “it is important to know how to do analysis rigorously and ‘by hand’ first, in order to truly appreciate the more modern, intuitive and abstract approach to analysis that one uses at the graduate level and beyond.” And in this blog, you also mentioned “the “rigorous” stage, in which one is now taught that in order to do maths “properly”, one needs to work and think in a much more precise and formal manner (e.g. re-doing calculus by using epsilons and deltas all over the place).” I think I am quite worried about this and maybe I misunderstand your ideas. I even doubt that what does TRUE mean in mathematics.

Why do you say “these sorts of fine distinctions, while important in logic, are only of “secondary” importance in analysis”? Why don’t I need to worry too much about the logic text when learning analysis, instead I simply need “make sense”? How should I understand “make sense” and “do it rigorously”? I know that LOGIC (or precisely mathematical logic?) is quite different a topic from analysis. But isn’t it true that all the argument we make in analysis is basic on logic? As I understand, everything is true in mathematics because it is “logically true”. Then when I want to do analysis rigorously, I always struggle with every detail logically.

As I understand so far, learned from the book, one need to “use” or “obey” logic when one does analysis instead of worrying about understanding the issues such as “what is logic?” and “how does it function?”. Does a mathematician need a very deep understanding in the area of logic or at least mathematical logic for understanding and doing math?

Because of worrying about such problems, it is difficult to tell when I can claim that “I am doing analysis rigorously”.

26 December, 2010 at 3:47 pm

Terence TaoTo perform analysis (or any other branch of mathematics) rigorously, you should use logically correct reasoning, such as that given in Appendix A of my textbook.

However, unless one is studying logic itself, it is not necessary to be able to formally analyse these rules of logical reasoning in order to be able to use them properly; for the purposes of doing rigorous mathematics, it is enough that these laws of logical reasoning make sense to you, so that you will use them correctly. The formal study of the laws of logic and their foundations is itself an interesting topic, but one which is of secondary importance for the purpose of performing rigorous analysis. (And if the laws of logic do not already make sense to you on an intuitive level, it is unlikely that a formal treatment of these laws is going to be the right way to resolve this conceptual difficulty. Indeed, as mentioned in that appendix, I deliberately refrained from formally listing the laws of logic in full, as this can in fact

get in the wayof obtaining a firm and intuitive grasp of how mathematical logic works, especially to students who are being exposed to formal mathematical reasoning for the first time.)To give an analogy: in order to write English prose properly, one needs to know the laws of grammar well enough that they make sense to you; but one does not need to know the laws of linguistics that generate these grammars, unless one is studying linguistics. This is not to say that linguistics is an uninteresting topic, but it is of secondary importance for the purpose of writing good English prose, which is ultimately more concerned with communicating the subject matter than it is with perfect grammar, though the latter should not be ignored.

27 December, 2010 at 3:30 pm

AnonymousDear Prof. Tao,

I noticed that you mentioned the concept “vacuously true” in the appendix more than once. In the “Universal quantifiers” section, there is an example:

. I was wondering how should one prove it? It seems that one can prove it as (a) multiply by 2 the "inequality 3<x<2", then one gets 6<2x<4; or (b) If , then . Since this is a false statement, is automatically true.(False implies true) How should one prove it? Is the statement "1=2 for all 3<x<2" also true? Is it the same thing as ?

How can a vacuously true statement can still be useful in an argument? Is there any concrete examples? I noticed that when defining the mathematical structures like metric space, vector space, topological space, etc., one always claims that "for a non-empty set…". Is it because one has to avoid something "vacuously true"? The empty set seems trivial, though, is it harmful to include it in the mathematical structure?

28 December, 2010 at 10:33 pm

AnonymousI tried to answer the question. I think the main idea is in (b), i.e., “false implies false”(rather than “false implies true”).

28 December, 2010 at 11:05 pm

AnonymousAnd for the second question, I myself think it is beyond the scope of mathematics. (It may be too boring to be a mathematical question) And nobody will be interested in the structures of an empty set.

28 December, 2010 at 11:21 pm

AnonymousAn excellent example for the usefulness of vacuously true statement, if I am right, is in “mathematical induction”.

29 December, 2010 at 3:37 am

J.P. McCarthyLet , and let be some property. To prove “, ” is true we simply have to show that for each element of that is true. If is empty then is essence we don’t have to check the truth of any of the elements of – there are none.

9 January, 2011 at 2:49 pm

AnonymousDear Prof. Tao,

For Exercise 2.2.2, if one uses induction on , the base case should be started from instead of . Is one allowed to do so or does one need to change Axiom 2.5 slightly? (But how can one change an “Axiom”?)

10 January, 2011 at 7:26 pm

YDear Prof. Tao,

In Section 3.1, I don’t understand why do you say the “is an element of” relation obeys the axiom of substitution. What I think is that “is an element of“ is just a property. And the one “OBEYS” the axiom of substitution should be “equality” that defined in the text instead of “is an element of” according to Section A.7.

18 January, 2011 at 8:53 pm

AnonymousAgain, about the “axiom of substitution”. In Section 3.3. “we observe that functions obey the axiom of substitution”. But in the appendix A.7, the substitution axiom is “Given any two objects x and y of the same type, if x=y, then f(x)=f(y) for all functions or operations f. These two argument are just like a “circle”. I am quite confused here.

21 January, 2011 at 5:25 am

Terence TaoThis is because we are continually upgrading our formal system by defining new operations and functions to add to the system. Each time we define a new operation (e.g. defining addition on the rational numbers, or multiplication on the real numbers, etc.) or introduce a new function, we need to check that the axiom of substitution remains true under this expanded system (this is usually known as checking that the operation or function is “well defined”). [One also needs to verify that the axiom of substitution is not destroyed whenever one identifies two mathematical objects together, as for instance was done in constructing the integers, rationals, and reals as equivalence classes of formal differences, sums, or limits of other numbers that had been identified together.]

Set-theoretic functions, by their nature, automatically obey the axiom of substitution, as observed in Section 3.3. But operations that are not defined via a set-theoretic function may fail the axiom, and so should not be placed inside one’s formal system of reasoning. For instance, the operation of taking the “numerator” of a rational number is not well-defined; 1/2 and 2/4 are equal as rational numbers, but have different numerators (1 and 2 respectively). [However, the operation of taking the numerator of a rational number after reducing to lowest terms is well defined, adopting the convention that the denominator of a rational in lowest terms is always a positive integer.]

11 January, 2011 at 5:25 am

observerHi Terry,

What is the difference between irreducible and prime numbers?

23 January, 2011 at 10:50 am

KVVHello,

I have a question on exercise 10.1.1 on page 256. Do we have to assume that the point is contained in both and ? I am unsure if we can apply definition 10.1.1 if we do not make such an assumption.

Thanks,

KVV

24 January, 2011 at 4:15 am

Terence TaoYes, needs to be contained in Y also; I’ve added an erratum to reflect this.

1 March, 2011 at 1:43 am

YeeweyDear Professor,

I got a problem when I read a Universiy text book for the first time when I was a fresh man many years ago. Where did those definitions come from ? How people designed them? etc.

I am not criticizing. Recently, I realize that the process of writing a text book is very similar to that of applying a Fourier transform to time series: the ‘essential’ ingredients remain but one can not know when these ingredients were created. Indeed, the time series of literature contain both knowledge and time labels. Many textbooks, however, omit the time labels.

I am wonder if you have thought about such problem? Or, could you please give some comments to the issue.

Thanks indeed!

Yeewey

2 March, 2011 at 2:11 pm

SandraDear Terry

I read the works on various author about Korovkin Theory but it seem there is not much information about what is Korovkin theory, etc. Can you recommend any reference materials that I can refer to for more indepth learning into this theory?

Thanks

Sandra

12 March, 2011 at 10:06 am

YIn the case of real number in this book, when where and , one has . Considering the continuous linear operator between two Banach space, say, . Can one still get the similar result? In this case, I think, becomes “ are invertible for all ” and “ is also invertible”. It seems that in the operator case the counterpart of the routine for “bounded away from zero” is not clear here.

12 March, 2011 at 11:00 am

Terence TaoIf one uses the operator norm topology, then it is indeed true that if a sequence of invertible linear operators converges in operator norm to an invertible linear operator , then the inverses converge in operator norm to . The point is that if one writes , then the converge in operator norm to zero, and so by Neumann series, exists and converges in operator norm to 1 for large enough n. Since , the claim follows.

12 March, 2011 at 11:39 am

AnonymousIf one drops the assumption that are Banach space, and only assume that they are normed vector space, Neumann series may fail to exist. However, if at the same time assume additionally that are continuous for all , which is not mentioned in your comment, can this still be true?

12 March, 2011 at 1:28 pm

Terence TaoIn my previous comment, the operators $T_n, T$ (and their inverses) should be understood to be continuous.

It is also possible to proceed without Neumann series (and without using completeness hypothesis), using the observation that if is small in operator norm (say, with norm less than 1/2) and is

a prioriknown to be invertible, then the inverse must also be close to the identity.17 March, 2011 at 9:20 am

AnonymousYou mentioned in the chapter 3 that formally, “sequence” is a function . But the definition of “limit of the sequence” and “limiting values of functions” do not look the same. The later one uses the “open set”. But what’s the “open set” for the former one? Is there a uniform definition for these two kinds of limit?

17 March, 2011 at 9:54 am

AnonymousIf can be identified as a “point” in the underlying space, say, or , where or , hopefully one can get the uniform definition? After all, “…, such that ,…” and “… such that for all …"look similar. Can one say something like "?

17 March, 2011 at 10:44 am

Terence TaoYes. This is discussed in Example 11 and Remark 6 of

https://terrytao.wordpress.com/2009/01/30/254a-notes-8-a-quick-review-of-point-set-topology/

12 April, 2011 at 10:22 pm

AnonymousDear Prof. Tao,

I don’t quite understand why the range of f needs to match the domain of g when one defines the composition(Definition 3.3.10) . Is this a convention? It seems that everything can be well-defined even range()\neq domain().

12 April, 2011 at 10:26 pm

AnonymousOf course with the assumption in the book that and $g:Y\to Z$.

Oops, how silly is my question. The range of and $f(X)$ are not the same in your book.

12 April, 2011 at 10:31 pm

AnonymousThe reason I concern about the concepts here is that I was wondering if the inverse of Exercise 3.37 is true: If is invertible, can one say that both and are invertible?

12 April, 2011 at 10:40 pm

AnonymousWell, at least when the cardinality of X, Y, Z is finite, the answer should be yes. And generally one can deduce that both and are 1-1 and is necessarily onto. But it seems that does not need to be onto.

18 April, 2011 at 8:25 am

AnonymousDear Prof. Tao,

In the section 8.5 “Ordered sets”, when you define the partially ordered sets, you use the word “relation”, and just “describe” it? Is there a formal definition of it? I think [this] (http://en.wikipedia.org/wiki/Relation_%28mathematics%29) is what you are talking about in the book, right?

Can one say that a relation on is actually a function according to your description in your book? Since you have already defined what is a “function”, then this should make sense.

18 April, 2011 at 9:23 am

Terence TaoFor the purposes of my book, the only relevant thing about a relation is what it does, which is described in the parenthetical comment in Definition 8.5.1: if x and y are in X, then is either a true or a false statement. One can model a relation, as you say, as a function from to a two-valued set such as , or , or one can also model a relation as a subset of as is given in the Wikipedia definition. (The connection between a relation , and a model of that relation, is analogous to the relationship between a function , and the graph of that function.)

However, as this text does not study the theory of relations

per se, we do not need to fix precisely what set-theoretic model one would use to model the relation concept, and can instead proceed on anad hocbasis, so that whenever one actually needs to apply, say, Zorn’s lemma, one verifies that the relation supplied by the application (e.g. subset inclusion) obeys the relevant properties required of a relation (in this case, that one can attempt to compare any two elements in the domain). The situation is similar to that of ordered pairs, discussed in Section 3.5, where again the exact set-theoretic model of an ordered pair is not particularly relevant for applications.18 April, 2011 at 9:55 am

AnonymousDear Prof. Tao,

can we have a function which is continuous only at a given points? we have usual Euclidean topology on the real line. I am able to find an example for , but for general case I am not.

Thanks

18 April, 2011 at 2:01 pm

juanLet be your points. Consider the polynomial . Now consider the function given by if and 0 otherwise. It is continuous only at the roots of , exactly the points .

19 April, 2011 at 11:03 am

AnonymousThanks Juan, that is great.

but now, let’s think about the following question :Can we find a function which is differentiable only at one point?one given specific point?

19 April, 2011 at 1:36 pm

juanConsider given by if and 0 otherwise. It’s not even continuous if , let alone differentiable. If , we can bound the limit using that $0 \leq f(x) \leq x^2$, by sandwich theorem, the limit exists (and is equal to 0), so the function is differentiable at that point. To construct a function differentiable only at a given , just “move” the function where needed.

19 April, 2011 at 2:50 pm

AnonymousJuan thanks. you are an excellent analyst.

18 April, 2011 at 3:35 pm

AnonymousDear Prof. Tao,

I don’t quite understand the “good set” in your proof for the lemma 8.5.14.

Why do you need to specify the property that for all ? Isn't it automatically true?

"we can thus assign a strict upper bound to each well-ordered subset of which has as its minimal element" But this assignment is not necessarily unique, how do we know when ? Doesn't it depend on the assignment ?

19 April, 2011 at 9:10 am

Terence TaoNo, the statement is not automatically true. Imagine for instance that with the usual ordering, and . Then the above claim would fail for . ( is certainly an upper bound for , but it is not necessarily the upper bound selected by .)

The property of being good does indeed depend on the choice of s. Given a set Y, the property that for all may or may not be true, depending on what s is (and indeed s will, in general, not be unique). In the proof, one fixes s first, and then considers sets that are good relative to that choice of s.

23 November, 2013 at 7:12 pm

JackAs I understand, is a choice function defined on

. When a good set is defined, it is well-ordered and contains as its minimal element, and obeys some property which is defined using . But how do we know that is a set in , which is the domain of the choice function ? (How do we know this set has a strict upper bound?)

24 November, 2013 at 6:35 am

Terence TaoBy hypothess of the argument, all well-ordered set with minimal element has a strict upper bound.

Also, is synonymous with (see Axiom 3.11)

29 December, 2013 at 8:58 pm

JackThe key construction of the proof is the “good sets”. But it seems rather artificial to me. What would be a possible motivation of this construction? You mentioned in the “intuition part” of the proof that these are the “partially completed” sets. In what sense are they “partially completed”?

23 November, 2013 at 7:21 pm

JackAnd for the definition of , do you mean instead?

24 April, 2011 at 10:52 am

AnonymousWhy can’t I see my previous comment? And when I repost it, it is said to be duplicated.

28 April, 2011 at 11:17 am

AnonymousI don’t understand the proof for proposition 7.2.12(alternating series test.) As long as we have for all , we can deduce that . Then for any , . How can one get ” the sequence is eventually -steady” instead of -steady?

28 April, 2011 at 11:56 am

Terence TaoSince for all , one has for all .

22 May, 2011 at 2:18 pm

YuProf. Tao,

As you said in the book, the proof for Lemma 9.6.3, which uses the Heine-Borel theorem is an indirect proof. I am wondering is there a direct proof for this lemma? Or does it have to be proved indirectly?

23 May, 2011 at 8:37 am

AnonymousDear Prof. Tao,

In p.290, you mentioned that there are ways to resolve the ambiguities of the notation of , most notably by introducing the notion of the differentiation along vector fields. But this is beyond the scope of this text. Could you recommend any references for this topic? Since the result of the direct search on Google for “differentiation along vector fields” is unsatisfied.

5 June, 2011 at 10:09 am

ckDear Prof. Tao,

I am using your text to learn analysis and I enjoy it very much. But I am puzzled by exercise 2.2.2.

I found a solution from internet as follow. Rephrase lemma 2.2.2 as:

Let P(n) be the statement: If n is not equal to 0, then there is a natural number b such that b++=n.

Proof: P(0) is true vacuously.

Suppose P(n) is true. Then P(n++) is the statement: If n++ is not equal to 0, then there is a natural number b such that b++=n++. This is true since we can take b=n.

Is this proof correct?

If it is so, I can imitate it to prove something like If n is not equal to 0 (positive), then n>n+1.

Proof: P(0) is true vacuously.

Suppose P(n) is true. Then n>n+1. So n=n+1+m for some positive number m. It follows that n++=(n+1+m)++=(n++)+1+m, hence n++>(n++)+1. So P(n++) is true. What’s wrong here?

5 June, 2011 at 1:33 pm

Terence TaoThe step “Then n>n+1” is invalid when n=0, by your definition of P(n).

14 June, 2011 at 7:13 am

AnonymousDear Prof. Tao,

I find that in definition 9.1.5, you use , while in definition 9.3.3, you use . I set myself as an exercise to prove that and are the same for both of these two cases. And I think it should true, isn't it? Generally, what's the principle or advantages to use one of them instead of another?

15 June, 2011 at 6:49 am

ChrisDear Professor Tao

With reference to the errata above, I have few queries.

Given:

1) Pg 70 5th Line of proof of Lemma 3.6.9 $1 \leq i \leq N$ should be $1 \leq i \leq n$ but in my book the 5th line is already correct! $i \in \mathbb{N} : 1 \leq i \leq n$

Please clarify.

Pg 96: A space is missing between “Proposition 4.2.4” and “allows”

My comment: Althought the page number is wrong, I am able to locate to the correct Pg at Pg 83 but I can’t see any missing space between “Proposition 4.2.4” and “allows”. In fact, I don’t see the word “allows”. Please clarify

Then:

P197: Exercise 8.3.2, f should be an injection rather than…….

My comment: May I know which of the text should be replaced? I am also confused by when you mention about definition of g. Can you provide the exact place in the text where it should replace?

15 June, 2011 at 9:30 am

Terence TaoThere were some issues with page numbering which should be fixed now (all errata to the second (hardback) edition should refer to the hardback page numbering).

16 June, 2011 at 7:00 am

ChrisDear Professor Tao

With respect to

Pg 70, 5th line of proof of Lemma 3.6.9, should it be 4th line? Please clarify

Pg 110 –> Should this be Pg 111 and the second edition seem to have this corrected. Please clarify.

Pg 168 –> This should be Pg 164.

[Corrected, thanks – T.]1 July, 2011 at 7:08 pm

AnonymousDear Prof. Tao,

is there a elementary proof of the fact that every convex function is continuous? is that correct?

Thanks

11 July, 2011 at 6:40 am

AnonymousMy Analysis I translate as Chinese book in which the proof of 8.5.14 that I think it is wrong in Chinese,who can give me the original English proof.Or give me a link to download Analysis I ,thank you.

14 July, 2011 at 1:31 am

studentOn errata, “•p. 108, proof of Lemma 5.3.18” should be “•p. 124, proof of Lemma 5.3.15” for the first edition.

21 July, 2011 at 1:44 pm

JackDear Prof. Tao,

I don’t know whether I can follow you hint for exercise 11.1.3(vol.I 2nd edition). For proving the statement in the hint, I think one has to prove it by contradiction. In other words, assume that is not strictly less than , then it follows that . But , it has to be of the form or for some . Am I right? Do you have a better proof?

22 July, 2011 at 8:45 am

Terence TaoThat proof would work. Alternatively, one can split into cases depending on whether is an open interval , a closed interval , a half-open interval , etc.

21 July, 2011 at 2:15 pm

JackI think there is a typo in p. 271(Vol.I 2nd edition).

Remarik 11.2.2 “then a function which is constant on can have only one constant value”—>”then a function which is constant on can have only one constant value”

[Corrected, thanks. -T]21 July, 2011 at 4:39 pm

JackFor the exercise 11.2.3, one finally needs to show that . Geometrically this is very obvious. But How can I prove it “in words”?

22 July, 2011 at 8:45 am

Terence TaoTry proving it in the special case when P consists of just a single interval J. The general case can be deduced from this case by an induction argument.

23 July, 2011 at 11:27 am

ChrisDear Professor Tao

Seeing no response from you, I am reposting this message for your kind consideration. Can I request you to put up errata to your books as in PDF form just like almost all other book’s authors did? When printed, the mathematical typings are clearer this way as compared when printed directly from blog as all mathematical typings turn blurry that way.

I hope you could take this into your consideration and put up a PDF. Thank You.

23 July, 2011 at 12:56 pm

JackHe has already given you the response. Maybe you didn’t noticed. :-)

13 November, 2011 at 11:05 pm

UlrichTerence, here some minor typos and questions (Analysis 1 – Second Edition)

Prop. 2.2.14 : second line: .. , and Let -> and let

p. 50, first line: h(2n+3)=h(2n+2) -> h(2n+3)=2n+2 (I guess)

Def. 4.3.4 : I guess, that is always rational, but this is not mentioned.

Prop. 4.4.1: I have not found the word “interspersing” in my dictionary, but interspersion. The same meaning?

Rem. 10.1.14: “Leibniz” without “t”

Question:Axiom 2.2 (p.17) is defining n++ as “is also a natural number”. Shouldn’t this be ” is also a unique natural number”? It is never mentioned, that the increment operator is a function.

I am using the book as textbook in my beginner “Analysis 1” lecture. Unfortunately, we had serious problems to get the book on time from Hindustan Book Agency, because they just print “on demand”. I don’t have a solution for this, but it is not a very commod situation for the students.

The same happens with the “Measure Theory” book. Still not available by Amazon as they told me.

[Corrections added, thanks. The uniqueness of is implicit from the axiom of substitution (Appendix A.7), but I’ll make a note to make this more clear. Interspersing is the present participle of “intersperse”. I have the first few chapters of Analysis I (and all of measure theory) online which can be used as a stopgap solution while the books are being ordered. -T]15 November, 2011 at 9:37 am

UlrichTerence, thanks for the remark. Maybe you can also add that Paul Cohen passed away in 2007 (p. 196).

What I miss is that between two real numbers you always find a non-rational real number, which is “Prop. 5.4.14+” telling not only the rational numbers are order dense in the real numbers but also the non-rational real numbers. Since this can be proved like Prop. 5.4.14 (Exerc. 5.4.5) it can be added there. Or have I overlocked something?

Regards

Ulrich

[I’ve added a suggested exercise along these lines to Section 5.5 (one needs to know the existence of at least one irrational number, such as , to get started.) -T]16 November, 2011 at 6:49 am

kI don’t understand the wrongness of one answer in the Java quiz about logic on the course homepage. The question was “Suppose one wishes to prove that ‘if all X are Y, then all Z are W'”. I answered ‘All Y are X, and all Z are W’, and it was marked as incorrect. Now that is very confusing to me: isn’t everything that is true trivially implied by everything?

16 November, 2011 at 6:50 am

kOh, I forgot the last part of the question. The complete question was “Suppose one wishes to prove that ‘if all X are Y, then all Z are W’. To do this, it would suffice to show that…”

16 November, 2011 at 8:45 am

Terence TaoThe (incorrect) answer you mention is “All Y are X, and all W are Z”, not “All Y are X, and all Z are W”.

18 November, 2011 at 9:47 am

AnonymousHow strange is that! I have clicked again through the quiz, and I am looking at the answers for the 20th time now, and I still see the option “All Y are X, and all Z are W” (which is marked as incorrect). I do not see an option “All Y are X, and all W are Z”.

18 November, 2011 at 11:15 am

Terence TaoAh, you seem to be using the old (java-based) version of the quiz, which is no longer being actively maintained. I have now corrected that particular answer, but the most recent versions of these quizzes can be found at

http://scherk.pbwiki.com/

18 November, 2011 at 12:45 pm

AnonymousAh, Thanks!

16 November, 2011 at 10:56 pm

UlrichOne more about Analysis 1: In Definition 6.2.6 (c) I suggest to mention, that in the definition the minus is the set-theoretic minus and not the minus in the number system $\mathbb{R}$ (at least this was my interpretation).

Ulrich

16 November, 2011 at 10:57 pm

Ulrich..

20 November, 2011 at 8:50 am

UlrichAnalysis 1, p.126, Prop. 6.1.4 (Proof, 6th line from below): you are refering Prop. 5.4.14. But what is really needed is Proposition 5.4.12 (which has been proved and is enough for this case).

Ulrich

[Added, thanks – T.]21 November, 2011 at 2:32 am

UlrichTerence, you are denoting the extended real numbers by , which bis normally the units of . Isn’t it better to denote by the extended real numbers, which is then also in sync with the closure symbol?

Ulrich

21 November, 2011 at 9:17 am

Terence TaoI’ve avoided using to denote the extended real line in this text in order to reduce confusion (the closure of in the real line is again ). Of course, once the reader is comfortable with the idea that closures depend on the ambient space as well as the set being closed up, there is nothing wrong with the notation, but I prefer to avoid it initially.

I also (mildly) prefer to use rather than for the group of units of a ring , as it gives slightly more emphasis to the multiplicative structure (while freeing up the asterisk symbol for other uses). But this is not a strong preference.

21 November, 2011 at 10:38 am

UlrichTerence, thanks for the comment and the clarification. The question came up during a discussion, because other books are using the bar-sign (including Bourbaki). But this is really not an issue.

Ulrich

27 November, 2011 at 3:53 am

AnonymousTerence, two minor typos:

Errata: p.108 Lemma 5.3.18 -> Lemma 5.3.15

Book p.144: Cor. 6.4.14: line 4: ” .. for all ” should be ” .. for all ”

Ulrich

[Corrected, thanks – T.]3 December, 2011 at 10:01 pm

DavidTrivial typo:

Analysis 1, p.186, Ex. 8.1.4. “….sequence f(0), f(1),…f(n) “. The last term should be f(n-1).

[Corrected, thanks – T.]12 December, 2011 at 5:52 am

ugrohTerence, in Thm. 7.5.1(b) you are talking about “… conditionally divergent”. Since this has not be defined (at least I have not found it), could you please clarify this? For my understanding, a series can either approach or or can be oscillating finitely or infinitely.

Thanks

Ulrich

12 December, 2011 at 8:45 am

Terence TaoYes, “conditionally divergent” is a confusing term, and “not conditionally convergent” should be substituted instead (and similarly “absolutely divergent” should be “not absolutely convergent”).

28 December, 2011 at 9:20 am

UlrichTerence, I have a question regarding Def. 9.3.6: If f is not defined for

, then the definition seems to be not completely correct. I guess, that one has to add the condition , that is should be a cluster point. Or do I understand the definition wrong?

Ulrich

28 December, 2011 at 9:32 am

Terence TaoOops, f needs to be restricted to E in the definition (i.e. the restriction of f to E is eps-close to L near x_0). In the case when x_0 lies in E, this forces L to equal f(x_0). (See also Remark 9.3.7.)

29 December, 2011 at 12:07 am

UlrichTerence, isn’t it better to have in the definition? Then its is clearer (at last for me) what limit means and e.g. what the difference to continuous is.

Another point: In Prop. 9.4.7 (c) you are writing . All the time you have instead . It is not serious, but it is maybe bemusing for newbies (I hope bemusing is the right word).

29 December, 2011 at 11:37 am

Terence TaoWell, that is certainly the more prevalent notational convention regarding limits, but I dislike it due to the lack of transitivity. With the usual definition of limit (in which is not allowed to be equal to ), the statements and do not imply . Using (the notation in my text) instead makes the lack of transitivity more apparent. (I prefer to interpret as “ whenever “, not as “ whenever and “.)

For Proposition 9.4.7, probably the best solution (given that it is used elsewhere in the text) is to add a fourth item to the list of equivalences in which one has non-strict inequality instead of strict inequality.

30 December, 2011 at 3:02 am

UlrichOK, I understand the intention. By the way: Def. 9.3.6 is on page 220 and Prop. 9.4.7 is on page 228 (just for the errata list to remove the ???).

Thanks and a good start into 2012.

Ulrich

[Corrected, thanks – T.]18 January, 2012 at 1:16 am

ugrohTerence, in the proof of Prop. 9.5.3 (p.232) you are pointing to Proposition 9.4.7 (first line). But at this stage of the proof you can’t use this proposition.

Ulrich

18 January, 2012 at 10:15 am

Terence TaoThe proposition is being applied to the restriction of f to the subdomain . I’ll add a note in the errata to clarify this.

18 January, 2012 at 4:30 am

ugrohTerence, I think one can shorten the proof of Prop. 9.6.7 (p. 236). First, as in the book one get . Now suppose there is no x where the sup will be attained. Then we consider the function

which is continuous, thus bounded on the interval. Now take ; then we get $ y $ such that

or or

, thus is unbounded.

Ulrich (hoping, this is correct)

18 January, 2012 at 10:24 am

Terence TaoYes, this works (though if one unpacks the proof that continuous functions are bounded, we see that this is essentially the same proof).

25 January, 2012 at 5:52 am

ugrohTerence, I am asking for Thm, 10.4.2. on page 261/262: I don’t see why

has to be continuous, which you use in the proof (line 7/8), in the case is differentiable at . I don’t have a counterexample for this, but I can’t prove this either. If is strictly monotone increasing/decreasing, then everything is ok (and applicable to the exercises).

Ulrich

25 January, 2012 at 10:50 am

Terence TaoThe continuity of f^{-1} at y_0 is a hypothesis of the theorem. (It can be shown that if f^{-1} exists at all, then it will be continuous at y_0 if one makes the additional hypothesis that f is continuous near y_0, by using the non-zero derivative and the intermediate value theorem. But with the hypotheses stated in that theorem, one can permute the values of f around on some dense set (e.g. the rationals) to show that f^{-1} is not necessarily continuous if one only assumes the remaining hypotheses of Theorem 10.4.2.)

25 January, 2012 at 11:31 pm

ugrohTerence, it was my lack of English. I thought your sentence in the theorem is a conclusion. I have to add a “and” after …. at , and …

Sorry for this.

Ulrich

26 January, 2012 at 9:24 am

ckDear Prof. Tao,

I am working on your text, trying exericse 5.4.3 about the integer part of a real number. I wonder if my solution is correct and if there is a better way to do it. My answer is as follow:

For existence let . Then there exists a natural number such that for all . Then by Prop. 4.4.3 (c), for all . By Prop 4.4.1, we have for some integer . Hence, for all . By Cor 5.4.10, . So either or or .

For uniqueness, if and . If , then wlog, we can assume . Then so we have . A contradiction.

Thanks for your help.

24 February, 2012 at 11:28 pm

Proving every positive natural number has unique predecessor | Help me on math , homework, math activities[…] am independently working through Tao’s Analysis I, and one of the exercises is to prove that every positive natural number has a unique predecessor. […]

1 March, 2012 at 11:27 am

AnonymousProfessor Terry Tao grateful if you give me information on how conceguir your book online because I am Peruvian and I have no access to buy so much presencial.se agredeceria him.

RENATO.

21 March, 2012 at 2:36 pm

Audrey PadillaOn pg. 115, exercise 5.4.7, it seems as though if epsilon < 0, than we should be trying to show that x<y + epsilon. (I understand that <= allows for strictly =0 to clear up this ambiguity? I am a analysis student, so please understand if I am incorrect.

21 March, 2012 at 2:47 pm

Audrey PadillaI’m sorry, my post is not correct. I meant to say epsilon > 0. Also, although less than or equal to allows for strictly less than, would it not be more clear to specify? Also, if epsilon was greater than or equal to zero, than I can see why we can show that x may be equal to y + epsilon. If epsilon is positive, and x<=y, then addind epsilon to y means that x<y+epsilon. If x=y, x cannot be equal to y+e, so why are we trying to show x<=y+e? Might there be a misprint?

21 March, 2012 at 6:35 pm

Terence TaoIt is true that one can replace by in Exercise 5.4.7 and still obtain a true statement. However the formulation with is slightly more useful in applications in which one is trying to prove an inequality of the form by first proving that holds for all (see Trick 2 of https://terrytao.wordpress.com/2010/10/21/245a-problem-solving-strategies/ for more discussion of this method). Note that proving is slightly easier than proving . (In practice, though, the distinction between the two statements is close to negligible.)

24 March, 2012 at 1:25 pm

AudreyThank you Prof. Tao. Also, the link has already proven to be of great help; your time and effort is much appreciated!

27 March, 2012 at 1:13 am

Provide the download of analysis IFor those who need the digital edition of Analysis I,you can download free from http://ishare.iask.sina.com.cn/f/18853777.html?from=like.

For those who do not need this book ,please ignore this comment.

29 March, 2012 at 6:59 am

Luqing YeDear professor Tao,

In exercise 8.5.19,

……we say that is an initial segment of if there exists an such that (so in particular ）……

I think should be changed into .

Please excuse me if I am wrong.

[Correction added, thanks – T.]30 March, 2012 at 12:51 am

Luqing YeDear professor Tao,

This book is relatively basic but emphasis a lot on rigourous.So please do not mind me asking a question concerning the rigourous of this book.I hope this question is not dumb.

In set theory,you have axiom 3.2:The existence of .

After introducing this axiom,you prove that is unique.After doing that,you have Lemma 3.1.6,this lemma says one can always pick out an element from a non-emptyset.

But before lemma 3.1.6,you hadn’t introduce an axiom that ensure the existence of non-emptyset……So I think lemma 3.1.6 should come up a bit later in this book.

30 March, 2012 at 8:57 am

Terence TaoLemma 3.1.6 is still valid even in the absence of non-empty sets; it simply becomes a vacuous truth. (Generally speaking, I prefer to state these foundational lemmas as soon as one has enough axioms to prove them, in order to emphasise which axioms they actually depend on.)

30 March, 2012 at 4:49 am

Luqing YeDear professor Tao,

Another question concerning rigorous .I hope my question is not a bad question.

You mentioned that the axiom of substitution works for specification.But it seems that you didn’t mention that the axiom of substitution works for the axiom of replacement.Maybe you think that you don’t need to mention a similar thing twice,but I think you’d better mention that……

30 March, 2012 at 8:59 am

Terence TaoSee the last sentence of Remark 3.1.24.

3 April, 2012 at 5:40 am

Luqing YeDear professor Tao,

Thanks for your reply.

Exercise 8.5.20 is a good exercise,readers such as I can understand this exercise well,and can prove it with the help of your hints.But I think there is a small problem in this exercise.This problem is so small that can not harm this exercise.Even so,I think it is neccessary for me to come up with this problem.

is a collection of subsets of ,what if ?If ,then the property “all the elements of intersect at least one element of ” will be not true.So I think you’d better mention that .

Maybe you have already realized that,I hope this is not a boring question.

[Correction added, thanks – T.]4 April, 2012 at 7:07 am

Luqing YeDear professor Tao,

I want to ask for advice,I will do exactly what you advise me to do.

This book let me do a lot of trival exercise ,for example,Exercise 7.1.1 ask me to prove lemma 7.1.4,but I think lemma 7.1.4 is so obvious that I do not need to write the proof down.

I even think that I don’t need to think about lemma 7.1.4,because that is so obvious,if some one let me prove lemma 7.1.4, I can prove it with no doubt.But if someone do not ask me to prove that ,I will definitely not even think about such trival thing.

So I want to know your attitude towards this,whether I should write my proof down……

Maybe many readers have same confusion,your advice will be appreciated.

4 April, 2012 at 7:43 am

Terence TaoFor a foundational course such as this, it is important to make sure that you can in fact supply a rigorous proof for any statement which is intuitively obvious to you; this point is also discussed at the end of https://terrytao.wordpress.com/career-advice/there%E2%80%99s-more-to-mathematics-than-rigour-and-proofs/ . In the specific case of Lemma 7.1.4 (basic facts about finite series), it is important to realise why “obvious” manipulations about finite series are true because such manipulations are not always true for infinite series (particularly if they are not absolutely convergent, or even conditionally convergent). In particular, it is important to realise that manipulation of finite series is ultimately justified through the axiom of induction (to make rigorous statements such as “use the associative law n times”), which is not directly available for the infinite series setting.

4 April, 2012 at 8:39 pm

Anonymous+1 for “it is important to make sure that you can in fact supply a rigorous proof for any statement which is intuitively obvious to you”. This has also been discussed in the preface of your textbook. :-)

6 April, 2012 at 4:44 am

Luqing YeDear professor Tao,

Lemma 7.1.13 say

But it seems that you hadn’t defined what is before.You just say that it is double finite series——finite series of finite series.I can’t figure out what does “finite series of finite series ” exactly mean.

6 April, 2012 at 6:51 am

Luqing YeI understand it now. is a finite set of cardinality ,there is a bijection from to . is a function such that .So . :-)

11 April, 2012 at 4:55 am

Luqing Yelemma 7.5.2:

Here is a similar result compared with lemma 7.5.2:

Let be a sequence,then .

In fact,I worked on lemma 7.5.2 for a long time but was not able to prove it myself until I come up with this result.As every one knows,there is similarity between + and ().So I just

change in lemma 7.5.2 into +,then get the result.The result is far more intuitive to me,because I am more used to deal with +.After proving this result,I prove lemma 7.5.2 successfully. :-)

13 April, 2012 at 10:02 pm

Luqing YeIn definition 8.2.4,When is uncountable,you defined as .(Lemma 8.2.5 ensured that is well defined.)

Can I define in another way by using the well ordering principle(Every set has at least a well ordering) and the principle of strong induction(proposition 8.5.10) ?

13 May, 2012 at 5:41 am

Sabyasachi MukherjeeDear Dr.Tao,

I am a high school student studying Analysis I.I have studied till chapter II and I am intrigued and absolutely delighted with the level of rigour However, .I noticed how other real analysis texts introduced some set theory and a bit of set-point topology before starting out.

Your book does not do that till volume II.I am rather interested in knowing your perspective on this and the reason your unusual choice of the sequence of topics.And yes,thank you for writing that text.It is quite accessible even to high school student like me!

13 May, 2012 at 11:34 am

Terence TaoActually, in my text I cover set theory as early as Volume I, Chapter 3 (with infinite sets being deferred to Volume I, Chapter 8).

General point-set topology is introduced in Volume II, Chapter 13, but is not emphasised in the text, which focuses more on the more concrete theory of metric spaces (Chapter 12) which suffices for many applications. This is partly due to the course divisions at UCLA (point set topology is taught initially in a separate undergraduate class from undergraduate real analysis, and covered in more depth in the graduate level analysis course), and partly for reasons of focus; the text is not meant to cover all of mathematics or even all of analysis, but is instead concentrating on what is needed for the foundations of undergraduate real analysis. (Thus, for instance, linear algebra is covered only very briefly, in Chapter 17; for similar reasons, I eventually decided not to cover integration of differential forms, as it required too much of a digression into geometry, and my hope was that any student who managed to get through my texts would then be experienced enough to be able to handle the analytic aspects of this material from other textbooks without too much difficulty.)

13 May, 2012 at 11:52 am

Sabyasachi MukherjeeDear Dr.Tao,

Thank you for clarifying.Indeed I was referring to set-point topology being introduced in volume II. Once again, thank you.You may be happy to know that here in India, a lot of high school students with Olympiad background are reading your Analysis books on their own as the two books are easily available.However, I am sorry to say that other books written by you are not really easily available;their imported editions need to be ordered at very high prices.Is there a way to address the problem?

14 May, 2012 at 8:40 am

shubhodip@Sabyasachi Mukherjee: This is not quite the case; for example ”Additive Combinatorics” is easily available. May be Analysis I and II are *more* easily available here because it is published by Hindustan Book Agency.

24 June, 2012 at 11:59 pm

abellongI’m studying Chapter 3: Set theory. I have some question: Axioms 3.1 says a set is also an object. So, is there anything that is not an object? I found no definitions of object there. Isn’t it necessary to define it first?

25 June, 2012 at 8:06 am

Terence TaoIn an axiomatic system, primitive concepts (in this case, the concept of a set and the concept of an object) do not need to be formally defined before the axioms are stated; see Remark 2.1.14. But one can give informal definitions. In this case, an object is anything which can be manipulated mathematically (which means, among other things, that they are subject to the axioms of equality, see Section A.7), and in particular can be part of a set (this is formalised in Axiom 3.3). Axiom 3.1 says that sets are a type of object, while Axiom 3.7 says that numbers are a type of object as well. However, the axiom system also permits other types of objects, and to some extent it is a matter of taste and convenience which objects one wishes to use in one’s set theory (see Remark 3.1.3). For instance, if one wishes to apply set theory to Euclidean geometry, then it may make sense to consider points and lines as primitive objects (this is very subtly different from the approach in Cartesian geometry, in which points are viewed as elements of and lines are viewed as subsets of , though for most purposes the two viewpoints are equivalent).

26 June, 2012 at 2:55 pm

Jack“primitive concepts do not need to be formally defined ‘before’ the axioms are stated”. Can I understand your words as “primitive concepts do not need to be formally defined”? After all, I can’t find a “formal” definition of “object” in the book but only the informal one, which has been mentioned in your answer. It seems that “object” is never defined formally at all.

I remember that you said in your linear algebra notes that for a mathematical concept (linear space in that notes), what is important is not “what it is” but “what it can do”. Can I say this is the same for the concept of set here?

26 June, 2012 at 3:14 pm

Terence TaoYes, if one is working in a purely axiomatic fashion, then the primitive concepts in one’s axiom system never need to be defined. (One can of course introduce definitions for non-primitive concepts in terms of primitive ones, though.)

Once the axiom system is in place, though, one can also start constructing

modelsof that system, by providing interpretations of the various primitive objects and operations of that system. For instance, one can construct a model of Peano arithmetic inside set theory by constructing a set to be designated as the natural numbers, and also constructing a successor operation on that set, and then verifying that all the axioms of Peano arithmetic are satisfied for this model. Note though that it is certainly possible for a single axiomatic system to have multiple models. (For instance, each group is a model for the axioms of group theory, and there are of course an infinite number of groups out there, which are typically not isomorphic to each other.)28 June, 2012 at 1:02 am

R-S integralDear Mr. Tao,

I saw two definitions of R-S integral,one kind is in your book and Rudin’s book,where is an inreasing function on ,the other kind is R-S integral in Apostol’s book,where there is no limitation to .I don’t know which definition I should follow,I can’t learn both because my time is very limited(I can’t finish my study task at school,and nearly drop out).Thanks for any piece of advice!

—-A sad Chinese undergraduate

7 July, 2012 at 6:36 am

AnonymousDear Dr Tao,

While I understand problem solving(in the conventional sense of the term) is not the real focus of Analysis I and II, it appears that if you had included some additional ,harder, exercises at the back of your books (a collection of say, 20-25 problems each) it would have been probably even better to those who are self-studying give one a sense of what one has managed to learn.I hope at some point you think of including something of that nature.Thank you.

9 July, 2012 at 7:03 am

AnonymousIn the proof of theorem 6.4.18 you set out to prove that every Cuachy sequence is convergent:

… Let a_n be a Cauchy sequence. We know from Cor. 6.1.17 that the sequence a_n is bounded…

But Cor. 6.1.17 states that every convergent sequence is bounded. At this point a_n might still be a non-convergent Cauchy sequence, so 6.1.17 doesn’t suffice to show that a_n is bounded.

9 July, 2012 at 8:46 am

Terence TaoOops, one should be using Lemma 5.1.15 here (or more precisely, its counterpart for reals instead of rationals) instead of Corollary 6.1.17.

13 July, 2012 at 6:47 am

ckDear Prof Tao,

Here are some misprints that I found in chapter 1 (2ed).

P2. Line 12: “can you add” should be “Can you add”

P9. Line 5: But the right-hand side is , delete “the right-hand side is”

P10. Line 2 in the formula, interchange x and y.

Lastly, in example 1.2.4, I suggest using the change of variable instead of . Then one doesn’t need to concern about the positive and negative infinities.

28 September, 2012 at 5:13 pm

Vineet NairDear Prof. Tao,

Analysis I, 2nd ed., pg. 24, 6th line from Def. 2.2.1. It says “From our discussion of recursion in the previous section we see that we have defined n + m for every integer n.” Shouldn’t it be “every natural number n”?

[Correction added, thanks – T.]8 October, 2012 at 5:58 am

Luqing YeDear prof.Tao,

I finally finished learning the construction of the real number system with the help of this very rigorous book.But a big problem left:What is a real number?In your book you say a real number is , is a Cauchy sequence of rational numbers.But I think is nothing,it is just a notion which somewhat relates to .A few sections later,you prove that is actually .But I think the definition of comes from ,so one should not explain by using .So what is a real number still remains a big problem for me……

8 October, 2012 at 8:55 am

Luqing YeI search google and find your google buzz essay https://profiles.google.com/114134834346472219368/buzz/RarPutThCJv .After seeing that essay I understand all……My only doubt is that why you didn’t define a real number as a equivalence class of Cauchy sequence of rational numbers in your text book.

8 October, 2012 at 9:09 am

Luqing YeThat is only a step away…..It seems that you replace the equivalence class by the strange notion .This makes me very uncomfortable because this notion seems meaningless,though it would eventually be replaced by ,but I think that is a “fake replace”……Because the concept of is based on .So a step away makes your text book introduce me a new object .I don’t like new objects.New objects,the fewer,the better.

8 October, 2012 at 12:04 pm

Terence TaoDear Luqing,

I think you are approaching mathematical foundations from a constructive viewpoint (focusing on what objects such as real numbers actually “are”) rather than from an axiomatic one (focusing on what

propertiesthese objects have). The distinction between the two perspectives is discussed in Remark 2.1.14 of my book. While the constructive viewpoint is initially more appealing conceptually, and is certainly of importance in foundations of mathematics, it turns out in thepracticeof mathematics that the axiomatic approach is much more flexible and powerful. In the end, once one leaves the foundational or logical aspects of mathematics, it doesn’t matter so much exactly which construction of the real numbers one takes as a model (whether it be equivalence classes of Cauchy sequences of rationals, formal limits of the same Cauchy sequences, Dedekind cuts of rationals, or whatever), so long as one can verify that this model obeys the basic axioms of the real numbers (e.g. the list in http://en.wikipedia.org/wiki/Real_number#Axiomatic_approach ). In particular, one is free to choose between a minimalist construction in a pure set theory in which one only works with constructions (such as tuples and equivalence classes) that were already constructed within the language of set theory, or a richer construction using an impure extension of set theory in which one adds additional formal symbols such as LIM. Both choices are equally valid for most mathematical purposes (the impure set theory with additional formal symbols is a conservative extension of pure set theory); I chose the latter for my book because it conceptually aligns the construction of the real numbers more closely with the way one usuallythinksof real numbers in practice – namely, as quantities that can be approximated to arbitrary precision by rationals (or terminating decimals). (I also discuss the distinction between minimalist approaches to foundations, and rich approaches to foundations, at http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/30231#30231 .)Incidentally, I adopt a similar approach in previous sections of my text in constructing the integers out of formal differences of natural numbers, and rationals out of formal quotients of integers. One could, if one wished, instead adopt a “minimalist” philosophy and construct the integers as equivalence classes of pairs of natural numbers, and the rationals as equivalence classes of pairs of rationals; this is a logically equivalent approach, but is further removed from one’s conceptual intuition about what integers or rationals should actually be.

8 October, 2012 at 2:39 pm

GilYoung Cheong@Prof. Tao: “It turns out in the practice of mathematics that the axiomatic approach is much more flexible and powerful” — is there any way to see why it’s “more” powerful? I always wonder if one write down several axioms for a a structure (or model) if we have to give an example to provide existence of such structure. But then again, as you point out in your lecture note, we do not really have “THE” foundation of mathematics (though I am just taking words from other people on this claim). As a student, I would like to know how powerful axiomatic approaches are, as it may affect future course choices, although it may sound like a too greedy question.

8 October, 2012 at 7:56 pm

Luqing YeThanks for your reply.Now I understand……That’s different philosophy.In your text book you construct real numbers,rational numbers,integers ,from a not minimalist approach,that is ,object oriented constructing ……In the past I regarded such approach as ugly approach….now I know….Thanks.

8 October, 2012 at 11:08 am

GilYoung CheongI don’t know if my comment can help, but I like how Prof. Tao introduces real numbers, so let me comment in case you want to read what I think.

The objects are the equivalence classes of the rational Cauchy sequences and he defines arithmetic on these objects inheriting the arithmetic of rational number that he has already created (this notion must be clear given that you know how to construct rationals from integers).

After the construction of these objects (which we call real numbers) with arithmetic (ordered field structure). And in these objects we get isomorphic copy of rational numbers, which does not hurt to be assume as “THE” rational numbers.

BUT, this is not the end unlike rationals!

He then explains how you get the least-upper-bound property (equivalently the greatest-lower-bound property), which is not seen in rationals but plays an important role in calculus (or even in elementary algebra and geometry). See how he proves it and copy it down with your own notations; there is a lot to learn in the proof.

(As far as I remember, Archimedean property was used and this simple property is REALLY important to understand what real numbers are. I might be wrong though.)

Finally, on the question on , note that both are actually real number although it is not clear the left-hand side exists. But the left-hand side should exist due to the completeness of real numbers and Prof. Tao gives a nice proof about in his note (it is a really fun proof too if you visualize what the objects in the proof do). Note that the proof is not very trivial despite the simplicity of statement.

Although I haven’t gone through details, it gave me a bit of intuition on the reason that the least-upper-bound property and the completeness are equivalent. It is a nice thing to think about when you (or I) have extra time for long and constructive proofs. The citation on the fact follows.

http://en.wikipedia.org/wiki/Completeness_of_the_real_numbers

What are “the” real numbers? : real numbers are any ordered field that contains rational numbers as a subfield and has the least-upper-bound property. It captures the idea of continuum (from the LUB property, or completeness) and we can prove the existence of them as we can construct it, so we can use it. Although I have not gone through the proof yet, it is a fact that however you construct real numbers, they are “essentially same.” (Another good thing to do when you (or I) have time.)

It is shocking that numbers can be understood in this perspective of sets, but after all, they are what we think they are (i.e., they have properties that we want). If I were you, I would not worry too much about it although I would keep thinking about it.

Why do we bother? : Well, at some point, you really want to deal with delicate part of functions defined on real numbers. Often, understanding very little details plays an important role on dealing those matters. (Due to lack of my expertise, I cannot come up with good examples.)

Hope this helps.

8 October, 2012 at 8:21 pm

Luqing YeIn the second paragraph of your comment,you say that “The objects are the equivalence classes of the rational Cauchy sequences “,but I don’t think prof.Tao pointed out this in his book.I think Prof.Tao thinks is just no more than a new object.We don’t care about what is ,we only need to know that it has all the property of a real number.

8 October, 2012 at 9:29 pm

GilYoung CheongThanks for the correction.

I should have said “basically equivalence classes” rather than saying they are “the” equivalence classes. When I wrote my own note using his lecture note, I used the equivalence classes and that’s why I was thinking that what he wrote were equivalence classes.

But I still think that formal limits are nothing more than representatives of equivalence classes. If you read the note, you recognize that each LIM(a_n) ties all the Cauchy sequences that is related to (a_n).

I still don’t see what is so different or more about it. I guess this is basically the question that I asked him (Prof Tao) that how could this be “more powerful.” Maybe it might be difficult for me to see it only with elementary things that I am used to, but it is very tough to imagine. Hope this can provide me (or any others) more motivations to study logic.

8 October, 2012 at 9:33 pm

GilYoung CheongThe part – “Why do we bother?” should not be considered to be of importance especially all we need are axiomatic properties. I guess I bothered because I thought the construction was nice.

14 October, 2012 at 1:33 am

Matthis LehmkühlerDear Prof. Tao,

In the proof of Proposition 11.7.1. (p. 290 in Second Edition, Hardcover) there is a little mistake in the formula because logically can’t be equal to as an interval can’t be equal to a number. So one simply has to replace by (see Theorem 11.1.13).

[Corrected, thanks – T.]1 January, 2013 at 3:37 pm

JackThe term “paradox” is used in this book and appears in many notes in this blog (e.g. Russell’s paradox and the Banach-Tarski paradox are mentioned lots of times). I’m a little confused about this term. According to [this article in wikipedia](http://en.wikipedia.org/wiki/Paradox#Logical_paradox), it has different meanings in different context. In this book, it seems to be used as “logical contradiction”, while in other places, it means something not true or counter-intuitive (e.g. in the preface). How should one understand which meaning is used?

1 January, 2013 at 6:19 pm

Terence TaoFrom context. (Not every word in a mathematical discussion is a formally defined technical term.)

24 February, 2013 at 12:26 am

Luqing YeDear Prof.Tao,

In this book you denote the successor of a natural number by ,indeference to modern computer language.

I only know C++ has this feature.Is this language C++?I learn C++ for four months when I am a freshman only because this book mentions ++…

Or are there any other language have this feature?

It seems that you know everything…..This makes me very amazing…

24 February, 2013 at 6:47 am

Terence Taohttp://en.wikipedia.org/wiki/Increment_and_decrement_operators

3 March, 2013 at 8:10 pm

MeWould anyone here be willing to sell–or donate or share–their copies of the second editions of Analysis I and Analysis II textbooks? I can only check them out on reserve at my university’s library, so I only get them for so long.

3 March, 2013 at 9:17 pm

Share bookAnalysis I http://ishare.iask.sina.com.cn/f/34822682.html?sudaref=www.baidu.com&retcode=0

Analysis II http://ishare.iask.sina.com.cn/f/34822684.html

Only for non-commercial,personal use.I share,I happy.Hope nobody blame me.

14 March, 2013 at 12:30 pm

HolgerDear Prof. Tao and readers,

I’m a happy student of your two volume Analysis series.

They are a great resource of independently doing math on one’s own.

I’d like to ask a question about exercise 5.4.3:

“Show that for every real number x there is exactly one integer N such that

N <= x < N + 1."

I don't want a spoiler, just a light hint, if I'm thinking in the right

direction. :-)

I could think of a proof by constructing a Cauchy sequence which is

equivalent to x whose elements are all less than 1-close, and going from there.

e.g. using that such a n exists for every rational number.

But is it really necessary to work on such a low level?

Something tells me (and maybe it's kidding me), that it should be enough to use

the archimedian property together with some of the ordered field facts.

I also tried a proof by Contradiction, but assuming

truth of the negated statement doesn't bring something nice to work with either.

Thanks for the great passion with which you teach math and motivate newcomers

like me,

Holger

14 March, 2013 at 4:43 pm

Luqing YeProve it by induction and contradiction.Firstly,the uniqueness of such N is easy to prove.We just need to prove existence.If such N does not exists,then for a natural number ,we have (why?),so ,so ,…,so by induction ,for all natrual number ,,this is absurd.

15 March, 2013 at 12:45 am

HolgerHi Luqing Ye,

thanks for your reply, but …

I said NO spoiler! ;-)

Now it’s trivial (the implication trick from the proof that sqrt 2 not being

rational + the archimedian property).

Thanks for the big hint,

Holger

15 March, 2013 at 2:11 am

Luqing YeDear Holger,

It seems my hint is too big…

When faced with such a problem,I just draw a number line and think “this property is obvious”.But this book requirs a rigourous proof,so I have to write a formal proof down.

Although we need a formal proof,but we should also make use of the vivid picture,because it helps us to prove,a vivid picture lead us to a rigourous proof,this vivid picture is your “right direction”.

15 March, 2013 at 4:28 am

HolgerDear Luqing Ye

you are absolutely right.

And in fact, I started out the same, imagining the real line with x and the integers around it.

That’s how I initially came to the idea of proving it by contradiction. (I didn’t go far enough though, as I just tried negating the whole claim).

I proved it rigorously right after your post, as it taught me an important lesson:

One has to introduce quantified variables in the right order, but besides that

they are independent of each other.

(Where you introduced x and only then went to contradict the \exists n)

So, thanks alot for sharing the lesson!

Holger

24 April, 2013 at 12:34 pm

NastassjaDear Professor Tao,

I’m an incoming student at UCLA preparing with Analysis I. I have a question on Exercise 5.4.8, proving that if , then .

Towards I contradiction I suppose . Following your hint, I know there is a rational such that . Writing , the inequalities imply , so for some , since the nonnegative reals are closed. Similarly, , so for some . Then for some and .

I'm not sure how to reach a contradiction from this as suggested in the hint. I'd appreciate any elaboration. Thanks.

-Nastassja

25 April, 2013 at 12:24 pm

Terence TaoThe assertions and are not merely true for some and , but are true for all sufficiently large and . Also, it may be helpful to use two rationals rather than just one, to give yourself a bit more room.

25 April, 2013 at 8:05 pm

NastassjaDear Professor Tao,

Thank you for taking the time to answer my question.

29 May, 2013 at 9:22 am

HolgerDear Nastassja and Prof. Tao,

after your question & response I’m unsure about my own approach. :-)

As the claim is a implication, I used contraposition. This gave me a simpler consequent to prove (I think).

Just to make sure my reasoning isn’t too garbage, this are the “milestones” I came across:

The contraposition of the claim is

.

Then I found a q between and .

If one reads the proposition of the non-negative reals are closed as a contraposition, then the for which pops right out of it.

Somehow I feel I’m still missing something.

Thanks for reading,

Holger

8 September, 2013 at 2:12 am

IdnI did just that, and I think it’s correct. I hope that if it’s not, Prof Tao could spare some time to suggest any errors we might have.

29 June, 2013 at 6:22 am

HolgerDear Prof. Tao and peers,

I’m trying to prove Lemma 5.6.6 (a) just from the points that are mentioned in the hints (and preceding text).

The bit which drives the square -root of 2 proof seems to cry for the binomial theorem in the n-th root case. But as the theorem isn’t even mentioned in the text, I don’t want to use it.

When I can’t go from upwards I thought I might go downwards from

to

depending on .

But the only thing I manage to find is

for an , which I try to get smaller without success since about a week. :-)

Am I totally off track?

Is my assumption correct, that I should be able to find the key idea more or less directly from the proposition on bouding of reals by rationals?

Thanks for reading,

Holger

29 June, 2013 at 7:12 am

Luqing YeDear Holger,

I don’t know whether you are totally off track or not,I just give you a tiny hint,this is what I did when I was proving lemma 5.6.6(a):

Combine with the definition of ,if ,Then find a such that ,where .so this will contradict the definition of .I suggest that you do not find exactly ,just prove that such exsit will be enough.

I have to say this hint is not detailed enough,there is still some technical detail,I suggest you notice the formula:

$\displaystyle latex a^n-b^n=(a-b)\cdot \Delta$,where is a complex stuff.

29 June, 2013 at 8:48 am

HolgerDear Luqing Ye,

thanks very much for your answer!

The hint is small enough this time, as I still have to think about it. ;-)

As I still have a slight hope that I can do the proof without a big sum, I’ll check whether I can infer properties for from being positive.

But if you as a much more experienced math student than me had to pull out a I think I can abandon the idea of finding a simpler proof.

Thank you once again,

Holger

29 June, 2013 at 9:01 am

Luqing YeAlthough in this book it seems that Mr.Tao tries to avoid big sum and construct every thing from the foundation, in general I do not avoid big sum because I know these big sum can be proved by mathematical induction.Mathematical induction itself is from simple to complex.

BTW,my proof is not likely to be unique or elegant,think freely yourself…

29 June, 2013 at 9:58 am

HolgerHi Luqing Ye,

you are probably right.

But one of the things I like most about Prof. Taos Analysis Text is that it’s completely non-circular.

In this faith I take it as warning sign when I feel I have to invoke a big result from somewhere else or when my proof would be end up more advanced/complex than the general tone of the current chapter.

Holger

29 June, 2013 at 8:08 pm

Luqing YeBut those things such as binomial formula or are not big result.They depend on very minimal foundations such as the four arithmetic operations of real numbers ,they looks complex only because they use mathematical induction.So use them will not result in circular augument.

24 September, 2013 at 5:09 pm

José Antonio Lara BenítezHi everyone is my first time here and I wondering about this lemma. I know that the easier way to prove is by the binomial theorem which is not difficult at all but maybe there is a more fundamental or simple way to do it. My question is, with respect to the lemma 5.6.6 is there a way to do it without the extra effort to prove the binomial theorem (which is mentioned until the chapter 7) ? Thanks in advance. I would really appreciate any help.

22 July, 2013 at 9:20 pm

Giancarlo Mantero Astroza.Hi everyone,

i’m really confused about the equality axioms for the natural numbers and his operations, cause in all the others types of numbers (integers, rationals and reals), the equality axioms are verified, and also for his operations (well-defined), but in natural numbers there’s no any mention about this, for example what happen with: “if a=b, then a++=b++” ? (the operation ‘++’ is well-defined ?), or “a=a”, “if a=b, then b=a”, “if a=b and b=c, then a=c”, “if ‘a’

is a natural number and a=b, then ‘b’ is a natural number” ? (natural numbers obey the equality axioms ?). Someone can help me plis ?, i’m very frustrated with this.

Thanks in advance!

PD: maybe we need to create a forum for discuss the book lol, it would be great.

Giancarlo.

22 July, 2013 at 10:29 pm

Luqing YeDear Giancarlo,

I do have some sympathy with your confusion,cause when I was learning the first several chapters of this book,my brain is also full of all kinds of such problems as “well defined”,”the equality axioms” …

Here is my opinion on your question:

As a matter of fact,the theory of natural number can be developed based on the theory of sets,i.e,every natural number can be defined constructively as ordinals,which is a special type of sets.But it appears that the author tries to avoid such definition and instead takes an axiomatic view point,i.e,the author defines natural number axiomaticly,using Peano’s five axioms to describe the property of natural numbers.

So the auther regard the theory of natural numbers as a foundation,and all the other theories such as integer theory,rational number theory,real number theory are based on the natural number theory.

So in such context,I think there is no need to verify the four equality axioms(reflecxive,symetry,transitive,substitution),as they are all trival,i.e,these properties can be verifed directly from Peano’s five axioms.

However,let’s imagine,if the author makes the theory of natural numbers based on the theory of sets,i.e,if the author define 0 as ,1 as ,…,then the verification of the well-definedness of the equality of natural numbers,and the ++ operation becomes a must ,so as to ensure ourselves that these definition do not conflict the rules of set theory.

Maybe my opinion is naive,I appreciate any comments.

23 July, 2013 at 1:43 am

HolgerBuenas Giancarlo:

I too was very worried about equality between natural numbers for a while. It subsided when I found out, that in formal logic two terms are assumed to be equal if they are syntactically (structually) the same.

i.e. foo = foo, because it’s the same string.

It turns out, as I just discovered, that the reflexivity, symmetry and transitivity axioms are indeed part of the peano axioms for natural numbers:

http://en.wikipedia.org/wiki/Peano_axioms#The_axioms

Just in case not only the natural numbers are new to you, but doing proofs in general, the thing that really boost my understanding about the axiomatic approach and how to do proofs, was proving most of chapter 2 in coq

(http://coq.inria.fr/).

The instant feedback was very instructive, especially when one is plagued with concerns about the validity of each single step of ones proofs. :-)

Holger

PS: Yeah, an analysis I and II fan forum would be great!

10 August, 2013 at 4:47 am

Luqing YeDear Prof.Tao,

I think in the proof of proposition 7.1.8,

should be .

I found this typo because someone told me that he could’t understand this formula…

[Erratum added, thanks – T.]10 August, 2013 at 5:32 am

Luqing YeIn addition, I think “almost every” in this proof should be replaced by .

Because is in ,but what we can actually add are real numbers,that is ,.

27 August, 2013 at 6:03 pm

AnonymousIn the language of set theory, how should one write the set of all real numbers in Definition 5.3.1? seems incorrect since those which are equivalent are identified as one real number.

27 August, 2013 at 9:13 pm

Terence Taohttp://en.wikipedia.org/wiki/Quotient_space

28 August, 2013 at 6:31 am

JackI think you are talking about quotient sets (http://www.proofwiki.org/wiki/Definition:Quotient_Set) since there is no topological structure (yet) here?

28 August, 2013 at 9:03 am

AnonymousBy an algebraic approach, one can also construct using rational numbers: . What’s the fundamental difference between and the constructed by the equivalent rational Cauchy sequences? Both of them satisfies the equation . But it seems no hope to define limits in the world of .

20 September, 2013 at 4:51 am

AnonymousIn Exercise 8.1.10, I don’t fully understand what “explicitly” mean. For example, if one gives a bijection , can be counted as an answer. Does one have to give a “formula” for ?

Let . Then one can order this set so that the elements in it is strictly increasing. Then we have a bijection such that $\latex \phi$ is increasing. Can such be counted as an “explicit” function?

22 September, 2013 at 7:38 am

Terence TaoThe term ‘explicit’ is not completely precise in mathematics, but to me at least, I would say that an object has been explicitly constructed if the recipe given for is unambiguous (in particular, different mathematicians following the recipe should create the same object at the end) and clearly gives an object of the correct class (in this case, a function from to ). So using an inverse of an explicit bijection would be considered an explicit function, so long as it has been demonstrated (as opposed to merely asserted) that is indeed a bijection.

21 September, 2013 at 9:08 pm

CkI borrowed this book from library and learned a lot from it. Eventually I decided to invest my money on this book. The content is best, but the publisher is worst. The binding of the book I received is so tight that it makes me hard to read because it is hard to open the book flat without damaging the binding. And I even found a large hole on one page. I don’t understand why Prof Tao chose to publish with this publisher. If there will be a third edition in the future, publish it with AMS, like most of your books, or even better, publish with Springer.

22 September, 2013 at 9:05 am

CkDear Prof Tao,

I found the following errata from your book (2nd edition).

Page 15 middle: Guiseppe Peano should be Giuseppe Peano.

Page 37 Example 3.1.10: so is singleton set is a set should be so is singleton set a set.

Page 54 Defintion 3.3.17: direct image f(X) has not yet been defined.

Moreover, I have some comments about chapter 3. Defintion 3.1.4 defines the equality of two sets by saying that two sets are equal iff they consist of the same elements. Then you mention in the paragraph after example 3.1.5 that from this defintion, one can verify that this notion of equality of sets satisfies the axioms of equality, in particular the axiom of substitution: if x belongs to A and A=B, then x belongs to B.

I read through the whole chapter 3 and thought very hard on this defintion. After that, I have the following difficulties.

Supopse x and y are objects and A is a set. If x=y, then x belongs to A iff y belongs to A. This is by axiom of substitution. Is this right? So I suppose that at the beginning of the theory, we have already had the notion of equality of objects as primitive concept and this notion of equality satisfies all the axioms of equality. Now by axiom 3.1, sets are objects, so we should have the notion of equality of sets before we did anything. What I am thinking is we don’t need to define the equality of two set and the claim “if x belongs to A and A=B, then x belongs to B” is a priori true by axiom of substitution for objects. One has this by default rather than need to verify this. Of course, to take this approach, one needs to postulate the axiom of extensionality in order to have another direction of implication of defintion 3.1.4. Am I correct?

Another question is about Remark 3.5.8 and Exercise 3.5.2, in which you used two different defintions for Cartesian product. In remark 3.5.8, you used the power set axiom to form a set and then used the axiom of spectification to restrict the set to the Cartesian product, the elements of which are functions with the same range (the union of all X_i) but not necessarily surjective. In exercise 3.5.2, you defined the Cartesian product the other way round by first defining an n-tuple as some surjective function and then defining the collection of all those surjective functions as Cartesian product. So are they set theoretically the same (clearly the elements are different) and which one defintion should we employ?

Finally, I would like to tell you that your analysis I and II are really great. I think they are even better than baby Rudin. I have no difficulties with rudin’s book. Reading Rudin’s book, I can understand all his proofs, but reading your book and following your hints, I can write down the proofs myself, which makes me feel that I really understand the theorems.

23 September, 2013 at 2:08 pm

Terence TaoThanks for the corrections and comments!

Regarding equality, what is going on here is that I am allowing the ability to

redefinebasic concepts such as equality as part of the process of building foundations; this is commonly used to identify objects which were previously not considered equal. For instance, the natural number is identified with the integer , the integer is identified with the rational number , and so forth. One can strengthen the notion of equality at any time, provided that one checks that the axioms of equality remain satisfied when doing so. So the way the axioms of set theory are set up in the book, it may be initially that one could have two distinct sets with the same elements, e.g. {1,2,3} and {1,3,2}, but they become identified with each other once Definition 3.1.4 is in place. Of course, as you say, one can also go down the route of adding the axiom of extensionality to the axioms of set theory, which also works.Regarding Remark 3.5.8, there is an additional step missing in that remark, namely one has to take each of the functions with the stated property and then restrict the range of each of these functions to make them surjective (using the axiom of replacement). I’ll add an erratum to this effect.

23 September, 2013 at 9:31 pm

CkProf Tao, thanks for your reply and clarifications!

7 October, 2013 at 9:24 pm

AmyAt the bottom of page 26, there is a note about Proposition 2.1.16.

It says the definition of function will not be circular, as the concept of a function does not require the Peano axioms.

￼ Proposition 2.1.16. is Recursive definitions.

But I think even if the concept of a function requires the Peano axioms, this will not be circular. Because firstly we define a natural number, and then we use Peano axioms is construct a function. Finally we proof the Proposition 2.1.16. The clue is just like: Peano axiom-> natural numer; Peano axiom-> function; natural number & function-> Proposition 2.1.16. Thus, no circle would occur.

29 October, 2013 at 3:01 pm

AnonymousThe last inequality in Lemma 7.5.2 says that

.

Can one actually come up with an example of the strict inequality?

31 October, 2013 at 6:17 am

JackThe whole two volumes only mention “improper integral” in Remark 11.3.6.

When I learn your book for 245ABC, I don’t see this topic either. Is it because the improper integral can be completely done by the Lebesgue integration theory or is it of “not much” importance in analysis (compared to those topics covered in the book)?

31 October, 2013 at 7:12 am

Terence TaoSee Remark 1.2.18 of “An introduction to measure theory”.

11 November, 2013 at 1:10 am

AnonymousI’m a bit perplexed about the definition of suprema in the extended reals from the course notes from your UCLA webpage. I’m not sure if these definitions appear as given below in the text.

—

Definition. Let be a subset of , and be a real number. We say that is an upper bound for , iff we have for every element in .

Definition. Let be a subset of , and be a real number. We say that is a least upper bound for iff (a) is an upper bound for , and also (b) any other upper bound for must be larger than or equal to .

Definition. Let be a subset of , and be a real number. Then we define the supremum sup() or least upper bound of by the following rule.

(a) If is contained in (i.e. and are not elements of ), then we let sup() be defined as in Week 2 notes.

(b) If contains , then we set sup() .

(c) If does not contain but does contain , then we set sup() sup() (which is a subset of and so falls under case (a)).

—

Are we assuming here that in the case (a), the of the definitions of upper bounds and least upper bounds is not restricted to ? Otherwise, it seems that the given definition would fail to assign suprema to empty set as well as nonempty subsets of which are unbounded above in (but which do not contain ).

Moreover, it seems that a slight modification of the original definitions (wherein we let and let ) would suffice without the need to separate the definition into cases. Am I overlooking something?

11 November, 2013 at 8:32 am

Terence TaoYes, in case (a), the supremum or infimum can be (this is done explicitly in Definition 5.5.10 of the text, and I think is also done in the original lecture notes). The alternate definition of the supremum you mention is equivalent to the existing one; this is Theorem 6.2.11 (or Exercise 6.2.2) in the text, and I think it is also in the original notes.

11 November, 2013 at 12:21 pm

AnonymousWow, thanks for the quick response. The original notes actually did not do this, but after typing up the previous post and seeing the definitions side-by-side, I realized that was probably the situation.

19 November, 2013 at 4:01 pm

AnonymousShouldn’t Definition 9.3.6 (convergence of functions at a point) consider whether the restriction of to (rather than itself) is -close to near for the rephrasing that follows to be equivalent?

[Correction added, thanks – T.]19 November, 2013 at 7:51 pm

AnonymousOh, wait. I think I might have somehow overlooked the errata listing for this definition (9.3.6) from the second edition, which likely means the error is not present in the third. Apologies if this is the case.

[No problem, I’ve removed the duplicate erratum – T.]6 December, 2013 at 12:37 pm

AnonymousWhen I had tried to understand how Axiom 2.5 limits the set of natural numbers with the elements 0,1,2… , I did the following proposition, it may be helpful i think,

Proposition:

1) x≠0

2) x≠n then x≠n++

If both 1 and 2 are true, then x isn’t a natural number.

Proof: Suppose to the contrary that 1 and 2 are true for an x=A, and A is a natural number. This means the property P(x)=(1 and 2)(x) is true for x=A. Now consider the property P'(x)=(1 and 2)'(x)=(1′ or 2′)(x).

P'(x)= [(x=0) or (there is an n such that x≠n and x=n++)]

P'(0) is true. (0=0)

Suppose that P'(x) is true. Then x=0 or (x≠n and x=n++ for some n)

a)If x=0 then x++=0++≠0 so (x++≠0 and x++=0++) is true. Then for this case, P'(x++) is true.

b)If x≠n and x=n++ for some n, then x++≠n++ and x++=(n++)++

Then P'(x++) is also true for this case.

Hence P'(x) is true for all natural numbers x by Axiom 2.5 Then since by our assumption A is a natural number, P'(A) should be true. But we know that P(A) is true. Then both P(A) and P'(A) should be true, which is a contradiction.

15 December, 2013 at 9:28 am

AnonymousDear Prof. Dr. Tao

I didn’t understand the following two points:

1)On Axiom 3.8 (universal specification), we assume that for each x, P(x) is either true or false. (thus, not both.) Then since the property P(x)=”x is an element of x” does not satisfy this (it is both true and false), then this means that Axiom 3.8 doesn’t speak anything about P(x). Then how P(x)=”x is an element of x” can contradict with Axiom 3.8?

2) Let x is an object but not a set. Let A={x , A}. Then how A contradicts with regularity axiom?

Thanks for all your helps.

15 December, 2013 at 2:57 pm

Terence Tao1. For any object x and set A, the statement

“ is an element of ” is going to be either true or false (but not both); since sets are objects, this means in particular that for any set x, “ is an element of ” is also going to be true or false (but not both). (Actually, assuming the axiom of regularity, this statement will always be false.)

Assuming the axiom of universal specification, one can run the Russell paradox argument to construct a sentence of the form “ is an element of ” which is simultaneously true and false, which contradicts what I just said; this is why the axiom of universal specification cannot be assumed.

2. A itself does not directly contradict the axiom of regularity, but the set {A} does.

8 January, 2014 at 12:44 am

José Antonio Lara BenítezDear Dr. Tao: I think in the exercise 3.6.8, the set A should be nonempty.

[Correction added, thanks – T.]11 January, 2014 at 3:11 am

Frieder SimonDear Prof. Tao,

there seems to be a slight inconsistency between the definitions (5.5.10 and 6.2.6) for suprema of sets in resp. . This amounts to whether a definition is [case A] a “metamathematical” one, i.e. giving a new name to a mathematical statement (which is an assertion *about* mathematical objects) or [case B] a “true, mathematical” one, i.e. giving a new name to an mathematical object itself. I know that this is hair-splitting and apologize in advance, but wanted to point it out in case other readers were wondering the same.

In Definition 5.5.10 the supremum equals if the sets were non-empty and had no upper bound. In Remark 5.5.11 it was emphasized that was just an abbreviation for the statement that had no upper bound, since is a meaningless symbol, so it is of category [A]; then Definition 6.2.6 was anticipated, where this symbol would have some meaning attached – so in light of this Remark one would expect that in Definition 6.2.6 isn’t an abbreviation for some mathematical statement any more, but a “true” mathematical definition, of category [B], since is now a mathematical object. But that definition doesn’t live up to this expectation, since the case for sets with no upper bound is reduced to Definition 5.5.10, so is again of category [A]. Thus we essentially have two meanings for : As a meaningsless symbol, that doesn’t exist in it’s own right and serves only to indicate that abbreviates something and as an object that has a mathematical existence.

11 January, 2014 at 9:03 am

Terence TaoAt the point in the text when Definition 5.5.10 are introduced, and are interpreted as “

at presentmeaningless” symbols (as remarked in Remark 5.5.11), because they have notnecessarilyacquired the status of a mathematical object (so that they can be used to form sets, be compared with other objects via the equality symbol, and so forth.) However, once the extended real line is introduced in Definition 6.2.1, these symbols acquire additional meaning; they are now confirmed to have the status of a mathematical object, and set-theoretic operations and the equality operation may be applied to them.To put it another way, at the time Definition 5.5.10 is introduced, it is left unspecified whether the statement is a purely syntactical one, or is defining as a mathematical operation, because this specification does not need to be made explicit until Section 6.2. For any of the discussion up to that point, one is free to interpret the definition either way (but the text itself is agnostic on this issue). But from Section 6.2 onwards, and are confirmed to be mathematical objects, and so the latter interpretation is in force from that point.

13 January, 2014 at 6:26 pm

José Antonio Lara BenítezDear professor Terence Tao, first of all let me give you a congratulation for your extraordinary book. On the other hand, I think in the exercise 8.5.16 instead of should be . I have also trouble with the last part.

Show that the maximal ordering of are precisely the total ordering. Using Zorn’s lemma show that given any partial ordering of there exists a total ordering such that is coarser than

We have to show that the set is non-empty. If then where denote the empty relation. In case that , then it contains relation.

In either case . Now let be a totally ordered subset of we have to show that there is an upper bound. Let be defined as follows, we say that for such that .

We have to show is well-defined. Suppose and there exists such that and but . Since both are in either or . Without loss of generality suppose that , then implies and since so which contradicts our assumption that . Hence is well-defined. Also is not difficult to show that is a partial ordering of .

Clearly is an upper bound for because whenever for would imply , which means . Thus by Zorn's lemma contains at least one maximal element which we call . We claim that is a total ordering of . Suppose for the sake of contradiction that is not a total ordering of . Thus there must be a pair for which neither nor . Thus we can define for all and we set . Thus is a partial ordering of , i.e., and also , note that the implication is vacuously true, which is a contradiction. Hence is a total ordering of .

I hope you could have time to answer it. If we create the maximal ordering as above what guarantee us that any given is comparable to ? in the sense of . Thanks for all your helps.

13 January, 2014 at 6:48 pm

Terence TaoThanks for the correction! Regarding the last part of the question, the trick is not to apply Zorn’s lemma to the entire class P (which, as you noted, may give a “useless” total ordering), but to instead apply that lemma to just the subclass consisting of those orderings that are coarser than .

13 January, 2014 at 8:49 pm

José Antonio Lara BenítezThanks for your help Dr. Tao. But is not necessary that the subclass also contain the orderings for which is coarser than? I mean, if we suppose as given. Define the subclass as the collection of all such that either or .

3 February, 2014 at 11:14 pm

José Antonio Lara BenítezDear professor Terence Tao ; I think in the exercise 9.1.15 it should be add it that and also in the example 9.3.21 should be instead of . Similarly for .

Jose

[Erratum added, thanks – T.]4 February, 2014 at 5:52 pm

José Antonio Lara BenítezDear Dr. Tao, I also have serious problems to follow your hint of the proposition 9.4.11, I think one has to prove it by contradiction but the argument is not simple for the general case when . If you have time, I’d appreciate any elaboration. Thanks.

Jose

4 February, 2014 at 6:34 pm

Terence TaoTo deduce the general case of from the special case , use the law .

5 February, 2014 at 12:10 am

Giancarlo ManteroIs it possible to prove that if ‘x’ is an object, and ‘x = y’, then ‘y’ is an object too (to construct for example, the set {1}, instead the set {0++}) ?

7 February, 2014 at 7:14 am

Motorel MaI think this proposition don’t need a proof ,because only between the same kind of objects can we define “=”.

7 February, 2014 at 7:36 am

Motorel MaDear professor tao,

I am an Chinese student and reading this book. My English is not very well. I want to ask you a question about the natural number’s addition. Because the equality of natural number obeys the substitution axiom ,then we can assert “if m=n, then m++=n++”. But the proposition “if m=n, then m+a=n+a” can be concluded not only by the substitution axiom but also by the principle of mathematical induction. Are the two proof effective? Is the substitution axiom always right about any operation of natural numbers as long as the operation is well-defined?

7 February, 2014 at 8:40 am

Luqing Ye老兄,

我认为对于自然数不必验证等号定义的合理性（也就是说那四条法则都不必验证），因为对于自然数来说，【相等】是一种最原始的符号而已，自动地满足了四条相等法则.

而对于整数，有理数等等，则需要验证等号定义的合理性。比方说，整数是用自然数来定义的，对于整数定义了整数相等的同时，我们当然默认了它必须符合整数相等的四条法则（等价关系+一条替换公理，在没定义整数的其它运算之前，替换公理还暂时不必验证），此时我们需要检查，对于整数相等的这四条法则有没有和整数的定义矛盾。

注意此时替换法则还没验证，因为我们还没有对于整数定义任何运算！然后，当我们定义整数加法这种运算的时候，我们需要验证整数相等的替换法则对于整数加法是否适用.

不知道我表达的有没有让你明白。欢迎指教。我英语也很烂，所以就不写英文了，反正咋俩都认识。你明白了的话，就用英语回复一句I understand now 吧.

7 February, 2014 at 8:57 am

Luqing YeHere is the English version.

Dear Motorel Ma,

I think there is no need to check that the four laws of equality must hold for natural numbers because to natural numbers,the notion of equality is no more than a symbol that follow the four laws AUTOMATICALLY.

But to integers and rational numbers,etc,we have to check whether the definition of equality is well defined.For example,integers are defined based on the natural numbers,so when we defined that two integers are equal,we must check that the four laws for the equality of integers do not conflict the definition of integers.Be aware that before we define any operation for integers,we do not need to check the substitution axiom.

After we defined the addition of integers,then it is the time to check THE SUBSTITUTION AXIOM FOR ADDTION OF INTEGERS,ie,if ,then ,etc.

Maybe I am wrong,please ask me in English when you do not understand me or have any disagreement.

8 February, 2014 at 5:48 am

Motorel Ma我明白你的意思啦。我是在想，对于自然数，它的相等满足那四条法则，那么对于命题“如果m=n，那么m+a=n+a”可以直接由替换公理得到，但同时也可以由数学归纳法证明得到。那么这两种途径是否都是有效的？我们在随便定义一种关于自然数的运算后，替换公理立刻成立吗？虽然我们只定义了加法乘法和指数运算。唉，叶兄啊，我英语太烂，连你都没看明白我的意思，那么陶哲轩肯定也看不明白啦。所以我干脆说中文好了，你帮我翻译吧。

8 February, 2014 at 10:00 am

Motorel MaHere is the English version

I know what you mean. I was thinking, for natural Numbers, it’s equality satisfy the four principles, then for the proposition “if m = n, m + a + a = n” can be directly by the substitution axiom, but at the same time, it can be obtained by mathematical induction proof. Then the two approaches are effective? We are literally defined after an operation on natural Numbers, substitution axiom hold immediately? Although we only defines the addition multiplication and exponentiation .

8 February, 2014 at 10:04 am

Motorel Ma“if m = n, m + a + a = n” should be “if m = n, m + a = n + a”.

8 February, 2014 at 10:35 am

Luqing YeFor natural numbers ,you prove by induction?I wander how you did that.You can provide me the detail next time when you reply.In my opinion,it is not needed,I think if you prove it by induction,that would be a fake proof.

Also,I think when we define any operation on natural numbers,there is

no need to check the substitution principle,for the reason we check

the substitution principle for operations of integers is that we want to check whether that operation of integers has any conflict with the definition of the equality of integers,but to natural numbers there is no such problem because the equality of two natural numbers is just a primitive notion,it do not based on something else.

对于自然数 你用归纳法证明 ?我想知道你是怎么做的.反正我觉得吧,这里根本不必用归纳法,即便用了也会是无效的.

我认为当我们随便定义了自然数的一种运算后,替换公理是可以直接用的.因为我们之所以要对整数之类的检验替换公理,是因为我们要检查定义在整数上的运算是不是和整数相等的定义矛盾.而对于自然数来说,其相等是一种原始概念,不必检验.

8 February, 2014 at 8:33 pm

Motorel Ma证明很简单啊，对a 进行归纳，基础情形m+0=m=n=n+0，假设对某个自然数a成立m+a=n+a，则m+(a++)=(m+a)++=(n+a)++=n+(a++)，当然这个证明本身用到了自然数的相等对“++”的替换公理，我有点难以接受自然数相等的替换公理对于任何自然数的运算都成立

8 March, 2014 at 9:26 am

JackIs the converse of Zorn’s lemma true?

[The answer to this question will depend quite heavily on exactly what one means by “converse to Zorn’s lemma”. I suggest as an instructive exercise to try to write out an explicit formulation of such a converse, and then trying to answer your question yourself. -T.]8 March, 2014 at 11:04 am

F r i e d e r S i m o nDear Prof. Tao,

the use of “=” in Axiom 3.3 (Singleton sets and pair sets) seems problematic to me (or at least I’m having trouble understanding its use in light of a logical perspective regarding your set theory): When you say “For every object , we have if and only if ” it isn’t clear how to interpret the “=”.

As far as I can see, you seem to be implicitly working with a set theory that is formalized in a first-order language without equality, since in Definition 3.1.4 it was explicitly defined (and not axiomatically stated) how to read “=” and there are also later definitions, as the one in 3.3.7, where equality for functions is established.

Since there may be objects aren’t sets, as mentioned on p.39, Definition 3.1.4 doesn’t cover the expression and one should therefore first define what “=” means for arbitrary objects. But that does not seem feasable (at least to me), because objects can be very different things (numbers, functions etc.) and I can’t see how one could cover all types of objects in a single definition (which, if this is true, would also motivate the exclusion of “=” from the “first-order language” in which your set theory is set up).

So the problem remains how is to be understood in Axiom 3.3. A possible fix could maybe consist of modifying to only mean the equality for those objects, which are defined at some point in the text (like Definition 3.3.7 for function equality), but that doesn’t seem very elegant and also forces one to move all those Definitions before Axiom 3.3 which seems impractical.

(Side remark: A similar issue arises with Definition 3.3.1 resp. 3.3.7 at resp. which again could be arbitrary objects – and also at various other places, like in Example 3.1.31 on p. 49 where .)

8 March, 2014 at 3:07 pm

Terence TaoAs my text is an analysis text rather than a logic text, I deliberately chose not to be excessively formal with regards to the set-theoretic foundations of mathematics here, which I am regarding as part of the metatheory of analysis, rather than part of the theory. If one were to formalise the metatheory implicitly used here, though, it would be a set theory as a language with equality, but for which the axioms are assumed to hold impredicatively (and dynamically): I prefer not to specify in advance exactly what types of objects exist in the mathematical universe (because I want to reserve the right to introduce new types of objects (e.g. functions, integers, real numbers, ordered pairs, etc.) in later sections), but as each new class of objects is admitted (and the notion of equality defined for each new class of objects), the set theoretic axioms are assumed to apply to this new enlarged universe.

From a strictly logical viewpoint, this is thus a slightly stronger set of set-theoretic assumptions than simply assuming the existence of a single, static set-theoretic universe obeying the ZFC axioms. Of course, one could work entirely with the latter assumptions, either by interpreting all new objects strictly within the given set theory (e.g. using the Kuratowski construction to define ordered pairs, using the graph of a function to define a function, using equivalence classes to define integers, rationals and reals, and so forth); alternatively, every time one wishes to introduce a new type of object, one could pass from the existing mathematical language to a conservative extension that includes this new type of object, after taking the effort to carefully check that the extension is indeed conservative. However, either of these approaches, while more “economical” in the sense of requiring a smaller and simpler set of logical assumptions, would require a lot more attention given to set-theoretic and logical formalism, which would distract from the core mission of the text, which is to teach real analysis, and would also give the misleading impression that analytic objects such as integers or real numbers must necessarily be interpreted in set-theoretic terms (e.g. as equivalence classes of Cauchy sequences) in order to be able to properly manipulate them mathematically. So my philosophy here is to work in a looser metamathematical framework in which there is “always enough set theory” around to do mathematics (even when one dynamically expands the language of mathematics by introducing new objects and definitions), but one does not focus on the precise formal setup of this set theory, which is operating more as a metatheory here to the underlying mathematics than as the theory itself.

EDIT: This mathoverflow answer by Andreas Blass is a good summary of the approach to metamathematics that I identify with (and implicitly use in this text), albeit without the emphasis on the dynamic nature of that metatheory.

9 March, 2014 at 1:02 am

F r i e d e r S i m o nThis, together with the reference you linked, was very enlightening. Thank you very much.

14 March, 2014 at 9:16 am

M. ScorseseDear Dr. Tao, I’m having problems understanding Definition 3.3.1 of functions, since it seems to me to be too vague.

The concrete reasons why I’m dubious about this definition is, that

1) the definition doesn’t contain a precise description what $f(x)$ really is. We define the object $f$ by using $f(x)$, but we don’t know given an object $f$ how to interpret $f(x)$.

2) we have very vew axioms that tell us how to generate new objects (in contrast to the existance of sets for which we have numerous axioms that tell us how to build new ones from already existing ones). Essentially we only know that every nonempty set contains an object (The single-choice Lemma 3.1.6).

Therefore: Given sets $X,Y$ and a property $P(\cdot,\cdot)$ such that for every $x\in X$ there exists only one $y\in Y$ such that $P(x,y)$ holds, how can I prove that exactly one object $f$ exists such that $P(x,y)$ is true whenever $y=f(x)$ is true ?

Could you please help me out ? Thanks.

14 March, 2014 at 11:19 am

Terence TaoThe axiom that generates function objects is Axiom 3.10, which in particular guarantees that every function f defined from Definition 3.3.1 is in fact an object in the set theoretic universe. (If you like, Definition 3.3.1 associates a “formal object” f to each property P(x,y) obeying the vertical line test, and Axiom 3.10 then upgrades this formal object to a set-theoretic object.)

17 March, 2014 at 2:55 pm

José Antonio Lara BenítezDear Dr. Tao I think there is a mistake in the page 275 after the proof of the Intermediate value theorem: “Note that if is not assumed to be continuous, then the intermediate value theorem no longer applies.”

But the function for and is non-continuous on , in particular the limit at $0$ doesn’t exist, but satisfy the IVT.

17 March, 2014 at 6:04 pm

Terence TaoThe sentence is technically correct as it stands – the way the intermediate value theorem is stated, continuity is an explicit hypothesis and so the theorem is not applicable for discontinuous functions even if the conclusion ends up being true anyway. But perhaps the more useful sentence to assert here is “Note that if is not assumed to be continuous, then the conclusion of the intermediate value theorem is not necessarily true.”.

9 June, 2014 at 11:03 pm

AnonymousDear Dr.Tao. in Proposition 10.3.3, if i change all [a,b] to (a,b), does the conclusion remain valid?In fact, i think the close interval version imply the open interval one.

[Yes – T.]12 June, 2014 at 7:40 pm

AnonymousDear Dr.Tao. In Exercise 11.6.5. Use Proposition 11.6.4 to prove Corollary 11.6.5, we should evaluate integral 1/x^p over [1,N], but without Theorem 11.9.4 (Second Fundamental Theorem of Calculus). how can we do that? Exercise 11.6.5 in wrong place or you have other consideration. Thank you.

12 June, 2014 at 10:28 pm

Terence TaoHmm, that’s odd. I don’t recall now why I placed that question there, but the simplest thing to do is to permit the use of Theorem 11.9.4 for that exercise, since that theorem does not use Corollary 11.6.5 in its proof.

15 June, 2014 at 5:10 am

AnonymousIf I understand it correctly, the two UCLA-courses Math 131A and B make up for what most universities simply call “Real Analysis.” If I wish to use your analysis-books as a supplement to Rudin in my Real Analysis-course (I’m a student, not a lecturer), should I go ahead and buy both? For what its worth, here is the course description:

“After completed course, the students are expected to be able to

– Describe the basic differences between the rational and the real numbers.

– Understand and perform simple proofs

– Answer question concerning uniform convergence of concrete numerical sequences and series

– Give the definition of concepts related to metric spaces, such as continuity, compactness, completeness and connectedness

– Give the essence of the proof of Stone-Weierstrass’ theorem, the contraction theorem as well as the existence of convergent subsequences using equicontinuity.

17 June, 2014 at 11:38 pm

AnonymousDear Dr. Tao ,i have such a understanding for 11.8 The Riemann-Stieltjes integral :

if a is NOT continuous in I , Lemma 11.8.4 is NOT true. we CAN NOT get something like Proposition 11.2.13 for P.c. Riemann-Stieltjes integral. finally we fail to define Riemann-Stieltjes integral, so a’s continuty is a KEY point for Riemann-Stieltjes integral. you touch this point (a’s discontinuous) in the last paragraph. but i think what you try to tell us is quite different from what i got.

especially for this point: if f and a are both discontinuous at the same point, then Riemann-Stieltjes integral is unlikely to be defined .

please correct me about this.

thank you.

18 June, 2014 at 12:02 am

Anonymousplease IGNORE this post. there is Riemann–Stieltjes Integrals with α a Step Function . http://www.math.ubc.ca/~feldman/m321/step.pdf

13 July, 2014 at 9:08 pm

cristHian Gz. (gcca)Dear Dr. Tao, I am using the second edition. On pg. 76, Exercise 3.5.12, 10 line: . I think should be .

[Correction added, thanks – T.]23 July, 2014 at 10:00 am

Gonzales Castillo Cristhian A.Dear Professor Tao,

using the 2nd Edition, on page 79, there is a subtlety:

7th line of proof of Lemma 3.6.9, it says: “Now define the function to “.

Perhaps you mean:

from to

or

[Correction added, thanks – T.]26 July, 2014 at 8:08 pm

Duodaa.comDear Professor Tao, I think there is little mistake on page 231, exercise 8.4.3. The Partition Principle (PP), which states that A can be mapped onto B (or B is empty) if and only if B can be injected into A. Whether (or not ) PP implies Axiom of Choice is a problem still open. But “the converse” part in exercise 8.4.3 is a such problem.

[Oops, I had neglected to add an additional condition on f, namely that it was a right inverse for g. -T.]27 July, 2014 at 12:41 am

Duodaa.comMaybe, somebody can close this problem as an exercise. :-)

24 August, 2014 at 5:22 pm

SiquanDear Prof. Tao

I am using your week notes to self learn Analysis 1. There is one thing in the proof of Theorem 27 (Least Upper Bound Property) in “week 2 note” that I don’t understand.

The note says that “there exits some natural number i with 0<= i <=k such that x0+i/n is an upper bound for E, while x0+(i-1)/n is not an upper bound for E …….. then it is easy to use induction to show that x0+i/n is not an upper bound for E for any 0<= i <=k". I really don't understand this point. What is the induction hypothesis? How should we condider x0+(i-1)/n is not an upper bound for E? Would you please tell me the specific process of this induction proof? Thank you.

Best regards

Siquan

3 November, 2014 at 5:41 pm

Gonzales Castillo Cristhian A.Dear Prof. Tao, below the proof of Proposition 6.4.12, a right parenthesis is missing after “(providing that abs are finite”.

Also, the Example 6.6.3 says . I think “” is missing, so .

[Corrections added, thanks -T.]15 November, 2014 at 12:25 pm

Gonzales Castillo Cristhian A.Dear Prof. Tao, I found the following errata (in second edition):

– In Lemma 7.1.4(c) a dot is missing after in the end of line. So

– In the proof of Proposition 7.4.1, 2th line, it says: “We know that the sequences and “. I think it should be “ and ” (replacing sub-index by and by ).

– In the proof of Proposition 7.4.3, in the last paragraph, the last sentence says: “Thus is -close to “. I think it should be “-close to ” ( by ).

– The Corollary 8.1.9 says: “”Let be a countable set , and let be a function. Then is at most countable”. It think it should be “Then is at most countable” (replacing by ).

[Corrections added, thanks, although I do believe Corollary 8.1.9 is correct as stated. -T.]17 November, 2014 at 6:16 am

RituparnaDear Dr. Tao,

I really find your book very refreshing in aspect to how it deals with Analysis in general, which is different from other books. But, I came across some things which I thought was needed to clarify..

In Chapter-2, pg-15 of your book Analysis I you write – “The natural numbers {0,1,2,…}”. I have got a little confused as I learnt in school that whole numbers start from 0 while natural numbers start from 1.

Also in pg-16, you write – “Why is 123.4444… a real number, while …444.321 is not?”. I maybe wrong but, shouldn’t both the numbers be real?

Hoping to clear the ambiguites

Thank you…

17 November, 2014 at 7:24 am

Terence TaoDifferent authors have different conventions as to whether to include zero as a natural number. For instance, as a general rule, number theorists and algebraists, who are more interested in the multiplicative structure of the natural numbers, tend to exclude zero, whereas set theorists, analysts, and combinatorialists, who are more interested in the ability of natural numbers to count the cardinality of sets, tend to include zero. See http://en.wikipedia.org/wiki/Natural_number

Real numbers have to be bounded by some finite natural number (by the Archimedean property of the reals), and so their decimal expansion cannot continue indefinitely to the left; 4, 44, and 444 are real numbers, but the infinite string …444 is not. (Such a string can be viewed as a different type of number, namely a 10-adic number, but that is another story.)

18 November, 2014 at 8:48 am

RituparnaDear Dr. Tao,

Thank you so much for clarifying my queries…

23 November, 2014 at 2:48 am

AnonymousDear Prof Tao,

In the new 3rd edition of analysis i, all the cross reference to analysis ii become wrong now since the new analysis ii starts from chapter 1 rather than chapter 12. For example, on page 7, on the line before example 1.2.7, see theorem 11.50.1 should be theorem 8.5.1 of analysis ii.

[Ugh. Yes, we’ll have to have this fixed in the next revision of the ms. -T.]25 December, 2014 at 12:48 pm

Gonzales Castillo Cristhian A.I found in Second Edition:

Theorem 8.3.1 (Cantor’s theorem): The end of the proof says “But if , then “. It must be “then “.

[Actually, the argument works as stated: since , and , we have . Of course, we also have as well – this is where the contradiction is coming from! -T.]Sorry for two things:

– The incorrect last errata. I must be sleepy and did not analyze the statement. :-(

– Right now, I’m in Chapter 9. I saw two another erratas in Chapter 8 but I not remember.

27 December, 2014 at 2:22 am

lorenzoDear prof. Tao,

I’m having trouble with Exercise 2.2.2 (numbering of the third edition), which asks to prove the following Lemma 2.2.10: “Let a be a positive number. The there exists exactly one natural number b such that b++=a”. This exercise is proposed before a notion of order is defined and therefore before one has the technique of strong induction available. The hint following the problem says: “use induction” but it seems to me that the lemma is calling for an induction on a (not on b) and so I don’t understand how we can use “ordinary” induction (Axiom 2.5) to prove it (since Axiom 2.5 has P(0) as the base case, not P(1)).

I’ve also tried induction on b, but I got nowhere so far.

Could you explain to me where I go wrong?

Thank you.

Best regards,

lorenzo.

27 December, 2014 at 7:22 am

Terence TaoYou can modify the statement of the exercise to an equivalent statement which makes sense for . (The statement will probably look something like “For any natural number , either , or else there exists exactly one…”.)

29 December, 2014 at 3:48 pm

Gonzales Castillo Cristhian A.Dear Prof. Tao, in Excercise 8.5.18 of Second Edition:

1. a “)” is missing at the end of “(i.e. there is no other totally ordered subset of which contains “; and

2. the first word of the last sentence is “Tthus”.

[Correction added, thanks -T.]11 January, 2015 at 7:49 am

AnonymousConsider the real-valued function defined as and the monotone increasing function . Then is Riamann-Stieltjes integrable according to the definition in the book. But according to the first definition in the wikipedia article Riemann-Stieltjes integral, it is not. What’s going wrong?

11 January, 2015 at 10:15 am

Terence TaoThe Riemann-Stieltjes integral in my text corresponds to the generalised Riemann-Stieltjes integral in the wikipedia article.

11 January, 2015 at 1:04 pm

AnonymousIs the definition in this textbook also equivalent to the Darboux sum definition, which is in Rudin’s

Principles of Mathematical Analysis?Is the Lebesgue-Stieltjes integral in the Math 245a notes 6 an extension (Exercise 16 in that notes) for all sorts of definitions in the link above (so that people don’t bother deal with the issues of different definitions of the R-S integrals) ?

9 February, 2015 at 2:15 pm

lorenzoDear prof. Tao,

I’m having trouble with Exercise 3.1.11 (numbering of the third edition), which asks to prove that the Axiom of Replacement (Axiom 3.6) implies the Axiom of Specification (Axiom 3.5).

Intuitively I think that this proof boils down to proving that if one requires that y (in Axiom 3.6) also belongs to A then one obtains the same statement of Axiom 3.5, but I don’t know how to show this rigorously (if it ever is correct).

Could you give me an advice about how to proceed?

Thank you.

Best regards,

lorenzo

9 February, 2015 at 3:53 pm

Terence TaoIn Axiom 3.6, you can select the property to be whatever you wish (as long as for each in there is at most one satisfying ), so if you want to put a condition such as in your choice of , you are free to do so.

11 March, 2015 at 9:22 am

XianjinHi, professor. For theorem 8.5.10 (the strong induction), should we add some basic conditions such as $P\left(a\right)$ is true, $P\left(b\right)$ is true,…,and then form the induction? Otherwise, suppose $P\left(n\right)$ is false for any $n \in X$, we get vacuously true, then we get $\forall n \in X, P\left(n\right) is true$. Although the proposition is still true, but this makes no sense.

11 March, 2015 at 9:41 am

Terence TaoVacuously true implications are still true, and can be useful (because there are fewer conditions one needs to check if one wishes to use the theorem). See Appendix A.2.

13 March, 2015 at 2:11 pm

The importance of sketching. The best way to learn is to write. | thoughtsonthecommontoad[…] a few books based on his blog; I found Compactness and Contradiction particularly good. [His two volume treatment of Analysis is […]

13 March, 2015 at 2:38 pm

XianjinFor problem 8.5.20, should we change the condition to ? Since if , then we cannot find a subset , which can satisfy the condition.

[The most recent edition of the book should have the condition . -T.]13 March, 2015 at 7:47 pm

Luqing Yeand are same notations.

16 March, 2015 at 7:26 am

RituparnaDear Dr. Tao,

In pg-63, Exercise 3.3.1 says – “if and are functions such that and , then .” I was wondering how we define the domain and range of here since the range of is while the domain of is . Is it some error here or am I missing something?

Thanks.

[There is a typo; see the errata for p.55 of the second edition. -T.]3 April, 2015 at 2:38 am

lorenzoDear prof. Tao,

I’m stuck on Exercise 3.4.7 (numbering of the third edition), which asks to show that the collection of all partial functions from a set X to a set Y is itself a set.

The only progress I’ve been able to make is realizing that if somehow I could “build” the set $latex{Y_1^{X_1}, Y_2^{X_2},…}$ (where the and are elements of and respectively) then the $UA$ would be (I think) the desired set.

To build that set I suspect one could use the Axiom of Replacement but all my attempts in doing so have been unsuccessful.

Could you give me an advice about how to proceed?

Thank you.

Best regards,

lorenzo

3 April, 2015 at 6:35 am

Terence TaoYes, this is how you should proceed. You can begin by showing that the set of pairs , with and , is a set. Then you can use the axiom of replacement with a predicate to capture the assertion that a given set is equal to .

3 April, 2015 at 1:54 pm

lorenzoBy pair you mean pair set in the sense of Axiom 3.3? I’m asking this because the notion of (ordered) pair of objects is not introduced until Chapter 3.5, which comes after the Exercise we are talking about, so I’m a bit confused about what you mean with the notation .

Best regards,

lorenzo.

3 April, 2015 at 8:59 pm

Terence TaoIf one wishes to avoid the use of ordered pairs, one can use the axiom of replacement twice rather than once, first to collect all the partial functions with a fixed domain and arbitrary range (or vice versa), and then again to allow to vary (the axiom of union will be useful here too).

19 April, 2015 at 6:18 am

XianjinFor theorem 10.4.2. Is it possible to remove the condition that is continuous at ? If is differentiable at , , then we can suppose , then by using intermediate theorem, we can show that near , is increasing, so exists and continuous at

19 April, 2015 at 11:59 am

Terence Taodoes not imply that is increasing. For instance, if for (and , then has a positive derivative at 0, but is not invertible (or increasing) near 0.

19 April, 2015 at 1:35 pm

XianjinBut what if when we already know exits and is differentiable at , can we know that near ?

[No; see the counterexample in my previous comment. -T.]25 April, 2015 at 12:16 pm

XianjinDear Prof Tao. In the last paragraph of the section 11.8, you say that if are interrupted at the same points, the Riemann-Stieltjes integral may not be defined. Can you give an example why it is not true? Thanks very much!

1 May, 2015 at 5:13 am

Ti GongDear Prof. Tao,

I only have the first edition. Below the proof of Proposition 6.4.12, you say that “Parts (c) and (d) of Proposition 6.4.12 say, in particular, that … “. But I think it may be “Parts (d) and (e) of Proposition 6.4.12 say, …” Thank you.

Best regards,

Ti

[Corrected, thanks – T.]2 May, 2015 at 8:00 pm

ttigongDear Prof. Tao, in the last sentence of the proof of Theorem B.2.2, you say that “Since already has a positive integer decimal representation by Theorem B.1.4, …”. But I think it may be “Since is a natural number, it can be 0, or has a positive integer decimal representation by Theorem B.1.4, …” Thank you.

Best regards,

Ti

6 May, 2015 at 5:04 am

lorenzoDear prof. Tao,

I’m stuck on Exercise 3.2.1 (numbering of the third edition), which asks to prove that the axiom of universal specification, Axiom 3.8, if assumed to be true would imply Axioms 3.2, 3.3, 3.4, 3.5 and 3.6.

I proved the statement for Axioms 3.2, 3.3, 3.4 and 3.5 but I’m having trouble doing the same with Axiom 3.6, because this last one uses a statement of two variables $P(x,y)$ to build a new set (out of an existing one), while Axiom 3.8 uses a statement of one variable, $P(x)$ to do the same (but starting “from scratch”), so I don’t see how I could derive the first from the second.

Could you give me an advice about how to proceed?

Thank you.

Best regards,

lorenzo

6 May, 2015 at 7:54 am

Terence TaoWhile the statement (or more precisely, the predicate) depends on two variables , the predicate “ is true for some ” depends only on (see the discussion on free and bound variables in Appendix A.4).

10 May, 2015 at 10:16 am

Diego G.MDear professor Tao.

I have a question about the absolute value.

Is there some special reason why you have decided to put the theory about the absolute value in the chapter 4. Integers and rationals ?

10 May, 2015 at 12:20 pm

Terence TaoThis is needed to define Cauchy sequences in Section 5.1.

12 May, 2015 at 10:39 am

Diego G.MBut what i don’t understand is why in other books different from yours, the absolute value is explained inside the real numbers ?

Do you think that it’s better to explain it from the point of view of rational numbers due to the fact that they already include integers, natural numbers.. ? (and also due to the fact that if we considere the absolute value from the point of view of real numbers, the only difference respect rational numbers are the irrational numbers. Why don’t you include irrational numbers? Maybe because they have not accuracy enought to be applied the absolute value ?

Thanks you for spend your time in this. It’s maybe a nonsense. I hope not.

12 May, 2015 at 12:17 pm

Terence TaoAbsolute values for the reals are defined in Definition 5.4.5. This supersedes the notion of absolute value for the rationals, defined in Definition 4.2.8. However, we temporarily need the rational notion of absolute value in order to define the notion of a Cauchy sequence of rationals, used here to define the reals. In some other texts, the reals are constructed by other means (e.g. Dedekind cuts), in which case one does not need to define absolute values for the rationals separately; however I find the Cauchy sequence definition to be conceptually superior and so used this route for my text.

15 May, 2015 at 8:38 am

lorenzoDear prof. Tao,

I have a question regarding proofs about natural numbers:

If one has to prove a property regarding natural numbers, should one always use induction?

For example, if one wanted to prove the statement ” *”.

Is the latter approach (i.e.proving an implication involving natural numbers “directly”) acceptable or should one always use induction to prove such statements? (Perhaps this is a silly question, but I’m in doubt since I always see proofs involving natural numbers carried out by induction).

Thank you for your time.

Best regards,

lorenzo.

*I have left out some details of this proof for brevity since I have only used it as an example to better explain myself.

15 May, 2015 at 10:53 am

Terence TaoInitially, one only has the five Peano axioms to work with, one of which is the powerful axiom of induction, and so it is natural that one will often rely on induction to prove many of the basic results about numbers (e.g. the commutative, associative, and distributive laws). As one progresses with the subject, though, one starts to obtain many other powerful results (e.g. the fundamental theorem of arithmetic, or various ways to relate sums to integrals) that can be used instead; many of these results are proven using induction at some point, but one does not need to explicitly invoke induction in order to

applysuch results.Basically, induction should not be a tool to be applied blindly, but only when it has a chance of being useful: one should only attempt to prove by induction if there is some link between and (or between and for some smaller values of ) which could potentially be exploited. For instance, if trying to prove a formula for , induction on would be a plausible strategy since there is a clear relationship between and $1^2+2^2+\dots+(n+1)^2$. On the other hand, if one were to try to prove non-existence of natural number solutions to for , induction on does not appear to be a viable strategy since there is no obvious relationship between the solvability of and of . (However, due to the obvious relationship between solvability of and , one can at least use induction to reduce to the case when is prime.)

A good test case for students is the Euclidean algorithm (Proposition 2.3.9 in my book). It can be proven by induction, but if one applies induction blindly one is unlikely to be able to prove it – one has to have some understanding of *why* one would attempt to use induction in order to be able to use induction effectively to establish that the Euclidean algorithm works.

p.s. to use < and > in wordpress, use < and > ; see https://terrytao.wordpress.com/about/

15 May, 2015 at 9:10 am

lorenzoSomething is definitely wrong with WordPress but I hope you nonetheless see the point of my question: is it possible to prove rigorously properties of natural numbers without using induction?

21 May, 2015 at 8:10 am

Meng DongIn the first edition:

p. 176, line 4 from bottom: Shouldn’t “Lemma 6.5.3” be “Theorem 6.1.19(e)”?

[One can use either of these results to establish the claim. For instance, one can apply Lemma 6.5.3 with replaced by . -T]2 June, 2015 at 8:19 am

lorenzoDear prof. Tao,

in Exercise 4.1.1 you ask to prove that the definition of equality on the integers is reflexive and symmetric; my question is: do these properties (and also the others listed in Appendix A.7) hold also for natural numbers? The answer should (obviously?) be yes and I’ve looked in the text for an answer but I haven’t found any.

P.S. The error on pag. 70 listed in the errata for the second edition hasn’t been corrected in the third edition: the remark about n-1 has been added but in the fourth line of Lemma 3.6.9 there is still an N and not an n (at least in my copy of the book)

[Correction added, thanks. The laws for equality for the natural numbers are implicitly included within Assumption 2.6 where the natural numbers are introduced to the mathematical universe by assumption (rather than by definition). -T.]7 June, 2015 at 12:10 am

D.GarciaDear prof. Tao,

I’m a student who has already take ‘calculus’. Now, i’m going to start ‘analysis’. I found two interesting books in the library:

-Your book: Analysis I ( I think that it’s a very clear book)

– Walter Rudin’s principle of mathematical analysis ( it’s said that it’s a classic in this field: https://notendur.hi.is/vae11/%C3%9Eekking/principles_of_mathematical_analysis_walter_rudin.pdf . Although this book it’s not as clear as yours, i like it because it’s quite hard and it means that i will spend some hours thinking about how to solve every single problem.

I have always used to study from two different books because i like to have different point of views. It doesn’t mean that i don’t trust your book. Indeed, i have decided to follow your book’s structure since i think it’s quite neat.

The reason why i’m writing you is because i would like to know your opinion about the fact of studing from these two books at once. The Rudin’s book contains a chapter about ‘Basic topology’ and some concepts such as ‘metric spaces…’ which are used very often. In this book, it’s said that topology it’s important to understand the next chapters of the book.

-Are both books compatible to be studied at the same time ?

– How would you arrange the Rudin’s book contents and your book’s ones such that one could study from these two books at once ?

(I’m trying to find a common structure between these two books but i’m not able.)

Greetings from Spain! Thank you in advance.

30 July, 2015 at 10:08 am

Gabriel SalmerónHi Terry Tao! I’m reading your “Analysis I” book, which I’m really enjoying :).

Besides some typos that don’t deserve further comments, I found a mistake that, I think, could be interesting:

*Example 3.3.22* The function f:N -> N\{0} defined by f(n) := n++ is a bijection (in fact, this is simply restating axioms 2.2, 2.3 and 2.4)…

Let’s remember. Axiom 2.2 says every succesor of a natural is a natural. Axiom 2.3 says that 0 is not a succesor. These two axioms indeed guarantee that f is well defined in the given domain and range.

Axiom 2.4 guarantees that f is injective.

But you cannot prove that f is surjective based only on those axioms, cos if we ‘double the naturals’ so we unite the usual {0,1,2,3….} to another set {0bis,1bis,2bis,3bis,…} in which 0bis++=1bis, 1bis++=2bis, 2bis++=3bis and so on, we end up with a set that satisfies axioms 2.2, 2.3, and 2.4, but 0bis is not the succesor of any other natural, so it’s not covered by f.

It’s also interesting to note that if we added an axiom saying: ‘every nonzero natural is succesor of another natural’ then f would be bijective, but we still might not have the naturals as we know them cos we still lack the axiom of induction.

Sorry for my bad english and for not knowing how to insert LateX here (tried to eliminate my comment from the other section but couldn’t).

[Correction added, thanks – T.]30 July, 2015 at 10:51 am

Gabriel SalmerónSorry for not seeing the instructions to enter LateX just below my eyes, in the previous comment.

I see there’s an attempt to exhoust errate, so I probably I should mention that in example 1.2.12, near the bottom of page 11, you are using L’Hopital’s rule to get: But if I’m not missing something, the subtrahend should be cos, isn’t the derivative of equal to ?

[What edition of the text are you using? All editions I am aware of have rather than . -T.]30 July, 2015 at 11:12 am

Gabriel SalmerónOh, I’m reading through the sample chapters I downloaded from this very page. I din’t bother to buy an edition before reading the sample chapters. Hope you don’t mind.

20 August, 2015 at 5:30 am

lorenzoDear prof. Tao,

in the proof of Theorem 5.5.9 (Existence of least upper bound) you say that “By the Archimedean property again, there exists another integer such that " . I don't understand how you have applied the Archimedean property to conclude as much. Could you explain it in detail?

Thank you.

Best regards,

lorenzo.

20 August, 2015 at 10:36 am

Terence TaoThe inequality is equivalent to , and the Archimedean property guarantees the existence of an integer multiple of greater than .