Analysis I

Last updated: Sep 27, 2021

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 to older versions than the corrected third edition can be found here.

— Errata to the corrected third edition (hardback) —

Page 1: On the final line, $-2$ should be in math mode.
Page 7: In Example 1.2.6, Theorem 19.5.1 should be “Theorem 7.5.1 of Analysis II”.
Page 8: In Example 1.2.7, “Exercise 13.2.9” should be “Exercise 2.2.9 of Analysis II”. In Example 1.2.8, “Proposition 14.3.3” should be “Proposition 3.3.3 of Analysis II”. In Example 1.2.9, “Theorem 14.6.1” should be “Theorem 3.6.1 of Analysis II”.
Page 9: In Example 1.2.10, “Theorem 14.7.1” should be “Theorem 3.7.1 of Analysis II”.
Page 11: In the final line, the comma before “For instance” should be a period.
Page 14: “without even aware” should be “without even being aware”.
Page 17: In Definition 2.1.3, add “This convention is actually an oversimplification. To see how to properly merge the usual decimal notation for numbers with the natural numbers given by the Peano axioms, see Appendix B.”
Page 19: After Proposition 2.1.8: “Axioms 2.1 and 2.2” should be “Axioms 2.3 and 2.4”.
Page 20: In the proof of Proposition 2.1.11, the period should be inside the parentheses in both parentheticals. Also, Proposition 2.1.11 should more accurately be called Proposition Template 2.1.11.
Page 23, first paragraph: delete a right parenthesis in $f_3(a_3))$ .
Page 27: In the final sentence of Definition 2.2.7, the period should be inside the parentheses. In proposition 2.2.8, “ $a$ is positive” should be “ $a$ is a positive natural number”.
Page 29: Add Exercise 2.2.8: “Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m)$ is true, $P(m+\!+)$ is true. Show that if $P(n)$ is true, then $P(m)$ is true for all $m \geq n$ . This principle is sometimes referred to as the principle of induction starting from the base case $n$ “.
Page 31: “Euclidean algorithm” should be “Euclid’s division lemma”.
Page 39: in the sentence before Proposition 3.1.18, the word Proposition should not be capitalised.
Page 41: In the paragraph after Example 3.1.22, the final right parenthesis should be deleted.
Page 45: at the end of the section, add “Formally, one can refer to ${\bf N}$ as “the set of natural numbers”, but we will often abbreviate this to “the natural numbers” for short. We will adopt similar abbreviations later in the text; for instance the set of integers ${\bf Z}$ will often be abbreviated to “the integers”.”
Page 47: In “In $\Omega$ did contain itself, then by definition”, add “of $\Omega$ “. After “On the other hand, if $\Omega$ did not contain itself,” add “then by definition of $P$ “, and after “and hence”, add “by definition of $\Omega$ “.
Page 48: In the third to last sentence of Exercise 3.2.3, the period should be inside the parenthesis.
Page 49: “unique object $f(x)$ ” should be “unique object $f(x) \in Y$ “, and similarly “exactly one $y$ ” should be “exactly one $y \in {\bf N}$ “.
Page 49+: change all occurrences of “range” to “codomain” (including in the index). Before Exercise 3.3.2, add the following paragraph: “Implicit in the above definition is the assumption that whenever one is given two sets $X, Y$ and a property $P$ obeying the vertical line test, one can form a function object. Strictly speaking, this assumption of the existence of the function as a mathematical object should be stated as an explicit axiom; however we will not do so here, as it turns out to be redundant. (More precisely, in view of Exercise 3.5.10 below, it is always possible to encode a function $f$ as an ordered triple $(X, Y, \{ (x,f(x)): x \in X \})$ consisting of the domain, codomain, and graph of the function, which gives a way to build functions as objects using the operations provided by the preceding axioms.)”
Page 51: Replace the first sentence of Definition 3.3.7 with “Two functions $f: X \to Y$ , $g: X' \to Y'$ are said to be equal if and only if they have the same domain and codomain (i.e., $X=X'$ and $Y=Y'$ ), and $f(x)=g(x)$ for <I>all</I> $x \in X$ .” Then add afterwards: “According to this definition, two functions that have different domains or different codomains are, strictly speaking, distinct functions. However, when it is safe to do so without causing confusion, it is sometimes useful to “abuse notation” by identifying together functions of different domains or codomains if their values agree on their common domain of definition; this is analogous to the practice of “overloading” an operator in software engineering. See the discussion [in the errata] after Definition 9.4.1 for an instance of this.”
Page 52: In Example 3.3.9, replace “an arbitrary set $X$ ” with “a given set $X$ “. Similarly, in Exercise 3.3.3 on page 55, replace “the empty function” with “the empty function into a given set $X$ “.
Page 56: After Definition 3.4.1, replace “a challenge to the reader” with “an exercise to the reader”. In Definition 3.4.1, “ $S$ is a set in $X$ ” should be “latex S$ is a subset of $X$ “.
Page 62: Replace Remark 3.5.5 with “One can show that the Cartesian product $X \times Y$ is indeed a set; see Exercise 3.5.1.”
Page 65: Split Exercise 3.5.1 into three parts. Part (a) encompasses the first definition of an ordered pair; part (b) encompasses the “additional challenge” of the second definition. Then add a part (c): “Show that regardless of the definition of ordered pair, the Cartesian product $X \times Y$ is a set. (Hint: first use the axiom of replacement to show that for any $x \in X$ , the set $\{ (x,y): y \in Y \}$ is a set, then apply the axioms of replacement and union.)”. In Exercise 3.5.2, add the following comment: “(Technically, this construction of ordered $n$ -tuple is not compatible with the construction of ordered pair in Exercise 3.5.1, but this does not cause a difficulty in practice; for instance, one can use the definition of an ordered $2$ -tuple here to replace the construction in Exercise 3.5.1, or one can make a rather pedantic distinction between an ordered $2$ -tuple and an ordered pair in one’s mathematical arguments.)”
Page 66: In Exercise 3.5.3, replace “obey” with “are consistent with”, and at the end add “in the sense that if these axioms of equality are already assumed to hold for the individual components $x,y$ of an ordered pair $(x,y)$ , then they hold for an ordered pair itself”. Similarly replace “This obeys” with “This is consistent with” in Definition 3.5.1 on page 62.
Page 67: In Exercise 3.5.12, add “Let $X$ be an arbitrary set” after the first sentence, and let $f$ be a function from ${\mathbf N} \times X$ to $X$ rather than from ${\mathbf N} \times {\mathbf N}$ to ${\mathbf N}$ ; also $c$ should be an element of $X$ rather than a natural number. This generalisation will help for instance in establishing Exercise 3.5.13.
Page 68: In the first paragraph, the period should be inside the parenthetical; similarly in Example 3.6.2.
Page 71: The proof of Theorem 3.6.12 can be replaced by the following, after the first sentence: ” By Lemma 3.6.9, ${\bf N} \backslash \{0\}$ would then have cardinality $n-1$ . But ${\bf N}$ has equal cardinality with ${\bf N} \backslash \{0\}$ (using $x \mapsto x+1$ as the bijection), hence $n = n-1$ , which gives the desired contradiction. Then in Exercise 3.6.3, add “use this exercise to give an alternate proof of Theorem 3.6.12 that does not use Lemma 3.6.9.”.
Page 73: In Exercise 3.6.8, add the hypothesis that $A$ is non-empty.
Page 77: “negative times positive equals positive” should be “negative times positive equals negative”. Change “we call $-n$ a negative integer“, to “we call $n$ a positive integer and $-n$ a negative integer“.
Page 89: In the first paragraph, insert “Note that when $n=1$ , the definition of $x^{-1}$ provided by Definition 4.3.11 coincides with the reciprocal of $x$ defined previously, so there is no incompatibility of notation caused by this new definition.”
Page 94, bottom: “see Exercise 12.4.8” should be “see Exercise 1.4.8 of Analysis II”.
Page 97: In Example 5.1.10, “1-steady” should be “0.1-steady”, “0.1-steady” should be “0.01-steady”, and “0.01-steady” should be “0.001-steady”.
Page 104: In the proof of Lemma 5.3.7, after the mention of 0-closeness, add “(where we extend the notion of $\varepsilon$ -closeness to include $\varepsilon=0$ in the obvious fashion)”, and after Proposition 4.3.7, add “(extended to cover the 0-close case)”.
Page 113: In the second paragraph of the proof of Proposition 5.4.8, add “Suppose that $x>y$ ” after the first sentence.
Page 122: Before Lemma 5.6.6: “ $n^{th}$ root” should be $n^{th}$ roots”. In (e), add “Here $k$ ranges over the positive integers”, and after “decreasing”, add “(i.e., $x^{1/k} < x^{1/l}$ whenever $l>k$ )”. One can also replace $x<1$ by $0 < x < 1$ for clarity.
Page 123, near top: “is the following cancellation law” should be “is another proof of the cancellation law from Proposition 4.3.12(c) and Proposition 5.6.3”.
Page 124: In Lemma 5.6.9, add “(f) $(xy)^q = x^q y^q$ .”
Page 130: Before Corollary 6.1.17, “we see have” should be “we have”.
Page 131: In Exercise 6.1.6, $LIM$ should be $\mathrm{LIM}$ .
Page 134: In the paragraph after Definition 6.2.6, add right parenthesis after “greatest lower bound of $E$ “.
Page 138: In the second paragraph of Section 6.4, $-1$ should be in math mode (three instances). After $xL=L$ in the proof of Proposition 6.3.10, add “(here we use Exercise 6.1.3.)”.
Page 140: In the first paragraph, $-1$ should be in math mode.
Page 143, penultimate paragraph: add right parenthesis after “ $L^+$ and $L^-$ are finite”.
Page 144: In Remark 6.4.16, “allows to compute” should be “allows one to compute”.
Page 147: “(see Chapter 1)” should be “(see Chapter 1 of Analysis II)”.
Page 148: In the first sentence of Section 6.6, replace $(a_n)_{n=1}^\infty$ to $(a_n)_{n=m}^\infty$ . After Definition 6.6.1, add “More generally, we say that $(b_n)_{n=m'}^\infty$ is a subsequence of $(a_n)_{n=m}^\infty$ if there exists a strictly increasing function $f: \{ n \in {\bf N}: n \geq m'\} \to \{ n \in {\bf N}: n \geq m\}$ such that $b_n = a_{f(n)}$ for all $n \in {\bf N}$ .”.
Page 153: Just before Proposition 6.7.3, “Section 6.7” should be “Section 5.6”.
Page 157: At the end of Definition 7.1.6, add the sentence “In some cases we would like to define the sum $\sum_{x \in X} f(x)$ when $f: Y \to {\bf R}$ is defined on a larger set $Y$ than $X$ . In such cases we use exactly the same definition as is given above.”
Page 161: In Remark 7.1.12, change “the rule will fail” to “the rule may fail”.
Page 163: In the proof of Corollary 7.1.14, the function $h$ should be replaced with its inverse (thus $h: Y \times X \to X \times Y$ is defined by $h(y,x) := (x,y)$ . In Exercise 7.1.5, “Exercise 19.2.11” should be “Exercise 7.2.11 of Analysis II“.
Page 166: In Remark 7.2.11 add “We caution however that in most other texts, the terminology “conditional convergence” is meant in this latter sense (that is, of a series that converges but does not converge absolutely).
Page 172: In Corollary 7.3.7, $q$ can be taken to be a real number instead of rational, provided we mention Proposition 6.7.3 next to each mention of Lemma 5.6.9.
Page 175: A space should be inserted before the (why?) before the first display.
Page 176: In Exercise 7.4.1, add “What happens if we assume $f$ is merely one-to-one, rather than increasing?”. Add a new Exercise 7.4.2.: “Obtain an alternate proof of Proposition 7.4.3 using Proposition 7.4.1, Proposition 7.2.14, and expressing $a_n$ as the difference of $a_n+|a_n|$ and $|a_n|$ . (This proof is due to Will Ballard.)”
Page 177: In beginning of proof of Theorem 7.5.1, add “By Proposition 7.2.14(c), we may assume without loss of generality that $m \geq 1$ (in particulaar $|a_n|^{1/n}$ is well-defined for any $n \geq m$ ).”.
Page 178: In the proof of Lemma 7.5.2, after selecting $N$ , add “without loss of generality we may assume that $N \geq 1$ “. (This is needed in order to take n^th roots later in the proof.) One can also replace $|a_n|^{1/n} \geq 1$ and $|a_n| \geq 1$ with $|a_n|^{1/n} > 1$ and $|a_n| > 1$ respectively.
Page 186: In Exercise 8.1.4, Proposition 8.1.5 should be Corollary 8.1.6.
Page 187, After Definition 8.2.1, the parenthetical “(and Proposition 3.6.4)” may be deleted.
Page 188: In the final paragraph, after the invocation of Proposition 6.3.8, “convergent for each $m$ ” should be “convergent for each $n$ “.
Page 189, middle: in “Why? use induction”, “use” should be capitalised.
Page 190: In the remark after Lemma 8.2.5, “countable set” should be “at most countable set”.
Page 193: In Exercise 8.2.6, both summations $\sum_{m=N}^\infty$ should instead be $\sum_{m=0}^N$ .
Page 198: In Example 8.4.2, replace “the same set” with “essentially the same set (in the sense that there is a canonical bijection between the two sets)”.
Page 203: In Definition 8.5.8, “every non-empty subset of $Y$ has a minimal element $\mathrm{min}(Y)$ ” should be “every non-empty subset $Z$ of $Y$ has a minimal element $\mathrm{min}(Z)$ “.
Page 203: In Proposition 8.5.10, “Prove that $P(n)$ is true” should be “Then $P(n)$ is true”.
Page 204: Before “Let us define a special class….”, add “Henceforth we fix a single such strict upper bound function $s$ “.
Page 205: The assertion that $Y_\infty \cup \{ s(Y_\infty)\}$ is good requires more explanation. Replace “Thus this set is good, and must therefore be contained in $Y_\infty$ ” with : “We now claim that $Y_\infty \cup \{ s(Y_\infty)\}$ is good. By the preceding discussion, it suffices to show that $x = s( \{ y \in Y_\infty \cup \{ s(Y_\infty)\}: y < x \}$ when $x \in (Y_\infty \cup \{s(Y_\infty)\}) \backslash \{x_0\}$ . If $x = s(Y_\infty)$ this is clear since $\{ y \in Y_\infty \cup \{ s(Y_\infty)\}: y < x \} = Y_\infty$ in this case. If instead $x \in Y_\infty$ , then $x \in Y$ for some good $Y$ . Then the set $\{ y \in Y_\infty \cup \{ s(Y_\infty)\}: y < x \}$ is equal to $\{ y \in Y: y < x \}$ (why? use the previous observation that every element of $Y' \backslash Y$ is an upper bound for $x$ for every good $Y'$ ), and the claim then follows since $Y$ is good. By definition of $Y_\infty$ , we conclude that the good set $Y_\infty \cup \{ s(Y_\infty)\}$ is contained in $Y_\infty$ “. In the statement of Lemma 8.5.15, add “non-empty” before “totally ordered subset”.
Page 206: Remove the parenthetical “(also called the principle of transfinite induction)” (as well as the index reference), and in Exercise 8.5.15 use “Zorn’s lemma” in place of “principle of transfinite induction”. In Exercise 8.5.6, “every element of $x$ ” should be “every element of $X$ “.
Page 208: In Exercise 8.5.18, “Tthus” should be “Thus”. In Exercise 8.5.16, “total orderings of $P$ ” should be “total orderings of $X$ “.
Page 215: Exercise 9.1.1 should be moved to be after Exercise 9.1.6, as the most natural proof of the former exercise uses the latter.
Page 216: In Exercise 9.1.8, add the hypothesis that $I$ is non-empty. In Exercise 9.1.9, delete the hypothesis that $x$ be a real number.
Page 221: At the end of Remark 9.3.7, $\lim_{x \in x_0; x \in E \backslash \{x_0\}}$ should be $\lim_{x \to x_0; x \in E \backslash \{x_0\}}$ .
Page 222: Replace the second sentence of proof of Proposition 9.3.14 by “Let $(a_n)_{n=0}^\infty$ be an arbitrary sequence of elements in $E$ that converges to $x_0$ .”
Page 223: Near bottom, in “Why? use induction”, “use” should be capitalised.
Page 224: In Example 9.3.17, (why) should be (why?). In Example 9.3.16, “drop the set $X$ ” should be “drop the set $E$ “, and change $\lim_{x \in x_0; x \in X}$ to $\lim_{x \to x_0; x \in X}$ .
Page 225: In Example 9.3.20, all occurrences of ${\bf R} - \{1\}$ should be ${\bf R} - \{-1\}$ .
Page 226: After Definition 9.4.1, add “We also extend these notions to functions $f: X \to Y$ that take values in a subset $Y$ of ${\bf R}$ , by identifying such functions (by abuse of notation) with the function $\tilde f: X \to {\bf R}$ that agrees everywhere with $f$ (so $\tilde f(x) = f(x)$ for all $x \in X$ ) but where the codomain has been enlarged from $Y$ to ${\bf R}$ .
Page 230: In Exercise 9.4.1, “six equivalences” should be “six implications”. “Exercise 4.25.10” should be “Exercise 4.25.10 of Analysis II“.
Page 231: In the second paragraph after Example 9.5.2, Proposition 9.4.7 should be 9.3.9. In Example 9.5.2, all occurrences of $x_0$ should be $0$ . In the sentence starting “Similarly, if $(b_n)_{n=0}^\infty$ …”, all occurrences of $a_n$ should be $b_n$ .
Page 232: In the proof of Proposition 9.5.3, in the parenthetical (Why? the reason…), “the” should be capitalised. Proposition 9.4.7 should be replaced by Definition 9.3.6 and Definition 9.3.3.
Page 233-234: In Definition 9.6.1, replace “if” with “iff” in both occurrences.
Page 235: In Definition 9.6.5, replace “Let …” with “Let $X$ be a subset of ${\bf R}$ , and let …”.
Page 237: Add Exercise 9.6.2: If $f,g: X \to {\bf R}$ are bounded functions, show that $f+g, f-g$ , and $f \cdot g$ are also bounded functions. If we furthermore assume that $g(x) \neq 0$ for all $x \in X$ , is it true that $f/g$ is bounded? Prove this or give a counterexample.”
Page 248: Remark 9.9.17 is incorrect. The last sentence can be replaced with “Note in particular that Lemma 9.6.3 follows from combining Proposition 9.9.15 and Theorem 9.9.16.”
Page 252: In the third display of Example 10.1.6, both occurrences of $(0,\infty)$ should be $(-\infty,0)$ .
Page 253: In the paragraph before Corollary 10.1.12, after “and the above definition”, add “, as well as the fact that a function is automatically continuous at every isolated point of its domain”.
Page 256: In Exercise 10.1.1, $Y \subset X$ should be $Y \subseteq X$ , and “also limit point” should be “also a limit point”.
Page 257: In Definition 10.2.1, replace “Let …” with “Let $X$ be a subset of ${\bf R}$ , and let …”. In Example 10.2.3, delete the final use of “local”. In Remark 10.2.5, $\subset$ should be $\subseteq$ .
Page 259: In Exercise 10.2.4, delete the reference to Corollary 10.1.12.
Page 260: In Exercise 10.3.5, $X \subset {\bf R}$ should be $X \subseteq {\bf R}$ .
Page 261: In Lemma 10.4.1 and Theorem 10.4.2, add the hypotheses that $X, Y \subseteq {\bf R}$ , and that $x_0,y_0$ are limit points of $X,Y$ respectively.
Page 262. In the parenthetical ending in “$latex f^{-1} is a bijection”, a period should be added.
Page 263: In Exercise 10.4.1(a), Proposition 9.8.3 can be replaced by Proposition 9.4.11.
Page 264: In Proposition 10.5.2, the hypothesis that $f,g$ be differentiable on $[a,b]$ may be weakened to being continuous on $[a,b]$ and differentiable on $(a,b]$ , with $g'$ only assumed to be non-zero on $(a,b]$ rather than $[a,b]$ . In the second paragraph of the proof “converges to $x$ ” should be “converges to $a$ “.
Page 265: In Exercise 10.5.2, Exercise 1.2.12 should be Example 1.2.12.
Page 266: “Riemann-Steiltjes” should be “Riemann-Stieltjes”.
Page 267: In Definition 11.1.1, add “ $X$ is nonempty and” before “the following property is true”, and delete the mention of the empty set in Example 11.1.3. In Lemma 11.1.4, replace “connected” by “either connected or empty”. (The reason for these changes is to be consistent with the notion of connectedness used in Analysis II and in other standard texts. -T.)
In the start of Appendix A.1, “relations between them (addition, equality, differentiation, etc.)” should be “operations between them (addition, multiplication, differentiation, etc.) and relations between them (equality, inequality, etc.)”.
Page 276: In the proof of Lemma 11.3.3, the final inequality should involve $f$ on the RHS rather than $g$ .
Page 280: In Remark 11.4.2, add “We also observe from Theorem 11.4.1(h) and Remark 11.3.8 that if $f: [a,b] \to {\bf R}$ is Riemann integrable on a closed interval $[a,b]$ , then $\int_{[a,b]} f = \int_{(a,b]} f = \int_{[a,b)} f = \int_{(a,b)} f$ .
Page 282: In Corollary 11.4.4, replace” $|f| = f_+ - f_-$ ” by “ $|f|$ , defined by $|f|(x) := |f(x)|$ “, and add at the end “(To prove the last part, observe that $|f| = f_+ - f_-$ .)”
Page 283: In the penultimate display, $(\overline{f_+}-\underline{f_-}(x))$ should be $(\overline{f_+}-\underline{f_-})(x)$ .
Page 284: Exercise 11.4.2 should be moved to Section 11.5, since it uses Corollary 11.5.2.
Page 288: In Exercise 11.5.1, (h) should be (g).
Page 291: In the paragraph before Definition 11.8.1, remove the sentences after “defined as follows”. In Definition 11.8.1, add the hypothesis that $\alpha$ be monotone increasing, and $X$ be an interval that is closed in the sense of Definition 9.1.15, and alter the definition of $\alpha[I]$ as follows. (i) If $I$ is empty, set $\alpha[I]:=0$ . (ii) If $I = \{a\}$ is a point, set $\alpha[\{a\}] := \lim_{x \to a^+; x \in X} \alpha(x) - \lim_{x \to a^-; x \in X} \alpha(x)$ , with the convention that $\lim_{x \to a^+; x \in X} \alpha(x)$ (resp. $\lim_{x \to a^-; x \in X} \alpha(x)$ ) is $\alpha(a)$ when $a$ is the right (resp. left) endpoint of $X$ . (iii) If $I = (a,b)$ , set $\alpha[(a,b)] := \lim_{x \to b^-; x \in X} \alpha(x) - \lim_{x \to a^+; x \in X} \alpha(x)$ . (iv) If $I = [a,b)$ , $(a,b]$ , or $[a,b]$ , set $\alpha[I]$ equal to $\alpha(\{a\}) + \alpha((a,b))$ , $\alpha((a,b)) + \alpha(\{b\})$ , or $\alpha(\{a\}) + \alpha((a,b)) + \alpha(\{b\})$ respectively. After the definition, note that in the special case when $\alpha$ is continuous, the definition of $\alpha[I]$ for $I = (a,b), [a,b), (a,b], [a,b]$ simplifies to $\alpha[I] = \alpha(b) - \alpha(a)$ , and in this case one can extend the definition to functions $\alpha$ that are continuous but not necessarily monotone increasing. In Example 11.8.2, restrict the domain of $\alpha$ to $[0,+\infty)$ , and delete the example of $\alpha[(-3,-2)]$ .
Page 292: In Example 11.8.6, restrict the domain of $\alpha$ to $[0,+\infty)$ . In Lemma 11.8.4 and Definition 11.8.5, add the condition that $X$ be an interval that is closed, and $\alpha$ be monotone increasing or continuous.
Page 293: After Example 11.8.7, delete the sentence “Up until now, our function… could have been arbitrary.”, and replace “defined on a domain” with “defined on an interval that is closed” (two occurrences).
Page 294: The hint in Exercise 11.8.5 is no longer needed in view of other corrections and may be deleted.
Page 295: In the proof of Theorem 11.9.1, after the penultimate display $|F(y)-F(x)| \leq M|x-y|$ , one can replace the rest of the proof of continuity of $F$ with “This implies that $F$ is uniformly continuous (in fact it is Lipschitz continuous, see Exercise 10.2.6), hence continuous.”
Page 297: In Definition 11.9.3, replace “all $x \in I$ ” with “all limit points $x$ of $I$ “. In the proof of Theorem 11.9.4, insert at the beginning “The claim is trivial when $b=a$ , so assume $b > a$ , so in particular all points of $[a,b]$ are limit points.”. When invoking Lemma 11.8.4, add “(noting from Proposition 10.1.10 that $F$ is continuous)”.
Page 298: After the assertion $F[J] = F(d)-F(c)$ , add “Note that $F$ , being differentiable, is continuous, so we may use the simplified formula for the $F$ -length as opposed to the more complicated one in Definition 11.8.1.”
Page 299: In Exercise 11.9.1, $q$ should lie in ${\bf Q} \cap (0,1)$ rather than ${\bf Q} \cap [0,1]$ . In Exercise 11.9.3, $x_0$ should lie in $(a,b)$ rather than $[a,b]$ . In the hint for Exercise 11.9.2, add “(or Proposition 10.3.3)” after “Corollary 10.2.9”.
Page 300: In the proof of Theorem 11.10.2, Theorem 11.2.16(h) should be Theorem 11.4.1(h).
Page 310: in the last line, “all logicallly equivalent” should be “all logically equivalent”.
Page 311: In Exercise A.1.2, the period should be inside the parentheses.
Page 327: In the proof of Proposition A.6.2, $0 \leq y \leq x$ may be improved to $0 < y < x$ ; similarly for the first line of page 328. Also, the “mean value theorem” may be given a reference as Corollary 10.2.9.
Page 329: At the end of Appendix A.7, add “We will use the notation $X \equiv Y$ to indicate that a mathematical object $X$ is being identified with a mathematical object $Y$ .”
Page 334: In the last paragraph of the proof of Theorem B.1.4, “the number $a_n \dots a_0$ has only one decimal representation” should be “the number $a_n \dots a_1$ has only one decimal representation”.

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 Adam, James Ameril, Paulo Argolo, José Antonio Lara Benítez, Dingjun Bian, Philip Blagoveschensky, Tai-Danae Bradley, Brian, Eduardo Buscicchio, Matheus Silva Costa, Gonzales Castillo Cristhian, Ck, William Deng, Kevin Doran, Lorenzo Dragani, Evangelos Georgiadis, Elie Goudout, Ti Gong, Cyao Gramm, Christian Gz., Ulrich Groh, Minyoung Jeong, Erik Koelink, Brett Lane, David Latorre, Matthis Lehmkühler, Bin Li, Percy Li, Ming Li, Mufei Li, Manoranjan Majji, Mercedes Mata, Simon Mayer, Pieter Naaijkens, Vineet Nair, Cristina Pereyra, Huaying Qiu, David Radnell, Tim Reijnders, Issa Rice, Eric Rodriguez, Pieter Roffelsen, Luke Rogers, Feras Saad, Gabriel Salmerón, Vijay Sarthak, Leopold Schlicht, Marc Schoolderman, Rainer aus dem Spring, SkysubO, Sundar, Karim Taha, Chaitanya Tappu, Winston Tsai, Kent Van Vels, Andrew Verras, Daan Wanrooy, John Waters, Yandong Xiao, Hongjiang Ye, Luqing Ye, Muhammad Atif Zaheer, and the students of Math 401/501 and Math 402/502 at the University of New Mexico for corrections.

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3 July, 2021 at 11:22 am

Darshan Pillay

Dear Professor Tao,

I am currently using the 3rd corrected edition of Analysis 1 and I am having some trouble understanding Definition 9.3.6 (Convergence of functions at a point) and also Definition 9.3.3 (Local epsilon-closeness). In particular when we set X to be the empty set then X has no adherent points and I am not sure how to make sense of Definition of 9.3.3 and Definition 9.3.6 in this case. In the Definition 9.3.3 do we impose that X is non-empty and in Definition 9.3.6 do we leave the limit undefined when X is empty?

Kind Regards,
Darshan Pillay

3 July, 2021 at 11:26 am

Darshan Pillay

in Definition 9.3.6 do we also require that E be non-empty?

6 July, 2021 at 12:34 am

Terence Tao

The non-emptiness of $X$ or $E$ is a consequence of the hypotheses of the definition (since, as you say, there would otherwise not be any adherent points), but does not need to be explicitly stated; there is no way in which Definition 9.3.6 could be applied to an empty set since one could not produce the adherent point $x_0$ that is part of the concept being defined.

6 July, 2021 at 6:53 am

yalikes

Dear Professor Tao,

I am reading the 3rd edition of Analysis 1, and I have a question about the definition of $\alpha$ -length in Riemann-Stieltjes integral(in the Errata to the third edition, definition 11.8.1),

Do we need function $\alpha$ to be bounded or domain of $\alpha$ X to be closed interval?
you see, if $I := (0,1), X:=(0,1)$ and $\alpha(x)=1/(1-x)$

$\alpha[I]$ does not define.

This means for some $\alpha$ , $\alpha$ -length didn’t define for some bounded interval I. length of interval always have a definition, but $\alpha$
doesn’t. Is this intended?

I am confusing about this.
(English is not my native language, sorry).

[Fair enough, I have now required $X$ to be closed. -T]

13 July, 2021 at 10:57 am

William Deng

If $X$ is now required to be closed, then would the domain $[0,+\infty)$ in Examples $11.8.2$ and $11.8.6$ , or the domain $\textbf R=(-\infty,+\infty)$ in Examples $11.8.3$ and $11.8.7$ be considered closed? They wouldn’t from the perspective of Definition $9.1.1$ , but would be relatively closed with respect to $\textbf R$ (as in Section $1.3$ of Analysis II).

[Here we are using the notion of closed set from Definition 9.1.15; I’ve updated the errata accordingly. -T]

11 July, 2021 at 12:17 am

Anonymous

Dear prof tao.

in Proposition 5.5.12, the book says “Since x^2 >2, we can choose 0<e2 , and thus (x-e)^2 >2”

how can we choose an e??

13 July, 2021 at 12:50 pm

William Deng

In Exercise $8.5.6$ , after the map $f(x):=(x)$ is defined, it says “that sends every element of $x$ to its order ideal” when it should say “that sends every element of $X$ to its order ideal”. In Exercise $8.5.16$ , there is a part that read “Show that the maximal elements of $P$ are precisely the total orderings of $P$ ” when it should instead be “the total orderings of $X$ “.

14 July, 2021 at 12:07 pm

Terry, this page used to load all the comments at once and one can easily search among them conveniently. Now it loads only a few. Do you know a way to search the comments conveniently (e.g. find all the comments by “Terence Tao”)?

16 July, 2021 at 9:35 am

William Deng

In Definition $8.5.1$ , don’t we technically also need to know the nature of the equality relation which has been defined on the objects of $X$ in order to check anti-symmetry for instance? So instead of viewing a poset as just a pair $(X,\leq_X)$ , we should view it as $(X,\leq_X,=_X)$ , where $=_X$ is the equality relation which has been defined on the elements of $X$ ?

[All first-class mathematical objects are understood to have an equality relation assigned to them. -T]

20 July, 2021 at 10:20 am

William Deng

In Zorn’s lemma (Lemma $8.5.15$ ), is the assumption that $X\neq\emptyset$ really necessary? I believe the other hypothesis that every chain in $X$ has an upper bound ( $\textbf{in}$ $X$ ) is sufficient to ensure that $X\neq\emptyset$ .

[Technically yes, but in practice one only verifies the upper bound property for non-empty totally ordered sets when invoking Zorn’s lemma and then one needs the additional hypothesis that $X$ is non-empty, which is in analogy with the base case of mathematical induction. I’ve added an erratum to reflect this. -T]

27 July, 2021 at 10:26 pm

Anonymous

Dear Professor Tao,
According to Example 8.4.2,if I is a set of the form \{i\in:1≤i≤n\},then the infinite cartesian products is the same set as the set n-fold cartesian product defined in Definition 3.5.7.However,from Exercise 3.5.2 we can kwon ordered n-tuple in n-fold cartesian product must be a surjective function.But the functions in the infinite cartesian products are not necessarily surjective.Why the two set are the same?
(English is not my native language, hope you can understand,sorry!)
Thank you for reading my comment!

[Fair enough, the more precise statement is that $\prod_{i=1}^n X_i$ and $\prod_{\alpha \in \{i \in {\bf N}: 1 \leq i \leq n\}} X_\alpha$ can be canonically identified through a one-to-one correspondence, rather than being literally equal, though in most cases one can safely “abuse notation” by viewing the two sets as the same. (The point being that a function $x : \{i \in {\bf N}: 1 \leq i \leq n\} \to \bigcup_\alpha X_\alpha$ in the latter set can be uniquely identified with a surjective function $x' : \{i \in {\bf N}: 1 \leq i \leq n\} \to X$ in the former set by taking $X$ to be the range of $x$ . -T]

3 August, 2021 at 9:27 pm

Anonymous

Dear Prof Tao,
In the page 100, the book says “If we are to define the real numbers from the Cauchy sequences of rationals as limits of Cauchy sequences, we have to know when two Cauchy sequences of rationals give the same limit, without first defining a real number(since that would be circular).

I can’t understand what argument would be circular.

a circular argument would be like this.

“why is A?” “because B” “then why is B?” “because A”

but I can’t find what argument would be circular in this 100p paragraph.

please help!

4 August, 2021 at 1:05 pm

solveallx

I don’t think the inferences are circular, but rather, the definition would be circular. A circular argument goes the way that you say, but a circular definition would go something like this:

“What is A? It is a thing which we can define in terms of B. But what is B? It is a thing that we can define in terms of A.”

Here the challenge is to define the equality of real numbers, assuming that we define a real number to be any Cauchy sequence of rational numbers. If x, y are two real numbers, how should we define x=y? We cannot say that this equation holds so long as x and y converge to the same real number, because if we did, then we would be defining the real numbers by reference to the real numbers. That would be a circular definition.

4 August, 2021 at 4:47 am

William Deng

Dear Professor Tao
I believe that due to the revisions that were made to Definition $11.8.1$ , the hint in Exercise $11.8.5$ is now obsolete since one can directly find piecewise constant functions majorizing and minorizing $f$ whose (piecewise constant) Riemann-Stieltjes integral is precisely $2f(0)$ . For instance, if $f:[-1,1]\to\textbf R$ , being continuous on a closed interval, is bounded by some real number $M$ , then the functions $f^+,f^-:[-1,1]\to\textbf R$ defined by $f^{\pm}(0):=f(0)$ and $f^{\pm}(x):=\pm M$ for $x\neq 0$ majorize and minorize $f$ respectively and are both piecewise constant with respect to the partition $\{[-1,0),\{0\},(0,1]\}$ of $[-1,1]$ , and in both cases we have

$\displaystyle p.c.\int_{[-1,1]}f^{\pm}d\mathrm{sgn}=2f(0)$

[Erratum added, thanks – T.]

4 August, 2021 at 12:56 pm

solveallx

Dear professor Tao, in your development of Fubini’s theorem of infinite sums, you show that if a sum over a countable set is absolutely convergent then it equals the iterated sum. (By the way, I searched FAR and wide to find someone who proves this, and of about 30 texts that I looked at, yours was one of two that contained such a proof! So I deeply thank you for this!)

However, I am wondering if the converse is true (if the iterated series converges then the sum over the terms is absolutely convergent). I mention this in part because I wonder if you would enjoy including this in your text … and of course selfishly, because I’d enjoy seeing a proof. :)

Anyway, if you ever do include it in a future version I’ll be happy to see it.

Thank you again for the text.

Best wishes,
Adam Frank

[Convergence of the iterated series is not sufficient; see Example 1.2.5. But absolute convergence of the iterated series suffices; this follows from the Fubini-Tonelli theorem, which is covered in my measure theory text. -T]

4 August, 2021 at 12:57 pm

solveallx

(Note: I have not found the converse in ANY of the texts that I’ve searched.)

17 August, 2021 at 11:34 pm

guanyuminghe

Dear professor Tao,
in your introduction of Cartesian products (finite or infinite), you first give definitions, then state their existence by the finite choice lemma and the axiom of choice.

I noticed that in these two statements, you only say that the cartesian product set is non-empty. I wonder how can this non-emptiness imply that the sets contain exactly the elements that the definitions give.

I appreciate your response about this text.

20 August, 2021 at 9:48 pm

guanyuminghe

Dear professor Tao,
I have a question about the proof of Proposition 8.1.5, in which you construct an infinite sequence by $a_n:=\min\{x \in X: x \ne a_m \forall m < n\}$ .

I wonder if this construction requires the axiom of choice or a similar assertion. It is to select an infinite sequence from an infinite set. By induction we can know that for every finite $n$ , there is a finite sequence. However, can we really show that the entire sequence, the function $f$ is defined on the entire $\mathbb{N}$ ?

I know that by the well-ordering principle, every $a_n$ is well-defined, but, the problem is, it is trying to define infinitely.

By the way, if it really requires the axiom of choice, I don't know how to apply the axiom since that the choices we made are all definitive.

Thank you for reading my comment.

22 August, 2021 at 1:12 pm

Terence Tao

The axiom of choice is not required here. For each $N$ , one can define the partial function $\{1,\dots,N\} \to \{a_1,\dots,a_N\}$ , or equivalently define the partial graph $\{ (1,a_1),\dots,(N,a_N) \}$ . Applying the axioms of union (and replacement), one can then define the complete graph $\{ (1,a_1), (2,a_2), \dots \}$ , and this gives the full function $n \mapsto a_n$ .

21 August, 2021 at 4:38 am

Anonymous

Dear prof Tao.
hello professor
Is equality a relation?
some books are saying that the set $latex\left \{ \left ( x,x \right )|x\in S \right \}$ is equality relation for any given set S.
But I think this definition already implies what’s the same and what’s different.

Is it correct definition of equality?

21 August, 2021 at 9:38 am

solveallx

Yes, equality is a relation. The definition you gave does not use the notion of equality, so it’s not a circular definition. What it uses is the ability of a quantifier to pick out any x in the domain and use it (repeatedly) to state something about it.

22 August, 2021 at 2:00 am

Anonymous

if you think like that, then can you prove that this equality relation is an equivalence relation only using the definition which I gave ? {(x, x)|x $\ in S$ }

22 August, 2021 at 1:39 pm

solveallx

Yep. For all x: (x,x) is in the set. And if (x,y) is in the set then (x,y)=(x,x)=(y,x) so the relation is symmetric. And so on.

22 August, 2021 at 8:41 pm

Anonymous

I said only using the definition. you did like this. (x,y)=(x,x), -> x=x, y=x -> x=x, x=y -> (y,x)=(x,x). but the equality signs in here ” x=x, y=x -> x=x, x=y ” is not what we defined. it is something that we should already know to define the set {(x, x)|x in S}.

22 August, 2021 at 5:01 pm

Anonymous

dear professor tao, i wonder about realation between Proposition 5.4.14 and Definition 5.4.5. how can i understand about it?

4 September, 2021 at 1:45 am

johnhtwan

Dear Professor Tao,

In Analysis I Ch3, is “Equality of set” an axiom or definition? Coz in my hardcover book it’s an axiom, while in electronic copy it’s a definition. Thank you so much!

John

21 September, 2021 at 7:27 am

Atom

Professor.

I sent an errata up to Chapter 6 of Analysis I (in a LaTeX pdf) on your email: tao@math.ucla.edu. But I guess that since you are a very busy person, I thought of bringing that to your notice here. (seeing that you take out time for comments here).

For your reference, I sent the email on September 8 (11:25 pm IST), entitled “Errata for Analysis I”.

(Feel free to delete this comment.)

30 September, 2021 at 7:19 am

Anonymous

Dear professor Tao,

In the Errata it said that in Exercise 8.4.3, “there exists an injection $f: A \to B$ ; in other words…” should be “there exists an injection $f: A \to B$ with $g \circ f: A \to A$ the identity map; in particular…”. (This is needed in order to establish the converse part of the question.)
But I seem to solve the converse part of the question without this errata and I could not find any mistake in the proof. My proof is as follows. Could you please check whether it is valid?
Suppose for arbitrary sets $A,B$ such that there exists a surjection $g: B \to A$ , there then exists an injection $f: A \to B$ . We need to show that the axiom of choice is true.
Let $I$ be a set, and for each $\alpha \in I$ let $X_{\alpha}$ be a non-empty set. Suppose that all the sets $X_{\alpha}$ are disjoint from each other. Then for each $\alpha \in I$ , we define a function $g_{\alpha}: X_{\alpha} \to \{\alpha\}$ by $g_{\alpha}(x):= \alpha$ .As $X_{\alpha}$ is a non-empty set for each $\alpha \in I$ , we can see that $g_{\alpha}$ is a surjection for each $\alpha \in I$ . So there exists an injection $f_{\alpha}: {\alpha} \to X_{\alpha}$ for each $\alpha \in I$ . Now if we define set $Y :=\{ f_{\alpha}(\alpha): \alpha \in I \}$ ,we will have $\#(Y \cap X_{\alpha})=1$ for all $\alpha \in I$ . By Exercise 8.4.2 we can know that the axiom of choice is true.
English is not my native language, hope you can understand, sorry!
Thank you for reading my comment!

3 October, 2021 at 4:46 pm

Terence Tao

In order to select a single $f_\alpha$ for each $\alpha \in I$ , you have to invoke the axiom of choice, which is what you are trying to prove.

4 October, 2021 at 2:12 am

Woody Young

Dear Professor Tao, I’d like to refer to Definition 6.2.6 in your book Analysis I for Supremum of sets of extended reals. Suppose $E=A\cup\{+\infty\}$ where $A\subset\mathbb{R}$ and $A\neq\emptyset$ . If $A$ is not bounded below, $\inf(E)$ doesn’t exist. Can I therefore suggest two small changes to Definition 6.2.6: [1] append to rule (c) “… provided that $\sup(E\setminus\{-\infty\})$ exists or supremum of E is not defined. [2] Append “… provided that $\sup(-E)$ exists if rule (c) is applicable or infinmum of E is not defined.” to the definition of infinmum at the bottom of 6.2.6? Please correct me if I am wrong. (By the way, thank you very much for writing the book. :-))

4 October, 2021 at 8:03 am

Terence Tao

By the analogue of Definition 5.5.10 for infima, the infimum of a set that is not bounded below is $-\infty$ by definition.

4 October, 2021 at 7:37 pm

Woody Young

Thanks, Prof. I just read Definition 5.5.10 and realized that the two cases were already covered. Thanks for your response.

13 October, 2021 at 4:23 am

Anonymous

Dear Professor Tao,
In Exercise 8.5.19,we need prove that the maximal elements of Ω are precisely the well-orderings (X,≤) of X.However,whether Ω has a maximal element is unknown.So I think it is better to change it into:if Ω has maximal elements,then the maximal elements of Ω are precisely the well-orderings (X,≤) of X.Exercise 8.5.16 also needs the same change.
What’s your opinion about this?
Thank you for reading my comment!

13 October, 2021 at 4:55 am

Anonymous

I think maybe I need a further explanation.The question here is that whether X has at least one well-ordering is unknown.If it has one,then it is easy to prove the maximal elements of Ω are precisely the well-orderings (X,≤) of X.If we only know that Ω has maximal elements,but do not know whether X has at least one well-ordering,then we should suppose for sake of contradictionthe that the maximal element is not the well-orderings (X,≤) of X,then we can find a bigger one,a contradiction.So if Ω has maximal elements,then the well-orderings (X,≤) of X exist,and the maximal elements of Ω are precisely the well-orderings (X,≤) of X.
English is not my native language,hope you can understand,sorry!

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