**Notational convention:** As in Notes 2, I will colour a statement red in this post if it assumes the axiom of choice. We will, of course, rely on every other axiom of Zermelo-Frankel set theory here (and in the rest of the course).

In this course we will often need to iterate some sort of operation “infinitely many times” (e.g. to create a infinite basis by choosing one basis element at a time). In order to do this rigorously, we will rely on *Zorn’s lemma*:

Zorn’s Lemma.Let be a non-empty partially ordered set, with the property that every chain (i.e. a totally ordered set) in X has an upper bound. Then X contains a maximal element (i.e. an element with no larger element).

Indeed, we have used this lemma several times already in previous notes. Given the other standard axioms of set theory, this lemma is logically equivalent to

Axiom of choice.Let X be a set, and let be a collection of non-empty subsets of X. Then there exists a choice function , i.e. a function such that for all .

One implication is easy:

**Proof of axiom of choice using Zorn’s lemma.** Define a *partial choice function* to be a pair , where is a subset of and is a choice function for . We can partially order the collection of partial choice functions by writing if and f” extends f’. The collection of partial choice functions is non-empty (since it contains the pair consisting of the empty set and the empty function), and it is easy to see that any chain of partial choice functions has an upper bound (formed by gluing all the partial choices together). Hence, by Zorn’s lemma, there is a maximal partial choice function . But the domain of this function must be all of , since otherwise one could enlarge by a single set A and extend to A by choosing a single element of A. (One does not need the axiom of choice to make a single choice, or finitely many choices; it is only when making infinitely many choices that the axiom becomes necessary.) The claim follows.

In the rest of these notes I would like to supply the reverse implication, using the machinery of well-ordered sets. Instead of giving the shortest or slickest proof of Zorn’s lemma here, I would like to take the opportunity to place the lemma in the context of several related topics, such as ordinals and transfinite induction, noting that much of this material is in fact independent of the axiom of choice. The material here is standard, but for the purposes of this course one may simply take Zorn’s lemma as a “black box” and not worry about the proof, so this material is optional.

– Well-ordered sets –

To prove Zorn’s lemma, we first need to strengthen the notion of a totally ordered set.

Definition 1.Awell-ordered setis a totally ordered set such that every non-empty subset A of X has a minimal element . Two well-ordered sets X, Y areisomorphicif there is an order isomorphism between them, i.e. a bijection which is monotone ( whenever ).

**Example 1.** The natural numbers are well-ordered (this is the well-ordering principle), as is any finite totally ordered set (including the empty set), but the integers, rationals, or reals are not well-ordered.

**Example 2.** Any subset of a well-ordered set is again well-ordered. In particular, if a, b are two elements of a well-ordered set, then intervals such as , , etc. are also well-ordered.

**Example 3.** If X is a well-ordered set, then the ordered set , defined by adjoining a new element to and declaring it to be larger than all the elements of X, is also well-ordered. More generally, if X and Y are well-ordered sets, then the ordered set , defined as the disjoint union of X and Y, with any element of Y declared to be larger than any element of X, is also well-ordered. Observe that the operation is associative (up to isomorphism), but not commutative in general: for instance, is not isomorphic to .

**Example 4.** If X, Y are well-ordered sets, then the ordered set , defined as the Cartesian product with the lexicographical ordering (thus if , or if and ), is again a well-ordered set. Again, this operation is associative (up to isomorphism) but not commutative. Note that we have one-sided distributivity: is isomorphic to , but is not isomorphic to in general.

**Remark 1.** The axiom of choice is trivially true in the case when X is well-ordered, since one can take to be the choice function. Thus, the axiom of choice follows from the well-ordering theorem (every set has at least one well-ordering). Conversely, we will be able to deduce the well-ordering theorem from Zorn’s lemma (and hence from the axiom of choice): see Exercise 11 below.

One of the reasons that well-ordered sets are useful is that one can perform induction on them. This is easiest to describe for the principle of strong induction:

**Exercise 1. **(Strong induction on well-ordered sets) Let X be a well-ordered set, and let be a property of elements of X. Suppose that whenever is such that is true for all , then is true. Then is true for every . This is called the *principle of strong induction*. Conversely, show that a totally ordered set X enjoys the principle of strong induction if and only if it is well-ordered. (For partially ordered sets, the corresponding notion is that of being well-founded.)

To describe the analogue of the ordinary principle of induction for well-ordered sets, we need some more notation. Given a subset A of a non-empty well-ordered set X, we define the *supremum* of A to be the least upper bound

(1)

of A (thus for instance the supremum of the empty set is ). If , we define the *successor* by the formula

. (2)

We have the following Peano-type axioms:

**Exercise 2. ** If x is an element of a non-empty well-ordered set X, show that exactly one of the following statements hold:

- (Limit case) .
- (Successor case) for some .

In particular, is not the successor of any element in X.

**Exercise 3.** Show that if x, y are elements of a well-ordered set such that , then x=y.

**Exercise 4.** (Transfinite induction for well-ordered sets) Let X be a non-empty well-ordered set, and let be a property of elements of X. Suppose that

- (Base case) is true.
- (Successor case) If and is true, then is true.
- (Limit case) If and is true for all , then is true. [Note that this subsumes the base case.]

Then is true for all .

**Remark 2.** The usual Peano axioms for succession are the special case of Exercises 2-4 in which the limit case of Exercise 2 only occurs for (which is denoted 0), and the successor function never attains . With these additional axioms, X is necessarily isomorphic to .

Now we introduce two more key concepts.

Definition 2.Aninitial segmentof a well-ordered set X is a subset Y of X such that for all (i.e. whenever y lies in Y, all elements of X that are less than y also lie in Y).A

morphismfrom one well-ordered set X to another Y is a map which is strictly monotone (thus whenever ) and such that is an initial segment of Y.

**Example 5.** The only morphism from to is the inclusion map. There is no morphism from to .

**Remark 3.** With this notion of a morphism, the class of well-ordered sets becomes a category.

We can identify the initial segments of X with elements of :

**Exercise 5. **Let X be a non-empty well-ordered set. Show that every initial segment I of X is of the form for exactly one .

**Exercise 6.** Show that an arbitrary union or arbitrary intersection of initial segments is again an initial segment.

**Exercise 7. ** Let be a morphism. Show that maps initial segments of X to initial segments of Y. If is such that x’ is the successor of x, show that is the successor of .

As Example 5 suggests, there are very few morphisms between well-ordered sets. Indeed, we have

Proposition 1.(Uniqueness of morphisms) Given two well-ordered sets X and Y, there is at most one morphism from X and Y.

**Proof. ** Suppose we have two morphisms , . By using transfinite induction (Exercise 4) and Exercise 7, one can show that agree on for every ; setting gives the claim.

**Exercise 8.** (Schroder-Bernstein theorem for well-ordered sets) Show that two well-ordered sets X, Y are isomorphic if and only if there is a morphism from X to Y, and a morphism from Y to X.

We can complement the uniqueness in Proposition 1 with existence:

Proposition 2.(Existence of morphisms) Given two well-ordered sets X and Y, there is either a morphism from X to Y or a morphism from Y to X.

**Proof. ** Call an element *good* if there is a morphism from to Y, thus is good. If is good, then we are done. From uniqueness we see that if every element in a set A is good, then the supremum is also good. Applying transfinite induction (Exercise 4), we thus see that we are done unless there exists a good such that is not good. By Exercise 5, for some . If then we could extend the morphism to by mapping a to b, contradicting the fact that is not good; thus and so is surjective. It is then easy to check that exists and is a morphism from Y to X, and the claim follows.

**Remark 4.** Formally, Proposition 1, Exercise 8, and Proposition 2 tell us that the collection of all well-ordered sets, modulo isomorphism, is totally ordered by declaring one well-ordered set X to be at least as large as another Y when there is a morphism from Y to X. However, this is not quite the case, because the collection of well-ordered sets is only a class rather than a set. Indeed, as we shall soon see, this is not a technicality, but is in fact a fundamental fact about well-ordered sets that lies at the heart of Zorn’s lemma. (From Russell’s paradox we know that the notions of class and set are necessarily distinct.)

– Ordinals –

As we learn very early on in our mathematics education, a finite set of a certain cardinality (e.g. a set ) can be put in one-to-one correspondence with a “standard” set of the same cardinality (e.g. the set ); two finite sets have the same cardinality if and only if they correspond to the same “standard” set ). (The same fact is true for infinite sets; see Exercise 12 below.) Similarly, we would like to place every well-ordered set in a “standard” form. This motivates

Definition 3.Arepresentation of the well-ordered setsis an assignment of a well-ordered set to every well-ordered set X such that

- is isomorphic to X for every well-ordered set X. (In particular, if and are equal, then X and Y are isomorphic.)
- If there exists a morphism from X to Y, then is a subset of (and the order structure on is induced from that on . (In particular, if X and Y are isomorphic, then and are equal.)

**Remark 5.** In the language of category theory, a representation is a covariant functor from the category of well-ordered sets to itself which turns all morphisms into inclusions, and which is naturally isomorphic to the identity functor.

**Remark 6.** Because the collection of all well-ordered sets is a class rather than a set, is not actually a function (it is sometimes referred to as a *class function*).

It turns out that several representations of the well-ordered sets exist. The most commonly used one is that of the ordinals, defined by von Neumann as follows.

Definition 4.An ordinal is a well-ordered set with the property that for all . (In particular, each element of is also a subset of , and the strict order relation on is identical to the set membership relation .)

**Example 5.** For each natural number , define the ordinal number recursively by setting and for all , thus for instance

(3)

and so forth. (Of course, to be compatible with the English language conventions for ordinals, we should write instead of , etc., but let us ignore this discrepancy.) One can easily check by induction that is an ordinal for every n. Furthermore, if we define , then is also an ordinal. (In the foundations of set theory, this construction, together with the axiom of infinity, is sometimes used to *define* the natural numbers (so that for all natural numbers n), although this construction can lead to some conceptually strange-looking consequences that blur the distinction between numbers and sets, such as and .)

The fundamental theorem about ordinals is

Theorem 1.

- Given any two ordinals , one is a subset of the other (and the order structure on is induced from that on ).
- Every well-ordered set X is isomorphic to exactly one ordinal .
In particular, is a representation of the well-ordered sets.

**Proof. ** We first prove 1. From Proposition 2 and symmetry, we may assume that there is a morphism from to . By strong induction (Exercise 1) and Definition 4, we see that for all , and so is the inclusion map from into . The claim follows.

Now we prove 2. If uniqueness failed, then we would have two distinct ordinals that are isomorphic to each other, but as one ordinal is a subset of the other, this would contradict Proposition 1 (the inclusion morphism is not an isomorphism); so it suffices to prove existence.

We use transfinite induction. It suffices to show that for every , that is isomorphic to an ordinal (which we know to be unique). This is of course true in the base case . To handle the successor case , we set , which is easily verified to be an ordinal isomorphic to . To handle the limit case , we take all the ordinals associated to elements in and take their union (here we rely crucially on the axiom schema of replacement and the axiom of union); by use of part 1 one can show that this union is an ordinal isomorphic to a as required.

**Remark 7. ** Operations on well-ordered sets, such as the sum and product defined in Examples 3, 4, induce corresponding operations on ordinals, leading to ordinal arithmetic, which we will not discuss here. (Note that the convention for which order multiplication proceeds in is swapped in some of the literature, thus would be the ordinal of rather than .)

**Exercise 9. **(Ordinals are themselves well-ordered) Let be a non-empty class of ordinals. Show that there is a least ordinal in this class, which is a subset of all the other ordinals in this class. In particular, this shows that any set of ordinals is well-ordered by set inclusion.

**Remark 8.** Because of Exercise 9, we can meaningfully talk about “the least ordinal obeying property P”, as soon as we can exhibit at least one ordinal with that property P. For instance, once one can demonstrate the existence of an uncountable ordinal (which follows from Exercise 11 below), one can talk about the least uncountable ordinal.

**Exercise 10.** (Transfinite induction for ordinals) Let be a property pertaining to ordinals . Suppose that

- (Base case) is true.
- (Successor case) If for some ordinal , and is true, then is true.
- (Limit case) If , and is true for all , then is true.

Show that is true for every ordinal .

Now we show a fundamental fact, that the well-ordered sets are just too “numerous” to all fit inside a single set, even modulo isomorphism.

Theorem 2.There does not exist a set A and a representation of the well-ordered sets such that for all well-ordered sets X.

**Proof.** By Theorem 1, any two distinct ordinals are non-isomorphic, and so get mapped under to a different element of A. Thus we can identify the class of ordinals with a subset of A, and so the class of ordinals is in fact a set. In particular, by the axiom of union, we may take the union of all the ordinals, which one can verify to be another ordinal . But then is another ordinal, which implies that , which contradicts the axiom of foundation.

**Remark 9.** It is also possible to prove Theorem 2 without the theory of ordinals, or the axiom of foundation. One first observes (by transfinite induction) that given two well-ordered sets X, X’, one of the sets is a subset of the other. Because of this, one can show that the union S of all the (where X ranges over all well-ordered sets) is well-defined (because the form a subset of A) and well-ordered. Now we look at the well-ordered set ; by Proposition 1, it is not isomorphic to any subset of S, but is necessarily contained in S, a contradiction.

**Remark 10.** The same argument also shows that there is no representation of the ordinals inside a given set; the ordinals are “too big” to be placed in anything other than a class.

– Zorn’s lemma –

Now we can prove Zorn’s lemma. The key proposition is

Proposition 2.Let X be a partially ordered set, and let be the set of all well-ordered sets in X. Then there does not exist a function such that is a strict upper bound for C (i.e. for all ) for all well-ordered .

**Proof. **Suppose for contradiction that there existed X and g with the above properties. Then, given any well-ordered set Y, we claim that there exists exactly one isomorphism from Y to a well-ordered set in X such that for all . Indeed, the uniqueness and existence can both be established by a transfinite induction that we leave as an exercise. (Informally, is what one gets by “applying g Y times, starting with the empty set”.) From uniqueness we see that whenever Y and Y’ are isomorphic, and another transfinite induction shows that whenever Y is a subset of Y’. Thus is a representation of the ordinals. But this contradicts Theorem 2.

**Remark 11.** One can use transfinite induction on ordinals rather than well-ordered sets if one wishes here, using Remark 10 in place of Theorem 2.

**Proof of Zorn’s lemma.** Suppose for contradiction that one had a non-empty partially ordered set X without maximal elements, such that every chain had an upper bound. As there are no maximal elements, every element in X must be bounded by a strictly larger element in X, and so every chain in fact has a strict upper bound; in particular every well-ordered set has a strict upper bound. Applying the axiom of choice, we may thus find a choice function from the space of well-ordered sets in X to X, that maps every such set to a strict upper bound. But this contradicts Proposition 2.

**Remark 12.** It is important for Zorn’s lemma that X is a set, rather than a class. Consider for instance the class of all ordinals. Every chain of ordinals has an upper bound (namely, the union of the ordinals in that chain), and the class is certainly non-empty, but there is no maximal ordinal. (Compare also Theorem 1 and Theorem 2.)

**Remark 13.** It is also important that every chain have an upper bound, and not just countable chains. Indeed, the collection of countable subsets of an uncountable set (such as ) is non-empty, and every countable chain has an upper bound, but there is no maximal element.

**Remark 14. ** The above argument shows that the hypothesis of Zorn’s lemma can be relaxed slightly; one does not need every chain to have an upper bound, merely every well-ordered set needs to have one. But I do not know of any application in which this apparently stronger version of Zorn’s lemma dramatically simplifies an argument. (In practice, either Zorn’s lemma can be applied routinely, or it fails utterly to be applicable at all.)

**Exercise 11.** Use Zorn’s lemma to establish the well-ordering theorem (every set has at least one well-ordering).

**Remark 15.** By the above exercise, can be well-ordered. However, if one drops the axiom of choice from the axioms of set theory, one can no longer prove that is well-ordered. Indeed, given a well-ordering of , it is not difficult (using Remark 1) to remove the axiom of choice from the Banach-Tarski constructions in Notes 2, and thus obtain constructions of non-measurable subsets of . But a deep theorem of Solovay gives a model of set theory (without the axiom of choice) in which every set of reals is measurable.

**Exercise 12.** Define a (von Neumann) cardinal to be an ordinal with the property that all smaller ordinals have strictly lesser cardinality (i.e. cannot be placed in one-to-one correspondence with ). Show that every set can be placed in one-to-one correspondence with exactly one cardinal. (This gives a representation of the category of sets, similar to how gives a representation of well-ordered sets.)

It seems appropriate to close these notes with a quote from Jerry Bona:

“The Axiom of Choice isobviouslytrue, the well-ordering principleobviously false, and who can tell aboutZorn’s Lemma?”

[*Update*, Jan 29: some more remarks added.]

[*Update*, Jan 30: more remarks added.]

[*Update*, Jan 31: See also "How to use Zorn's lemma", by Tim Gowers.]

## 29 comments

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28 January, 2009 at 3:28 pm

AnonymousIn example 4, when describing the well-ordering on the lexicographic ordering, I think you meant y <= y’.

[Corrected, thanks - T.]28 January, 2009 at 6:44 pm

EricI think it’s worth remarking on how a Zorn’s Lemma proof can be turned into a transfinite induction proof, as there’s a very simple algorithm. When using Zorn’s Lemma, there are two steps. First, you must prove that chains are bounded in your poset, and second you must show that a maximal element has the properties you desire. These correspond exactly to the limit and successor steps of constructing your object by transfinite induction, respectively. For example, to construct a basis for a vector space, you need that a union of a chain of linearly independent sets is linearly independent and that a maximal linearly independent set spans and is hence a basis. By transfinite induction you can construct a basis by taking unions at limit steps and adding new elements at successor steps if you don’t yet have a basis, which is always possible since not being a basis implies that what you have is non-maximal.

31 January, 2009 at 2:27 pm

kung fu pandaDear Mr Tao, Dear People,

First of all, thank you so much for all the articles on your blog and especially for this one.

I am trying to prove countable choice using dependent choice. Could someone correct me please? Thank you.

I restate dependent choice (D.C): If is a binary relation on a nonempty set and if for every a belonging to there exists a belonging to such that then there is a sequence in such that for all belonging to .

Countable choice (C.C): Every countable family of nonempty sets has a choice function.

D.C implies C.C.

proof:

Let be a family of nonempty sets and let be the set of all choice functions on some . We use the inclusion for the binary relation.

The set is nonempty because for each , smaller than , we can find a choice function on the elements of . the elements of are countable subsets so we can "choose" the least elements of the elements of the ‘s.

Secondly, for every choice function of , does there exist a choice function of such that is included in . is a subset of the product of the , elements of the ‘s and so one can find a subset of this product. Using D.C, there is a sequence in such that each is included in for all . This means that one can find a choice function on

Thank you in advance for you suggestions and corrections.

11 February, 2009 at 3:44 pm

245B, Notes 10: Compactness in topological spaces « What’s new[...] Show that every filter is contained in at least one ultrafilter. (Hint: use Zorn’s lemma, see Notes 7.) Exercise 5 A collection of subsets of has the finite intersection property if every finite [...]

16 February, 2009 at 7:28 am

实分析0-10 « Liu Xiaochuan’s Weblog[...] 第七节整节讲选择公理，我看得真是爽死了。要知道，我在很长时间里都希望找一个这方面的材料好好的学一下。本节中包括的内容有：良序集，序数，Zorn引理，超限归纳以及他们和选择公理之间一些互推关系。众所周知，上面这些多半等价，本节短短几页把其中关系写的清清楚楚。 [...]

5 November, 2009 at 12:48 am

The “no self-defeating object” argument « What’s new[...] with the theory of ordinals) can be used to give a quick proof of Zorn’s lemma: see these lecture notes of mine for further discussion. Notice the similarity between this argument and the proof of Proposition 1. [...]

25 September, 2010 at 10:58 pm

245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems « What’s new[...] algebra) (This exercise requires familiarity with the theory of ordinals, which is reviewed here. Recall that we are assuming the axiom of choice throughout this course.) Let be a collection of [...]

22 February, 2011 at 1:11 pm

AnonymousIn exercise 10, I think the succesor case should say: if $\alpha=\beta\cup\{\beta\}$ or something similar instead of $\alpha={\beta,\{\beta\}\}$ (the latter is, in most cases, not even a transitive set).

[Corrected, thanks - T.]10 March, 2012 at 5:39 pm

Luqing YeIn example3,the link of “disjoint union” directs me to 【well ordering principle】in wikipedia.The correct link: http://en.wikipedia.org/wiki/Disjoint_union

[Corrected, thanks - T.]18 March, 2012 at 3:49 am

Luqing YeDear professor Tao ,

In definition 4,I think you should check whether is well defined,i.e,if is a set such that ,,then is a well ordered set.

[The requirement that be well-ordered is part of Definition 4, and is not a consequence of the remainder of the definition. -T.]20 March, 2012 at 5:32 am

Luqing YeDear professor Tao,

Suppose \alpha is an ordinal which has at least two elements.It is easy to verify that the minimal element of is .I wonder whether the minimal element of is ……I can not prove it ,maybe it is not true….

20 March, 2012 at 10:01 am

Luqing YeWell….Now I understand this is trivally true.Let be the minimal element of ,Let be the minimal element of .The only element that belongs to is ,so ……What a stupid question.

Even though this question is stupid,it makes me fully understand that the first several elements of is .This makes ordinals more intuitive for me.

By the way,I want to say that this blog is brilliant,I’ll try my best to make some constructive comments(though sometimes my comments may seem stupid) which is relevant to the posts.I hope you will not mind my comments taking up too much space.Thanks.

21 March, 2012 at 5:46 pm

PanDear professor Tao,

I have a question about the note above. In the definition of supremum of a subset A of a well-ordering set X, it does not dependent on the choice of A or just a typo on the definition? Thanks

[That was a typo,now corrected, thanks - T.]22 March, 2012 at 5:26 am

Luqing YeHere is an another conclusion compared to Theorem 1.1:

Given any two ordinals and ,,then or .

Proof:By symmetry,Suppose .Now let’s see and .If ,then .If ,then must belong to ,otherwise, ,which is a contradiction to .

But I think this is not as useful as Theorem 1.1,because it assumes ,what a pity. :-)

24 March, 2012 at 9:13 pm

Luqing YePropsition 2 is essenially same with Lemma 8.5.14 in your text book (I read the Chinese version,because I have no access to the English version).Some time ago,I read the proof of Lemma 8.5.14 in ,but I feel rather uncomfortable after reading the proof,because I think in that proof,”the good subset of X”(I don’t know whether I translate it right or not,in Chinese,that is called 偏序集X的好子集) is not well motivated.Now after reading this post,I understand that “the good subset of X ” is essentially “the subset of X which is isomorphic to an ordinal”.

27 March, 2012 at 8:10 pm

叶卢庆(Luqing Ye)I have a hypothesis,I can’t prove it ,maybe it is not true.

Definition:An ordinal is a successor ordinal iff there exists an ordinal such that .

Now there exists an ordinal ,, is a successor ordinal,prove that is a natural number.

I don’t know whether it is true……

27 March, 2012 at 8:46 pm

叶卢庆(Luqing Ye)As soon as I write it down,I realize my stupidness.Let .Done.So the hypothesis is wrong :-(

19 May, 2012 at 8:08 am

questa123456About Remark 1, which says that if is a well-ordered set, and , then existence of a choice function is established by choosing a minimal element in the respective subset, without AC and in fact illustrating AC.

My question is, for a general , why can’t I just say I choose a “random” element from the respective set such that ? or alternatively I couldn’t give you a pre-specified rule about choosing from , but if you give to me now, I can choose a from ? By this way, the AC seems redundant for me…

So why a uniform rule, such as , is OK, but some some rule unspecified ex-ante is not OK?

In a similar line, we don’t need the Axiom of Choice to show the existence of a function , because we have .

19 May, 2012 at 9:04 am

Terence TaoOne has to make a distinction between “lazily evaluated” functions, of the type you propose, and the set-theoretic functions which occur in standard set theory (and in particular in Zermelo-Frankel set theory), defined as sets of ordered pairs. A lazily evaluated function can mimic a set-theoretic function to some extent, but only as long as one is only interested in the values of that function on a finite number of points (this is related to how the axiom of finite choice can already be deduced from the other axioms of set theory, but the axioms of countable or unrestricted choice cannot). However, many operations in set theory (such as invocation of the axiom schema of replacement) require one to manipulate an infinite number of values of the function at once, and this cannot be done with lazily evaluated functions without allowing for an infinite number of operations to be applied, which is a very dangerous step to allow as it opens the door for any number of paradoxes (e.g. the Ross-Littlewood paradox). The solution to all this in traditional set theory is to forget about lazily evaluated functions and infinite numbers of operations, and instead work exclusively with finite numbers of operations applied to set-theoretic functions. (One can then rigorously recover the intuitive idea of applying “infinitely many operations” by tools such as transfinite induction, but only as long as one works within the confines of set theory.)

However, there are other models of set theory in which lazy evaluation would be permitted, in which case the axiom of choice can be replaced with various “adaptive oracles”. This would no longer be traditional set theory, but can still be conceptually useful for some purposes. I discuss this perspective in this previous blog post.

20 May, 2012 at 6:00 am

questa123456Dear Professor Tao, thank you. When I read books on Set Theory, these things were not discussed explicitly, and they confuse me a lot. Your answer really makes me think more carefully about these issues. I think I need more time to digest the contents you mentioned, but they really help!

Thank you again.

25 January, 2013 at 8:51 pm

Luqing YeI found some of Escher’s drawings have a lot to do with Zorn’s lemma and the axiom of regurarity and russel’s paradox.For example,one of his picture is “A young man who is seeing a picture is in that picture”,another example is this picture:http://upfile.chinaschool.org.cn/image/2011/10/19/20111019093448_12375.jpg

16 May, 2013 at 12:28 pm

Arjun JainMy question is in regards to Remark 15. I am new to studying Mathematics properly, and so don’t know the precise meaning of measurability, but has anyone actually constructed a well ordering of the real number set? After reading about the well ordering theorem from Halmos’s book, i tried well- ordering different sets, and was stuck at the reals, and the internet was also not of much help here.

16 May, 2013 at 9:43 pm

Luqing YeThe well ordering principle says any set can be well ordered,so can be well ordered.

The construction of the well ordering of the is quite easy,but the construction is based on Zorn’s lemma(Or well ordering principle,they are equivalent).

You choose a real numebr every second,and spend your entire life choosing real numbers,and simultaneously arranged them one after one in order.Obviously your life is not enough,because your life is consisted of finite number of seconds.But Zorn’s lemma suppose you are a god,so you can live FOREVER.So if you are a god,then you will finally managed to choosing all the real numbers and arranged them in order.

Then is well ordered.So only god can see intuitively that a well ordering of actually exists,we are human beings ,so we can only see logically the a well ordering of exsits.

16 May, 2013 at 10:11 pm

Arjun JainSorry, but that doesn’t seem to be convincing to me. Maybe, a more precise answer?

16 May, 2013 at 10:55 pm

Luqing YeIn fact,the construction of the well ordering of is highly undetermined.It is more like a existence proof instead of a constructive proof,because the construction is essentially based on the Axiom of set(AC is equivalent to Zorn’s lemma).

As for the detailed proof of the well ordering of ,simply use the Well ordering principle.So it is not interesting to prove that can be well ordered,instead,it is interesting to prove the well ordering principle by axiom of choice.

As for the detailed proof,please see some set theory book,(or this post),especially order theory,such as ordinal numbers,well ordered sets.

16 May, 2013 at 11:54 pm

Luqing YeAfter some thought,now let me explain more precisely.

When a set is finite,it is easy to verify by mathematical induction that you can choose finitely many times,and and simultaneously arranged the elements of this finite set one after one in order.

But in the case of when the set is infinite,in order to well order this set,you should choose infinite many times and arranged them in well order,in the specific case of real number system,you should choose number of times and simultaneously arranged real numbers one after one in order,where represents the cardinality of .It is in the case undetermined infinite choice that axiom of choice must be used.

8 March, 2014 at 10:49 am

JackThis might be a dumb question: It takes me a while to realize that the statement of the Axiom of Choice in this note is equivalent to the one you gave in your Analysis I, in which you use an index set instead of using —“the collection of non-empty subsets of . Is there a particular reason that you use different forms of the statement?

8 March, 2014 at 3:35 pm

Terence TaoThis particular formulation of the axiom of choice is well adapted to the approach to proving Zorn’s lemma given in this blog post. Other formulations of the axiom of choice may be slightly more convenient for some other purposes, but ultimately it is largely a matter of taste which formulation to use for a given context. (In general, I believe it is better mathematical practice to switch freely between different equivalent formulations of a mathematical concept or statement, in order to best fit the context, than to try to stick dogmatically to a single “best” formulation regardless of context out of a misguided desire to be “consistent”. See for instance Thurston’s discussion on the concept of differentiation in Section 2 of this essay.)

8 October, 2014 at 7:45 am

MarceloGreat article! In Remark 7 you refer to Exercises 3 and 4, but the reference should be to Examples 3 and 4 instead. Also, there is an Example 1 appearing between Examples 4 and 5.

[Corrected, thanks - T.]