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I thought many results in calculus need axiom choice. For example, I thought one needs AC to prove that a bounded sequence in the real line has a convergent subsequence. Recently I was taught that one only needs mathematical induction to prove it. So here are my questions.

Can most results in calculus be proved without AC? If yes, what are the exceptions, to name a few?

Obviously a theorem which uses Zorn's lemma most likely does need AC. So please exclude obvious ones.

Edit By "without AC", I mean without any form of AC, i.e. countable or not, dependent or not. In other words, within ZF.

Edit One of the motivations of my question is as follows. People often unconciously use AC to prove theorems. And it often turns out that their uses of AC are unnecessary. For example, an infinite subset of a compact metric space has a limit point. In his book "Principles of mathematical analysis", Rudin proves this by choosing a suitable neighborhood of every point of the space. He uses AC here, though he doesn't say so. However, since a compact metric space is separable, you can avoid AC to prove this theorem.

Edit I'll make the above statement clearer. You can even avoid countable AC to prove the above theorem. In other words, you can prove it within ZF.

Edit I'll make my questions clearer and more specific. By calculus, I mean classical analysis in Euclidean spaces. Take, for example, Rudin's "Principles of mathematical analysis". Can all the results in this book be proved within ZF? If not, what are the exceptions?

Makoto Kato
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    The result that every bounded sequence in $\Bbb R$ has a convergent subsequence holds in the absence of AC. Zorn's lemma is equivalent to AC, so there's no 'most likely' about it! – Brian M. Scott May 06 '12 at 09:01
  • What if it doesn't need Zorn's lemma in the first place? It's very unlikely, but you can't say it's absolutely necessary without checking the proof. – Makoto Kato May 06 '12 at 09:19
  • Ah, I misunderstood: I thought that you were saying that needing Zorn's lemma 'most likely' meant that it needed AC, when in in fact in that case it would certainly need AC. – Brian M. Scott May 06 '12 at 09:22
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    You are wrong. It is not "often" that AC is unneeded, it is often that the full power of AC is unneeded. Most modern mathematics require some fragment of AC to behave nicely, except for very explicit cases. – Asaf Karagila May 06 '12 at 21:56
  • You repeat the same thing over and over: "prove it in ZF". You don't really specify what. It is also hard to give general examples. Sure, polynomials are still continuous and all that; the derivative of $\sin x$ is still $\cos x$; and exponents are still similar. But now you have like three possible kinds of compactness and two forms of continuity. This is a vast topic, maybe the blog post and the books I linked should prove a good starting point. – Asaf Karagila May 06 '12 at 23:14
  • @Karagila You wrote "It is not "often" that AC is unneeded,". I'm talking about calculus, which means classical analysis in Euclidean spaces. – Makoto Kato May 07 '12 at 01:57
  • @Karagila You wrote "You repeat the same thing over and over". I only repeated twice. I have no idea what's wrong with making my statements clearer. Nothing would be lost by doing so, though nothing might be gained. – Makoto Kato May 07 '12 at 02:06
  • @Makoto: My first name is Asaf. Furthermore, if you wish my user to receive a notification please use Asaf or AsafKaragila, not just Karagila. Secondly, classical analysis is still a pretty big subject. Prior to the formulation of AC many people used things like DC or countable choice without noticing. Later it was shown that such use was sometimes essential (e.g. continuity; Arzela-Ascoli for general families of functions on the real line). – Asaf Karagila May 07 '12 at 06:12
  • (Cont'd) no one is going to do the work for you. If you want to find out what can be done in ZF, start from Tao's blog and Herrlich's book. He has a section about unneeded choice too. Both are very readable. I put quite a lot of effort into my answer, and you simply "clarified" your answer by repeating something that was clear to me from the beginning. You are asking how would one go about doing analysis in $\mathbb R^n$ in ZF, and which theorems remain valid there. My answer to such a general question is: some theorems, and some particular cases of theorems remain true. – Asaf Karagila May 07 '12 at 06:15
  • @Asaf So your answer to my question "Can most results in calculus be proved within ZF?" is no? – Makoto Kato May 07 '12 at 14:43
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    No, my answer is that it depends greatly on the type of results you want. If they are very general then no. Otherwise maybe. – Asaf Karagila May 07 '12 at 14:59
  • @Asaf How about Rudin's "Principles of mathematical analysis"? Can most results in this book be proved within ZF? – Makoto Kato May 07 '12 at 15:16
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    Again, this is a way too general question. Some of the results require no choice whatsoever, other require it for most cases, some you can remove choice if you limit yourself to particular functions or subspaces. One can write books and not give you a complete answer. The reason my answer below is so general is that your question is simply too broad. I can relate and I can understand the wish to have an answer, but the question "can most freshman calculus be done without choice?" has no well-defined and reasonable answer except "Yes, but also not really, except sometimes." – Asaf Karagila May 07 '12 at 17:47
  • @Asaf What's so general? One only needs to pick up all the theorems in the Rudin's book which cannot be proven in ZF. – Makoto Kato May 08 '12 at 01:40
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    I am very sorry, but asking about an entire book is not specific at all. If you want such answer I suggest you study about set theory and foundations, read the references I linked below, then sit and write a book of your own. Otherwise this is just not going to work. – Asaf Karagila May 08 '12 at 04:42
  • @Asaf There are not so many theorems which are proved using AC in the Rudin's book. Checking all those theorems may not be easy but obviously not too hard for the knowlegeable people. – Makoto Kato May 08 '12 at 05:00
  • I don't see any person with knowledge in this field sitting through a field and checking. Furthermore it is never immediate that no choice is involved and it is even less immediate if unnecessary choice is involved. – Asaf Karagila May 08 '12 at 05:01
  • I tend to think there are only a few theorems outside of ZF if any in the Rudin's book contrary to the Asaf's opinion. – Makoto Kato May 08 '12 at 05:06
  • @Asaf Just because you can't answer it does not necessarily mean others can't. I can wait. – Makoto Kato May 08 '12 at 05:11
  • I can, I just don't see why I should do it for someone else. I would gladly help if you had asked about one theorem. Go ahead, wait. – Asaf Karagila May 08 '12 at 05:14
  • @Asaf You seem to have a different opinion from mine regarding the rate of non-ZF theorems in the Rudin's book. I think it's a good reason for you to show us you are actually right. – Makoto Kato May 08 '12 at 07:05
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    No, I think that going through the entire Rudin book and pointing out "This theorem uses choice, this one does not, and that one can be modified into a choiceless proof while this one can be reduced to a choiceless proof in the case of $C^1$ functions" is not what I should be doing. This is a great chance for you to learn about axiom of choice and foundational things and prove that it is "easy" for an expert to spot a choiceless proof and a theorem which can be proved without any choice (and then prove it too). Go ahead, become an expert and show me how easy it is. – Asaf Karagila May 08 '12 at 07:10
  • @Asaf Then how did you come up with the conclusion that I was wrong? See the first comment of yours. – Makoto Kato May 08 '12 at 09:41
  • Because most theorems are stated in ZFC and therefore can be stated in a rather broad and general way. Without the axiom of choice you may lose a bit, but it is often not a big part of the theorem which gets lost. It is enough, however, to justify the use of choice. So yes, most theorems do require some choice, but several relatively minor changes can be made to make the theorem true in ZF. – Asaf Karagila May 08 '12 at 10:15
  • @Asaf How do you know most theorems in the Rudin's book do require some choice? – Makoto Kato May 08 '12 at 10:57
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    I secretly read it, but now I just want to make your squiggle for the information. Seriously, though, I don't. I just know what are the standard theorems in freshman calculus and I know a few things about the axiom of choice. My guess is that most books (and especially the book which is assumed as the standard textbook) will talk about these theorems in the fashion they are often presented. I have to ask, why do you want to know about Rudin's book without the axiom of choice? It seems as though there is some agenda behind this request... – Asaf Karagila May 08 '12 at 11:00
  • @Asaf Someone told me almost all theorems in elementary calculus can be proven in ZF. I didn't believe him. He showed me how the theorem above on a bounded sequence can be proven using only induction. It was an eye opener. I want to make sure what he said. Besides I prefer proofs that don't use AC. They are down-to-earth. – Makoto Kato May 08 '12 at 11:28
  • Then you should ask that someone. You should also study for yourself and figure things out on your own. Continuity at a point is no longer equivalent without some choice; the different form of compactness are no longer equivalent without any choice; now theorems which make use of compactness need to be modified and theorems which make use of continuity need to be modified. The Arzela-Ascoli theorem fails; measurability related questions are totally out of the question; almost anything which you prove by induction need to be carefully double-checked. But sure, polynomials still play nice. – Asaf Karagila May 08 '12 at 11:34
  • @Asaf I got a nice question badge on this one. It means not a few people are interested. Do you have some hidden agenda which makes you against my question? What do you mean by "continuity at a point is ..."? – Makoto Kato May 08 '12 at 11:59
  • Yes. This is why I put so much into my answer and this is why I keep this discussion with you. Because I really don't want you to have an answer. On the other hand, $f\colon A\to\mathbb R$ where $A\subseteq\mathbb R$ is continuous at $x_0$ if for every $\varepsilon>0$ there is $\delta>0$ such that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\varepsilon$. To show that this is the same as to say $f(\lim x_n)=\lim f(x_n)$ for $\lim x_n=x_0$ requires some choice to be present. So is the equivalence between sequential compactness (every sequence has a convergent subsequence) and compactness too. – Asaf Karagila May 08 '12 at 12:04
  • @Asaf I don't know why you don't want me to get an answer. I think Herrlich's The Axiom of Choice is helpfull. It'd be nicer if there is a book whose main subject is similar to my question, i.e. how far one can go without AC in calculus or something like that. – Makoto Kato May 09 '12 at 09:39
  • I was being sarcastic when I said that I don't want you to have an answer. See that you are writing "I would like a book to answer my question". I am sorry but I have no intentions of ever writing this book for you. Generally speaking, if you ask a question whose answer should be roughly a book then this is not a good question for this website. – Asaf Karagila May 09 '12 at 09:43
  • @Asaf You write as if I'm asking the question to you only. You wrote as if I repeated "within ZF" only for you. Of course not. I'm not asking you to write a book for me, either.

    "Generally speaking, if you ask a question whose answer should be roughly a book then this is not a good question for this website."

    Why not?

     – Makoto Kato May 09 '12 at 15:40
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    You should read the FAQ perhaps. – Asaf Karagila May 09 '12 at 16:30
  • Okay can I delete my question? Joke aside, I don't understand why not. Why asking about a book which solves a certain question is not good on this website? For example, why a question like "do you know a good book on AC?" is not good? – Makoto Kato May 09 '12 at 20:05
  • Reasonable questions should have reasonable answers too. Your question is good, but broad. I replied with an answer which was equally broad. To your edits I replied with a comment since you did not even bother to reply my answer and it seemed to me that you are dissatisfied with it. From there this ridiculously long discussion began in the comments. To your question "Do you know a good book on AC" I already replied with several references, included a blog post about analysis which deals with the difference between hard and soft analysis which is related to your question. (Cont.) – Asaf Karagila May 09 '12 at 23:19
  • (...) However your edit which asks "Is Rudin's book true in ZF?" is a question that to answer it would require an immense amount of time, as it would require to overview every theorem and definition and every proof. It would require to simply write a complete book on the topic. This question is not well-suited for this website, no one will write a book just for you due to a request on this website. This much I can guarantee you. Except, of course, yourself. If you are willing to do that. – Asaf Karagila May 09 '12 at 23:19
  • @Asaf How do you know there is no book which can answer my question if not perfectly? – Makoto Kato May 11 '12 at 00:37
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    Because in the past 15 months my life literally revolved around the axiom of choice and its usage in mathematics - in some sense this is what I do for a living. Of course I cannot be sure, but I have gone through most common references as well uncommon references. I don't think that there is a book about the axiom of choice which my hands did not flip through at one point or another. – Asaf Karagila May 11 '12 at 06:29
  • @Asaf I'm not in a hurry. I can wait. – Makoto Kato May 11 '12 at 14:34
  • I am not really sure what you are waiting for. Someone that will write such book? I honestly doubt it will happen soon. Books take years to write and some time to get published. Good luck with that. – Asaf Karagila May 11 '12 at 19:34
  • @Asaf Not necessarily so. Someone may recommend a book or paper which can answer my question if not perfectly. – Makoto Kato May 12 '12 at 22:14

1 Answers1

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To do classical analysis (i.e. things in $\mathbb R^n$) you don't need the whole axiom of choice. Mostly you would need the axiom of countable choice, and rarely the stronger principle of dependent choice.

For example, the proof that continuity by $\varepsilon$-$\delta$ is equivalent to continuity by sequences (at a given point $x$) requires the axiom of countable choice.

We often use definitions by induction to define a certain sequence, the induction itself tells us that if we have defined $a_n$ we can define $a_{n+1}$ - so for a given finite number we can define a sequence longer that this number. We are usually interested, though, in the case where there is an infinite sequence, this is exactly where the axiom of choice comes into play. It tells us that there is such sequence and that we can use it.

However not everything requires the axiom of choice. For example the Heine-Borel theorem that states that closed and bounded intervals are compact (every open cover has a finite sub-cover) requires no choice whatsoever. Similarly a continuous function everywhere is sequentially continuous everywhere and vice versa (while requiring continuity at a single point at a time needs some choice).

Do note that general theorems about general functions/sequences/sets usually require some choice, on the other particular and very specific cases may be proved without it.

Some words to address the second edit: It is true that most of the people simply work in ZFC, and why shouldn't they? It's a very comfortable system and it saves you the need to verify certain things (e.g. if you want to use countable choice you have to keep checking that the families you choose from are countable). On the other hand it is also very true that most results people use require far less than the full axiom of choice, at least in elementary analysis.

Regardless to the above, once you get to functional analysis then the axiom of choice become more apparent (e.g. assuming the Ultrafilter lemma + Krein-Milman theorem implies AC in full). It is natural, too, that the larger your sets become the more choice you will need, and once you start talking about classes of spaces (e.g. all Hilbert spaces, or all locally-convex spaces) instead of a specific space - the more choice you will need to cover the general case.

Similarly in the "small" cases you can get away without choice (or with very very little) if you have a very specific case at hand. If, on the other hand, you want to prove a general theorem you might need to use some choice principle directly.

Doing analysis without any choice whatsoever is very hard and very limiting. I suppose it can be challenging and fun to people who find pleasure in that, much as I find pleasure in studying the possible structure of the universe without the axiom of choice. My suggestion is to examine your internal drive for this question: if you merely wish to learn on the axiom of choice and what is possible without it, I suggest picking up some books about mathematics and the axiom of choice:

  1. Herrlich, H. The Axiom of Choice (Springer, 2006)
  2. Moore, G. H. Zermelo's Axiom of Choice (Dover Publications, 2012)
  3. Fremlin, D. H. Measure Theory, vol. 5 (Torres Fremlin, 2000)
  4. Schechter, E. Handbook of Analysis and Its Foundations (Academic Press, 1997).

If on the other hand you simply wish to reject the axiom of choice, and you have decided to work without it, I am certain there are references suited for that, alas I am unfamiliar with any of them.

Further reading:

  1. Continuity and the Axiom of Choice
  2. Foundation for analysis without axiom of choice?
  3. Terry Tao's blog: Soft analysis, hard analysis, and the finite convergence principle.
Community
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Asaf Karagila
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    Can you point to an example where $DC$ is needed and $AC(\omega)$'s not enough? – t.b. May 06 '12 at 09:21
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    @t.b.: Fubini's theorem. – Asaf Karagila May 06 '12 at 09:27
  • Can you be more specific? Fremlin does not mention Fubini's theorem (for finite products) as a source of problems. – t.b. May 06 '12 at 09:41
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    @t.b.: Hmmmm, this seems to be folklore, then. I always heard that Fubini needs DC. Either way, Baire Category is equivalent to DC in the general case. – Asaf Karagila May 06 '12 at 09:49
  • Okay, thanks. That's good enough an example for me :) – t.b. May 06 '12 at 09:52
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    @t.b. another example is the Ascoli-Arzelà Theorem, which requires dependent choice to assert the existence of the diagonal sequence. – Emilio Ferrucci May 06 '12 at 12:52
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    @Emilio: Actually, depending on the form you want Arzelà-Ascoli to have, you can get away with countable choice: see here. The usual phenomenon: if something seems to require some form of choice on the nose, you can't be sure that this form is really needed. In particular, applications of DC can very often be replaced by a judicious use of countable choice (that's basically the reason for my question). – t.b. May 06 '12 at 14:38
  • @t.b. Interesting... someday I will have to find out how that proof is carried out. I don't think there is an easy way to adapt the proof I know so that it only uses $AC_{\omega}$, but I may be wrong. – Emilio Ferrucci May 06 '12 at 14:54
  • @t.b.: I would imagine that there might be a Borel related result needing the full power of DC (or at least more than countable choice) to generate some sequence of codes in one way or another. – Asaf Karagila May 06 '12 at 21:33
  • @t.b.: Feel free to edit the bibliography to whatever format you see fit! :-) – Asaf Karagila May 06 '12 at 22:26
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    Hi, Asaf. I remember that the definition of an abstract $C^\infty$ manifold included a "maximal atlas" of coordinate charts, and concluding at some early point that a beefy slice of AC was required for that. http://en.wikipedia.org/wiki/Manifold#Atlases If I misunderstood, it was also a long time ago. The OP may be asking only about Euclidean space, of course. That's life. – Will Jagy May 06 '12 at 23:17
  • @Will: The OP asks about Euclidean spaces, I also know nothing about manifolds and atlases and I prefer sticking to things I can later explain (at least partial explanations). This is certainly not one of them. – Asaf Karagila May 07 '12 at 06:36
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    @WillJagy Dear Asaf and Will. I think that we do not need the axiom of choice in order to prove the existence of a maximal atlas of smooth charts in any class of equivalence (by compatibility) of atlases of smooth charts on a topological manifold. Infact, if I don't misunderstand, it is just the union of the elements of the class. Bye. – agt May 07 '12 at 08:25
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    @WillJagy: George Lowther asserts here, Mariano goes into a bit more detail here and Zhen Lin gives a proof here that existence of maximal atlas does not require choice. – Willie Wong May 08 '12 at 09:16
  • @WillieWong, I was young, they said it wasn't addictive, and who would be hurt anyway? – Will Jagy May 08 '12 at 18:56
  • @AsafKaragila I think proving sequentially continuity implies continuity requires countable choice... – YuiTo Cheng Jan 12 '19 at 17:22
  • @YuiToCheng: Yes, https://math.stackexchange.com/questions/126010/continuity-and-the-axiom-of-choice – Asaf Karagila Jan 12 '19 at 17:32
  • @AsafKaragila But what you wrote seems implicitly implying it requires no choice – YuiTo Cheng Jan 12 '19 at 17:35
  • @YuiToCheng: I literally wrote in the second sentence of my answer that it does. – Asaf Karagila Jan 12 '19 at 17:36
  • @AsafKaragila I meant the fourth paragraph "Similarly..." – YuiTo Cheng Jan 12 '19 at 17:38
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    @YuiToCheng: Yes, because if a function is continuous everywhere then choice is not needed. (Again, see the link, specifically Brian's answer there.) And of course this only applies for the case where you're talking about $\Bbb R^n$, not in general for metric spaces, because that's the context of this question. – Asaf Karagila Jan 12 '19 at 17:42