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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?

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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!2012-05-06
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    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.2012-05-06
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    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.2012-05-06
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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.2012-05-06
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    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.2012-05-06
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    @Karagila You wrote "It is not "often" that AC is unneeded,". I'm talking about calculus, which means classical analysis in Euclidean spaces.2012-05-07
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    @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.2012-05-07
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    @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).2012-05-07
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    (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.2012-05-07
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    @Asaf So your answer to my question "Can most results in calculus be proved within ZF?" is no?2012-05-07
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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.2012-05-07
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    @Asaf How about Rudin's "Principles of mathematical analysis"? Can most results in this book be proved within ZF?2012-05-07
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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."2012-05-07
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    @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.2012-05-08
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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.2012-05-08
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    @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.2012-05-08
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    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.2012-05-08
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    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.2012-05-08
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    @Asaf Just because you can't answer it does not necessarily mean others can't. I can wait.2012-05-08
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    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.2012-05-08
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    @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.2012-05-08
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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.2012-05-08
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    @Asaf Then how did you come up with the conclusion that I was wrong? See the first comment of yours.2012-05-08
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    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.2012-05-08
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    @Asaf How do you know most theorems in the Rudin's book do require some choice?2012-05-08
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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...2012-05-08
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    @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.2012-05-08
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    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.2012-05-08
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    @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 ..."?2012-05-08
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    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.2012-05-08
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    @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.2012-05-09
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    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.2012-05-09
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    @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?2012-05-09
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    [You should read the FAQ](http://math.stackexchange.com/faq#dontask) perhaps.2012-05-09
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    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?2012-05-09
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    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.)2012-05-09
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    (...) 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.2012-05-09
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    @Asaf How do you know there is no book which can answer my question if not perfectly?2012-05-11
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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.2012-05-11
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    @Asaf I'm not in a hurry. I can wait.2012-05-11
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    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.2012-05-11
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    @Asaf Not necessarily so. Someone may recommend a book or paper which can answer my question if not perfectly.2012-05-12

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