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This is inspired by this discussion. I see that the debates about the necessity of the axiom of choice in this or that statement are still ongoing. In this regard, I became interested in whether there are textbooks that describe what mathematical disciplines turn into when the axiom of choice is consistently excluded from them.

It seems to me that, theoretically, nothing prevents a logician from writing a book that describes algebra or topology or analysis in some axiomatic set theory without the axiom of choice, for example, in ZF.

So my question is

are there texts describing what remains of analysis (or algebra, or topology) if it is built consistently without the axoim of choice, for example, in ZF (not in ZFC)?

I think such a text would be very helpful, because in my observation, people who argue about the application of the axiom of choice tend to focus on particulars without seeing the big picture, and moreover, having a rather vague idea of the subject.

For example, I have met analysts who believe that in analysis the axiom of choice appears only in a few statements, such as the Hahn-Banach theorem, and if you do not use them, you can consider your conscience "unstained by its application".

People on this thread have already explained to me that some things in analysis can be preserved, in particular (forgive my ignorance), it was a surprise to me that the theory of real numbers is preserved in some form (although, as far as I understand, the classical theorems of mathematical analysis mostly disappear). And apparently, something can be preserved in algebra and topology. I would be terribly interested to look at this picture.

I remember Boris Kushner's book on "Constructive mathematical analysis", but it's about something else, about a variant of intuitionistic mathematics, where "constructiveness", as the adepts understand it, is woven into the system of axioms of logic, not set theory. Is there any overview or a book explaining what is preserved in mathematics when it is built in ZF?

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    $\begingroup$ This is way too open ended. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 20, 2022 at 11:49
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    $\begingroup$ Yes, but you're asking "Hey, did anyone ever redo all of mathematics without choice?" and the answer is obviously not. There are some books, like Schecther's Handbook of Analysis and its Foundations which discuss some aspects of, well, analysis; Fremlin's Measure Theory has a part at the end of vol. 5 dedicated to, well, measure theory; and surely similar books exist in other fields. Herrlich has The Axiom of Choice which contains many possible disasters that may occur without choice throughout mathematics. But none of those is even remotely complete in any sense of the word. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 20, 2022 at 12:35
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    $\begingroup$ @SergeiAkbarov: It is very not true. No. It might be wise, before starting with mathematics without AC, to learn a bit of wet theory without AC. For that Jech's Axiom of Choice book is pretty decent. Azriel Levy is known to keep track of it, so I imagine that his book in Basic Set Theory would be good as well. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 20, 2022 at 18:11
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    $\begingroup$ @SergeiAkbarov: you confidently made a host of claims about how all sorts of mathematics "breaks down" and "nothing can be done" without the axiom of choice, which is simply false. I would urge a more cautious approach that doesn't irk people who actually know what can be done without AC and will simply ignore you as they're tired fighting this particular combination of ignorance and arrogance, driven by unchecked confirmation bias. $\endgroup$
    – Andrej Bauer
    Commented Sep 21, 2022 at 6:33
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    $\begingroup$ Minor comment, but no one's brought it up yet: Absoluteness results from set theory imply that AC cannot be necessary to prove sufficiently simple statements. For example, Shoenfield's absoluteness theorem implies that if you can prove, using AC, a statement of the form "for all $X$ there is a $Y$ such that $\varphi(X,Y)$" where $X$ and $Y$ range over sets of integers and $\varphi$ only quantifies over integers, then you can prove it without using AC. Many theorems about countable objects can be put in this form—my favorite example here is Hindman's theorem—so they're all preserved in only ZF. $\endgroup$
    – Julia Williams
    Commented Sep 22, 2022 at 21:51

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I agree with Asaf Karagila that the question as literally stated is a bit too sprawling, but you might want to start with Simpson's book, Subsystems of Second-Order Arithmetic. Its goals aren't the same as yours, but along the way, the book shows how a large chunk of "countable mathematics" (whatever that means), including analysis, can be developed on the basis of second-order arithmetic (or even in $\mathsf{ACA}_0$), and a fortiori on the basis of ZF. Part of the trick is to avoid "overly general" statements of certain theorems; if you fix certain choices in advance, then you don't have to invoke AC to make those choices on your behalf later, and you can prove versions of (for example) Bolzano–Weierstrass that suffice for applications.

You might also want to check out another MO answer of mine which briefly discusses some other potentially relevant books. Again, they don't have exactly the same goals that you have, but the work they do may be relevant. Here's a quote from Bishop which gives some of the flavor:

Applications of the axiom of choice in classical mathematics either are irrelevant or are combined with a sweeping use of the principle of omniscience. The axiom of choice is used to extract elements from equivalence classes where they should never have been put in the first place. For example, a real number should not be defined as an equivalence class of Cauchy sequences of rational numbers; there is no need to drag in the equivalence classes.

It's harder to avoid AC when it comes to "uncountable mathematics" (whatever that means). In another MO question, I asked whether Hahn–Banach for $\ell^\infty$ implies the existence of a non-measurable set. The answer is apparently not well known, and perhaps is not known at all. From this experience, I infer that analysts by and large have not even bothered to figure out in detail how much functional analysis can be carried out just on the basis of ZF+DC, let alone ZF on its own.

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  • $\begingroup$ Tim, "on the basis of ZF" - does this mean that Simpson turns ZF somehow into a second order theory? $\endgroup$
    – Sergei Akbarov
    Commented Sep 20, 2022 at 19:23
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    $\begingroup$ I would say it this way: second-order arithmetic is interpretable in ZF. The theorems of second-order arithmetic, so interpreted, are theorems of ZF. $\endgroup$
    – Timothy Chow
    Commented Sep 20, 2022 at 22:30
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    $\begingroup$ @SergeiAkbarov The terminology "second-order arithmetic" may be slightly misleading; it's really a two-sorted theory that is still based on first-order logic. One sort is natural numbers and the other sort is sets of natural numbers. It's called second-order because you can quantify over sets of natural numbers and hence over real numbers, but $\mathbb{R}$, thought of as the set of all subsets of $\mathbb{N}$, is not something you can refer to directly. Chapter 1 of Simpson's book is available online; I refer you to it for more details. $\endgroup$
    – Timothy Chow
    Commented Sep 21, 2022 at 12:08
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    $\begingroup$ @SergeiAkbarov Correct. But I think the point is that if you're serious about getting rid of AC, then the way to go about it is not to begin with classical mathematics, hunt around for uses of AC, and see if you can pull out AC like a Jenga piece without causing the whole tower to collapse. The way to do it is to rebuild mathematics from the ground up. Since nobody seems to have done exactly what you were hoping for (building math using ZF), the closest thing is to see how various people have developed large portions of mathematics without using AC. $\endgroup$
    – Timothy Chow
    Commented Sep 21, 2022 at 13:18
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    $\begingroup$ Tim, I understand what you mean, but before doing what you say, the first step for me would be an attempt to see what happens if I pull out the brick that everyone is trying to pull out. $\endgroup$
    – Sergei Akbarov
    Commented Sep 21, 2022 at 13:50
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One of the standard texts which presents functional analysis only based on ZF+DC is the monograph (consisting of 3 volumes) Henry G. Garnir, Marc de Wilde, and Jean Schmets, Analyse Fonctionnelle.

Also in most of my monographs, in particular Topological Analysis, you will find many of the standard results of analysis and topology with explicit notes for which parts of the assertions more than ZF+DC is needed (and in a few cases also remarks when ZF alone is sufficient). Also in those of my monographs more related with integration and measure theory no use of anything more than ZF+DC is made unless explicitly mentioned. For nonstandard analysis the situation is different, although there are some recent papers that a certain internal nonstandard analysis can be carried out in ZF(+DC) as well.¹

In pure ZF (without DC) most of analysis is known to break down, in particular, it is almost impossible to do a reasonable measure or integration theory (as the real line might be a countable union of countable sets) or even topology (since sequential and topological definitions of a limit can differ already for functions of the real line).

Edit: ¹See e.g. Karel Hrbacek, Mikhail G. Katz, Infinitesimal analysis without the Axiom of Choice, Annals Pure Appl. Logic 172 (2021) (6)

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  • $\begingroup$ I'm not sure that "In pure ZF ... it is almost impossible to do a reasonable measure or integration theory" is quite correct. There are difficulties, to be sure, but Bishop already took a stab at developing measure theory without choice. See also Constructive algebraic integration theory without choice by Spitters and Pre-measure spaces and pre-integration spaces in predicative Bishop-Cheng measure theory by Petrakis and Zeuner. $\endgroup$
    – Timothy Chow
    Commented Sep 24, 2022 at 20:57
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    $\begingroup$ Without any choice what you get does not really resemble the classical setting. Lebesgue "measure" is not $\sigma$-additive, for instance. It is true you can "recover" more than one would expect. Fremlin has an interesting approach using "codes" of sets rather than the sets themselves. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 24, 2022 at 22:16
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    $\begingroup$ @TimothyChow ”almost impossible” was probably a bit too harsh, but you have to be extremely careful which definitions you choose. So there is not “the” integration theory anymore in ZF, but several such theories, usually preserving different properties (compared to the ZF+DC situation which has all properties and in which all approaches are equivalent, at least for the Lebesgue measure). $\endgroup$
    – Martin Väth
    Commented Sep 25, 2022 at 15:02
  • $\begingroup$ @MartinVäth I realize that I'm confused about your comment regarding nonstandard analysis. Shouldn't there be some kind of transfer principle that means that the apparent use of stronger axioms isn't really necessary? Admittedly, I'm not very familiar with nonstandard analysis. $\endgroup$
    – Timothy Chow
    Commented Aug 2, 2023 at 1:09
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    $\begingroup$ @TimothyChow: In nonstandard analysis of Robinson/Luxemburg, the transfer principle applies, roughly speaking, only to standard sentences; but there are more. You can “explicitly” define non-measurable sets there. For instance, let $f\colon[0,1)\times\mathbb N\to\{0,1\}$ be such that $f(x,n)$ is the $n$-th digit of the binary expansion of $x$. Then for any infinite hypernatural $h$ the set $\{x\in[0,1):{}^*f({}^*x,h)={}^*1\}$ is not Lebesgue measurable. It is surprising that this set cannot be defined in the restricted languages of internal set theory from the cited paper of Hrbacek-Katz. $\endgroup$
    – Martin Väth
    Commented Aug 5, 2023 at 21:27
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ZF cannot prove the local equivalence between "epsilon delta" continuity and "sequential" continuity for functions on the reals. A small fragment of (countable) AC does suffice for proving the equivalence.

In this light, most/many proofs in analysis would break down in ZF; a lot of work would be needed to sift through the details.

However, and in answer to your question, the aforementioned equivalence is provable in ZF for e.g. regulated functions. Thus, for the special case of regulated functions, no (extra) AC is needed anywhere.

A more general answer (based on recent research) is the following: it seems that countable AC can be avoided if we require that all functions under study are Baire 1 (or effectively Baire $n$).

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    $\begingroup$ It is very important to point out that this failure of equivalence is only when talking about "a single point", and a function that is everywhere sequentially continuous is also everywhere continuous. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 27, 2022 at 0:59
  • $\begingroup$ Hence the "local" equivalence I mentioned, but you are of course right. $\endgroup$
    – Sam Sanders
    Commented Sep 27, 2022 at 7:14
  • $\begingroup$ Just a reference for 'Baire 1/Baire n' functions, since I had to look it up: en.wikipedia.org/wiki/Baire_function $\endgroup$
    – David Roberts ♦
    Commented Nov 18, 2022 at 2:24
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Timothy Chow gave a fine answer in the context of classical mathematics. Here are some further sources for you to ponder. These not only work without choice, but also without excluded middle:

The moral of the story is that the folk tales that mathematicians tell about how choice-free mathematics is completely different, or even impoverished beyond recognition, are just folk tales. Of course, one has to be a bit more careful, but that is always the case when we generalize, as new phenomena arise.

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    $\begingroup$ Andrej, you misunderstand me all the way. I did not say that the orthodox foundations are better than others. I was trying to understand what happens if we remove AC from ZFC, this is not the same. $\endgroup$
    – Sergei Akbarov
    Commented Sep 21, 2022 at 14:54
  • $\begingroup$ Thanks for your remark, I edited the paragraph to address your question more directly. It's not easy to give a precise technical answer, though, because you're asking about "all math". $\endgroup$
    – Andrej Bauer
    Commented Sep 21, 2022 at 15:20
  • $\begingroup$ Just for my own education, I take it that Blakers-Massey for simplicial sets (not just the "synthetic" form of Blakers-Massey) can also be proved without AC or LEM? $\endgroup$
    – Timothy Chow
    Commented Sep 21, 2022 at 17:14
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    $\begingroup$ @TimothyChow: that's probably hard to figure out, as classic proofs are unlikely paying attenition to where and whether there are essential uses of AC and LEM. But already showing the basic categorical structure of Kan simplicial sets is tricky without LEM – this was a stumbling block for a long while in HoTT, and was resolved by passage to cubical sets. $\endgroup$
    – Andrej Bauer
    Commented Sep 21, 2022 at 20:43
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    $\begingroup$ @SamSanders: Bishop-style constructive analysis does indeed rely on countable choice (Dependent choice). Choice-free analysis uses Dedekind reals, and is less well-developed, although there is some amount of it. For instance, the HoTT books develops the basics without choice, and there are others. It's easier to do general topology without countable choice. Choice-free algebra has been studied quite a bit and it has its own set of challenges. $\endgroup$
    – Andrej Bauer
    Commented Sep 27, 2022 at 9:24

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