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To do calculus, we (presumably) need the real numbers (or perhaps some abstract complete metric space?).

When the real numbers are constructed using Cauchy sequences or Dedekind cuts, does this happen in "pure" ZF (without any choice axiom)?

In addition, are any choice axioms (Countable Choice, Dependent Choice, etc.) needed when:

  • defining/studying convergence of sequences of reals or functions $f:\mathbb{R} \longrightarrow \mathbb{R}$?
  • defining/studying continuity of functions $f:\mathbb{R} \longrightarrow \mathbb{R}$?
  • defining/studying derivatives of functions $f:\mathbb{R} \longrightarrow \mathbb{R}$?
  • constructing the Riemann/Lesbegue integral?
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    $\begingroup$ math.stackexchange.com/questions/103743/… $\endgroup$
    – Alessandro Codenotti
    Commented Nov 4, 2019 at 16:58
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    $\begingroup$ The usual proof that continuity is equivalent to sequential continuity uses countable choice. $\endgroup$
    – PhoemueX
    Commented Nov 4, 2019 at 17:59
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    $\begingroup$ You need DC for most of calculus, for specifically R, there are times were CC is enough, but without any choice, things become weird $\endgroup$
    – Holo
    Commented Nov 4, 2019 at 18:51
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    $\begingroup$ @ℋolo . I dk how much of measure theory can be salvaged without Choice. $\endgroup$
    – DanielWainfleet
    Commented Nov 4, 2019 at 20:53
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    $\begingroup$ @DanielWainfleet: The famed Fremlin's Measure Theory book has a very thorough section on using Borel codes for measure theory in ZF, it's just... terrible, though. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 5, 2019 at 6:40

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The Cauchy reals and Dedekind reals are isomorphic in "pure" ZF (although their isomorphism requires the Law of Excluded Middle, which means it doesn't hold generally in intuitionistic logic : this is a logic weaker than the one you're used to)

All the definitions you allude to below are choice-free; but certain characterizations require the axiom of choice.

For instance, "$f$ is continuous at $x$ if and only if for all sequences $(x_n)$ converging to $x$, $f(x_n)\to f(x)$" requires some amount of choice (it is usually proved with countable choice - and if you say that it holds for all metric spaces, you get full countable choice) - however, surprisingly, "$f$ is continuous if and only if for all sequences convergent $(x_n)$, $(f(x_n))$ converges" does not require more than ZF (a proof of that can be found in Herrlich's Axiom of choice)

More generally, when you try to characterize topological properties of functions with sequences, you'll often end up needing some form of choice : usually dependent choice is enough, and sometimes you can get away with countable choice.

Lebesgue measure theory uses some small amount of choice, the best being to just assume dependent choice : sometimes you can get away with just countable choice, but all in all dependent will end up being useful.

If you don't have any choice, then Lebesgue breaks down because $\mathbb R$ could be a countable union of countable sets, in which case it would have measure $0$ (so it wouldn't really be meaningful)

For the rest of calculus you would need a case-by-case analysis to determine if such and such result uses choice. As I explained above, in basic calculus the main culprit is dependent choice : if you assume that and not full choice you can get away with pretty much anything, and there are some times when you'll actually need it.

Actually, it can even be benefitial to not assume full choice, because (under some large cardinal assumptions if I recall correctly) the theory "ZF+dependent choice + all sets of reals are Lebesgue measurable" is consistent, so you can do maths in it without risks, and, well, dependent choice allows you to do all the basic calculus that you want while "all sets are Lebesgue measurable" smooths some points in measure theory.

However, dependent choice is not enough when you start encountering wilder beasts such as infinite dimensional vector spaces, and infinite products of spaces (where you often need stuff like the Hahn-Banach theorem and Tychonoff's theorem); but that's later in analysis.

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  • $\begingroup$ "$f$ is continuous if and only if for all sequences convergent $(x_n)$, $(f(x_n))$ converges" It seems like this, combined with "for a continuous function $f$, if $x_n\rightarrow x$, then $f(x_n)\rightarrow f(x)$." would prove the statement that requires choice. Does this statement itself require choice, or does the argument combining them require it? $\endgroup$
    – eyeballfrog
    Commented Nov 5, 2019 at 16:33
  • $\begingroup$ @eyeballfrog : how would it imply it ? If anything, a proof of the implication requires choice, because your statement is very basic and holds in way more general contexts $\endgroup$
    – Maxime Ramzi
    Commented Nov 5, 2019 at 16:42
  • $\begingroup$ Oh...right. I would need the reverse of that statement for that to work. $\endgroup$
    – eyeballfrog
    Commented Nov 5, 2019 at 17:00

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