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I'm trying to see if the following proof use any kind of choice beyond the choices ZF allows: Assume nested intervals theorem on real numbers. So shortly, if $I_n$ is a sequence of nested intervals whose length converges to $0$ (with closed intervals on R), then intersection of all of them is a singleton.

We want to show that every non-empty subset of R which is bounded from above has a supremum.

Now, let $A \subset$ R and let $a \in A$ and $r \in$ R be an upper bound for the set $A$. Now our first interval is $[a,r]$. If there is no element in $A$, which is bigger than $a$, then we are done and likewise, if there is no real number less than $r$ which is an upper bound, then we are done. So we can choose this two new numbers and build another interval.

Now, writing this proof here, I noticed that the length doesn't have to go to $0$, so this proof is not correct. I don't ask a correct proof, which would be I am sure a duplicate. I am asking if the process above requires AC or something alike.

aeyalcinoglu
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  • The nested intervals theorem (also known as Cantor's theorem) is provable in ZF. – Asaf Karagila Jan 27 '17 at 20:34
  • I know this is an old post, but for reference, if you make one choice after another where each choice depends on the previous ones, countably many times, then this uses dependent choice (DC) but in real analysis this can often be avoided as Asaf's answer shows, by using the rationals. In some cases, it cannot be totally avoided. For example, the equivalence of continuity at a point and the sequential criterion for continuity at that point cannot be proven in ZF but can be proven in ZF plus countable choice (CC). Note that CC is strictly weaker than DC, which is in turn strictly weaker than AC. – user21820 Aug 21 '20 at 19:15

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You can always choose these numbers to be rational numbers, since the rational numbers are dense in every non-singleton interval.

Since the rational numbers are countable, we can prove there is a nice choice function from the non-empty sets of rational numbers, so no choice is actually needed.

Asaf Karagila
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