# Heine's theorem and axiom of choice

I recently read a proof on Wikipedia of Heine's theorem here. I am using their notations.

The proof aims at showing that given two metric spaces $$M$$ and $$N$$ where $$M$$ is compact, if $$f$$ is a continuous mapping $$M\rightarrow N$$, then $$f$$ is uniformly continuous.

Part of the proof states that for every $$\epsilon>0$$ and every $$x$$, there exists a $$\delta_x$$ such that $$d_M(x,y)<\delta_x \Rightarrow d_N(f(x),f(y))<\epsilon$$, and then uses $$\delta_x$$ which is a mapping from $$M$$ to $$\mathbb R^+$$.

Am I right when wondering if the axiom of choice is implicitly assumed?

Going back to various proofs on compact sets, I have the feeling that some form of Axiom of Choice (either general or at least countable) must be assumed in quite a few proofs.

For instance, when showing that a sequence of values in a bounded interval in $$\mathbb R$$ has a converging subsequence, you need to choose recursively subsequences and sequence elements, so at least do a countable number of choices.

When I was taught that at university, I don't remember that was mentioned, so I am wondering if those technical details have been brushed under the carpet back then.

• Topologists (those that I know) are all firm believers in AC and apply it all the time, without even considering alternative arguments. See the classic paper Horrors of Topology without Choice by van Douwen. A famous quote by the topologist Mary-Ellen Rudin: "first we well-order everything in sight" (on how to start a topology proof). Aug 19, 2021 at 7:57

Since for each $$x$$ there is some $$\delta$$ that is a candidate for $$\delta_x$$, and moreover, if we make that $$\delta$$ smaller then we do not lose any generality either, we can simply take $$\delta_x$$ to be $$\frac1n$$ where $$n$$ is the least one for which $$\frac1n$$ works.
• @Chris We choose the smallest $n$ that works so we use the well-order on $\Bbb N$ to avoid AC. Aug 19, 2021 at 21:27