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The other day, I read something that made reference to a well-known result in topology: the fact that a compact Hausdorff space is normal. I took a moment to recall the proof.

Let $A$ and $B$ be closed subsets of $X$. Fix a point $b \in B$. For all $a \in A$, there exist disjoint open subsets $U_a$ and $V_a$ of $X$ such that $a \in U_a$ and $b \in V_b$. The collection $\{ U_a : a \in A \}$ is an open cover for $A$, and admits a finite refinement, so...

Then, all of a sudden, I spotted that I had invoked the axiom of choice! I found it a little irritating that the proof I was pulling from my memory was none other than the standard one I'd learnt from multiple books, yet none of these books had ever drawn my attention to the use of the axiom of choice. I'm not one of those people who objects to the axiom of choice, but I do like to know where it is being used!

Moreover, I was pretty sure that this result wasn't one of those famous theorems where the axiom choice is essential (e.g. Tychonoff, Baire category). Indeed, after a certain amount of reflection, I managed to reorganise the proof so as to eliminate the use of the axiom of choice.

Let $A$ and $B$ be closed subsets of $X$. Fix a point $b \in B$. The collection $$\{ U \text{ open} \subset X : \exists V \text{ open} \subset X \text{ s.t. } U \cap V = \emptyset \text{ and } b \in B\}$$ is an open cover for $A$, and admits a finite refinement, so...

I then became doubly irritated because nobody had ever mentioned this alternative proof to me.

This is not an isolated case. Last month, I answered a question on math.SE, invoking the axiom of choice without even realising I had done so. Another user kindly flagged what I had done and provided an alternative argument.

For the sake of mental hygiene, I'd love to compile a list of elementary results I'm likely to be familiar with, that are normally proved using the axiom of choice for the sake of readability, but can also be proved without the axiom of choice. I suspect the majority of these will be in point-set topology. Would you be able to help with suggestions or references?

Kenny Wong
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    This is so similar I don't know if it deserves a separate answer: the fact that if you have a basis $\mathcal{B}$ (with the definition in terms of existence of $B \in \mathcal{B}$ such that $x \in B \subseteq U$), then every open set is a union of a subfamily of $\mathcal{B}$. The natural proof to think of is to choose some such $B$ for every $x\in U$; while again, you can just take $\mathcal{B} \cap \mathcal{P}(U)$ as the subfamily which avoids a use of AC. – Daniel Schepler Apr 04 '23 at 16:24
  • @DanielSchepler That's a great example! I've always gone with the natural proof, and it has never occurred to me that AC was used. – Kenny Wong Apr 04 '23 at 16:29
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    A general pattern with compactness results is that you shouldn’t try to choose one open set per $x$; you should instead throw in all the open sets which have the property. – Mark Saving Apr 04 '23 at 16:37
  • @MarkSaving Yes, that pattern does indeed apply to both of the examples I've come up with. – Kenny Wong Apr 04 '23 at 16:38
  • @MarkSaving (and OP) Note that in that sort of case we can often make precise the idea that choice simplifies things by noting that the "all open sets" approach generally increases the number of open sets needed; see e.g. MSE/MO. – Noah Schweber Apr 04 '23 at 16:57
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    This question on [mathoverflow.se] might be of interest: Unnecessary uses of the axiom of choice. And I will mention the specific result pointed out here: Continuity and the Axiom of Choice (a proof in ZF about equivalence between two approaches to the global continuity for functions $\mathbb R\to\mathbb R$). – Martin Sleziak Apr 05 '23 at 09:55
  • I didn't know that $f$ sequentially continuous everywhere $\implies$ $f$ continuous everywhere doesn't require AC, even though $f$ sequentially continuous at $x$ $\implies$ $f$ continuous at $x$ requires AC. Interesting. In that example, the proof with AC is vastly more intuitive than the proof without. – Kenny Wong Apr 05 '23 at 10:02
  • Another one I recently found. You need to AC to prove that Noetherian rings are factorization rings in generality. But for $\mathbb Z$ and $k[x]$, I think you can avoid AC, since there are constructive ways of selecting a pair of proper divisors for a reducible element. – Kenny Wong Apr 09 '23 at 06:34

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