I went over the proof of Baire's Theorem for complete metric spaces (that a countable union of closed sets with empty interior has empty interior). In the proof, a decreasing nested sequence of closed balls is constructed. Then we choose a particular point from each of these balls and construct a sequence of those points.

I thought that we used axiom of choice when we chose these points. But then I read in multiple source that we actually don't use axiom of choice and that we use a weaker form, called the axiom of dependent choice. I looked up the definition of dependent choice:

Axiom of dependent choice (DC): Let $A$ be a non-empty set and $R \subseteq A \times A$ satisfy $\forall a \in A \; \exists b \in A: a R b$. Then there exists a sequence $(a_n)_n$ such that $a_n \; R \; a_{n+1} \; \forall n$.

The problem I have with this is that I don't understand how this axiom was used. First of all, what is the set $A$ and relation $R$ when we apply DC to prove Baire's Theorem? And secondly, could someone explain in some more details how this axiom is used.

  • 2
    $\begingroup$ I wrote a whole chapter about it: arxiv.org/abs/2010.15632 $\endgroup$
    – Asaf Karagila
    Jan 11, 2022 at 22:02
  • $\begingroup$ And this might also answer your question. $\endgroup$
    – Asaf Karagila
    Jan 11, 2022 at 22:04
  • $\begingroup$ @AsafKaragila is it possible to prove Baire's Theorem for locally compact hausdorff spaces without DC? If so, could you refer me to a proof that doesn't use it? (I know it isn't possible for complete metric spaces, because they are equivalent). $\endgroup$
    – Matthew
    Jan 11, 2022 at 22:32
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    $\begingroup$ Well, assuming BPI it is equivalent to to DC, so you can't. See mathoverflow.net/a/391946/7206 $\endgroup$
    – Asaf Karagila
    Jan 11, 2022 at 22:40
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    $\begingroup$ Can you share the exact proof that you saw? $\endgroup$ Jan 11, 2022 at 22:42

1 Answer 1


Define a sequence of points $x_n$ such that $x_{n+1}$ is within $2^{-n}$ of $x_n$. We're applying DC to the relation "$x_{n+1}$ is within $2^{-n}$ of $x_n$". (This answer uses some help from the text that Asaf Karagila linked in the comments, but I have not yet read its explanation of why DC is enough.)


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