# Compact metric spaces is second countable and axiom of countable choice

Why we need axiom of countable choice to prove following theorem: every compact metric spaces is second countable?

In which step it's "hidden"?

Thank you for any help.

• The "obvious" proof uses countable choice, Why we need countable choice is a harder question. – Andrés E. Caicedo Nov 19 '13 at 21:56
• @Andres, I'm knee deep in references... – Asaf Karagila Nov 19 '13 at 21:59

Usually the proof would go like this:

For every $n$ define $\mathcal U_n=\{B(x,\frac1n)\mid x\in X\}$, this is clearly an open cover so it has a finite subcover $\mathcal V_n$.

Finally we can show that $\bigcup\mathcal V_n$ is a basis for the topology.

Here used twice countable choice:

1. We chose a finite subcover for every $n$.
2. We took the union of countably many finite sets, each with more than one element. Then we claim that this union is countable.

Both things are a consequence of the axiom of countable choice, and cannot be proved in general without it.

In the following paper the authors show that for compact metric spaces the statement that the space is separable is equivalent to the statement that it is second-countable. This makes it easier to find a counterexample, as non-separable compact metric spaces are easier to come by.

Keremedis, Kyriakos; Tachtsis, Eleftherios. "Compact metric spaces and weak forms of the axiom of choice." MLQ Math. Log. Q. 47 (2001), no. 1, 117–128.

Models where these fail (i.e. there is a compact metric space which is not second-countable) are also given.

• Thank you! But in reason nr 2. if $\ \mathcal V_n = \{B(x_{n,i},\frac{1}{n}), i=1,...,d(n)\}$ for some function $d$ and if we take $f_{n}:\mathcal V_n \rightarrow \mathbb{N}$ such that $f_{n}(B(x_{n,i}))=i$ we cannot use this functions (injections) to prove that $\bigcup\mathcal V_n$ is countable? Is that because we have to know more about choosing $x_{n,i}$ and $d(n)$? – user63123 Nov 19 '13 at 22:46
• If you can uniformly enumerate the $\mathcal V_n$'s then you can indeed show that their union is countable. But generally I don't see any reason for that to be true. Think about $[0,1]$. How many real numbers can be chosen for a cover by intervals of length $\frac18$? Can you uniformly choose from these real numbers? – Asaf Karagila Nov 19 '13 at 22:52