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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.

Martin Sleziak
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user63123
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1 Answers1

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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.

Martin Sleziak
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Asaf Karagila
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  • 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
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    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
  • Why does the countability of a countable union of fine sets require (any form of) choice? Shouldn't this follow from the countability of $\mathbb{N} \times \mathbb{N}$? Does that also require choice? – Carlos Esparza Feb 15 '20 at 16:02
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    @CarlosEsparza: Because you need to choose injections from the finite sets into $\Bbb N$. The proof that $\Bbb{N\times N}$ is countable does not require us to make infinitely many choices. – Asaf Karagila Feb 15 '20 at 16:03
  • What is the meaning of “We chose a finite subcover for every n” ? Since it is compact, there is a finite subcover for each n, so why do we need make choice for this? – Tim Nov 29 '24 at 10:06
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    @Tim: Because if you're choosing one for each $n$, you're making infinitely many arbitrary choices. That's why. – Asaf Karagila Nov 29 '24 at 12:12
  • Thanks you very much. I still have a question. Is it correct that the two points you listed are proving $\cup V_n$ is countable. Because proving $\cup V_n$ is a base does not use axiom of countable choice. – Tim Nov 29 '24 at 12:29
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    @Tim: Yes, that is correct. The fact that $\bigcup\mathcal V_n$ is a basis is fine. But you need choice to define it in the first place and you need choice to prove that it is countable (because it might be that you can still somehow get the choice of finite subcovers, but not the countability of the union, for example). – Asaf Karagila Nov 29 '24 at 13:06
  • This helps me a lot and think, thanks you. We need AoCC so that ${V_1,V_2\cdots}$ is a set and using axiom of union, we can define $\cup V_n$. That's your point 1 does. Similarly, to prove $\cup V_n$ is countable, we have to construct a map that have to choose injections/surjections for each n. – Tim Nov 30 '24 at 05:04