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If every infinite subset has a limit point in a metric space $X$, then $X$ is separable (in ZF)

Yesterday, i posted this question and got an answer that 'Limit Point Compact⇒Separable' is unprovable in ZF.

As you can see, the proof in the link is done by the fact that there could be an infinite set which doesn't have a countable subset. I accepted the equivalence as an consequence of AC and continued my study.

However, most of spaces in classical analysis are Dedekind-infinite(i.e. $X≧|\aleph_0|$) as far as i know, and i have no idea how to prove this in ZF.(Since the way of my argument i tried today is exactly the same as the one i tried yesterday, that is, i can't make a countable choice).

I want to prove this if possible, or know whether it is unprovable. Help

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up vote 2 down vote accepted

Take the same space as JDH suggested, and embed it into $[0,1]$, now add to that the interval $[2,3]$.

This set is Dedekind infinite and limit point compact, but it is not separable for obvious reasons.


I couldn't find anywhere a particular account for the exact consistency strength needed for this sort of statement. However if there are no infinite D-finite sets then the assertion "Every infinite set has an accumulation point" is equivalent to that of "Every sequence has a convergent subsequence"

I have a hunch that this too is not enough to prove that $X$ is separable.

Generally speaking, though, there is a plethora of statements about metric spaces and compactness, Lindelof-ness and such, all which are equivalent to countable choice. I would not be surprised if this turns out to be equivalent to the axiom of countable choice as well, but I don't have much time nowadays to sit and tackle this problem on my own.

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Do you have any idea what statement $\psi$ makes '$\psi$+limit point compact ⇒ separable' possible? –  Katlus Aug 9 '12 at 9:27
    
I wouldn't be surprised if it is in fact equivalent to countable choice. If you have Herrlich's The Axiom of Choice it may have the answer. Otherwise you could wait a couple of hours until I am near my copy again. –  Asaf Karagila Aug 9 '12 at 9:32
    
I don't want to disturb you, but that book is not in the library i'm studying. Would you mind posting a short summary? –  Katlus Aug 9 '12 at 9:42
    
Are you assuming $X$ is a subspace of $\mathbb R$ or a general metric space? –  Asaf Karagila Aug 9 '12 at 9:55
    
I referred $X$ to a subspace of Dedekind-infinite metric space –  Katlus Aug 9 '12 at 10:03
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