# A metric space in which every infinite set has a limit point is separable

I am struggling with one problem. I need to show that if $X$ is a metric space in which every infinite subset has a limit point then $X$ is separable (has countable dense subset in other words).

I am trying to use the result I have proven prior to this problem, namely every separable metric space has a countable base (i.e. any open subset of the metric space can be expressed as a sub-collection of the countable collection of sets). I am not sure this is the right way, can anyone outline the proof? Thanks a lot in advance!

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Let $\langle X,d\rangle$ be a metric space in which each infinite subset has a limit point. For any $\epsilon>0$ an $\epsilon$-mesh in $X$ is a set $M\subseteq X$ such that $d(x,y)\ge\epsilon$ whenever $x$ and $y$ are distinct points of $M$. Every $\epsilon$-mesh in $X$ is finite, since an infinite $\epsilon$-mesh would be an infinite set with no limit point. Let $\mathscr{M}(\epsilon)$ be the family of all $\epsilon$-meshes in $X$, and consider the partial order $\langle \mathscr{M}(\epsilon),\subseteq\rangle$. This partial order must have a maximal element: if it did not have one, there would be an infinite ascending chain of $\epsilon$-meshes $M_0\subsetneq M_1\subsetneq M_2\subsetneq\dots$, and $\bigcup_n M_n$ would then be an infinite $\epsilon$-mesh. Let $M_\epsilon$ be a maximal $\epsilon$-mesh; I claim that $$X=\bigcup_{x\in M_\epsilon}B(x,\epsilon)\;,$$ where as usual $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$. That is, each point of $X$ is within $\epsilon$ of some point of $M_\epsilon$. To see this, suppose that $y\in X\setminus \bigcup\limits_{x\in M_\epsilon}B(x,\epsilon)$. Then $d(y,x)\ge\epsilon$ for every $x\in M_\epsilon$, and $M_\epsilon \cup \{x\}$ is therefore an $\epsilon$-mesh strictly containing $M_\epsilon$, contradicting the maximality of $M_\epsilon$.

Now for each $n\in\mathbb{N}$ let $M_n$ be a maximal $2^{-n}$-mesh, and let $$D=\bigcup_{n\in\mathbb{N}}M_n\;.$$ Each $M_n$ is finite, so $D$ is countable, and you should have no trouble showing that $D$ is dense in $X$.

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Brian, Can I possibly ask for a bit of clarification of the above proof? There is one bit of the proof where I'm struggling. Shouldn't the union of all e-meshes be infinite? Otherwise if its finite, wouldn't it mean there exist e-mesh, which you could 'slice' no more? Would the union of all the e-meshes (for every positive 'e') form an infinite set with a limit point in X? Can you possibly clarify your thinking of that part of the proof? HUGE thanks in advance, Leon –  Leon Dec 4 '11 at 0:07
@Leon: I’m not sure quite what you’re asking here. The union of all possible $\epsilon$-meshes is $X$, but the only union of $\epsilon$-meshes that we need to consider is $D$. $D$ actually can be finite, but in general it is indeed infinite; however, it’s only countably infinite, so if you can show that it’s dense in $X$, you’ve shown that $X$ is separable. To do this, let $x\in X$; for each $n\in\mathbb{N}$ there is some $x_n\in M_n$ such that $d(x,x_n)<2^{-n}$, and it should be clear that $\langle x_n:n\in\mathbb{N}\rangle\to x$. –  Brian M. Scott Dec 4 '11 at 0:41
@Leon: Or were you asking why $M_\epsilon$ is finite? That’s simply because (1) it’s an $\epsilon$-mesh, and (2) every $\epsilon$-mesh in $X$ is finite. –  Brian M. Scott Dec 4 '11 at 0:43

Since $X$ is limit point compact, it is totally bounded. That is, for every $\epsilon>0$, there is a finite cover of $X$ consisting of balls of radius $\epsilon$ (if not, one could construct a sequence in $X$ that has no limit point). For each positive integer $n$, let $A_n$ a finite cover of $X$ of open sets of radius $1/n$. Now consider $\cup C_n$, where $C_n$ is the set of centers of the elements in $A_n$.

This is essentially what Brian did, but I'll post it anyway. In a nutshell, you are showing: limit point compactness implies $X$ is totally bounded and a totally bounded space is seperable.

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