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!


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 $M_\epsilon \cup \{y\}$ 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 $X$.

  • $\begingroup$ 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 $\endgroup$ – Leon Dec 4 '11 at 0:07
  • $\begingroup$ @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 $\langle x_n:n\in\mathbb{N}\rangle\to x$. $\endgroup$ – Brian M. Scott Dec 4 '11 at 0:41
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Dec 4 '11 at 0:43
  • $\begingroup$ why $D$ is countably infinite. It should be finite because it is the union of finite $\epsilon-$ meshes. $\endgroup$ – Saikat Mar 25 '16 at 14:30
  • $\begingroup$ @Saikat: The union of countably infinitely many finite sets is countable, but it certainly need not be finite. After all, $\Bbb N$ is the union of the finite sets $\{n\}$ for $n\in\Bbb N$. $\endgroup$ – Brian M. Scott Mar 25 '16 at 14:32

A proof based on Rudin's Hints (Page 45, Qn 24)

Step 1: Fix $ \delta >0$, and pick $x_{1}\in X$. Having chosen $x_{1},...,x_{j}\in X$, choose $x_{j+1}\in X$, if possible, so that $d(x_{i},x_{j+1})\geq \delta $ for $i =1,...,j.$ This process must stop after a finite number of steps, otherwise $x_{i}\in X, i\in N$ is an infinite set in X, so it should have a limit point, say $x\in X$. Then any neighborhood of $x$ with radius less than $\frac{\delta}{2}$ contain at most one term of the sequence (remember, any two distinct terms of the sequence are of atleast $\delta$ distance). A contradiction.

Thus X can be covered by finitely many neighborhoods of radius $\delta$.

Step 2: Take $\delta = \frac{1}{n}$ ($n = 1,2,3,...$). Let $\{x_{n_{1}},...x_{n_{k(n)}}\}$ be the finite set obtained from step 1 corresponding to $\delta=\frac{1}{n}$. Let $D = \cup_{n=1}^{\infty} \{x_{n_{1}},...x_{n_{k(n)}}\}$. Then D is countable.\Next we prove D is dense in X which will prove the result.

If $D=X$ nothing to prove, otherwise let $x\in X \setminus D$ and take an $\epsilon$ - neighborhood of $x$. Choose n such that $\frac{1}{n}<\epsilon$. Neighborhoods of $x_{n_{1}},...x_{n_{k(n)}}$ with radius $\frac{1}{n}$ will cover X. So $x$ will be in one of such neighborhoods, say neighborhood of $x_{n_{i}}$, hence $d(x,x_{n_{i}})<\frac{1}{n}<\epsilon$. Thus $x$ is a limit point of D.


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.


Proof sketch:

It is well known that the hypothesis is equivalent with compactness of your space.

For every $n\geq 1$, consider the open cover

$$\{B_X(x,1/n)\mid x \in X\}$$

and extract finite subcovers. Choose the centers of the balls from the finite subcovers and put them together in a set (this requires choice).

The obtained set is at most countable and you can show that it is dense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.