2
$\begingroup$

I am trying to prove the following statement: Prove that if (X,d) is a compact metric space, then X must be separable.

Where separable means the following: We say a topological space is separable if it has a countable dense subset.

I think the strategy behind this would be to use the fact that a compact metric space is sequentially compact, meaning that every sequence in X has a convergent sub sequence in X (use this to prove the limit point p to which such sub sequence converges to is also in X. This making X dense).

I am uncertain of how to prove the countability of such set. I was thinking about using the set:

{{B(x,d) | x is an element of X}}

Extracting a finite subcover and constructing a subset contained in X by removing a particular open ball Bi(x,d). I am not sure as to whether or not the balls are countable sets themselves.

I really appreciate you taking the time to read through and help :)

$\endgroup$
4
$\begingroup$

Looking at covers by open balls is a good idea, but not in quite the manner that you seem to have in mind. For $n\in\Bbb Z^+$ let

$$\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}\;;$$

this is an open cover of $X$, so it has a finite subcover $\mathscr{V}_n$. This means that there is a finite $F_n\subseteq X$ such that

$$\mathscr{V}_n=\left\{B\left(x,\frac1n\right):x\in F_n\right\}\;;$$

Now see what you can do with the set $D=\bigcup_{n\in\Bbb Z^+}F_n$.

$\endgroup$

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.