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How to prove that Every compact metric space is separable$?$

Thanks in advance!!

Martin Sleziak
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Aang
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2 Answers2

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Hint: Consider the countable family of coverings $\mathcal U_n=\left\{B\left(x,\frac1n\right)\mid x\in X\right\}$ for all $n$, use compactness to distill a countable family of points, and show it is dense.

Asaf Karagila
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  • Thanks for the hint, i got it. But, i feel it difficult to find this kind of open covering. There are a lot of other open coverings ; is there any trick for how to know which one would help? – Aang Sep 04 '12 at 06:16
  • @Avatar: All of them. – Asaf Karagila Sep 04 '12 at 06:20
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    The easiest open set in a metric space are balls; the easiest way to ensure that all points are covered is to use all points as centers; the easiest way to ensure best "granularity" is to allow arbitrarily small radii. – Hagen von Eitzen Sep 04 '12 at 06:22
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    @Asaf Karagila: You exhibit a countable set of coverings, whic might be confusing. The straightforward covering ${B(x,r)|x \in X, r>0}$ should suffice. – Hagen von Eitzen Sep 04 '12 at 06:24
  • @Avatar: But why would you care about other open covers? All you want is a countable dense set, and Asaf’s argument gives it to you. – Brian M. Scott Sep 04 '12 at 06:25
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    @Hagen: A single cover isn’t sufficient. You need to use a sequence of covers to get finite sets with granularities approaching $0$. – Brian M. Scott Sep 04 '12 at 06:26
  • D'oh. I was thinking of something completely different – Hagen von Eitzen Sep 04 '12 at 06:29
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Recall that every infinite subset $E$ of a compact metric space $X$ has a limit point in $X$. Now define a recursive process as follows: Fix some $\delta > 0$ and pick an $x_1 \in X$. Now having defined $x_1,\ldots,x_n$, if possible choose $x_{n+1}$ such that $d(x_{n+1},x_i) \geq \delta$ for all $1 \leq i \leq n$. Now we claim that this process must terminate. For suppose it does not. Consider $$E = \{x_1,x_2,x_3,\ldots \}.$$

Now by assumption of $X$ being compact, $E$ has a limit point $a \in X$ say. But then $B_\frac{\delta}{2}(a)$ can only contain at most one point of $E$, a contradiction. It follows that $X$ can be covered by finitely many neighbourhoods of radius $\delta$ about some points $x_1,\ldots,x_n$. Now what happens if you consider all balls of radius $\frac{1}{m}$ about each $x_i$ for $1 \leq i \leq n$? This will be countable, but why will it be dense in $X$? $m$ by the way runs through all positive integers.

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    Just a minor quibble: $B_\delta (a)$ could contain more than one point of $E$. $B_{\delta / 2} (a)$ on the other hand... – user642796 Sep 04 '12 at 11:16
  • @ArthurFischer I always get mixed up with the radius and diameter :D –  Sep 04 '12 at 13:07
  • The $n$ depends on the $m$. We get different points for each radius $1/m$. – Stefan Hamcke Feb 09 '14 at 22:59
  • @ArthurFischer

    Why does "every infinite sub-set $E$ of a compact metric space $X$ has a limit point in $X$"?

     – Eric_ Oct 19 '14 at 18:30
  • @Eric_ Every infinite subset E having a limit point in E is one possible definition of compactness for metric spaces. It is equivalent to sequential compactness. – R R Feb 05 '15 at 05:46