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A metric space is called separable if it contains a countable dense subset.

I have no idea how to go about proving this. What sort of things should I understand to do this. What would a sample proof look like?

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  • $\begingroup$ Do you know what it means for a set to be dense? $\endgroup$ – graydad Oct 13 '14 at 16:19
  • $\begingroup$ Do you know why $\mathbb R$ is separable? $\endgroup$ – azarel Oct 13 '14 at 16:19
  • $\begingroup$ Think rational... $\endgroup$ – copper.hat Oct 13 '14 at 16:20
  • $\begingroup$ Yeah, for a set to be dense, for every a in our set A, that point A will include a point from our metric space X. $\endgroup$ – J.r Oct 13 '14 at 16:21
  • $\begingroup$ @J.r That seems a little off to me. The definition I learned is that a subset $A \subset X$ is dense if $\overline A = X$. That is, the closure of $A$ gives you $X$. $\endgroup$ – graydad Oct 13 '14 at 16:24
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Well, you're looking for something that's countable and dense. The natural thing to look at would be $\mathbb Q$. In this case, look at $\mathbb Q ^K$. It's countable as it's a finite crossproduct of countable sets. To show it's dense, you have to show that every point in $\mathbb R ^k$ is a limit point of $\mathbb Q^k$. For that, use a sequence approach (Decimal approximations is the natural one)

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Since this is also a question with "general-topology" tag:

Theorem: Let X and Y be separable topological spaces. Then $X \times Y$ is separable.

Proof: There are two countable sets $A \subseteq X $ and $B \subseteq Y$ such that $\overline A=X$ and $\overline B =Y$. Since A and B are countable $A \times B$ is countable too.

For an arbitrary $(x,y) \in X \times Y $ holds: For every open neighborhoods $U,V$ of $x$ and $y$ the intersections $U\cap A$ and $V\cap B$ are not empty, hence the intersection $U \times V \cap A \times B$ is not empty too.

From this it follows that the intersection of $A \times B$ with every open neighborhood containing $(x,y)$ is non-empty.

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Hint: Explore some properties of $\Bbb{Q}^k$

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