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I am hoping to prove the following:

$A \subset X$ is such that if for any $x ∈ X$ and $\epsilon > 0$ there exists $y ∈ A$ such that $d(x, y) < \epsilon$

is equivalent to saying for any $B \subset X, B$ nonempty and open, then $B \cap A$ is nonempty.

I have proved the reverse, but having some trouble with this direction. I think I need to use open balls and show that their intersection contains a point, but I'm confused by the specifics.

Johner
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  • Hint: Every open set in a metric space under the topology induced by said metric can be expressed as an (arbitrary) union of balls (of various epsilons). Use this to fact to acquire an $\epsilon >0$ and $x\in B$ and apply the premise to this situation. – Justin Benfield Oct 22 '16 at 16:13
  • Can I choose epsilon to be anything? I have a proof that works if epsilon is greater than the radius of another ball, but I'm not so sure I can assume this @JustinBenfield – Johner Oct 22 '16 at 16:19
  • Well, what you get is that since $B$ is nonempty, there is an $x$ in $B$ (and since $B\subset X$, $x\in X$). Now since $B$ is open, it is a union of open balls, choose any $x\in B$ , then there is an $\epsilon$-ball containing it that is entirely within $B$ (why?) This meets the requirements of the hypothesis, thus you can invoke the hypothesis to conclude that $\exists y\in A$, which is also in said ball, and this in $B$, therefore $B\cap A$ is nonempty. – Justin Benfield Oct 22 '16 at 16:28

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Suppose $B$ is open and non empty, choose $x \in B$ then $B(x,\epsilon) \subset B$ for some $\epsilon>0$ since $B$ is open. By hypothesis, there is some $y \in A $ such that $d(y,x) < \epsilon$ and hence $y \in B$. So $A \cap B $ is non empty.

copper.hat
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