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Let $X$ be a subset of a metric space M. Prove that $X$ is a closed subset of M if and only if whenever x is a point in M such that $B_{\varepsilon} \cap X \ne \varnothing$ for every $\varepsilon > 0$ , then x $\in X$.

I've been working on this question for awhile. I understand that intuitively it makes sense. However, I can't quite figure out how to go about it. I have tried using the fact that $X'$ will be closed. And thus for all x' $\in X'$ there is an $\varepsilon$ > 0 such that $B_{\varepsilon} \subset X'$.

Conversely, I figure since x $\in X$ whenever the ball intersects with $X$, then it must be closed.

Am I on the right track here?

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    What is $X'$ in your question? – 2017-02-21
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    I'm not sure what you mean with the notation $X'$, and it would also be helpful to know which definition of closed you are using (either the complement of an open set or a set containing all its limit points), because the proof would be slightly different for each. – 2017-02-21
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    I'm sorry, in this course we were taught to use X' as the notation for the complement of X, that is what it represents. Could we not prove it either way? I just need it to be proved. This is simply what I thought so far. @B.Mehta – 2017-02-21

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First assume $X$ is closed in $M$. Then $X'$ is open, meaning that for any $y \in X'$, there is an $\epsilon>0$ such that $B_\epsilon(y) \subset X'$.

We want to show that for $x \in M$, if $B_\epsilon (x) \cap X \neq \varnothing$ for any $\epsilon>0$, then $x \in X$. To this end, suppose that $x \notin X$ and aim to show that $B_\epsilon (x) \cap X = \varnothing$ for some $\epsilon>0$, the contrapositive. But $x \notin X$ means exactly that $x \in X'$, so there is an $\epsilon>0$ such that $B_\epsilon(x) \subset X'$. Now for this $\epsilon$, $B_\epsilon (x)$ is contained entirely within $X'$, but we know $X$ and $X'$ are disjoint, so $B_\epsilon (x) \cap X = \varnothing$ must be true, and so we are done.

Using those ideas, it shouldn't be too difficult to prove the converse - work as much as you can with $X'$ since the definition of $X$ closed is in terms of its complement, and use the idea that if $U \subset Y$, then $U \cap Y' =\varnothing$.

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    Excellent, very helpful. I am working on the converse, but I am unsure if I started it the right way, or maybe if I just need to keep going. Here is what I've done: We want to show that X' is open, so for all *y* $\in X'$ there exists $B_\varepsilon (y)$ such that $B_\varepsilon (y) \subset X'$. This implies that $B_\varepsilon \cap X = \varnothing$. – 2017-02-21
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    Obviously we need to use that which we are given, but I am a little confused how to do that. – 2017-02-21