Proving that $X$ is a closed subset if and only if whenever $B_\epsilon (x) \cap X \neq \emptyset $ for every $\epsilon >0$, then $x \in X$.

Question

Suppose $X$ is a subset of a metric space $M$. I would like to prove that $X$ is a closed subset of $M$ if and only if whenever $x$ is a point in $M$ such that $B_\epsilon (x) \cap X \neq \emptyset $ for every $\epsilon >0$, then $x \in X$.

I did one direction (the easier one), but am having trouble coming up with a rigorous proof for the other direction. Any help would be graciously appreciated!

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Definitions I am using:

If every limit point of $X$ belongs to $X$, I say that $X$ is closed. Also, $X$ is open if and only if $X^\mathsf{c}$ is closed.

Let $M$ be a metric space and $X$ be a subset of $M$. We say that a point $x$ in $M$ is a limit point of $X$ if there is a sequence $x_n$ such that $x_n \in X$ for every positive integer $n$ and limit as n goes to infinity of $x_n = x$.

Answer 1

HINT: Construct a sequence tending to $x$ by considering different values of $\epsilon$, like $\frac{1}{n}$ for all $n$ and finding a point, say $x_n$, in $B_\epsilon (x) \cap X$. And the apply definitions. This will work to prove both sides of the double implication.

Answer 2

I'll try to prove the reverse-implication. But my knowledge on metric spaces is not great. I've mostly been confined to $\Bbb R^n$ but I'm fairly certain the following should help.

Consider the complement of $X$ in $M$, say $C(X)$. Let $x \in C(X)$. Due to our condition there is a $\epsilon \gt 0$ such that $B_{\epsilon}(x) \cap X = \emptyset$. That is $B_{\epsilon}(x) \subseteq C(X)$. We have just proved that $C(X)$ is open and hence $X$ is closed.