Let $X$ be a metric space with a metric $d$.
Why does $x\in \overline A \iff \forall \epsilon>0, A\cap V_{\epsilon} (x) \neq \emptyset \tag{1}$
?
There are many posts that take it as no-need-to-be-proven result, however I haven't found a proof, which is probably quite basic.
I know the definition of a limit point, but it is difficult for me to derive $(1)$ from it.
My definition of the closure is $\overline A = A \cup A'$ where $A'$ are the limit points of $A$.
A limit point of $A$ is a point $p$ such as for every neighborhood $V$ of $p$ there exists a point $x\in A$ such as $x\neq p$ and $x\in V$.
A neighborhood of $x$ is the set of $y$ such as $d(x,y)<\epsilon$ where $\epsilon \in \mathbb R^{+*}$.
PS: $V_{\epsilon} (x) $ is the set of $y$ such as $d(x,y)< \epsilon$.
