# An element a belongs to closure A iff there exist a sequence an belongs to A s.t an -> a

I am reading a proof of a theorem stating that:

Point a belongs to closure A iff there exist a sequence an belongs to A s.t an -> a

Why the last inequality prove that a belongs to closure A?

The important thing to realize is that $$\overline{A} = A \cup A' = A \cup \{\text{limit points of A}\}$$ and the statement 'for every $\epsilon>0$, $B_\epsilon(a) \cap A \neq \emptyset$' is exactly the statement that $a\in A \cup A'$.
This is because either $a\in A$, in which case we're done, or $a\notin A$, so the non-emptiness of $B_\epsilon(a) \cap A$ implies that there is some point in this intersection which isn't $a$, i.e. $(B_\epsilon(a)\setminus\{a\}) \cap A \neq \emptyset$. Since $\epsilon$ is arbitrary, this means that $a$ is a limit point of $A$.