Answer by student13 for $x\in \overline A \iff$ - how to derive definition of closure

I'm assuming that with the limit definition you mean that $\overline A$ is the collection of points $a$ such that a path $\{x_i\}_{i = 1}^\infty \subseteq A$ exists that converges to $a$.

The proof is actually quite easy.

1. if $a \in \overline A$, then such a path exists for $a$, then by limit definition, for all $\epsilon > 0$ we can find a point $x_i$ such that $d(a, x_i) < \epsilon$, hence the $V_\epsilon(a)$ has overlap with $A$
2. if every $V_\epsilon(a)$ has overlap with $A$, we can find points $x_i \in V_\frac{1}{i}(a) \cap A$. Now $d(x_i, a) < 1/i$, so $\{x_i\}_{i = 1}^\infty$ converges to $a$, and for all $x_i \in A$. Therefore $a \in \overline A$.

I hope this gives you an idea on why these definitions are equivalent.