Closure of closure of set equals closure of set Let $A\subseteq \mathbb{R}^n$. I want to prove that $cl(cl(A))\subseteq cl(A)$. Let $x\in cl(cl(A))$, then $x$ is adherent to $cl(A)$, which implies that $B(x;r)\cap cl(A)\neq\emptyset, \forall r>0$. How do I proceed from here? And do I really need to?
I'm guessing that it might be sufficient to argue as follows:
Let $x\in cl(cl(A))$, then $x$ is adherent to $cl(A)$, which implies that $x\in cl(A)$.
But would this be rigorous enough?

Suppose $x$ is an adherent point of $\operatorname{cl}(A)$, we want show $x$ is an adherent point of $A$. Suppose it were not. Then there is an open set $O$ (or open ball if you work in metric spaces) that contains $x$ and misses $A$, so $O \cap A = \emptyset$. Now for all $p \in O$, the set $O$ is a witness to the fact that $p \notin \operatorname{cl}(A)$ as well (it's an open neighbourhood of $p$ missing $A$). But then $O$ contains no points of $\operatorname{cl}(A)$ at all, showing that $x \notin \operatorname{cl}(\operatorname{cl}(A))$ at all. Contradiction. So $x$ is an adherent point of $A$.
So in metric spaces you need the minor lemma: $p \in B_r(x)$ then there exists $r' >0 $ such that $B_{r'}(p) \subseteq B_r(x)$( so $B_r(x)$ is an open set) to make the proof work with metric open balls (then you take this smaller ball as a witness for $p \in O = B_r(x)$). This is, however, a simple application of the triangle inequality.
