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?

Closure of a closed set is equal to itself. As $cl(A)$ is closed, therefore 
$$cl(cl(A))=cl(A)$$
