TITLE: Prove that $A$ is closed if and only if $A=\ cl(A)$
QUESTION [1 upvotes]: Prove that $A$ is closed if and only if $A=\ cl(A)$

So this is my attempt for an answer.
Pf-Suppose $A\subseteq X$ is closed.
In the backward direction, the proof is trivial since if $A=\ cl(A)\Rightarrow A$ is closed.
In the forward direction,
Since $\ cl(A)$ is the smallest closed subset that contains $A$ then by definition of the closure of a set, we obtain the trivial answer  $A\subseteq \ cl(A)$.
Since A is closed. let $x\in \ cl(A)$ $\Rightarrow \forall r>0,\space B(x;r)\cap A\neq \emptyset\space $. Now, fix $r>0\space$ so that   $\space B(x;r)\cap A\subseteq A\ \Rightarrow x\in A$ so we obtain we obtain $\ cl(A)\subseteq A$.
So we get $cl(A)=A$

REPLY [0 votes]: Well one direction is trivial, because the closure of a set is by definition a certain type of closed set.
The other direction is nearly as easy because of the type of closed set the closure is.  It's the smallest closed set containing $A$.  But there is no smaller set containing $A$ than $A$ itself.
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