Let \(X\) be topological space . A subset \(A\subseteq X\) is closed if and only if \(\bar{A} = A\).
Proof
Due to (0x677afc5e) we have \(A \subseteq \bar{A}\). But since \(A\) is closed we also have \(\bar{A}\subseteq A\).
On the contrary \(A\) is closed, since \(\bar{A}\) is closed.