Due to (0x677afc5e) we have \(A \subseteq \bar{A}\). Since \(A\) is closed we also have \(\bar{A} \subseteq A\).
On the other hand \(A\) is closed since \(\bar{A}\) is closed.
Due to (0x677afc5e) we have \(A \subseteq \bar{A}\). Since \(A\) is closed we also have \(\bar{A} \subseteq A\).
On the other hand \(A\) is closed since \(\bar{A}\) is closed.