Let \(X\) be a topological space and \(A\subseteq X\). A point is in \(\bar{A}\) if and only if every neighbourhood of it contains a point of \(A\).
Proof
If there is a neighbourhood such that it does not contain any point of \(A\), the point is in \(\Ext A\) (exterior ).
In the contrary if \(x\not\in \bar{A}\), it is in \(\Ext A\) and since it is open, there is a neighbourhood distinct to \(A\).