Show that the \(T_1\) axiom is equivalent to the condition that for each pair of points of \(X\), each has a neighborhood not containing the other.
Solution
Assume first that \(X\) is \(T_1\), so each singleton is closed. Given distinct \(x,y\in X\), the set \[ X-\{y\} \] is an open neighborhood of \(x\) not containing \(y\), and \(X-\{x\}\) is an open neighborhood of \(y\) not containing \(x\).
Conversely, suppose that for every pair \(x\ne y\), each has a neighborhood omitting the other. Fix \(x\in X\). For every \(y\ne x\), choose an open neighborhood \(U_y\) of \(y\) with \(x\notin U_y\). Then \[ X-\{x\}=\bigcup_{y\ne x}U_y \] is open. Thus \(\{x\}\) is closed for every \(x\), so \(X\) is \(T_1\).