Theorem 1   Let $X$ be a topological space. Then $X$ is a $T_1$ -space if and only if sets $\{x\}$ , $\{y\}$ are separated for all distinct $x,y\in X$ .

Proof. Suppose $X$ is a $T_1$ -space. Then every singleton is closed and if $x,y\in X$ are distinct, then \begin{eqnarray*} \{x\} \cap \overline{\{y\}} &=& \{x\} \cap \{y\} =\emptyset, \\ \overline{ \{x\} }\cap \{y\} &=& \{x\} \cap \{y\} =\emptyset, \end{eqnarray*}and $\{x\}$ , $\{y\}$ are separated. On the other hand, suppose that $\{x\} \cap \overline{\{y\}} =\emptyset$ for all $x\neq y$ . It follows that $\overline{\{y\}}=\{y\}$ , so $\{y\}$ is closed and $X$ is a $T_1$ -space.