> [!definition] > > ![[separations.png|400]] > > Let $(X, \topo)$ be a [[Topological Space|topological space]], then there are properties of the topology that guarantee the existence of [[Open Set|open sets]] that separate points or [[Closed Set|closed sets]] from each other. > - $T_0$: If $x \ne y$, then there exists an [[Open Set|open set]] containing $x$ but not $y$, *or* $y$ but not $x$. > - $T_1$: If $x \ne y$, then there exists an open set containing $y$ but not $x$. > - $T_2$ ([[Hausdorff Space|Hausdorff]]): If $x \ne y$, then there are disjoint open sets $U, V$ such that $x \in U$, $y \in V$. > - $T_3$ ([[Regular Space|Regular]]): A $T_1$ space where for any closed set $A \subset X$ and $x \not\in A$, there exists disjoint open sets $U, V$ such that $A \subset U$, $x \in V$. > - $T_4$ ([[Normal Space|Normal]]): A $T_1$ space where for any disjoint closed sets $A, B \subset X$, there exists disjoint open sets $U, V$ such that $A \subset U$, $B \subset V$. > [!theorem] > > Let $(X, \topo)$ be a topological space. $X$ is $T_1$ if and only if the singleton sets $\bracs{x}$ is closed for every $x \in X$. > > *Proof*. Suppose that $X$ is $T_1$. Let $x \in X$ and $y \ne x$. Then there exists an open set $U_y$ such that $y \in U_y$ but $x \not\in U_y$. This allows writing > $ > \bracs{x}^c = \bigcup_{y \ne x}U_y > $ > which is an open set since each $U_y$ is open. Therefore $\bracs{x}^c$ is closed. > > Suppose that every $\bracs{x}^c$ is closed. Then for any $x \ne y$, $t \in \bracs{x}^c$ and $x \in \bracs{y}^c$ are open. Therefore $X$ is $T_1$. > [!theorem] > > All $T_4$ spaces are $T_3$, all $T_3$ spaces are $T_2$. > > *Proof*. $\bracs{x}$ is a closed set.