T3 Regular Hausdorff Topological Spaces
Recall from the T0 Kolmogorov Topological Spaces page that a topological space $X$ is said to be a T_{0} space or a Kolmogorov space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that either $x \not \in V$ or $y \not \in U$.
On the T1 Fréchet Topological Spaces page we said that a topological space $X$ is said to be a T_{1} space or a Fréchet space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $x \not \in V$ and $y \not \in U$.
On the T2 Hausdorff Topological Spaces page we said that a topological space $X$ is said to be a T_{2} space or a Hausdorff space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.
Before we look at the next type of topological space, we will first need to give some definitions.
Definition: Let $X$ be a topological space and let $A$ and $B$ be disjoint subset of $X$, i.e., $A \cap B = \emptyset$. If $U$ and $V$ are open sets such that $A \subseteq U$, $B \subseteq V$, and $U \cap V = \emptyset$ then $U$ and $V$ are said to Separate $A$ and $B$. A topological space $X$ is said to be Regular if for every $x \in X$ and for every closed set $F \subseteq X$ not containing $x$ there exists open sets $U$ and $V$ which separate $\{ x \}$ and $F$. |
We will now define the next type of topological space called a T_{3} or Regular Hausdorff topological space.
Definition: Let $X$ be a topological space. Then $X$ is said to be a T_{3} Space or a Regular Hausdorff Space if $X$ is both a regular space and a T_{1} space. |
Theorem 1: Let $X$ be a topological space. If $X$ is a T_{3} space then $X$ is a T_{2} space. |
- Proof: Let $X$ be a T_{3} space. Then $X$ is both a regular space and a T_{1} space. Let $x, y \in X$, $x \neq y$. Since $X$ is a T_{1} space, every singleton set is closed. In other words, $\{ x \}$ and $\{ y \}$ are both closed. Since $X$ is a regular space, if we let $F = \{ y \}$ then there exists open sets $U$ of $\{ x \}$ and $V$ of $F = \{ y \}$ such that $U \cap V = \emptyset$. So $X$ is a T_{2} space. $\blacksquare$