T3 Regular Hausdorff Topological Spaces

Recall from the T0 Kolmogorov Topological Spaces page that a topological space $X$ is said to be a T₀ 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 T₁ 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 T₂ 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 T₃ or Regular Hausdorff topological space.

Definition: Let $X$ be a topological space. Then $X$ is said to be a T3 Space or a Regular Hausdorff Space if $X$ is both a regular space and a T1 space.

Theorem 1: Let $X$ be a topological space. If $X$ is a T3 space then $X$ is a T2 space.

• Proof: Let $X$ be a T₃ space. Then $X$ is both a regular space and a T₁ space. Let $x, y \in X$, $x \neq y$. Since $X$ is a T₁ 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 T₂ space. $\blacksquare$