T4 Normal Hausdorff Topological Spaces

# T4 Normal 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 $U \cap V = \emptyset$.

On the T3 Regular Hausdorff Topological Spaces page we saw that a topological space $X$ is said to be regular if for every $x \in X$ and for every closed set $F$ not containing $x$ there exists open sets $U$ and $V$ that separate $\{ x \}$ and $F$, i.e., $x \in U$ and $F \subseteq V$ where $U \cap V = \emptyset$. We then said that $X$ is a T3 space or a regular Hausdorff space if $X$ is both regular and T1

We will need to define another type of space before we can formally define a T4 space.

 Definition: A topological space $X$ is said to be Normal if for every pair of disjoint closed sets $E$ and $F$ of $X$ there exists open sets $U$ and $V$ such that $U \subseteq E$, $V \subseteq F$, and $U \cap V = \emptyset$.

We will now define what a T4 space is.

 Definition: Let $X$ be a topological space. Then $X$ is said to be a T4 Space or a Normal Hausdorff Space if $X$ is both a normal space and a T1 space.

As we will see in the following Theorem, every T4 space is a T3 space.

 Theorem 1: Let $X$ be a topological space. If $X$ is a T4 space then $X$ is a T3 space.
• Proof: Let $X$ be a T4 topological space. Then $X$ is both a normal space and a T1 space. We only need to show that $X$ is a regular space. Let $x \in X$ and let $F$ be a closed set in $X$ that does not contain $x$. Since $X$ is a T1 space, the singleton sets $\{ x \}$ are closed. So since $X$ is a regular space, there exists open sets $U$ and $V$ with $\{ x \} \subseteq U$ and $F \subseteq V$ such that $U \cap V = \emptyset$. Therefore $X$ is regular.
• Since $X$ is a regular space and is a T1 space we have that $X$ is a T3 space. $\blacksquare$