T4 Normal 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 $U \cap V = \emptyset$.
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 T_{3} space or a regular Hausdorff space if $X$ is both regular and T_{1}
We will need to define another type of space before we can formally define a T_{4} 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 T_{4} space is.
Definition: Let $X$ be a topological space. Then $X$ is said to be a T_{4} Space or a Normal Hausdorff Space if $X$ is both a normal space and a T_{1} space. |
As we will see in the following Theorem, every T_{4} space is a T_{3} space.
Theorem 1: Let $X$ be a topological space. If $X$ is a T_{4} space then $X$ is a T_{3} space. |
- Proof: Let $X$ be a T_{4} topological space. Then $X$ is both a normal space and a T_{1} 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 T_{1} 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 T_{1} space we have that $X$ is a T_{3} space. $\blacksquare$