Normal Space is T3 Space

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Theorem

Let $\left({S, \tau}\right)$ be a normal space.


Then $\left({S, \tau}\right)$ is also a $T_3$ space.


Corollary

Let $\left({S, \tau}\right)$ be a normal space.


Then $\left({S, \tau}\right)$ is also a regular space.


Proof

Let $T = \left({S, \tau}\right)$ be a normal space.

From the definition of normal space:

$\left({S, \tau}\right)$ is a $T_4$ space
$\left({S, \tau}\right)$ is a Fréchet ($T_1$) space.


Let $F$ be any closed set in $T$.

Let $y \in \complement_S \left({F}\right)$, that is, $y \in S$ such that $y \notin F$.

As $T$ is a Fréchet ($T_1$) space it follows from Equivalence of Definitions of $T_1$ Space that $\left\{{y}\right\}$ is closed.


As $T = \left({S, \tau}\right)$ is a normal space, we have that:

$\forall A, B \in \complement \left({\tau}\right), A \cap B = \varnothing: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.

But $F$ and $\left\{{y}\right\}$ are disjoint closed sets.

So:

$\forall F \subseteq S: \complement_S \left({F}\right) \in \tau, y \in \complement_S \left({F}\right): \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \varnothing$

which is precisely the definition of a $T_3$ space.

$\blacksquare$


Sources