Fully T4 Space is T4 Space

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Theorem

Let $T = \left({S, \tau}\right)$ be a fully $T_4$ space.

Then $T$ is a $T_4$ space.


Proof

Recall that $\left({S, \tau}\right)$ is a $T_4$ space if and only if:

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

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


Consider the open cover $\mathcal U = \left\{{\complement \left({A}\right), \complement \left({B}\right)}\right\}$.

Since $T$ is fully $T_4$, there exists a star refinement $\mathcal V$ of $\mathcal U$.

Define $\mathcal V_A$ and $\mathcal V_B$ by:

$\mathcal V_A := \left\{{V \in \mathcal V: A \cap V \ne \varnothing}\right\}$
$\mathcal V_B := \left\{{V \in \mathcal V: B \cap V \ne \varnothing}\right\}$

and subsequently $U_A$ and $U_B$ by:

$U_A := \bigcup \mathcal V_A$
$U_B := \bigcup \mathcal V_B$


Now given $a \in A$, since $\mathcal V$ is a cover, there exists $V \in \mathcal V$ such that $a \in V$.

But then $V \in \mathcal V_A$, so that:

$a \in U_A$

Hence $A \subseteq U_A$, and similarly $B \subseteq U_B$.


Aiming for a contradiction, suppose that $x \in U_A \cap U_B$.

Then there exist $V_A \in \mathcal V_A, V_B \in \mathcal V_B$ such that:

$x \in V_A, x \in V_B$

By definition of $\mathcal V_A$ and $\mathcal V_B$, there exist:

$a \in A \cap V_A$
$b \in B \cap V_B$


Now recall the definition of the star $x^*$ of $x$:

$x^* := \bigcup \left\{{V \in \mathcal V: x \in V}\right\}$

Since $T$ is fully $T_4$, either $x^* \subseteq \complement \left({A}\right)$ or $x^* \subseteq \complement \left({B}\right)$.

But $a \in V_A \subseteq x^*$, so $x^* \nsubseteq \complement \left({A}\right)$.

Similarly $b \in V_B \subseteq x^*$, so $x^* \nsubseteq \complement \left({B}\right)$.


From this contradiction it is inferred that $x \in U_A \cap U_B$ cannot exist.

That is:

$U_A \cap U_B = \varnothing$

and it follows that $T$ is a $T_4$ space.

$\blacksquare$


Sources