T5 Space is T4 Space

Theorem
Let $\left({S, \tau}\right)$ be a $T_5$ space.

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

Proof
Let $\left({S, \tau}\right)$ be a $T_5$ space.

From the definition of $T_5$ space:


 * $\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \varnothing: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $A^-$ is the closure of $A$ in $T$.

Let $C, D \subseteq S$ be disjoint sets which are closed in $T$.

Thus $C, D \in \complement \left({\tau}\right)$ from the definition of closed set.

From Topological Closure is Closed:
 * $C^- = C, D^- = D$

and so from $C \cap D = \varnothing$:
 * $C^- \cap D = C \cap D^- = \varnothing$

Thus from the definition of $T_5$ space:


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

which is precisely the definition of a $T_4$ space.