T5 Space is T4 Space

Theorem
Let $\struct {S, \tau}$ be a $T_5$ space.

Then $\struct {S, \tau}$ is also a $T_4$ space.

Proof
Let $\struct {S, \tau}$ be a $T_5$ space.

From the definition of $T_5$ space:


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

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

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

Thus $C, D \in \map \complement \tau$ from the definition of closed set.

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

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

Thus from the definition of $T_5$ space:


 * $\forall C, D \in \map \complement \tau, C \cap D = \O: \exists U, V \in \tau: C \subseteq U, D \subseteq V$

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