Equivalence of Definitions of T5 Space

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

Definition by Open Sets implies Definition by Closed Neighborhoods
Suppose $\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$.

Let $Y,A \subseteq S$.

Suppose $A \subseteq Y^\circ$ and $A^- \subseteq Y$.

Then:

We also have:

We have by hypothesis:
 * $\exists U, V \in \tau: A \subseteq U, \relcomp S Y \subseteq V, U \cap V = \O$

Then:

So we have demonstrated that there exists a closed neighborhood $\relcomp S V$ of $A$ contained in $Y$.

As $Y$, $A$ are arbitrary:


 * $\forall Y,A \subseteq S: (A \subseteq Y^\circ \wedge A^- \subseteq Y) \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$

Definition by Closed Neighborhoods implies Definition by Open Sets
Suppose $\forall Y,A \subseteq S: (A \subseteq Y^\circ \wedge A^- \subseteq Y) \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$.

Let $A,B \subseteq S$ be separated sets.

Then $A^- \cap B = A \cap B^- = \O$.

Then:

We also have:

We have by hypothesis that there exists a closed neighborhood $N$ of $A$ contained in $\relcomp S B$.

Then $\exists U \in \tau: A \subseteq U \subseteq N \subseteq \relcomp S B$.

Then we have:

Finally:

Therefore we have:

As $A$, $B$ are arbitrary,


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