Equivalence of Definitions of T5 Space

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

Definition by Open Sets
$T = \left({S, \tau}\right)$ is a $T_5$ space :


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

That is:
 * $\left({S, \tau}\right)$ is a $T_5$ space when for any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.

Definition by Closed Neighborhoods
$T = \left({S, \tau}\right)$ is a $T_5$ space :
 * each subset $Y$ of $S$ contains a closed neighborhood of each $A \subseteq Y^\circ$ for which $A^- \subseteq Y$.

Here, $Y^\circ$ denotes the interior of $Y$ and $A^-$ denotes the closure of $A$.

Definition by Closed Neighborhoods implies Definition by Open Sets
Suppose that each subset $Y \subseteq S$ contains a closed neighborhood of each $A \subseteq Y^\circ$ for which $A^- \subseteq Y$.

Let $Y \subseteq S$.

Let $B = \complement_S \left({Y}\right)$.

From Complement of Interior equals Closure of Complement:
 * $B^- \cap Y^\circ = \varnothing$

Hence:
 * $A \subseteq Y^\circ \implies A \cap B^- = \varnothing$

Also:
 * $A^- \subseteq Y \implies A^- \cap B = \varnothing$

demonstrating that $A$ and $B$ are separated.

Now let $N_A$ be a closed neighborhood of $A$ such that $N_A \subseteq Y$.

Then $\complement_S \left({N_A}\right)$ is an open neighborhood of $B$.

As $N_A$ is a (closed) neighborhood of $A$:
 * $\exists U \in \tau: A \subseteq U \subseteq N_A$

As $\complement_S \left({N_A}\right)$ is an open neighborhood of $B$:
 * $\exists V \in \tau: B \subseteq V \subseteq \complement_S \left({N_A}\right)$

From Intersection with Complement is Empty iff Subset it follows that:
 * $U \cap \complement_S \left({N_A}\right) = \varnothing$

and
 * $V \cap N_A = \varnothing$

from which it follows that:
 * $U \cap V = \varnothing$

and so the conditions for a $T_5$ space from open sets are fulfilled.