# Equivalence of Definitions of T5 Space

## Theorem

The following definitions of the concept of $T_5$ space are equivalent:

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

### Definition by Open Sets

$\left({S, \tau}\right)$ is a $T_5$ space if and only if:

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

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

$\left({S, \tau}\right)$ is a $T_5$ space if and only if:

every subset $Y \subseteq S$ contains a closed neighborhood of each $A \subseteq Y^\circ$ for which $A^- \subseteq Y$.

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

## Proof

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

$B^- \cap Y^\circ = \varnothing$

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.

$\blacksquare$