# Condition for Alexandroff Extension to be T2 Space

## Theorem

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

Let $p$ be a new element not in $S$.

Let $S^* := S \cup \left\{{p}\right\}$.

Let $T^* = \left({S^*, \tau^*}\right)$ be the Alexandroff extension on $S$.

Then $T^*$ is a $T_2$ (Hausdorff) space if and only if $T$ is a locally compact Hausdorff space.

## Proof

### Necessary Condition

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

Let $x, y \in S$.

Then as $T$ is Hausdorff, there exist two disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.

But by definition of the Alexandroff extension on $S$, $U$ and $V$ are also open sets of $T^*$.

That covers the case where both $x \ne p$ and $y \ne p$.

Without loss of generality, let $y = p$.

As $T$ is a locally compact Hausdorff space, $x$ has a compact neighborhood $N_x$.

By definition of neighborhood, there exists an open set $U$ of $T$ containing $x$.

By definition of the Alexandroff extension, $U$ is an open set of $T^*$.

By Compact Subspace of Hausdorff Space is Closed, $N_x$ is closed in $T$.

Let $V = S^* \setminus N_x$.

By definition of $V$, $x \notin V$ and $p \in V$.

By definition of the Alexandroff extension, $V$ is an open set of $T^*$.

By Intersection with Complement is Empty iff Subset:

- $U \cup V = \varnothing$

So we have two open sets $U, V$ of $T^*$ such that $x \in U, p \in V$ and $U \cup V = \varnothing$.

Thus by definition $T^*$ is a $T_2$ space.

$\Box$

### Sufficient Condition

Let $T^*$ be a $T_2$ space.

By definition of $T$ is a subspace of $T^*$.

By $T_2$ Property is Hereditary, it follows that $T$ is a $T_2$ space.

It remains to be shown that $T$ is also (weakly) locally compact.

Let $x \in S$.

As $T^*$ is a $T_2$ space, we can find.

- an open set $U$ of $T^*$ containing $x$

and:

- an open set $V$ of $T^*$ containing $p$

such that $U \cap V = \varnothing$.

Then by definition of the Alexandroff extension, $S^* \setminus V$ is a compact (closed) neighborhood of $x$.

Thus by definition $S$ is a locally compact Hausdorff space.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 34: \ 3$