The Hausdorff Property on One-Point Compactifications of a Topo. Sp.

# The Hausdorff Property on One-Point Compactifications of a Topological Space

Recall from the One-Point Compactification of a Topological Space page that if $(X, \tau)$ is a topological space that is not compact then a one-point compactification of $X$ is the topological space $(X_{\infty}, \tau')$ where $X_{\infty} = X \cup \{ \infty \}$ and:

(1)\begin{align} \quad \tau' = \tau \cup \{ (X \setminus C) \cup \{ \infty \} : C \: \mathrm{is \: compact \: and \: closed \: in \:} X \} \end{align}

We will now look at a nice theorem which tells us exactly when the one-point compactification $X_{\infty}$ of a topological space $X$ is Hausdorff.

Theorem 1: Let $X$ be a topological space. Then $X$ is both Hausdorff and locally compact if and only if $X_{\infty}$ is Hausdorff. |

**Proof:**$\Rightarrow$ Let $X$ be both Hausdorff and locally compact, and let $x, y \in X_{\infty}$ be two distinct points. There are two cases to consider.

**Case 1:**Suppose that $x, y \neq \infty$. Then since $X$ is Hausdorff, there exists open neighbourhoods in $X$, $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$. But every open set in $X$ is an open set in the one-point compactification $X_{\infty}$. In other words, there exists open neighbourhoods $U$ in $X_{\infty}$, $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.**Case 2:**Without loss of generality, suppose that $x = \infty$, and $y \not \infty$. Then take any open neighbourhood $V$ of $y$ for which $\overline{V}$ is compact ($V$ is also an open neighbourhood of $y$ in $X_{\infty}$) which can be done from the local compactness of $X$. Since $\overline{V}$ is both compact and closed, we have that $U = (X \setminus \overline{V}) \cup \{ \infty \}$ is open in $X_{\infty}$ and contains $\infty$, and $U \cap V = (X \setminus \bar{V}) \cap V = \emptyset$.

- From both cases above we see that $X_{\infty}$ is indeed Hausdorff.

- $\Leftarrow$ Suppose that $X_{\infty}$ is Hausdorff. We first show that $X$ is Hausdorff.

- By definition, $X \subset X_{\infty}$ is a subspace of $X_{\infty}$ and since $X_{\infty}$ is Hausdorff and since the Hausdorff property is hereditary, we must have that $X$ is Hausdorff.

- We now show that $X$ is locally compact. Let $x \in X$. Since $X_{\infty}$ is Hausdorff there exists open neighbourhoods $U$ of $x$ and $V$ of $\infty$ such that $U \cap V = \emptyset$. Then for some compact and closed set $C$ in $X$ we have that $V$ is of the form:

\begin{align} \quad V = (X \setminus C) \setminus \{ \infty \} \end{align}

- Then we must have that $U \subset C$, and since $C$ is closed in $X$ we must have that $\bar{U} \subset C$. Since $C$ is compact in $X$ and $\bar{U}$ is closed in $X$ we must have that $\bar{U}$ is compact in $X$. So, for every point $x \in X$ there exists a neighbourhood of $x$ that is compact. Therefore $X$ is locally compact. $\blacksquare$