Definition 5.20.1 (Locally Compact Hausdorff Space).label Let $X$ be a Hausdorff space, then the following are equivalent:
- (1)
For any $x \in X$, there exists $K \in \cn(x)$ compact.
- (2)
For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of compact sets.
- (3)
For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of relatively compact sets.
If the above holds, then $X$ is a locally compact Hausdorff (LCH) space.
Proof. (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact. By Proposition 5.16.4, $K$ is closed. Let $U \in \cn^{o}(x)$ be open, then $K \setminus U$ is closed. By Proposition 5.16.3, $K \setminus U$ is compact. Since $X$ is Hausdorff, for each $y \in K \setminus U$, there exists $U_{y} \in \cn(x)$ and $V_{y} \in \cn^{o}(y)$ open such that $U_{y} \cap V_{y} = \emptyset$. As $K \setminus U$ is compact, there exists a $L \subset K \setminus U$ finite such that $\bigcup_{y \in L}V_{y} \supset K \setminus U$. In which case, $K \setminus \bigcup_{y \in L}V_{y}$ is compact by Proposition 5.16.3, and a neighbourhood of $x$ because it contains $\bigcap_{y \in L}U_{y}$.
(2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By Proposition 5.16.4, $K$ is closed, so $\overline{K^o}\subset K$ is compact by Proposition 5.16.3.$\square$
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