Equivalent Conditions for Local Compactness of Hausdorff Spaces

Last updated Nov 1, 2022

Let $X$ be a Hausdorff Topological Space. Then The following are Equivalent:

1. $X$ is Locally Compact.
2. Each $p \in X$ has a Precompact neighborhood.
3. $X$ has a Topological Basis of Precompact Sets.

$(1 \Rightarrow 2)$: Let $p \in X$. Since $X$ is Locally Compact, there exists Open $U \subset X$ so that $p \in U$ and $U \subset K$ for some Compact $K \subset X$. Since $X$ is Hausdorff, we know $K$ is Closed because Compact Sets in Hausdorff Spaces are Closed. Then, by definition of Closure, $\text{cl}U \subset K$. Recall, Closed Subset of a Compact Set is Compact, so $U$ is Precompact. $\checkmark$

$(2 \Rightarrow 3)$: For each $p \in X$, let $U_{p} \subset X$ be a Precompact neighborhood of $p \in X$. Consider the collection: $$\mathcal{B} := {U_{p} \cap V : p \in X, V \subset X\text{ open}}.$$ $\mathcal{B}$ is a collection of Open Sets, because finite Set Intersection of Open sets is Open. $\mathcal{B}$ is also a collection of Precompact sets because for each $B \in \mathcal{B}$, there exists $p \in X$ so that $B \subset U_{p}$. Since $U_{p}$ is Precompact, we see $B$ is Precompact as well because Subset of a Precompact Set is Precompact. Finally, we show $\mathcal{B}$ is a Topological Basis for $X$. Suppose $V \subset X$ is Open. Then for each $p \in V$, $U_{p} \cap V \in \mathcal{B}$. Then we have $$\bigcup\limits_{p \in V} U_{p} \cap V = (\bigcup\limits_{p \in V}U_{p}) \cap V = V.$$ Therefore, $\mathcal{B}$ is a Topological Basis for $X$. $\checkmark$

$(3 \Rightarrow 1)$: Let $p \in X$ and let $\mathcal{B}$ be a Topological Basis of Precompact sets. Since $p \in X$ and $X$ is Open, there must be some $B \in \mathcal{B}$ such that $p \in B$. Because $B$ is Precompact, $B \subset \text{cl}B$, which is Compact. By definition of Topological Basis, $B$ is Open. Thus $X$ is Locally Compact. $\checkmark$ $\blacksquare$

1. Lee - Introduction to Smooth Manifolds - Exercise A.55