Equivalence of Definitions of Strongly Locally Compact Space

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

1 implies 2
Follows immediately from Topological Closure is Closed.

2 implies 1
Let $x \in S$

Let $U$ be an open neighborhood of $x$.

Let $K \subseteq S$ be a closed compact subspace with $U \subseteq K$.

By Topological Closure is Closed $\overline U$ is closed in $T$.

By Set Closure as Intersection of Closed Sets, the closure $\overline U$ of $U$ in $T$ satisfies $\overline U \subseteq K$.

By Closed Set in Closed Subspace, $\overline U$ is closed in $K$.

By Closed Subspace of Compact Space is Compact, $\overline U$ is compact.