Equivalence of Definitions of Locally Compact Hausdorff Space

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

1 implies 2
Let $x\in S$.

Let $K$ be a compact neighborhood of $x$.

Let $U$ be a neighborhood of $x$.

We have to show that $U$ contains a compact neighborhood of $x$.

By Neighborhood in Topological Subspace iff Intersection of Neighborhood and Subspace, $U\cap K$ is a neighborhood of $x$ in $K$.

By Neighborhood in Compact Hausdorff Space Contains Compact Neighborhood, there is a compact neighborhood $V$ of $x$ in $K$ such that $V \subset U \cap K$.

By:
 * Neighborhood in Neighborhood Subspace
 * $K$ is a neighborhood of $x$ in $X$

$V$ is a compact neighborhood of $x$ in $X$.

Because $V\subset U$ and $U$ was arbitrary, $x$ has a neighborhood basis of compact sets.

2 implies 1
Follows from Locally Compact Space is Weakly Locally Compact.