Locally Compact iff Open Neighborhood contains Compact Set

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

Then
 * $T$ is locally compact


 * $\forall X \in \tau, x \in X: \exists Y \subseteq S: x \in Y^\circ \land Y \subseteq X \land Y$ is compact

where $Y^\circ$ denotes the interior of $Y$.

Sufficient Condition
Let
 * $T$ is locally compact

Let $X \in \tau, x \in X$.

By Locally Compact iff Open Neighborhood contains Compact Open Neighborhood
 * $ \exists U \in \tau: a \in U \land U \subseteq X \land U$ is compact

By Interior of Open Set:
 * $U^\circ = U$

Thus
 * $\exists Y \subseteq S: x \in Y^\circ \land Y \subseteq X \land Y$ is compact

Necessary Condition
Let
 * $\forall X \in \tau, x \in X: \exists Y \subseteq S: x \in Y^\circ \land Y \subseteq X \land Y$ is compact