Locally Compact iff Open Neighborhood contains Compact Set
Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Then:
- $T$ is locally compact
- $\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
where $K^\circ$ denotes the interior of $K$.
Proof
Sufficient Condition
Let $T$ be locally compact.
Let $U \in \tau, x \in U$.
Since $T$ is locally compact there exists a neighborhood basis $\BB$ of $x$ consisting of compact sets.
From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$.
By definition of a neighborhood basis:
- $\exists K \in \BB : K \subseteq U$.
By definition of a neighborhood of $x$:
- $\exists V \in \tau : x \in V \subseteq K$
By definition of the interior of a subset then:
- $x \in V \subseteq K^\circ$
Thus
- $\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
$\Box$
Necessary Condition
Let
- $\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
Let $\BB$ be the set of all compact neighborhoods of $x$.
It is shown that $\BB$ is a neighborhood basis of $x$.
Let $N$ be a neighborhood of $x$.
By definition of a neighborhood of $x$:
- $\exists U \in \tau : x \in U \subseteq N$
By assumption:
- $\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
By definition of the interior of a subset $K^\circ$ is open.
By definition of a neighborhood then $K$ is a neighborhood of $x$.
Hence $K \in \BB$ and $K \subseteq U \subseteq N$.
As $N$ was arbitrary, then $B$ is a neighborhood basis of $x$ consisting of compact sets.
Sources
- Mizar article WAYBEL_3:def 9