Locally Compact iff Open Neighborhood contains Compact Set

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Theorem

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


Then

$T$ is locally compact

if and only if

$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ 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 $\mathcal B$ of $x$ consisting of compact sets.

From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$.

By the definition of a neighborhood basis:

$\exists K \in \mathcal B : K \subseteq U$.

By the definition of a neighborhood of $x$:

$\exists V \in \tau : x \in V \subseteq K$

By the 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 $\mathcal B$ be the set of all compact neighborhoods of $x$.

It is shown that $\mathcal B$ is a neighborhood basis of $x$.


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

By the 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 the definition of the interior of a subset $K^\circ$ is open.

By the definition of a neighborhood then $K$ is a neighborhood of $x$.

Hence $K \in \mathcal B$ and $K \subseteq U \subseteq N$.

As $N$ was arbitrary, then $B$ is a neighborhood basis of $x$ consisting of compact sets.

Sources