Discrete Space is Strongly Locally Compact
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Theorem
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is strongly locally compact.
Proof
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$ of $T$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure in $T$.
From Singleton Set in Discrete Space is Compact, we have that $\set x$ is compact in $T$.
Hence the result by definition of strongly locally compact.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $7$