Discrete Space is Paracompact
Jump to navigation
Jump to search
Theorem
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is paracompact.
Proof
Let $\VV$ be any open cover of $S$.
Consider the set $\CC$ of all singleton subsets of $S$:
- $\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover which is locally finite.
This result also shows that $\CC$ is the finest cover on $T$.
So $\CC$ is an open refinement of $\VV$ which is locally finite.
So $T$ is paracompact, by definition.
$\blacksquare$
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: $7$