Discrete Space has Open Locally Finite Cover

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Theorem

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


Consider the set $\CC$ of all singleton subsets of $S$:

$\CC := \set {\set x: x \in S}$

Then $\CC$ is an open cover of $T$ which is locally finite.


This cover is the finest cover on $S$.

That is, if $\VV$ is a cover of $T$, then $\CC$ is a refinement of $\VV$.


Proof

We have that:

$\forall x \in S: \exists \set x \in \CC: x \in \set x$

and so $\CC$ is a cover for $S$.

Then from Set in Discrete Topology is Clopen, it follows that $\CC$ is an open cover of $T$.

From Point in Discrete Space is Neighborhood, every point $x \in S$ has a neighborhood $\set x$.

This neighborhood $\set x$ intersects exactly one element of $\CC$, that is: $\set x$ itself.

As $1$ is a finite number, the result follows from definition of locally finite.


Now consider any cover $\VV$ of $S$.

By definition:

$\forall x \in S: \exists V \in \VV: x \in V$

That is:

$\forall x \in S: \exists V \in \VV: \set x \subseteq V$

That is, every element of $\CC$ is contained in some element of $\VV$.

Thus by definition, $\CC$ is a refinement of $\VV$.

$\blacksquare$


Sources