Discrete Space has Open Locally Finite Cover

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 $\VV$.