Discrete Uniformity generates Discrete Topology

Theorem
Let $S$ be a set.

Let $\mathcal U$ be the discrete uniformity on $S$.

Then the topology generated by $\mathcal U$ is the discrete topology on $S$.

The diagonal relation $\Delta_S$ generates the basis for this discrete topology.

Proof
From the construction, let $\vartheta \subseteq \mathcal P \left({S}\right)$ be a subset of the power set of $S$, created from $\mathcal U$ by:


 * $\vartheta := \left\{{u \left({x}\right): u \in \mathcal U, x \in X}\right\}$

where:
 * $\forall x \in X: u \left({x}\right) = \left\{{y: \left({x, y}\right) \in u}\right\}$

We need to show that $\vartheta$ is the discrete topology.

Consider $\Delta_S \in \mathcal U$.

Let $x \in S$.

Then the set:
 * $U_x := \left\{{y: \left({x, y}\right) \in \Delta_S}\right\}$

We have that: $\forall x \in S: U_x = \left\{{x}\right\} \in \mathcal U$

Thus we have:
 * $\mathcal B := \left\{{U_x: x \in S}\right\} = \left\{{\left\{{x}\right\}: x \in S}\right\}$

From Basis for Discrete Topology, we have that $\mathcal B$ is a basis for the discrete topology on $S$.