Discrete Uniformity is Uniformity

Theorem
Let $S$ be a set.

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

Then $\mathcal U$ is indeed a uniformity.

Proof
We examine the uniformity axioms in turn:


 * U1: $\forall u \in \mathcal U: \Delta_S \subseteq u$

This follows by definition of the discrete uniformity:
 * $\mathcal U := \left\{{u \subseteq S \times S: \Delta_S \subseteq u}\right\}$


 * U2: $\forall u, v \in \mathcal U: u \cap v \in \mathcal U$

We have that $\forall u, v \subseteq S \times S: \Delta_S \subseteq u, \Delta_S \subseteq v$

So $\Delta_S \subseteq u \cap v$ from Intersection Largest.

So $u \cap v \in \mathcal U$, and U2 holds.


 * U3: $u \in \mathcal U, u \subseteq v \subseteq S \times S \implies v \in \mathcal U$

Let $u \subseteq v \subseteq S \times S$.

We have that $\Delta_S \subseteq u$ and so $\Delta_S \subseteq v$ from Subsets Transitive.

So $v \in \mathcal U$ by definition of $\mathcal U$, and U3 holds.


 * U4: $\forall u \in \mathcal U: \exists v \in \mathcal U: v \circ v \subseteq u$ where $\circ$ is defined as:
 * $u \circ v := \left\{{\left({x, z}\right): \exists y \in S: \left({x, y}\right) \in v, \left({y, z}\right) \in u}\right\}$

We have by definition of $\mathcal U$ that $\Delta_S \in \mathcal U$.

We note from Diagonal Relation Equivalence that $\Delta_S$ is an equivalence relation and so by definition transitive.

Therefore from Transitive Relation contains Composite with Self we have that $\Delta_S \circ \Delta_S \subseteq \Delta_S$.

So:
 * $\forall u \in \mathcal U: \exists \Delta_S \in \mathcal U: \Delta_S \circ \Delta_S \subseteq \Delta_S \subseteq u$

and so U4 holds.


 * U5: $\forall u \in \mathcal U: \exists u^{-1} \in \mathcal U$ where $u^{-1}$ is defined as:
 * $u^{-1} := \left\{{\left({y, x}\right): \left({x, y}\right) \in u}\right\}$

Let $u \in \mathcal U$, so $\Delta_S \subseteq u \subseteq S \times S$.

Thus:
 * $\forall x \in u: \left({x, x}\right) \in u \implies \left({x, x}\right) \in u^{-1}$

and so $\Delta_S \subseteq u^{-1}$

Thus $u^{-1} \in \mathcal U$.

All of the uniformity axioms are satisfied, and $\mathcal U$ is demonstrated to be a uniformity.