# Discrete Uniformity is Uniformity

## Theorem

Let $S$ be a set.

Let $\UU$ be the discrete uniformity on $S$

Then $\UU$ is indeed a uniformity.

## Proof

We examine the uniformity axioms in turn:

$\text U 1$
$\forall u \in \UU: \Delta_S \subseteq u$

This follows by definition of the discrete uniformity:

$\UU := \set {u \subseteq S \times S: \Delta_S \subseteq u}$

$\text U 2$
$\forall u, v \in \UU: u \cap v \in \UU$

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 is Largest Subset.

So $u \cap v \in \UU$, and $\text U 2$ holds.

$\text U 3$
$u \in \UU, u \subseteq v \subseteq S \times S \implies v \in \UU$

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

We have that $\Delta_S \subseteq u$ and so $\Delta_S \subseteq v$ from Subset Relation is Transitive.

So $v \in \UU$ by definition of $\UU$, and $\text U 3$ holds.

$\text U 4$
$\forall u \in \UU: \exists v \in \UU: v \circ v \subseteq u$ where $\circ$ is defined as:
$u \circ v := \set {\tuple {x, z}: \exists y \in S: \tuple {x, y} \in v, \tuple {y, z} \in u}$

We have by definition of $\UU$ that $\Delta_S \in \UU$.

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

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

So:

$\forall u \in \UU: \exists \Delta_S \in \UU: \Delta_S \circ \Delta_S \subseteq \Delta_S \subseteq u$

and so $\text U 4$ holds.

$\text U 5$
$\forall u \in \UU: \exists u^{-1} \in \UU$ where $u^{-1}$ is defined as:
$u^{-1} := \set {\tuple {y, x}: \tuple {x, y} \in u}$

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

Thus:

$\forall x \in u: \tuple {x, x} \in u \implies \tuple {x, x} \in u^{-1}$

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

Thus $u^{-1} \in \UU$.

$\Box$

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

$\blacksquare$