Pseudometric Space generates Uniformity

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Theorem

Let $P = \struct {A, d}$ be a pseudometric space.

Let $\UU$ be the set of sets defined as:

$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$

where:

$\R_{>0}$ is the set of strictly positive real numbers
$u_\epsilon$ is defined as:
$u_\epsilon := \set {\tuple {x, y}: \map d {x, y} < \epsilon}$


Then $\UU$ is a uniformity on $A$ which generates a uniform space with the same topology as the topology induced by $d$.


Proof

We check whether the Uniformity Axioms are satisfied.


$\text U 1$

Let $x \in A$.

Then $\tuple {x, x} \in \Delta_A$.

Let $u \in \UU$.

From definition:

$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$

By Metric Space Axiom $(\text M 1)$:

$\map d {x, x} = 0 < \epsilon$

Hence we have $\tuple {x, x} \in u_\epsilon$.

By definition of subset, $\Delta_A \subseteq u_\epsilon \subseteq u$.

$\Box$


$\text U 2$

Let $u, v \in \UU$.

From definition:

$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$
$\exists \delta \in \R_{>0}: u_\delta \subseteq v$

Without loss of generality assume that $\delta \ge \epsilon$.

Let $\tuple {x, y} \in u_\epsilon$.

Then $\map d {x, y} < \epsilon \le \delta$.

Thus $\tuple {x, y} \in u_\delta$.

By definition of subset, $u_\epsilon \subseteq u_\delta$.

Hence:

\(\ds u_\epsilon\) \(=\) \(\ds u_\epsilon \cap u_\delta\) Intersection with Subset is Subset
\(\ds \) \(\subseteq\) \(\ds u \cap v\) Set Intersection Preserves Subsets
\(\ds \leadsto \ \ \) \(\ds u \cap v\) \(\in\) \(\ds \UU\)

$\Box$


$\text U 3$

Let $v \in A \times A$.

Suppose $u \in \UU$ and $u \subseteq v$.

From definition:

$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$

Thus $u_\epsilon \subseteq v$.

Therefore $v \in \UU$ as well.

$\Box$


$\text U 4$

Let $u \in \UU$.

From definition:

$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$

Let $v = u_{\epsilon / 2}$.

Since $u_{\epsilon / 2} \subseteq v$ and $\dfrac \epsilon 2 > 0$, $v \in \UU$.

From definition:

\(\ds v \circ v\) \(=\) \(\ds \set {\tuple {x, z}: \exists y \in A: \tuple {x, y} \in v, \tuple {y, z} \in v}\)
\(\ds \) \(=\) \(\ds \set {\tuple {x, z}: \exists y \in A: \map d {x, y} < \frac \epsilon 2, \map d {y, z} < \frac \epsilon 2}\)

Let $\tuple {x, z} \in v \circ v$.

Then:

$\exists y \in A: \map d {x, y} < \dfrac \epsilon 2, \map d {y, z} < \dfrac \epsilon 2$

By Metric Space Axiom $(\text M 2)$: Triangle Inequality:

\(\ds \map d {x, z}\) \(\le\) \(\ds \set {\tuple {x, z}: \exists y \in A: \tuple {x, y} \in v, \tuple {y, z} \in v}\)
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \frac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

Hence $\tuple {x, z} \in u_\epsilon \subseteq u$.

By definition of subset, $v \circ v \subseteq u$.

$\Box$


$\text U 5$

Let $u \in \UU$.

From definition:

$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$

Suppose $\tuple {b, a} \in u_\epsilon$.

Then $\map d {b, a} < \epsilon$.

By Metric Space Axiom $(\text M 3)$:

$\map d {a, b} = \map d {b, a} < \epsilon$

Hence $\tuple {a, b} \in u_\epsilon \subseteq u$.

Consider the set $u^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in u}$.

From the above, $\tuple {b, a} \in u^{-1}$.

By definition of subset, $u_\epsilon \subseteq u^{-1}$.

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

$\Box$


This needs considerable tedious hard slog to complete it.
In particular: It remains to show that the topology induced by the uniformity $\UU$ is the same as the one induced by the pseudometric $d$
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
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Sources

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