uniformities on a set form a complete lattice

Theorem.

The collection of uniformities on a given set ordered by set inclusion forms a complete lattice.

Proof.

Let $X$ be a set. Let $𝔘(X)$ denote the collection of uniformities on $X$. The coarsest uniformity on $X$ is $\{X\times X\}$, and the finest is the discrete uniformity:

$$\{S\subset X\times X:\mathrm{\Delta }(X)\subseteq S\}.$$

Hence $𝔘(X)$ is bounded. To show that $𝔘(X)$ is complete, we must prove that it has the least upper bound property.

Suppose ${\{{𝒰}_{\alpha }\}}_{\alpha \in I}$ is a nonempty family of uniformities on $X$. Let $ℬ$ consist of all finite intersections of elements of the ${𝒰}_{\alpha }$. Let us check that $ℬ$ is a fundamental system of entourages for a uniformity on $X$.

(B1) Let $S$, $T\in ℬ$. Each of $S$ and $T$ is a finite intersection of elements of the ${𝒰}_{\alpha }$, so their intersection is as well. Hence $S\cap T\in ℬ$.

(B2) Every element of $ℬ$ is a finite intersection of subsets of $X\times X$ containing $\mathrm{\Delta }(X)$. So every element of $ℬ$ contains the diagonal.

(B3) Let $S\in ℬ$. Without loss of generality, $S=S_{\alpha }\cap S_{\beta }$, where $S_{\alpha }\in {𝒰}_{\alpha }$ and $S_{\beta }\in {𝒰}_{\beta }$. Since $S_{\alpha }\in {𝒰}_{\alpha }$, $S_{\alpha }^{-1}\in {𝒰}_{\alpha }$. Similarly, $S_{\beta }^{-1}\in {𝒰}_{\beta }$. Since the process of taking the inverse of a relation commutes with taking finite intersections, ${(S_{\alpha }\cap S_{\beta })}^{-1}\in ℬ$.

(B4) Let $S\in ℬ$. Again suppose $S=S_{\alpha }\cap S_{\beta }$ with $S_{\alpha }\in {𝒰}_{\alpha }$ and $S_{\beta }\in {𝒰}_{\beta }$. Then there exist $T_{\alpha }\in {𝒰}_{\alpha }$ and $T_{\beta }\in {𝒰}_{\beta }$ such that$T_{\alpha }\circ T_{\alpha }\subseteq S_{\alpha }$ and $T_{\beta }\circ T_{\beta }\subseteq S_{\beta }$. The set $T=T_{\alpha }\cap T_{\beta }$ is in $𝒰$, and since $T\circ T$ is a subset of both $S_{\alpha }$ and $S_{\beta }$, it is a subset of $S$.

The fundamental system $ℬ$ generates a uniformity $𝒰$. By construction, $𝒰$ is an upper bound of the ${𝒰}_{\alpha }$. But any upper bound of the ${𝒰}_{\alpha }$ would have to contain all finite intersections of elements of the ${𝒰}_{\alpha }$. So $𝒰={sup}_{\alpha \in I}{𝒰}_{\alpha }$.∎

This theorem is useful because it allows us to assert the existence of the coarsest uniform space satisfying a particular property.

Corollary.

Let $X$ be a set and let ${\mathrm{\{}{Y}_{\alpha }\mathrm{\}}}_{\alpha \mathrm{\in }I}$ be a family of uniform spaces. Then for any family of functions $\mathrm{\{}{f}_{\alpha }\mathrm{:}X\mathrm{\to }{Y}_{\alpha }\mathrm{\}}$, there is a coarsest uniformity on $X$ making all the $f_{\alpha }$ uniformly continous.

The coarsest uniformity making a family of functions uniformly continuous is called the initial uniformity or weak uniformity.