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

Proof. Let $X$ be a set. Let $\mathfrak{U}(X)$ denote the collection of uniformities on $X$ . The coarsest uniformity on $X$ is $\{X\times X\}$ , and the finest is the discrete uniformity: $$ \{S\subset X\times X\colon \Delta(X)\subseteq S\}. $$ Hence $\mathfrak{U}(X)$ is bounded. To show that $\mathfrak{U}(X)$ is complete, we must prove that it has the least upper bound property.

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

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

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

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

(B4) Let $S\in\BB$ . Again suppose $S=S_{\alpha}\cap S_{\beta}$ with $S_{\alpha}\in\UU_{\alpha}$ and $S_{\beta}\in\UU_{\beta}$ . Then there exist $T_{\alpha}\in\UU_{\alpha}$ and $T_{\beta}\in\UU_{\beta}$ 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 $\UU$ , and since $T\circ T$ is a subset of both $S_{\alpha}$ and $S_{\beta}$ , it is a subset of $S$ .

The fundamental system $\BB$ generates a uniformity $\UU$ . By construction, $\UU$ is an upper bound of the $\UU_{\alpha}$ . But any upper bound of the $\UU_{\alpha}$ would have to contain all finite intersections of elements of the $\UU_{\alpha}$ . So $\UU=\sup_{\alpha\in I} \UU_{\alpha}$ .

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