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:

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

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

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

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

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

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