In this entry, we extend the usual definition of a uniformly continuous function between metric spaces to arbitrary uniform spaces.

Let $(X,\mathcal{U}),(Y,\mathcal{V})$ be uniform spaces (the second component is the uniformity on the first component). A function $f:X\to Y$ is said to be uniformly continuous if for any $V\in\mathcal{V}$ there is a $U\in \mathcal{U}$ such that for all $x\in X$ , $U[x]\subseteq f^{-1}(V[f(x)])$ .

Sometimes it is useful to use an alternative but equivalent version of uniform continuity of a function:

Proof. Suppose $f$ is uniformly continuous. Pick any $V\in \mathcal{V}$ . Then $U\in \mathcal{U}$ exists with $U[x]\subseteq f^{-1}(V[f(x)])$ for all $x\in X$ . If $(a,b)\in U$ , then $b\in U[a]\subseteq f^{-1}(V[f(a)])$ , or $f(b)\subseteq V[f(a)]$ , or $g(a,b)=(f(a),f(b))\in V$ . The converse is straightforward.

Remark. Note that we could have picked $U$ so the inclusion becomes an equality.

Proof. Let $A$ be open in $Y$ and set $B=f^{-1}(A)$ . Pick any $x\in B$ . Then $y=f(x)$ has a uniform neighborhood $V[y]\subseteq A$ . By the uniform continuity of $f$ , there is an entourage $U\in \mathcal{U}$ with $x\in U[x]\subseteq f^{-1}(V[y])\subseteq f^{-1}(A)=B$ .

Remark. The converse is not true, even in metric spaces.