uniform continuity

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

Let $(X,𝒰),(Y,𝒱)$ 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 $V\in 𝒱$ there is a $U\in 𝒰$ 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:

Proposition 1.

Suppose $f\mathrm{:}X\mathrm{\to }Y$ is a function and $g\mathrm{:}X\mathrm{\times }X\mathrm{\to }Y\mathrm{\times }Y$ is defined by $g\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}{x}_{\mathrm{2}}\mathrm{)}\mathrm{=}\mathrm{(}f\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{)}\mathrm{,}f\mathit{}\mathrm{(}{x}_{\mathrm{2}}\mathrm{)}\mathrm{)}$. Then $f$ is uniformly continuous iff for any $V\mathrm{\in }\mathrm{V}$, there is a $U\mathrm{\in }\mathrm{U}$ such that $U\mathrm{\subseteq }{g}^{\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}V\mathrm{)}$.

Proof.

Suppose $f$ is uniformly continuous. Pick any $V\in 𝒱$. Then $U\in 𝒰$ 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.

Proposition 2.

. If $f\mathrm{:}X\mathrm{\to }Y$ is uniformly continuous, then it is continuous under the uniform topologies of $X$ and $Y$.

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 $U\in 𝒰$ 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.