# uniform continuity

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 $x\in X$, $U[x]\subseteq f^{-1}(V[f(x)])$.

###### Proposition 1.

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

###### 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.

Title uniform continuity UniformContinuity 2013-03-22 16:43:15 2013-03-22 16:43:15 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 54E15 UniformlyContinuous UniformContinuityOverLocallyCompactQuantumGroupoids uniformly continuous