# proof that uniformly continuous is proximity continuous

Let $f:X\to Y$ be a uniformly continuous function from uniform spaces $X$ to $Y$ with uniformities $\mathcal{U}$ and $\mathcal{V}$ respectively. Let $\delta$ and $\epsilon$ be the proximities generated by (http://planetmath.org/UniformProximity) $\mathcal{U}$ and $\mathcal{V}$ respectively. It is known that $X$ and $Y$ are proximity spaces with proximities $\delta$ and $\epsilon$ respectively. Furthermore, we have the following:

###### Theorem 1.

$f:X\to Y$

###### Proof.

Let $A,B$ be any subsets of $X$ with $A\delta B$. We want to show that $f(A)\epsilon f(B)$, or equivalently,

 $V[f(A)]\cap V[f(B)]\neq\varnothing,$

for any $V\in\mathcal{V}$. Pick any $V\in\mathcal{V}$. Since $f$ is uniformly continuous, there is $U\in\mathcal{U}$ such that

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

for any $x\in X$. As a result,

 $U[A]\subseteq f^{-1}(V[f(A)]),$

which implies that

 $f(U[A])\subseteq V[f(A)].$

Similarly $f(U[B])\subseteq V[f(B)]$. Now, $A\delta B$ is equivalent to $U[A]\cap U[B]\neq\varnothing$, so we can pick

 $z\in U[A]\cap U[B].$

Then

 $f(z)\in f(U[A])\cap f(U[B])\subseteq V[f(A)]\cap V[f(B)],$

and therefore

 $V[f(A)]\cap V[f(B)]\neq\varnothing.$

This shows that $f$ is proximity continuous. ∎

Title proof that uniformly continuous is proximity continuous ProofThatUniformlyContinuousIsProximityContinuous 2013-03-22 18:07:50 2013-03-22 18:07:50 CWoo (3771) CWoo (3771) 6 CWoo (3771) Proof msc 54E15 msc 54C08 msc 54C05 msc 54E17 msc 54E05