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 $𝒰$ and $𝒱$ respectively. Let $\delta$ and $ϵ$ be the proximities generated by (http://planetmath.org/UniformProximity) $𝒰$ and $𝒱$ respectively. It is known that $X$ and $Y$ are proximity spaces with proximities $\delta$ and $ϵ$ respectively. Furthermore, we have the following:

Theorem 1.

$f:X\to Y$ is proximity continuous.

Proof.

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

$$V[f(A)]\cap V[f(B)]\ne \mathrm{\varnothing },$$

for any $V\in 𝒱$. Pick any $V\in 𝒱$. Since $f$ is uniformly continuous, there is $U\in 𝒰$ 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 $U[A]\cap U[B]\ne \mathrm{\varnothing }$, 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

$$V[f(A)]\cap V[f(B)]\ne \mathrm{\varnothing }.$$

This shows that $f$ is proximity continuous.∎