proof that uniformly continuous is proximity continuous


Let f:XY be a uniformly continuous function from uniform spaces X to Y with uniformities 𝒰 and 𝒱 respectively. Let δ 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 δ and ϵ respectively. Furthermore, we have the following:

Theorem 1.

f:XY is proximity continuous.

Proof.

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

V[f(A)]V[f(B)],

for any V𝒱. Pick any V𝒱. Since f is uniformly continuous, there is U𝒰 such that

U[x]f-1(V[f(x)]),

for any xX. As a result,

U[A]f-1(V[f(A)]),

which implies that

f(U[A])V[f(A)].

Similarly f(U[B])V[f(B)]. Now, AδB is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to U[A]U[B], so we can pick

zU[A]U[B].

Then

f(z)f(U[A])f(U[B])V[f(A)]V[f(B)],

and therefore

V[f(A)]V[f(B)].

This shows that f is proximity continuous. ∎

Title proof that uniformly continuous is proximity continuous
Canonical name ProofThatUniformlyContinuousIsProximityContinuous
Date of creation 2013-03-22 18:07:50
Last modified on 2013-03-22 18:07:50
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Proof
Classification msc 54E15
Classification msc 54C08
Classification msc 54C05
Classification msc 54E17
Classification msc 54E05