proof that uniformly continuous is proximity continuous
Let f:X→Y 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:X→Y 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 V∈𝒱. Pick any V∈𝒱. Since f is uniformly continuous, there is U∈𝒰 such that
U[x]⊆f-1(V[f(x)]), |
for any x∈X. 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 equivalent to U[A]∩U[B]≠∅, so we can pick
z∈U[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 |