uniform proximity is a proximity

In this entry, we want to show that a uniform proximity is, as expected, a proximity.

First, the following equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath characterizations of a uniform proximity is useful:

Lemma 1.

Let X be a uniform space with uniformity U, and A,B are subsets of X. Denote U[A] the image of A under UU:

{bX(a,b)U for some aA}.

The following are equivalent:

  1. 1.

    (A×B)U for all U𝒰

  2. 2.

    U[A]U[B] for all U𝒰

  3. 3.

    U[A]B for all U𝒰

If we define AδB iff the pair A,B satisfy any one of the above conditions for all U𝒰, we call δ the uniform proximity.


(12) Suppose (a,b)(A×B)U. Then bU[A]. Since U is reflexiveMathworldPlanetmathPlanetmathPlanetmathPlanetmath, (b,b)U, or bU[B]. This means bU[A]U[B].

(23) For any U𝒰, we can find V𝒰 such that VVU. So V=VΔVVW, where Δ is the diagonal relation (since V is reflexive). Set W=VV-1. By assumptionPlanetmathPlanetmath, there is cW[A]W[B] (and hence cU[A]U[B] as well). This means (a,c),(b,c)W for some aA and bB. Since W is symmetricPlanetmathPlanetmathPlanetmath, (c,b)WV, so that (a,b)=(a,c)(c,b)VU. This means that bU[A]. As a result, U[A]B.

(31) If bU[A]B, then there is aA such that (a,b)U, or (A×B)U. ∎

We want to prove the following:

Proposition 1.

The binary relationMathworldPlanetmath δ on P(X) defined by

AδB𝑖𝑓𝑓(A×B)U for all U𝒰

is a proximity on X.


We verify each of the axioms of a proximity relation:

  1. 1.

    if AB, then AδB:

    pick cAB, then (c,c)U since the diagonal relation ΔU for all U𝒰.

  2. 2.

    if AδB, then A and B:

    If AδB, then (A×B)U for every U𝒰, since no U is empty, there is (a,b)U such that (a,b)A×B, or A and B.

  3. 3.

    (symmetry) if AδB, then BδA:

    If AδB, then there is (aU,bU)(A×B)U-1 for every U𝒰, so (bU,aU)U, which implies (B×A)U, or BδA.

  4. 4.

    (A1A2)δB iff A1δB or A2δB:

    Since (A1A2)×B=(A1×B)(A2×B),

    iff (a,b)((A1×B)(A2×B))U=((A1×B)U)((A2×B)U)
    iff (a,b)(A1×B)U or (a,b)(A2×B)U.
  5. 5.

    AδB implies the existence of CP(X) with AδC and (X-C)δB, where AδB means (A,B)δ.

    First note that δ is symmetric because δ is. By assumption, there is U𝒰 such that U[A]U[B]= (second equivalent characterization of uniform proximity from lemma above). Set C=U[B]. Then U[A]C=. By the third equivalent condition of uniform proximity, AδC. Likewise, U[B](X-C)=U[B](X-U[B])=, so Bδ(X-C), or (X-C)δB.

This shows that δ is a proximity on X. ∎

Title uniform proximity is a proximity
Canonical name UniformProximityIsAProximity
Date of creation 2013-03-22 18:07:21
Last modified on 2013-03-22 18:07:21
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 7
Author CWoo (3771)
Entry type Derivation
Classification msc 54E17
Classification msc 54E05
Classification msc 54E15