uniform proximity is a proximity
In this entry, we want to show that a uniform proximity is, as expected, a proximity.
First, the following equivalent 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 U∈U:
{b∈X∣(a,b)∈U for some a∈A}. |
The following are equivalent:
-
1.
(A×B)∩U≠∅ for all U∈𝒰
-
2.
U[A]∩U[B]≠∅ for all U∈𝒰
-
3.
U[A]∩B≠∅ for all U∈𝒰
If we define AδB iff the pair A,B satisfy any one of the above conditions for all U∈𝒰, we call δ the uniform proximity.
Proof.
(1⇒2) Suppose (a,b)∈(A×B)∩U. Then b∈U[A]. Since U is reflexive, (b,b)∈U, or b∈U[B]. This means b∈U[A]∩U[B].
(2⇒3) For any U∈𝒰, we can find V∈𝒰 such that V∘V⊆U. So V=V∘Δ⊆V∘V⊆W, where Δ is the diagonal relation (since V is reflexive). Set W=V∩V-1. By assumption, there is c∈W[A]∩W[B] (and hence c∈U[A]∩U[B] as well). This means (a,c),(b,c)∈W for some a∈A and b∈B. Since W is symmetric
, (c,b)∈W⊆V, so that (a,b)=(a,c)∘(c,b)∈V⊆U. This means that b∈U[A]. As a result, U[A]∩B≠∅.
(3⇒1) If b∈U[A]∩B, then there is a∈A such that (a,b)∈U, or (A×B)∩U≠∅. ∎
We want to prove the following:
Proposition 1.
Proof.
We verify each of the axioms of a proximity relation:
-
1.
if A∩B≠∅, then AδB:
pick c∈A∩B, then (c,c)∈U since the diagonal relation Δ⊆U for all U∈𝒰.
-
2.
if AδB, then A≠∅ and B≠∅:
If AδB, then (A×B)∩U≠∅ for every U∈𝒰, since no U is empty, there is (a,b)∈U such that (a,b)∈A×B, or A≠∅ and B≠∅.
-
3.
(symmetry) if AδB, then BδA:
If AδB, then there is (aU,bU)∈(A×B)∩U-1 for every U∈𝒰, so (bU,aU)∈U, which implies (B×A)∩U≠∅, or BδA.
-
4.
(A1∪A2)δB iff A1δB or A2δB:
Since (A1∪A2)×B=(A1×B)∪(A2×B),
(a,b)∈((A1∪A2)×B)∩U 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.
Aδ′B implies the existence of C∈P(X) with Aδ′C and (X-C)δ′B, where Aδ′B means (A,B)∉δ.
First note that δ′ is symmetric because δ is. By assumption, there is U∈𝒰 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, Aδ′C. Likewise, U[B]∩(X-C)=U[B]∩(X-U[B])=∅, so Bδ′(X-C), or (X-C)δ′B.
This shows that δ is a proximity on X. ∎
Title | uniform proximity is a proximity |
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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 |