# 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 $\mathcal{U}$, and $A,B$ are subsets of $X$. Denote $U[A]$ the image of $A$ under $U\in\mathcal{U}$:

 $\{b\in X\mid(a,b)\in U\mbox{ for some }a\in A\}.$

The following are equivalent:

1. 1.

$(A\times B)\cap U\neq\varnothing$ for all $U\in\mathcal{U}$

2. 2.

$U[A]\cap U[B]\neq\varnothing$ for all $U\in\mathcal{U}$

3. 3.

$U[A]\cap B\neq\varnothing$ for all $U\in\mathcal{U}$

If we define $A\delta B$ iff the pair $A,B$ satisfy any one of the above conditions for all $U\in\mathcal{U}$, we call $\delta$ the uniform proximity.

###### Proof.

($1\Rightarrow 2$) Suppose $(a,b)\in(A\times B)\cap U$. Then $b\in U[A]$. Since $U$ is reflexive, $(b,b)\in U$, or $b\in U[B]$. This means $b\in U[A]\cap U[B]$.

($2\Rightarrow 3$) For any $U\in\mathcal{U}$, we can find $V\in\mathcal{U}$ such that $V\circ V\subseteq U$. So $V=V\circ\Delta\subseteq V\circ V\subseteq W$, where $\Delta$ is the diagonal relation (since $V$ is reflexive). Set $W=V\cap V^{-1}$. By assumption, there is $c\in W[A]\cap W[B]$ (and hence $c\in U[A]\cap U[B]$ as well). This means $(a,c),(b,c)\in W$ for some $a\in A$ and $b\in B$. Since $W$ is symmetric, $(c,b)\in W\subseteq V$, so that $(a,b)=(a,c)\circ(c,b)\in V\subseteq U$. This means that $b\in U[A]$. As a result, $U[A]\cap B\neq\varnothing$.

($3\Rightarrow 1$) If $b\in U[A]\cap B$, then there is $a\in A$ such that $(a,b)\in U$, or $(A\times B)\cap U\neq\varnothing$. ∎

We want to prove the following:

###### Proposition 1.

The binary relation $\delta$ on $P(X)$ defined by

 $A\delta B\quad\mbox{iff}\quad(A\times B)\cap U\neq\varnothing\mbox{ for all }U% \in\mathcal{U}$

is a proximity on $X$.

###### Proof.

We verify each of the axioms of a proximity relation:

1. 1.

if $A\cap B\neq\varnothing$, then $A\delta B$:

pick $c\in A\cap B$, then $(c,c)\in U$ since the diagonal relation $\Delta\subseteq U$ for all $U\in\mathcal{U}$.

2. 2.

if $A\delta B$, then $A\neq\varnothing$ and $B\neq\varnothing$:

If $A\delta B$, then $(A\times B)\cap U\neq\varnothing$ for every $U\in\mathcal{U}$, since no $U$ is empty, there is $(a,b)\in U$ such that $(a,b)\in A\times B$, or $A\neq\varnothing$ and $B\neq\varnothing$.

3. 3.

(symmetry) if $A\delta B$, then $B\delta A$:

If $A\delta B$, then there is $(a_{U},b_{U})\in(A\times B)\cap U^{-1}$ for every $U\in\mathcal{U}$, so $(b_{U},a_{U})\in U$, which implies $(B\times A)\cap U\neq\varnothing$, or $B\delta A$.

4. 4.

$(A_{1}\cup A_{2})\delta B$ iff $A_{1}\delta B$ or $A_{2}\delta B$:

Since $(A_{1}\cup A_{2})\times B=(A_{1}\times B)\cup(A_{2}\times B)$,

 $\displaystyle(a,b)\in\big{(}(A_{1}\cup A_{2})\times B\big{)}\cap U$ iff $\displaystyle(a,b)\in\big{(}(A_{1}\times B)\cup(A_{2}\times B)\big{)}\cap U=% \big{(}(A_{1}\times B)\cap U\big{)}\cup\big{(}(A_{2}\times B)\cap U\big{)}$ iff $\displaystyle(a,b)\in(A_{1}\times B)\cap U\mbox{ or }(a,b)\in(A_{2}\times B)% \cap U.$
5. 5.

$A\delta^{\prime}B$ implies the existence of $C\in P(X)$ with $A\delta^{\prime}C$ and $(X-C)\delta^{\prime}B$, where $A\delta^{\prime}B$ means $(A,B)\notin\delta$.

First note that $\delta^{\prime}$ is symmetric because $\delta$ is. By assumption, there is $U\in\mathcal{U}$ such that $U[A]\cap U[B]=\varnothing$ (second equivalent characterization of uniform proximity from lemma above). Set $C=U[B]$. Then $U[A]\cap C=\varnothing$. By the third equivalent condition of uniform proximity, $A\delta^{\prime}C$. Likewise, $U[B]\cap(X-C)=U[B]\cap(X-U[B])=\varnothing$, so $B\delta^{\prime}(X-C)$, or $(X-C)\delta^{\prime}B$.

This shows that $\delta$ is a proximity on $X$. ∎

Title uniform proximity is a proximity UniformProximityIsAProximity 2013-03-22 18:07:21 2013-03-22 18:07:21 CWoo (3771) CWoo (3771) 7 CWoo (3771) Derivation msc 54E17 msc 54E05 msc 54E15