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

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

The following are equivalent:

1. $(A\times B)\cap U\ne \mathrm{\varnothing }$ for all $U\in 𝒰$

2. $U[A]\cap U[B]\ne \mathrm{\varnothing }$ for all $U\in 𝒰$

3. $U[A]\cap B\ne \mathrm{\varnothing }$ for all $U\in 𝒰$

If we define $A\delta B$ iff the pair $A,B$ satisfy any one of the above conditions for all $U\in 𝒰$, 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 $U\in 𝒰$, we can find $V\in 𝒰$ such that $V\circ V\subseteq U$. So $V=V\circ \mathrm{\Delta }\subseteq V\circ V\subseteq W$, where $\mathrm{\Delta }$ 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, $U[A]\cap B\ne \mathrm{\varnothing }$.

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

We want to prove the following:

Proposition 1.

The binary relation $\delta$ on $P\mathit{}\mathrm{(}X\mathrm{)}$ defined by

$$A\delta B\mathit{\hspace{1em}}\text{𝑖𝑓𝑓}\mathit{\hspace{1em}}(A\times B)\cap U\ne \mathrm{\varnothing }\mathit{\text{ for all }}U\in 𝒰$$

is a proximity on $X$.

Proof.

We verify each of the axioms of a proximity relation:

1. if $A\cap B\ne \mathrm{\varnothing }$, then $A\delta B$:

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

2. if $A\delta B$, then $A\ne \mathrm{\varnothing }$ and $B\ne \mathrm{\varnothing }$:

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

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 $U\in 𝒰$, so $(b_U,a_U)\in U$, which implies $(B\times A)\cap U\ne \mathrm{\varnothing }$, or $B\delta A$.

4. $(A_1\cup A_2)\delta B$ iff $A_1\delta B$ or $A_2\delta B$:

   Since $(A_1\cup A_2)\times B=(A_1\times B)\cup (A_2\times B)$,

   $(a,b)\in \left((A_1\cup A_2)\times B\right)\cap U$

   iff$(a,b)\in \left((A_1\times B)\cup (A_2\times B)\right)\cap U=\left((A_1\times B)\cap U\right)\cup \left((A_2\times B)\cap U\right)$

   iff$(a,b)\in (A_1\times B)\cap U\text{ or }(a,b)\in (A_2\times B)\cap U.$

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

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