Theorem 1   Let $X$ be a set. The following are true.

• Let $\ll$ be defined as above. Define a new relation $\delta$ on $P(X)$ : $A\delta'B'$ iff $A\ll B$ . Then $\delta$ so defined is a proximity relation, turning $X$ into a proximity space.
• Conversely, let $(X,\delta)$ is a proximity space. Define a new relation $\ll$ on $P(X)$ : $A\ll B$ iff $A\delta'B'$ . Then $\ll$ satisfies the six properties above.

Proof. Suppose first that $X$ and $\ll$ are defined as above. We will verify the individual nearness relation axioms of $\delta$ by proving their contrapositives in each case, except the last axiom:

1. if $A\delta'B$ , then $A\ll B'$ , or $A\subseteq B'$ , so $A\cap B=\varnothing$ ;
2. suppose either $A=\varnothing$ or $B=\varnothing$ . In either case, $A\ll B'$ , which means $A\delta' B$ ;
3. if $A\delta' B$ , then $A\ll B'$ , so $B''\ll A'$ , or $B\ll A'$ , or $B\delta' A$ ;
4. if $A_1\delta' B$ and $A_2\delta' B$ , then $A_1\ll B$ and $A_2\ll B$ , so $(A_1\cup A_2)\ll B$ , or $(A_1\cup A_2)\delta' B$ ;
5. if $A\delta' B$ , then $A\ll B'$ . So there is $D\subseteq X$ with $A\ll D$ and $D\ll B'$ . Let $C=D'$ . Then $A\ll C'$ and $C'\ll B'$ , or $A\delta' C$ and $C'\delta' B$ .

Next, suppose $(X,\delta)$ is a proximity space. We now verify the six properties of $\ll$ above.

1. since $X\delta' \varnothing$ , $X\ll \varnothing'$ , or $X\ll X$ ;
2. suppose $A\delta' B'$ , then if $x\in A$ , we have $x\delta'B'$ , implying $x\cap B'=\varnothing$ , or $x\in B$ ;
3. if $A\ll B$ and $C\ll D$ , then $A\delta' B'$ and $C\delta' D'$ , which means $A\delta' (B'\cup D')$ and $C\delta' (B'\cup D')$ , which together imply $(A\cap C)\delta' (B'\cup D')$ , or $(A\cap C)\delta (B\cap D)'$ , or $A\cap C\ll B\cap D$ ;
4. if $A\ll B$ , then $A\delta' B'$ , so $B'\delta' A$ (as $\delta$ is symmetric, so is its complement), which is the same as $B'\delta' A''$ , or $B'\ll A'$ ;
5. if $A\delta D'$ , then $B\delta C'$ (since $A\subseteq B$ and $D'\subseteq C'$ ), so $B\ll' C$ , a contradiction;
6. if $A\ll B$ , then $A\delta'B'$ , so there is $D\subseteq X$ with $A\delta'D$ and $D'\delta'B'$ . Define $C=D'$ , then $A\ll C$ and $C\ll B$ , as desired.

This completes the proof.

Theorem 2   if $B$ is a proximal neighborhood of $A$ in a proximity space $(X,\delta)$ , then $B$ is a (topological) neighborhood of $A$ under the topology $\tau(\delta)$ induced by the proximity relation $\delta$ . In other words, if $A\ll B$ , then $A\subseteq B^{\circ}$ and $A^c\subseteq B$ , where $^{\circ}$ and $^c$ denote the interior and closure operators.

Proof. Since $A\delta'B'$ , then $x\delta'B'$ whenever $x\in A$ , which is the contrapositive of the statement: $x\in A'$ whenever $x\delta B'$ , which is equivalent to $B'^c\subseteq A'$ , or $A\subseteq B^{\circ}$ . Furthermore, if $x\notin B$ , then $x\in B'$ . But $A\delta'B'$ b assumption. This implies $x\delta' A$ , which means $x\notin A^c$ . Therefore $A^c\subseteq B$ .