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proximal neighborhood
Let $X$ be a set and $P(X)$ its power set. Let $\ll$ be a binary relation on $P(X)$ satisfying the
following conditions, for any $A,B\subseteq X$:
1. $X\ll X$,
2. $A\ll B$ implies $A\subseteq B$,
3. $A\ll B$ and $C\ll D$ imply $A\cap C\ll B\cap D$,
4. $A\ll B$ implies $B^{{\prime}}\ll A^{{\prime}}$ (${}^{{\prime}}$ is the complement operator)
5. $A\subseteq B\ll C\subseteq D$, then $A\ll D$, and
6. if $A\ll B$, then there is $C\subseteq X$, such that $A\ll C\ll B$.
By 1 and 4, it is easy to see that $\varnothing\ll\varnothing$. Also, 3 and 4 show that $A\cup C\ll B\cup D$ whenever $A\ll B$ and $C\ll D$. So $\ll$ is a topogenous order, which means $\ll$ is transitive and antisymmetric. Under this order relation, we say that $B$ is a proximal neighborhood of $A$ if $A\ll B$.
The reason why we call $B$ a “proximal” neighborhood is due to the following:
Theorem 1.
Let $X$ be a set. The following are true.

Let $\ll$ be defined as above. Define a new relation $\delta$ on $P(X)$: $A\delta^{{\prime}}B^{{\prime}}$ 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 $A\delta^{{\prime}}B^{{\prime}}$. 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 $A\delta^{{\prime}}B$, then $A\ll B^{{\prime}}$, or $A\subseteq B^{{\prime}}$, so $A\cap B=\varnothing$;
2. suppose either $A=\varnothing$ or $B=\varnothing$. In either case, $A\ll B^{{\prime}}$, which means $A\delta^{{\prime}}B$;
3. if $A\delta^{{\prime}}B$, then $A\ll B^{{\prime}}$, so $B^{{\prime\prime}}\ll A^{{\prime}}$, or $B\ll A^{{\prime}}$, or $B\delta^{{\prime}}A$;
4. if $A_{1}\delta^{{\prime}}B$ and $A_{2}\delta^{{\prime}}B$, then $A_{1}\ll B$ and $A_{2}\ll B$, so $(A_{1}\cup A_{2})\ll B$, or $(A_{1}\cup A_{2})\delta^{{\prime}}B$;
5. if $A\delta^{{\prime}}B$, then $A\ll B^{{\prime}}$. So there is $D\subseteq X$ with $A\ll D$ and $D\ll B^{{\prime}}$. Let $C=D^{{\prime}}$. Then $A\ll C^{{\prime}}$ and $C^{{\prime}}\ll B^{{\prime}}$, or $A\delta^{{\prime}}C$ and $C^{{\prime}}\delta^{{\prime}}B$.
Next, suppose $(X,\delta)$ is a proximity space. We now verify the six properties of $\ll$ above.
1. since $X\delta^{{\prime}}\varnothing$, $X\ll\varnothing^{{\prime}}$, or $X\ll X$;
2. suppose $A\delta^{{\prime}}B^{{\prime}}$, then if $x\in A$, we have $x\delta^{{\prime}}B^{{\prime}}$, implying $x\cap B^{{\prime}}=\varnothing$, or $x\in B$;
3. if $A\ll B$ and $C\ll D$, then $A\delta^{{\prime}}B^{{\prime}}$ and $C\delta^{{\prime}}D^{{\prime}}$, which means $A\delta^{{\prime}}(B^{{\prime}}\cup D^{{\prime}})$ and $C\delta^{{\prime}}(B^{{\prime}}\cup D^{{\prime}})$, which together imply $(A\cap C)\delta^{{\prime}}(B^{{\prime}}\cup D^{{\prime}})$, or $(A\cap C)\delta(B\cap D)^{{\prime}}$, or $A\cap C\ll B\cap D$;
4. if $A\ll B$, then $A\delta^{{\prime}}B^{{\prime}}$, so $B^{{\prime}}\delta^{{\prime}}A$ (as $\delta$ is symmetric, so is its complement), which is the same as $B^{{\prime}}\delta^{{\prime}}A^{{\prime\prime}}$, or $B^{{\prime}}\ll A^{{\prime}}$;
5. if $A\delta D^{{\prime}}$, then $B\delta C^{{\prime}}$ (since $A\subseteq B$ and $D^{{\prime}}\subseteq C^{{\prime}}$), so $B\ll^{{\prime}}C$, a contradiction;
6. if $A\ll B$, then $A\delta^{{\prime}}B^{{\prime}}$, so there is $D\subseteq X$ with $A\delta^{{\prime}}D$ and $D^{{\prime}}\delta^{{\prime}}B^{{\prime}}$. Define $C=D^{{\prime}}$, then $A\ll C$ and $C\ll B$, as desired.
This completes the proof. ∎
Because of the above, we see that a proximity space can be equivalently defined using the proximal neighborhood concept. To emphasize its relationship with $\delta$, a proximal neighborhood is also called a $\delta$neighbhorhood.
Furthermore, we have
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 $A\delta^{{\prime}}B^{{\prime}}$, then $x\delta^{{\prime}}B^{{\prime}}$ whenever $x\in A$, which is the contrapositive of the statement: $x\in A^{{\prime}}$ whenever $x\delta B^{{\prime}}$, which is equivalent to $B^{{\prime c}}\subseteq A^{{\prime}}$, or $A\subseteq B^{{\circ}}$. Furthermore, if $x\notin B$, then $x\in B^{{\prime}}$. But $A\delta^{{\prime}}B^{{\prime}}$ b assumption. This implies $x\delta^{{\prime}}A$, which means $x\notin A^{c}$. Therefore $A^{c}\subseteq B$. ∎
Remark. However, not every $\tau(\delta)$neighborhood is a $\delta$neighborhood.
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