# 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. 1.

$X\ll X$,

2. 2.

$A\ll B$ implies $A\subseteq B$,

3. 3.

$A\ll B$ and $C\ll D$ imply $A\cap C\ll B\cap D$,

4. 4.

$A\ll B$ implies $B^{\prime}\ll A^{\prime}$ (${}^{\prime}$ is the complement operator)

5. 5.

$A\subseteq B\ll C\subseteq D$, then $A\ll D$, and

6. 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 anti-symmetric. 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. 1.

if $A\delta^{\prime}B$, then $A\ll B^{\prime}$, or $A\subseteq B^{\prime}$, so $A\cap B=\varnothing$;

2. 2.

suppose either $A=\varnothing$ or $B=\varnothing$. In either case, $A\ll B^{\prime}$, which means $A\delta^{\prime}B$;

3. 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. 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. 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. 1.

since $X\delta^{\prime}\varnothing$, $X\ll\varnothing^{\prime}$, or $X\ll X$;

2. 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. 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. 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. 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. 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.

Title proximal neighborhood ProximalNeighborhood 2013-03-22 16:58:25 2013-03-22 16:58:25 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 54E05 proximity neighborhood $\delta$-neighborhood