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

2. suppose either $A=\mathrm{\varnothing }$ or $B=\mathrm{\varnothing }$. 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 $X{\delta }^{\prime }\mathrm{\varnothing }$, $X\ll {\mathrm{\varnothing }}^{\prime }$, or $X\ll X$;

2. suppose $A{\delta }^{\prime }{B}^{\prime }$, then if $x\in A$, we have $x{\delta }^{\prime }{B}^{\prime }$, implying $x\cap B^{\prime }=\mathrm{\varnothing }$, 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 $\mathrm{(}X\mathrm{,}\delta \mathrm{)}$, then $B$ is a (topological) neighborhood of $A$ under the topology $\tau \mathit{}\mathrm{(}\delta \mathrm{)}$ induced by the proximity relation $\delta$. In other words, if $A\mathrm{\ll }B$, then $A\mathrm{\subseteq }{B}^{\mathrm{\circ }}$ and $A^c\mathrm{\subseteq }B$, where ${}^{\mathrm{\circ }}$ 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
Canonical name	ProximalNeighborhood
Date of creation	2013-03-22 16:58:25
Last modified on	2013-03-22 16:58:25
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	7
Author	CWoo (3771)
Entry type	Definition
Classification	msc 54E05
Synonym	proximity neighborhood
Synonym	$\delta$-neighborhood