Prove or disprove a particular statement in proximity spaces

Question

A proximity space is a set $X$ with a relation $\sim$ between its subsets such that it satisfies the following axioms

1. $A\sim B\Rightarrow B\sim A$
2. $A\sim B\Rightarrow A,B\ne \varnothing$
3. $A\cap B\ne \varnothing\Rightarrow A\sim B$
4. $A\sim (B\cup C)\iff (A\sim B)\text{ or }(A\sim C)$
5. $A\not\sim B\Rightarrow \text{ There exists }E\subseteq X \text{ such that } A\not\sim E \text{ and } B\not\sim E^c$

I want to consider the following statement in a proximity space.

If $\{x\}\sim A$ and for all $y\in A, \{y\}\sim B$, then $\{x\}\sim B$.

Till now, I have looked at some examples of proximity spaces (two subsets are related if 1. they are both non-empty; 2. if they have non-trivial intersection; 3. if distance between the sets is 0 [in a metric space]) and this seems to be true in all these cases. But I can't really prove the statement, neither disprove it.

I have tried the following.

Suppose $\{x\}\not\sim B$. Then there is a $E$ such that $\{x\}\not\sim E$ and $B\not\sim E^c$. This implies that $x\in E^c$ and $B\subseteq E$. But then I am lost.

It can be proven that if there exists $x$ such that $A\sim\{x\}$ and $\{x\}\sim B$, then $A\sim B$. I can use this if I could show that $\{x\}\sim A$ implies there is $y\in A$ such that $\{x\}\sim \{y\}$. But this is not true in the above example number 3. If such a $y$ exists, then $d(x,y)=0$ implies $x=y$. But $\{x\}\sim A$ implies $x\in\bar{A}$ which implies $x\in A$. But then $A$ is a closed set. This is not necessarily true.

There are a couple of other approaches I tried but all led to dead ends pretty quickly.

Any help is appreciated.

Answer 1

Let $(X,\sim)$ be a proximity space, and suppose

• $\{x\}\sim A$.
• $\{a\}\sim B$ for all $a\in A$.

holds for some $x\in X$ and some $A,B\subseteq X$.

Claim:$\;\{x\}\sim B$.

Proof:

Suppose instead that $\{x\}\not\sim B$.

Let $E\subseteq X$ be such that $\{x\}\not\sim E$ and $B\not\sim E^c$.

Let $a\in A$.

If $a\in E^c$ then \begin{align*} & \{a\}\sim B \\[4pt] \implies\;& B\sim\{a\} \\[4pt] \implies\;& B\sim\{a\}\cup (E^c{\setminus}\{a\}) \\[4pt] \implies\;& B\sim E^c \\[4pt] \end{align*} contradiction.

It follows that $A\cap E^c={\large{\varnothing}}$, hence $A\subseteq E$.

But then \begin{align*} & \{x\}\sim A \\[4pt] \implies\;& \{x\}\sim A\cup (E{\setminus}A) \\[4pt] \implies\;& \{x\}\sim E \end{align*} contradiction, which completes the proof of the claim.