Closure induced by a proximity corresponds to a topology How to prove that the closure induced by a proximity corresponds to a topology?
 A: $\newcommand{\cl}{\operatorname{cl}}$Let $\langle X,\delta\rangle$ be a proximity space, and for $A\subseteq X$ define $\cl A=\big\{x\in X:\{x\}\delta A\big\}$.
By definition of a proximity we have:


*

*$P_0$: $A\delta B$ iff $B\delta A$.  

*$P_1$: $A\delta(B\cup C)$ iff $A\delta B$ or $A\delta C$.  

*$P_2$: $\{x\}\delta\{y\}$ iff $x=y$.  

*$P_3$: $\varnothing\bar\delta X$.  

*$P_4$: If $A\bar\delta B$, there are $C,D\subseteq X$ such that $A\bar\delta C$, $B\bar\delta D$, and $C\cup D=X$.


We need a few very basic consequences.


*

*It follows from $P_1$ that if $A\delta B$ and $B\subset C$, then $A\delta C$, since $B\cup C=C$. 

*This together with $P_0$ and $P_2$ implies that if $A\cap B\ne\varnothing$, then $A\delta B$: if $x\in A\cap B$, then $\{x\}\delta\{x\}$ by $P_2$, $\{x\}\delta A$ by (1), $A\delta\{x\}$ by $P_0$, and $A\delta B$ by another application of (1). 

*Finally, $\varnothing\bar\delta A$ for every $A\subseteq X$, by $P_3$ and (1).
With these basics established it’s easy to check that $\cl$ is a closure operator.


*

*$\cl\varnothing=\varnothing$, since $\varnothing\bar\delta A$ for each $A\subseteq X$.

*$\cl A\subseteq A$, since for each $x\in A$ we have $\{x\}\delta A$ by (2).

*That $\cl(A\cup B)=\cl A\cup\cl B$ is an immediate consequence of $P_1$.

*To prove that $\cl(\cl A)=\cl A$, it clearly suffices to show that $\cl A\supseteq\cl(\cl A)$, i.e., that if $x\notin\cl A$, then $x\notin\cl(\cl A)$. If $x\notin\cl A$, then $\{x\}\bar\delta A$, so by $P_4$ there are $C,D\subseteq X$ such that $\{x\}\bar\delta C$, $A\bar\delta D$, and $C\cup D=X$. Then $X\setminus D\subseteq C$, so by (1) we must have $\{x\}\bar\delta (X\setminus D)$ and hence $x\in D$. That is, $X\setminus\cl A\subseteq D$, so $\cl A\subseteq X\setminus D\subseteq C$. Now recall that if $x\notin\cl A$, then $\{x\}\bar\delta C$; thus, $\{x\}\bar\delta\cl A$ by (1), and hence $x\notin\cl(\cl A)$, as desired.
It’s well-known that any closure operator induces a topology.
A: I've found a proof here at PlanetMath.
