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How to prove that the closure induced by a proximity corresponds to a topology?

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2  
What's a proximity? –  Jason DeVito Oct 22 '12 at 12:36
    
@JasonDeVito: en.wikipedia.org/wiki/Proximity_space –  porton Oct 22 '12 at 12:39
1  
@Jason: See this article. –  Brian M. Scott Oct 22 '12 at 12:39
    
That'll teach me to google before I ask ;-) –  Jason DeVito Oct 22 '12 at 12:51
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@Jason: Judging by the answer, it seems that this should teach porton to Google before he asks. –  Asaf Karagila Oct 22 '12 at 13:59

2 Answers 2

up vote 3 down vote accepted

$\newcommand{\cl}{\operatorname{cl}}$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.

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

  2. 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).

  3. 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.

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

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

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

  4. 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.

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In the formula "cl A ⊆ cl(cl A)" subset and superset are confused. –  porton Oct 22 '12 at 14:45
    
@porton: Thanks; fixed. –  Brian M. Scott Oct 22 '12 at 14:48

I've found a proof in PlanetMath at http://planetmath.org/encyclopedia/Proximity.html

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