Given a subset $S$ of a finite metric space $F$ with a distance function $d(,)$ and a number $\delta > 0$ let $N_\delta(S) = \{x \in F| d(x,S)\ge \delta\}$. Is there a characterization of conditions necessary and sufficient for $N_\delta(N_\delta(S)) = S$ to hold?
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1$\begingroup$ How should one define curvature for a finite metric space? $\endgroup$– user74022Commented May 23, 2015 at 4:07
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$\begingroup$ $N_\delta(N_\delta(S))=S$ if and only if every point $x\not\in S$ is in a ball of radius $\delta$ disjoint from $S$. $\endgroup$– Yoav KallusCommented May 23, 2015 at 6:27
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$\begingroup$ Not sure I understand this statement -- clearly there are examples when the equality holds and there are points which are not in the ball of radius $\delta$. E.g. take $S = {x}$, $\delta < \inf_{y,z} d(y,z)$ so that $N_\delta(S) = F \setminus {x}$. $\endgroup$– user74022Commented May 23, 2015 at 13:24
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$\begingroup$ I did not say "the ball", but "a ball". In your example every singleton is a delta ball, and every point not in S is in such a ball. $\endgroup$– Yoav KallusCommented May 23, 2015 at 15:44
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1 Answer
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$N_\delta(N_\delta(S))=S$ if and only the complement of $S$ is a union of $\delta$-balls. Equivalently, it is the union of all $\delta$-balls disjoint from $S$.
If $y\in S$, then there is no point $z\in N_\delta(S)$ at distance $d(y,z)< \delta$. Therefore, it is also in $N_\delta(N_\delta(S))$. Therefore, if $N_\delta(N_\delta(S))\neq S$, then there is a point $y\in N_\delta(N_\delta(S))$, $y\not\in S$. We know $y$ is disjoint from all balls with centers in $N_\delta(S)$, but those are actually all the balls disjoint from $S$. So not every point outside $S$ is in a ball disjoint from $S$.
For the converse, assume there is a point $y\not\in S$ that is disjoint from all balls disjoint from $S$, then it is in $N_\delta(N_\delta(S))$ and so $N_\delta(N_\delta(S))\neq S$.
answered May 23, 2015 at 16:19