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A metric space $Y$ has the binary intersection property provided that whenever a collection of closed balls in $Y$ intersects pairwise, then there is a common intersection point.

Does the metric space $M$ of compact metric spaces under the Gromov-Hausdorff distance satisfy the binary intersection property?

The motivation is simple: I have a metric space $X$ with subspace $A$ and a Lipschitz map $f:A \to M$. I'd like to know if I can extend $f$ to all of $X$ without increasing the Lipschitz constant. It turns out (see Prop 1.4 here) that this binary intersection property is one of two hypotheses that must be satisfied by $M$ if it is to admit Lipschitz extensions for arbitrary metric space pairs $(X,A)$.

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Just a clarification: when you speak of two subsets U, V of M (the space of compact metric spaces) "intersecting", presumably you mean that there exist u in U and v in V such that d(u, v) = 0? –  Tom Leinster Nov 17 '12 at 13:48
    
Tom: that's correct. –  Vidit Nanda Nov 17 '12 at 13:55
    
Would the statement in the Motivation become more true if $A$ (or even $X$) is assumed to be compact? –  Pietro Majer Nov 19 '12 at 22:05
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3 Answers

up vote 11 down vote accepted

No, Let $B_n\in M$ be the $n$-dimensional Euclidean unit ball and $r=\frac12+\varepsilon$ where $\varepsilon=\frac1{100}$. Then the $r$-balls in $M$ centered at $B_n$ intersect pairwise. Indeed, for $m>n$ the $m$-dimensional Euclidean ball of radius 1/2 lies within Gromov-Hausdorff distance 1/2 from both $B_n$ and $B_m$ (as seen from their natural inclusion into $\mathbb R^m$).

However there is no compact metric space $K$ which stays within distance $1/2+\varepsilon$ from every $B_n$. Indeed, suppose the contrary, then there is a map $f_n:B_n\to K$ which distorts distances by at most $1+2\varepsilon$. But $B_n$ contains $2n$ points with pairwise distances $\sqrt 2$, hence the $f_n$-images of these points are separated by distances at least $\sqrt2-1-2\varepsilon>\frac1{10}$. Thus $K$ contains arbitrarily many $\frac1{10}$-separated points, hence it is not compact.

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Hi Tom!

Parenthetical remark: (I guess in this case the "closed balls" are not sets but classes, being defined by a metric condition; and "common intersection point" can be taken quite literally.)

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Hi Vin! (Before anyone pounces on Vin for leaving an answer that is really a comment, note that he's new and therefore doesn't have the reputation to leave comments yet.) What was behind my comment was that it doesn't necessarily make sense to talk about literal equality of metric spaces. But yes, I see your point: if we're talking about balls, it doesn't matter. –  Tom Leinster Nov 18 '12 at 2:44
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Sergei Ivanov showed that infinite number of balls can have problems with compactness, but there are more bad news. I can take 3 balls which intersects pairwise, but don't have any common intersection point, or non compact "intersection point".

Indeed, Let $m^1,m^2,m^3$ be subsets of $\mathbb{R}$, consisting of three points each. And distances between them are $12,7,5$ for $m^1$, $10,5,5$ for $m^2$, and $10,7,3$ for $m^3$. Then GH-distanses between $m^1, m^2, m^3$ are less or equal then 1. If intersection of balls with centers in $m^1, m^2, m^3$ and radiuses $\frac{1}{2}$ is not empty, then I can take one element $m^{o}$ from it.

There is metric space $m^{1} \cup m^{o}$ such that Hausdorff distances between $m^{1}$ and $m^{o}$ is less then $\frac{1}{2} + ε$. For each point $A_{i}$ from $m^{1}$ I take a point $B_{i}$ from $m^{o}$ such that $|A_i B_i|$ is less then $\frac{1}{2} + 2ε$. I will call $m^{oo}$ the metric space consisting of $B_1$, $B_2$, $B_3$.

GH-distanses between $m^{i}$ and $m^{oo}$ are less or equal then $\frac{1}{2} + 2ε$,then distances between points of $m^{oo}$ are $11, 6, 4$ plus some epsilons, but that contradicts triangle inequality.

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Thanks, Vladimir! –  Vidit Nanda Apr 1 '13 at 21:10
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