Does the metric space of compact metric spaces satisfy the binary intersection property? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T10:35:48Z http://mathoverflow.net/feeds/question/112677 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112677/does-the-metric-space-of-compact-metric-spaces-satisfy-the-binary-intersection-pr Does the metric space of compact metric spaces satisfy the binary intersection property? Vidit Nanda 2012-11-17T13:11:12Z 2013-03-23T15:42:52Z <p>A metric space $Y$ has the <em>binary intersection property</em> provided that whenever a collection of closed balls in $Y$ intersects pairwise, then there is a common intersection point.</p> <blockquote> <p>Does the metric space $M$ of compact metric spaces under the <a href="http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence#Gromov.E2.80.93Hausdorff_distance" rel="nofollow">Gromov-Hausdorff distance</a> satisfy the binary intersection property?</p> </blockquote> <p>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 <a href="http://books.google.com/books?id=lXZ95EKwjYUC&amp;pg=PA13&amp;lpg=PA13&amp;dq=metrically+convex+binary+intersection&amp;source=bl&amp;ots=SvQJYCRWp8&amp;sig=CmTnEUt1Zxt2-3FBmxf0rlufVbQ&amp;hl=en&amp;sa=X&amp;ei=bIqnUKLIE8LQmAWQuIDIDQ&amp;redir_esc=y#v=onepage&amp;q=metrically%2520convex%2520binary%2520intersection&amp;f=false" rel="nofollow">here</a>) 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 $(X,A)$.</p> http://mathoverflow.net/questions/112677/does-the-metric-space-of-compact-metric-spaces-satisfy-the-binary-intersection-pr/112700#112700 Answer by Sergei Ivanov for Does the metric space of compact metric spaces satisfy the binary intersection property? Sergei Ivanov 2012-11-17T18:10:29Z 2012-11-17T18:10:29Z <p>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 $\mathbb R^m$).</p> <p>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.</p> http://mathoverflow.net/questions/112677/does-the-metric-space-of-compact-metric-spaces-satisfy-the-binary-intersection-pr/112740#112740 Answer by Vin de Silva for Does the metric space of compact metric spaces satisfy the binary intersection property? Vin de Silva 2012-11-18T02:04:42Z 2012-11-18T02:04:42Z <p>Hi Tom!</p> <p>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.)</p> http://mathoverflow.net/questions/112677/does-the-metric-space-of-compact-metric-spaces-satisfy-the-binary-intersection-pr/125384#125384 Answer by Vladimir Zolotov for Does the metric space of compact metric spaces satisfy the binary intersection property? Vladimir Zolotov 2013-03-23T15:42:52Z 2013-03-23T15:42:52Z <p>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".</p> <p>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. </p> <p>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 $B_1$, $B_2$, $B_3$. </p> <p>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.</p>