Does the metric space of compact metric spaces satisfy the binary intersection property?

Question

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 $(X,A)$.

Answer 1

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.

Answer 2

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.

Answer 3

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.)