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.



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> Does the metric space $M$ of compact metric spaces under the [Gromov-Hausdorff distance](http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence#Gromov.E2.80.93Hausdorff_distance) satisfy the binary intersection property?
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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](http://books.google.com/books?id=lXZ95EKwjYUC&pg=PA13&lpg=PA13&dq=metrically+convex+binary+intersection&source=bl&ots=SvQJYCRWp8&sig=CmTnEUt1Zxt2-3FBmxf0rlufVbQ&hl=en&sa=X&ei=bIqnUKLIE8LQmAWQuIDIDQ&redir_esc=y#v=onepage&q=metrically%2520convex%2520binary%2520intersection&f=false)) 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 $(X,A)$.



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


