Let $H_1,H_2,\dots,H_n$ be compact and convex sets in $\mathbb{R}^n$ such that $\bigcap_{j=1}^n H_j$ has non-empty interior and for each $i=1,2,\dots,n$ there exist at least one element $x \in H_i$ such that $x \not \in \bigcup_{j\not = i} H_j$. Can $\bigcap_{j=1}^n \partial H_j$ be a single point?
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3$\begingroup$ Maybe this works, take $n$ balls $B_i$ tangent to a single point such that $B_i \subset B_j$ for $i < j$, let's say of radius $1/i$, this example doesn't satisfy the the last condition, to get this, modify the spheres by cutting them along the meridian in two half spheres, adding long cylinders of length $100i$ at the meridians and then gluing the half spheres back, so each $B_n$ looks like a big pill. $\endgroup$– shurtadosCommented Sep 12, 2013 at 22:56
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3 Answers
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In ${\mathbb R}^3$, try three solid cones with different heights and the same base radius, axis and vertex (the vertex being the intersection of the three boundaries).
answered Sep 13, 2013 at 4:37
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$\begingroup$ I was just about to post a similar answer. The same construction works for all $n \ge 1$. $\endgroup$ Commented Sep 13, 2013 at 5:20
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1$\begingroup$ @Tapio: Hum, for $n=2$, it seems to me that one creates another component in the intersection of boundaries. Also, for $n=1$, you have one convex and compact set, whose boundary must have two points. So, this example should answer $n\ge 3$, and only $n=2$ looks unsettled. $\endgroup$ Commented Sep 13, 2013 at 7:46
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$\begingroup$ @Benoît: You are right. I was not careful enough. I missed $n=n$ (among other things...). $\endgroup$ Commented Sep 13, 2013 at 9:07
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1$\begingroup$ In the case $n = 2$ the intersection of the two boundaries seems to always contain at least two points. $\endgroup$ Commented Sep 13, 2013 at 9:17
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$\begingroup$ That's excellent, thanks. I still feel there is something non-generic about this but clearly I need to work out what it is. $\endgroup$ Commented Sep 13, 2013 at 14:34
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Yes: think of three closed balls in $\mathbb{R}^3$, such that the intersection of the boundaries of the first two balls is a small circle tangent to the boundary of the third ball from inside.
answered Sep 12, 2013 at 21:51
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$\begingroup$ Thanks! But sorry, it turns out I can rule out this case: I have a condition stronger than the one I posted---did not realize I would needed as strong. Will edit right. $\endgroup$ Commented Sep 12, 2013 at 22:04
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Yes it is possible. Here is a generic example: take $n$ red points, and one blue point in $\mathbf{R}^n$ so that the convex hull of these $n+1$ points has non-empty interior. In this interior take a green point. Now let $H_i$ be the convex hull of the blue point, the green point and the $n-1$ red points remaining after removing the $i$-th, and you obtain an adequate configuration.
If you do not like colours, consider $\tilde H_i$ the half space defined by $x_i \ge 0$ and consider $H_i$ the intersection of $\tilde H_i$ with the cube consisting of points whose coordinates are between $-1$ and $0$.
answered Sep 13, 2013 at 12:52
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$\begingroup$ I'm confused by the colored example. Why isn't the line segment joining the blue and green points in the intersection of the boundaries? $\endgroup$ Commented Sep 13, 2013 at 14:35
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$\begingroup$ I don't seem to be able to make this one work. Won't all sets $\partial H_i$ share both the blue and green point? $\endgroup$ Commented Sep 13, 2013 at 14:40
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$\begingroup$ Oh, you are both right! Please assume that you do not like colours. :) I guess we can make this work by making the blue point also red and then removing 2 points each time, but points have to be choosed adequaltely so it is much less illuminating than the example with the half planes. $\endgroup$ Commented Sep 13, 2013 at 15:51