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If there are 3 intervals, such that any 2 of them intersect, then all 3 of them intersect.

For any 4 disks, if any 3 of them have a non empty intersection, then all 4 of them have a common intersection.

This seems to generalize to higher dimensions.

What is this property, that for some $n$, knowing all subset of size $n-1$ of $n$ shapes have non-empty intersection implies all the shapes have non-empty intersection?

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3 Answers

up vote 6 down vote accepted

Helly's theorem. It applies to convex shapes, not to all shapes.

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Your final statement looks a little too general: first you seem to have lost the concept of dimension in your general statement; and second there are shapes where it probably does not apply, such as two-dimensional crescents: here are four crescents where any three intersect (twice) but the four do not. One crescent is shaded, as are the places of triple intersection.

enter image description here

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You have already got the answers, but maybe the theorem of Knaster, Kuratowski and Mazurkiewicz will interest you a bit. Here you have a link to Wikipedia:

Knaster-Kuratowski-Mazurkiewicz lemma

It's similar to the hypothesis you have stated but more complicated in assumptions. It says that, given an $n$-simplex $S$ covered by closed sets $C_i$, where $i\in I:=\{1,\ldots,n\}$, such that for all $J\subseteq I$ the face of $S$ spanned by all the $e_i's$ where $i\in J$ is covered by all the sets $C_i$ where again $i\in J$, the intersection of all $C_i$ is nonempty: $\bigcap_{i=1}^n C_i\neq\emptyset$.

Under the link you will find more interesting informations about the theorem as well as the proof of it.

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