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Helly's theorem : Let $C_1,\ldots,C_n$, $n\geq d+1$, be convex sets in $\Bbb R^d$. Suppose every $d+1$ have a common intersection. Then they all have a common intersection.

I can find the proofs for $n\geq d+2$ in books, Wikipedia everywhere.

But don't able to find any proof for $n=d+1$ in nowhere. Anyone help me to prove this.

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    $\begingroup$ It is trivial for $n=d+1$. $\endgroup$
    – copper.hat
    Commented Jul 8 at 5:36
  • $\begingroup$ @copper.hat but how to see the trivial case. $\endgroup$
    – A. H.
    Commented Jul 8 at 5:50
  • $\begingroup$ In that case, the statement is basically if P is true then P is true, which is always the case. $\endgroup$
    – copper.hat
    Commented Jul 8 at 13:13

1 Answer 1

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Hint: Look at the statement of the theorem carefully with your particular case in mind. Substitute $n=d+1$ explicitly if that helps.

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  • $\begingroup$ A collection of 3 sets in the plane(𝑑=2, n=3 ), such that every 2 of them intersect, but all 3 of them do not intersect. Let 𝐴𝐵𝐶 be a non-degenerate triangle. Let the sets be the line segments 𝐴𝐵,𝐵𝐶,𝐶𝐴. Clearly, any 2 of them intersect (at a vertex), and the 3 of them do not intersect. I don't find any example, would you give me any example? $\endgroup$
    – A. H.
    Commented Jul 8 at 7:24
  • $\begingroup$ @A.H.A First off, please use regular letters and ideally also the math mode that is embedded in this site. I can only see rectangles where you presumably wrote $d$ and $ABC$. Second, how many of the sets does the theorem require to have non-empty intersection? Is your example, where any two segments meet, a valid counterexample? $\endgroup$
    – Arthur
    Commented Jul 8 at 8:08
  • $\begingroup$ Would you expand the answer by explaining more? Actually your hints not help me well. $\endgroup$
    – A. H.
    Commented Jul 8 at 11:05
  • $\begingroup$ Read the theorem carefully. In the case of $n = d+1$, the theorem says "If any $d+1$ of the $d+1$ sets $C_1, \ldots, C_{d+1}$ have non-empty intersection, then all $d+1$ of these sets have non-empty intersection." Even more concretely, in the plane $d = 2$ case, it says "If any 3 of your 3 sets have non-empty intersection, then all three sets have non-empty intersection." $\endgroup$
    – Arthur
    Commented Jul 8 at 11:08

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