Given the system of closed arcs from the same circle, with length smaller than half circumference of that circle. If every three arc from that system have non-empty intersection, then intersection of all arcs is non-empty. How can I prove this? How can Helly's theorem be applied to this?
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Is your system finite? – iamvegan Sep 03 '16 at 16:23
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I have no idea. I would presume that it's infinite. – Sep 03 '16 at 16:25
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If it is finite, according to Helly's theorem you have nonempty intersection. It is because, arc of a circle is a convex set in $\mathbb{R}^2$. For infinite case I have no idea. – iamvegan Sep 03 '16 at 16:28
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@iamvegan Arcs are not convex in $\Bbb R^2$. Working with their convex hull probably fixes it, though. – Akiva Weinberger Jun 21 '17 at 23:52
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Also, Helly's theorem works for infinite sets as long as they're all compact (in fact, it even works as long as at least one is compact and the rest are closed). Closed arcs are compact. – Akiva Weinberger Jun 21 '17 at 23:55
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I'm not sure if Helly's theorem can be directly applied, but the proof for this is practically the same. Since there is nothing to prove in the three-arc case, consider the four arcs $\{a_1,\,a_2,\,a_3,\,a_4\}$ on the circumference of the circle in $\mathbb{R}^2$. Since every three-arc intersection is given to be nonempty, pick a point in each intersection denoted by:
$$v_1 \in a_2 \cap a_3 \cap a_4 \qquad v_2 \in a_1 \cap a_3 \cap a_4\\ v_3 \in a_1 \cap a_2 \cap a_4 \qquad v_4 \in a_1 \cap a_2 \cap a_3 $$
So now we have four points along the circle. WLOG let $v_1,v_4$ be the two points farthest away and $v_2,v_3$ the other two. Define new arcs $\alpha_{14} = [v_1,v_4]$ and $\alpha_{23} = [v_2,v_3]$ connected in the direction (will always be the minor arc) so that intersection is preserved: $$\forall x \in \alpha_{14} \implies x \in a_2 \cap a_3$$ $$\forall y \in \alpha_{23} \implies y \in a_1 \cap a_4$$
Also note that by definition of our new arcs that $\alpha_{23} \subset \alpha_{14}$. Take a point $z \in \alpha_{23}$, then:
$$\begin{align*}z \in \alpha_{23}\\ z \in \alpha_{23} \cap \alpha_{14}\\ z \in \left(a_1 \cap a_4\right) \cap \left(a_2 \cap a_3\right) \\ z \in a_1 \cap a_4 \cap a_2 \cap a_3\\ z \in a_1 \cap a_2 \cap a_3 \cap a_4 \end{align*} $$
For any set of arcs greater than $4$ we can apply this nesting argument multiple times. If there are an odd number of arcs given, then $v_i$ picked that is in the middle must actually be in the intersection of all arcs.
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