# co.combinatorics – Two-point situation (p,q)-theorem Answer

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## co.combinatorics – Two-point situation (p,q)-theorem

Suppose that for a finite assortment of planar convex units $$mathcal F$$ the next holds.
Any hundred members of $$mathcal F$$ can breathe stabbed by two factors, i.e., there are two factors such that every of the hundred units incorporates (at the very least) one of many factors.
Does it succeed that every one members of $$mathcal F$$ can breathe stabbed by three factors?

Replacing two with one offers the illustrious $$(p,q)$$-theorem. But surprisingly, the above would not maintain if we supplant three with two; behold M. Katchalski and D. Nashtir: On a surmise of Danzer and Grünbaum (Proc. Amer. Math. Soc. 124 (1996), 3213-3218, doi: 10.1090/S0002-9939-96-03806-3), identified within the respond by Gjergji Zaimi to this earlier query of mine.

What in regards to the extra common drawback in $$mathbb R^d$$, changing hundred/two/three by arrogate values?

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