Suppose $A_1A_2A_3$ is a triangle and $U_1$, $U_2$, $U_3$ are open subsets in the plane such that their union convers the triangle (including its interior). Also suppose that $A_1A_2 \subseteq U_3$, $A_2A_3 \subseteq U_1$ and $A_1A_3 \subseteq U_2$. Is it true that the intersection of $U_i$'s is non-empty ?

  • $\begingroup$ My intuition is saying that you should be able to make some sort of argument that says if this were true, what you were covering would be homotopic to a circle and from there get a contradiction since what you're covering is simply connected, although at the moment I'm not totally sure how to make this precise. $\endgroup$
    – Stahl
    Jan 29, 2016 at 13:11

1 Answer 1


Yes. By (para)compactness of the triangle and Lemma 5.1.6 from Engelking’s “General Topology”, there exist closed subsets $V_i\subset U_i$ for each $i$ of the triangle such that their union covers the triangle. Put $C_1=V_3\cup A_1A_2\subset U_3$, $C_2=V_1\cup A_2A_3\subset U_1$, and $C_3=V_2\cup A_3A_1\subset U_2$. Then the union of $C_i$’s covers the triangle and by KKM lemma $\varnothing\ne\bigcap C_i\subset \bigcap U_i$.

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  • $\begingroup$ Thank you and I am trying to understand your solution. I believe scince we are dealing with only 3 sets here, maybe one can prove more elementary. $\endgroup$
    – T.KM
    Jan 31, 2016 at 7:43
  • $\begingroup$ @User3186 I am afraid that no, because the question has a flavor of KKM Lemma or Sperner’s lemma, which have problems with elementary proofs. $\endgroup$ Feb 1, 2016 at 19:34

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