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The entire 2-dimensional plane is covered by 3 sets: Blue, Green and Red. It is given that:

  • All sets are closed.
  • All sets are interior-disjoint (but may meet at their boundaries).
  • Blue is bounded.
  • Blue meets both Red and Green (i.e. their boundaries have non-empty intersection).

The following diagram illustrates the given facts (not the entire coloring):

enter image description here

I would like to prove that there is a point in which all 3 sets meet.

The intuition is that, if we start from the intersection of Blue and Green and move along the boundary of Blue, we must somewhere meet the Red boundary, since the entire plane is covered.

However, there are probably some cases in which this is not true. If this is so, what conditions do I need to add in order to make sure that the intuitive conclusion is true?

Erel Segal-Halevi
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2 Answers2

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I think the result does not hold. Suppose Red is the closed unit disk (centered at the origin); blue is the closed annulus centered at the origin with inner radius $1$ and outer radius $2$ (so blue completely surrounds red); and green consists of all points at distance greater than or equal to $2$ from the origin.

paw88789
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If blue has a connected boundary, then it holds. Because if there were no point of intersection of all three sets, you could split $\partial\mathrm{Blue}$ to $\partial\mathrm{Blue}\cap\mathrm{Red}$ and $\partial\mathrm{Blue}\cap\mathrm{Green}$ which would yield a decomposition of $\partial\mathrm{Blue}$ into two open and closed sets, contradicting its connectedness.

(I should probably explain better why these sets are open. For contradiction, assume that $\partial\mathrm{Blue}\cap\mathrm{Red}$ is not open in $\partial\mathrm{Blue}$. Then for some $x\in\partial\mathrm{Blue}\cap\mathrm{Red}$, each neighborhood of $x$ in $\partial\mathrm{Blue}$ contains a point not contained in $\mathrm{Red}$. However, each point in $\partial\mathrm{Blue}$ is contained in either Red or Green. So, it follows that each neighborhood of $x$ in $\partial\mathrm{Blue}$ contains a point in Green and the closedness of Green implies that $x$ is a contained in Green and hence in the triple-intersection, contradicting our assumption.)

Conversely, if $\partial\mathrm{Blue}$ is not connected, then it can be split into two sets $C_1$ and $C_2$ with positive distance. Let $U_1$ and $U_2$ be some disjoint neighborhoods of $C_1$, $C_2$ in $\mathbb{R}^2$ and define the color on $U_i\setminus\mathrm{Blue}$ to be red resp. green for $i=1$ resp. $2$. Then you choose any red-green coloring of $\mathbb{R}^2\setminus (\mathrm{Blue}\cup U_1\cup U_2)$ so that Green and Red are both closed and you get a coloring wihout a triple-intersection.

Some condition on when the boundary is connected can be found in this discussion. (It seems to me that it follows from that discussion that if Blue has a connected interior whose closure is Blue and Green and Red are both connected and unbounded, then this is sufficient for $\partial\mathrm{Blue}$ to be connected, but I'm not completely sure)

Community
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Peter Franek
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  • This makes sense. I think the relevant condition on Blue is that it should be simply-connected, but I am not sure. – Erel Segal-Halevi Dec 21 '14 at 11:36
  • @ErelSegalHalevi (I doubt that simple-connectedness is sufficient, but I removed my remark about the Warsaw circle, it's not a counter-example) – Peter Franek Dec 21 '14 at 11:58
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    I'm going to sound dumb, but I do not understand the part going "then each neighborhood of $x$ in $\mathbb{R}^2$ would contain points that are neither in Blue nor in Red". Can you elaborate, if possible? (thanks!) – Clement C. Dec 21 '14 at 13:47
  • @ClementC. Probably you agree that each point on $\partial\mathrm{Blue}$ is contained either in Green or in Blue. If some neighborhood $U$ of $x$ in $\mathbb{R}^2$ contains only points from Blue and Red, then ${U}\cap\partial\mathrm{Blue}\subseteq \partial\mathrm{Blue}\cap\mathrm{Red}$ is a neighborhood of $x$ in $\partial\mathrm{Blue}$. Or not? – Peter Franek Dec 21 '14 at 13:53
  • Indeed; but you want the contrapositive for the statement above, don't you? In this case, the conclusion would be "$U$ contains only points that are not in Blue$\cap$Red", wouldn't it? (instead of "$U$ contains only points that are neither in Blue nor Red"). (or did I miss something?) – Clement C. Dec 21 '14 at 14:02
  • @ClementC. "If some neighborhood $U$ contains only points such that each point of this neighborhood is either in Red or in Blue, then we have an open neighborhood of $x$ in $\partial\mathrm{Blue}$ contained in Red." That's equivalent to "If there were no such open neighborhood, then each neighborhood of $x$ would contain a point that is not in Blue and not in Red" Or not? I'm already a bit confused but still don't see a mistake.. – – Peter Franek Dec 21 '14 at 14:23
  • Your last formulations are equivalent (at least to me), yes. But in your previous post, is that what you mean? ("If some neighborhood $U$ of $x$ in $\mathbb{R}^2$ contains only points from Blue and Red, then"? ((for the last statement to be true -- after the 'then' I thought you needed $U$ to contain only points from Blue$\cap$Red: isn't it the case? ). – Clement C. Dec 21 '14 at 15:34
  • No, I meant that $U$ is contained in $\mathrm{Red}\cup\mathrm{Blue}$. Anyway, if you agree with the last formulation, is the logic in my "proof" correct? I hope it is :) – Peter Franek Dec 21 '14 at 15:43
  • Let us continue this discussion in chat. – Clement C. Dec 21 '14 at 15:47
  • @ClementC. I expanded this part, could you please check whether it is correct now? – Peter Franek Dec 21 '14 at 15:54
  • It looks correct to me — thanks! (also, I have to say: I believe was correct before as well, it just confused me). – Clement C. Dec 21 '14 at 16:01
  • Ok. And thanks for the comments, I'm making mistakes all the time, so it's good to check. – Peter Franek Dec 21 '14 at 16:03
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    Alternatively, and perhaps easier to see: $BB$, the boundary of blue, is closed in the plane, so its intersections with $R$ (Red) and $G$ (Green) are closed in $BB$ (as intersections of closed sets), non-empty (hypothesis), and disjoint (if we assume no triple intersection). Under these hypotheses, $BB \cap R$ is the complement in $BB$ of the closed set $BB \cap G$, hence open in $BB$, etc. – Andrew D. Hwang Dec 21 '14 at 16:33