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Note: I have simplified the question. I think trying to generalise this straight away is biting off more than I can chew.

If $\ A, B, C\ $ are pairwise not disjoint, compact, connected subsets of $\ \mathbb{R}^2\ $ with $\ A \cap B \cap C = \emptyset\ $ then $\ \left(A\cup B\cup C\right)^c\ $ is disconnected. Is this true? Example diagram:

enter image description here

Note that if we did not require $\ A, B, C\ $ to be compact, and therefore closed, then $\ A = B'(\ (-1,0);\ 2\ ), B = B'(\ (1,0);\ 2\ ), C = \overline{B'(\ (0,10),\ 10-\sqrt{3})}\ $ where $\ B'(\ x;\ r)\ $ means the open ball centred at $\ x\ $ with radius $\ r,\ $ is an example where $\ \left(A\cup B\cup C\right)^c\ $ is not disconnected.

Adam Rubinson
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  • Note that this property is true for $\mathbb{R}$ (when $k=1$). If you want to generalize, you most likely need to vary the number of sets and involve higher-dimensional holes. – Brian Moehring Jan 28 '22 at 17:01
  • Oh, because it's vacuously true when $k=1$ ? – Adam Rubinson Jan 28 '22 at 17:03
  • Note in higher dimensions we have more than one kind of "hole". Like in $\mathbb{R^3}$ there's the hole "through" a donut or punched piece of paper, or the hole "inside" a tennis ball or an inflated and tied balloon. Homology and cohomology theories give a way of describing these more generally. – aschepler Jan 28 '22 at 17:11
  • I suspect that whenever you have $k$ sets in $\mathbb{R}^n$, each homotopy-equivalent to a point, and the intersection of any $k-1$ of the sets is not empty but the intersection of all $k$ is empty, then their union is homotopy-equivalent to $S^{k-2}$. (Which implies the conditions are impossible if $k>n+1$.) – aschepler Jan 28 '22 at 17:23
  • It could have 2 holes, or infinitely many. Cut some open holes out of $A\setminus (B\cup C).$ You did not say $A,B,C$ were simply-connected – DanielWainfleet Jan 28 '22 at 18:40
  • @DanielWainfleet Yes, I meant, "at least one" hole. And I need to say that $A,B,C$ are simply connected also. You are right. I edited these into the question. – Adam Rubinson Jan 28 '22 at 19:10
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    Topologists have defined (early twentieth century) a topological invariant called “topological dimension” for all separable metric spaces. (There are at least 3 equivalent definitions, yielding the same positive integer value $n$ for a given space (or possibly $\infty$)). Brouwer showed that indeed $\dim(\Bbb R^k)=k$ for all $k$. – Henno Brandsma Jan 28 '22 at 19:39
  • @HennoBrandsma I'm not sure what that has got to do with the question? – Adam Rubinson Jan 28 '22 at 20:31
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    The most obvious invariant. – Henno Brandsma Jan 28 '22 at 20:33
  • I suggest, you pick up a textbook on algebraic topology and start reading. This will help you to be able to formalize your questions like the one you just asked (and avoid using "hole" terminology) and even to answer these questions yourself. – Moishe Kohan Jan 28 '22 at 20:44
  • @MoisheKohan OK, fair enough. – Adam Rubinson Jan 28 '22 at 21:52
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    One more thing: "genus $n$ means it has $n$ holes" is actually wrong in the context of your question. Yes, in the classification of compact oriented connected surfaces, this is true, but "hole" there is different from "hole" in the context of your question. This ambiguity of the notion of a "hole" is one reason topologists (mostly) avoid using this terminology, see here. – Moishe Kohan Jan 28 '22 at 22:00
  • I think better than “genus $ \geq 1$” would be to say that $\ (A \cup B \cup C)^c $ is disconnected. – Adam Rubinson Jan 29 '22 at 07:54
  • If ABC are additionally path connected then you could make a 'triangle' with vertices in relevant intersections. Then try to show that it cannot be shrunk to a point and from that conclude that the union has nontrivial fundamental group – Radost Waszkiewicz Jan 29 '22 at 20:29
  • Yes, the current question has positive answer but the proof that I know requires Alexander duality and Chech cohomology. I am not sure you are ready for this. – Moishe Kohan Jan 29 '22 at 20:52
  • @MoisheKohan I am definitely not ready for those things. But maybe there is a proof that doesn't require those things... – Adam Rubinson Jan 29 '22 at 21:28

1 Answers1

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As the subsets are connected and $X\cap Y\ne\emptyset, X,Y\in\{A,B,C\}$, then we can draw a continuous path between each set. Furthermore, the end points of each path can be connected.

This loop cannot be contracted to null as $A\cap B\cap C=\emptyset$, and therefore the loop's interior $I$ intersected with $(A\cup B\cup C)'$ is non-empty.

$I$ doesn't contain all of $(A\cup B\cup C)'$, because each set is compact, and therefore $(A\cup B\cup C)'$ is disonnected.

JMP
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    The problem has the following Jordan curve-like theorem as a special case – “the complement in the plane of the homeomorphic image of a circle is disconnected” – which I’m not convinced can be proved with a few lines of elementary-looking topology not involving specific properties of the plane. – Aphelli Sep 22 '22 at 10:43
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    Moreover, $A,B,C$ are only assumed to be connected, they are not necessarily arc-connected, so the existence of a non-trivial loop may be false. – Christophe Leuridan Sep 22 '22 at 20:29
  • I actually kind of like this argument and find it somewhat intuitive. If we replaced "connected" with "path connected" in the question, then would this answer be valid? – Adam Rubinson Sep 27 '22 at 13:24