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Let $\{A_i\}_{i\in I}$ be a family of subsets of $\Bbb R^2$ (where $I=\Bbb N$ or $\Bbb Z$; I don't know if it makes a difference) such that

  • $\bigcup_{i\in I} A_i=\Bbb R^2$
  • $i\ne j\implies A_i\cap A_j=\emptyset$
  • $A_i\ne\emptyset$
  • $A_i$ is connected
  • $A_i\cup A_{i+1}$ is connected

How often can it happen that $A_i\cup A_j$ is connected for $j\notin\{i-1,i,i+1\}$?

Definition. Let's say that an index $n\in I$ is infinitely linked/almost completely linked/completely linked if $A_n\cup A_i$ is connected for infinitely many/almost all/all $i\in I$.

One possibility configuration is to let the $A_i$ be vertical stripes, in which case no $n\in I$ is infinitely linked. With another configuration one can achieve that there exists exactly one $n\in I$ that is completely linked (let $A_n=\{(0,0)\}$ and all other $A_i$ suitable sectors of $\Bbb R^2\setminus\{(0,0)\}$). Can more than that be achieved?

  • I.e., are there configurations with more (two, three, arbitrarily many, infinitely many, almost all, all) completely linked indices?
  • Or maybe at least more almost completely linked indices?
  • For $I=\Bbb Z$, I can find configurations two infinitely linked indices. Are three or more infinitely linked indices possible?
  • Are two infinitely linked indices possible with $I=\Bbb N$?

Note that with $\Bbb R^3$ instead of $\Bbb R^2$ one can easily achieve that all indices are completely linked.

Hagen von Eitzen
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  • It doesn't matter whether the index set is $N$ or $Z$ as they are both countably infinite. You could just say " $A$ is a countably infinite cover of $R^2$ ... ("cover" means $ \cup A\supset R^2$)... of pair-wise disjoint non-empty connected subsets of $R^2$ , such that the union of any 2 of them is connected". I like the q. – DanielWainfleet Nov 05 '15 at 21:01
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    @user254665 If my condition about $A_i\cup A_{i+1}$ were important, it would matter. It is only by the (currently two) answers that all $A_i\cup A_j$ can be made connected, thus making the order type imposed irrelavant. – Hagen von Eitzen Nov 05 '15 at 21:09

2 Answers2

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Using the axiom of choice, you can partition $\mathbb{R}^2$ into $2^{\aleph_0}$ sets $A_i$ such that any union of the $A_i$ is connected. Indeed, there are only $2^{\aleph_0}$ uncountable subsets of $\mathbb{R}^2$ that are either open or closed and each of them has cardinality $2^{\aleph_0}$, so by a straightforward diagonalization argument (very similar to the argument here, for instance), you can partition $\mathbb{R}^2$ into $2^{\aleph_0}$ disjoint sets $A_i$ with the property that each of them intersects every uncountable open or closed subset of $\mathbb{R}^2$.

Now suppose some $A_i$ were disconnected. Then there are open subsets $U,V\subset\mathbb{R}^2$ such that $A_i\subset U\cup V$, $U\cap A_i\neq\emptyset$, $V\cap A_i\neq\emptyset$, and $U\cap V\cap A_i=\emptyset$. Since $A_i$ intersects every nonempty open set, $U\cap V$ must be empty. But then $U\cup V$ is disconnected, and hence $\mathbb{R}^2\setminus (U\cup V)$ is an uncountable closed set (as the complement of any countable subset of the plane is connected). So $A_i$ must intersect $\mathbb{R}^2\setminus (U\cup V)$, which is a contradiction. By the same argument, any union of the $A_i$ (indeed, any set containing any $A_i$) is also connected.

(This is essentially counterexample 124 in Counterexamples in Topology, though there they only construct two such sets $A_i$.)

Community
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Eric Wofsey
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  • Although it is probably elementary, how do you show that $R^2\backslash (U\cup V)$ is uncountable$ And the remark in brackets ("as the complement of any countable...") seems to me to be irrelevant. – DanielWainfleet Nov 05 '15 at 20:53
  • If $\mathbb{R}^2\setminus(U\cup V)$ were countable, then its complement, $U\cup V$, would be connected. I have added a reference for the parenthetical remark. – Eric Wofsey Nov 05 '15 at 21:28
  • Thanks for clearing that up. I expected it would be simple. – DanielWainfleet Nov 05 '15 at 21:42
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    “…by a simple diagonalisation argument…” Maybe elaborate on this a little? The simplest version I can see relies two fairly non-trivial facts: that there are only $2^{\aleph_0}$-many uncountable open or closed sets, so that they can be enumerated correctly (not hard to prove, but not obvious to state); and that any uncountable closed set must have cardinality $2^{\aleph_0}$, so that it does not get exhausted at any stage (not at all trivial or obvious, if you don’t already know it). – Peter LeFanu Lumsdaine Nov 07 '15 at 12:33
  • @PeterLeFanuLumsdaine: Yes, that's the argument I had in mind. I'll add some more detail. – Eric Wofsey Nov 07 '15 at 17:02
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In On a connected dense proper subgroup of $\mathbb R^2$ whose complement is connected, Ryuji Maehara constructs a connected dense proper subgroup of $\mathbb R^2$. If we take the collection of all cosets of this subgroup, we get a partition into (continuum many) disjoint connected dense subsets. It is standard that the union of any number of these is connected. In particular, we can lump them together into $\mathbb N$-many subsets which are still dense and connected, and the union of any two of these will be connected.

John Gowers
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  • So I take it that Maehara's group as constructed in the ref is of continuum index? – Hagen von Eitzen Nov 05 '15 at 18:57
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    @HagenvonEitzen I should have said. The group is constructed as the graph of an additive function, so it has continuum index. – John Gowers Nov 05 '15 at 19:06
  • @Donkey_2009. I take it that the additive group is ${(x,f(x) :x\in R}$ where $f$ is any additive nowhere-continuous function? – DanielWainfleet Nov 05 '15 at 21:09
  • @user254665 Or, what amounts to the same, any additive $f$ not of the form $f(x)=cx$ ... Dense and contimnuum index is clear, I'm still contemplating connected – Hagen von Eitzen Nov 05 '15 at 21:14
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    @user254665 I don't know that any additive nowhere-continuous $f$ will do. The $f$ given in the paper is constructed so that its graph intersects every closed subset of the plane whose projection on to the $x$-axis contains some open interval. – John Gowers Nov 05 '15 at 22:00
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    In fact, the additiveness of the function isn't necessary (since we can just take translates up and down). So maybe this answer is in fact just adding a whole lot of useless machinery to Eric's answer. I'll have a think, though. – John Gowers Nov 06 '15 at 00:35
  • I would love this answer more, esp. since I so much like non-continuous additive functions. I'll accept the other though because it is more self-contained ... – Hagen von Eitzen Nov 07 '15 at 11:56