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It turned out that How well connected can a (special) partition of $\Bbb R^2$ be? had a a few nice answers using continum-many pairwise disjoint dense connected sets. Connectedness is just weirder than one first thinks ...

So let's up the ante to path-connectedness:

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 - in the linked question the distinction was irrelevant as solutions with all indices "completely linked" were found even for $I\approx \Bbb R$) 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 path-connected
  • $A_i\cup A_{i+1}$ is path-connected

How often can it happen that $A_i\cup A_j$ is path-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 path-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.

EDIT: Only after seeing Crostul's solution (and how it can be adapted for $I=\Bbb Z$, I noticed that it is not possible to have three or more completely linked indices: Pick a point in each of the three sets and a point in each of three other sets. Then we can pick a path witnessing the path-connectedness of the respective unions (wlog.(!) these paths do not intersect) and obtain a solution to the famous gas-water-electricity problem (i.e., a planar embedding of $K_{3,3}$), which is impossible.

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Hagen von Eitzen
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  • In your edit, how do you get that WLOG the paths do not intersect? – Eric Wofsey Nov 07 '15 at 19:31
  • @EricWofsey First of all, a path in $A_i\cup A_j$ connecting $a\in A_i$ and $b\in A_j$ can be assumed/simplified to transition between the two parts only once (because $A_i,A_j$ are also path-connected). If these transition points are distinct, we need only untangle them within a path-connected set, which should be easy. If two or all three transition points from $A_i$ (we do not consider more paths from $A_i$!) coincide, this transition point must belong to $A_i$ and we may pick it as our vertex. – Hagen von Eitzen Nov 07 '15 at 22:16
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    It is not obvious to me how you can assume the path transitions only once. For instance, consider $A_1=[0,1]\times{0}\cup (\mathbb{Q}\cap[0,1])\times[0,1)$ and $A_2=[0,1]\times{1}\cup([0,1]\setminus\mathbb{Q})\times(0,1]$. In this case there does exist a path from $A_1$ to $A_2$ which only transitions once, but it's unclear how you would get such a path by "simplifying" a horizontal path that started at $(0,1/2)$ and ended at $(\alpha,1/2)$ for some irrational $\alpha$. – Eric Wofsey Nov 07 '15 at 22:26
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    @EricWofsey Good point. Seems my adding the "(!)" to the "wlog" was justified :) So let me try a different approach: Consider paths $\gamma_{ab}\colon[0,1]\to A_a\cup A_b$ from $a\in A_a$ to $b\in A_b$ and $\gamma_{ac}\colon[0,1]\to A_a\cup A_c$ from $a\in A_a$ to $c\in A_c$. Consider $f\colon [0,1]^2\to [0,\infty)$, $(s,t)\mapsto |\gamma_{ab}(s)-\gamma_{ac}(t)|$. This continuous function is positive on the top and the right edge of $[0,1]^2$. By compactness we find a top-right-most $(s_0,t_0)$ with $f(s_0,t_0)=0$. Then $\gamma_{ab}(s_0)=\gamma_{ac}(t_0)$ implies (to be continued) – Hagen von Eitzen Nov 07 '15 at 22:41
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    (continuation) that $\gamma_{ab}=\gamma_{ac}\in A_a$ and we can take this point instead of $a$. By top-right-most-ness, the chopped paths have no point in common except the start. Together, they form a path from $b$ via $a$ to $c$. Now add a third path $\gamma_{ad}$ in $A_a\cup A_d$. The set of $t\in[0,1]$ where $\gamma_{ad}(t)$ is on one of the other paths is closed and has a maximum $t_1$. We relocate $a$ again to $\gamma_{ad}(t_1)$ and adjust the paths accordingly (note that $\gamma_{ab}\cup \gamma_{ac}$ joins the old and new position of $a$ within $A_a$).(to be continued) – Hagen von Eitzen Nov 07 '15 at 22:50
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    (continued) Note that all adjustments took place within $A_a$ except that possibly some "detour" pieces of paths in $A_b,A_c,A_d$ got chopped away. With the same method we untangle the three paths originating at $b$, then $c$, etc. Fortunately, all vertices in the $K_{3,3}$ we build this way have only degree $3$. -- I hope this sketches a better justification for my "wlog". :) – Hagen von Eitzen Nov 07 '15 at 22:53
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    I'm almost convinced, but still puzzled by one thing. When you relocate $a$ to $\gamma_{ad}(t_1)$, let's say that $\gamma_{ad}(t_1)$ also lies on $\gamma_{ab}$. Then I assume you are redefining $\gamma_{ab}$ to just start at $\gamma_{ad}(t_1)$ instead of the old $a$. But what do you do with $\gamma_{ac}$? You could make it start with a path from $\gamma_{ad}(t_1)$ to the old $a$ within $A_a$, but how do you know you can choose such a path that doesn't intersect $\gamma_{ad}$ or $\gamma_{ab}$? – Eric Wofsey Nov 07 '15 at 23:00

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This picture represents a partition $\{ A_n \}_{n \ge 1}$ with $A_1$ and $A_2$ completely linked.

enter image description here

Crostul
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  • This also works for $I=\Bbb Z$ if we put $A_0, A_{-1},A_{-2},\ldots$ in a similar pattern on the lower side. – Hagen von Eitzen Nov 07 '15 at 19:07
  • This completely resolves the question regarding completely linked as well as almost completely linked indices (cf. my edit). Unless someone comes up with interesting insights regarding three or more inifninitely linked indices (or maybe even then) your answer deserves to be accepted. Allow me to wait a few hours anyway – Hagen von Eitzen Nov 07 '15 at 19:19