How well connected can a (special) partition of $\Bbb R^2$ be? 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.
 A: 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 $A_i$.)
A: 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.  
