common boundary points of connected sets

Question

suppose $A_1, A_2, A_3,\ldots$ are connected closed sets on $\mathbb R^2$ such that for $i, j=1, 2, 3, \ldots$ we have $\operatorname{int}(A_i) \cap \operatorname{int}(A_j)=\varnothing$ and for $i=1, 2, 3, \ldots$ we have $\operatorname{int}(A_i)\ne\varnothing.$

draw $A_1$

draw $A_2$ such that $A_1\cap A_2 \ne \varnothing.$

draw $A_3$ such that $A_1\cap A_2 \ne \varnothing.$ and $A_1\cap A_3 \ne \varnothing.$ and $A_2\cap A_3 \ne \varnothing.$

and so on.

my problem is how many number of sets can be drawn like this

Answer 1

If two states, $A$ and $B,$ share a boundary, then a road can go from the capital of $A$ to the capital of $B$ without passing through any states besides $A$ and $B$. Now try this with four states mapping the roads between capital cities, between $A$ and $B,$ between $A$ and $C,$ between $A$ and $D,$ between $B$ and $C,$ between $B$ and $D,$ and between $C$ and $D.$ $$ \begin{array}{cccccccc} A & \leftrightarrow & B & \nwarrow \\ \downarrow & \searrow & \downarrow & \uparrow \\ C & \leftrightarrow & D & \nearrow \\ & \searrow & \rightarrow \end{array} $$ This picture is crude but I hope you can see the road from $C$ to $B.$

A fifth capital city, if connected to $A,$ $B,$ and $C,$ could not reach $D$ without passing through another state.

So five is more than will fit in a plane in this way.