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Let $S=\{a_i\}$ be a countable set of bounded,connected and closed subsets of $\mathbb R^n$, each of nonzero area, such that any two $a_i$ and $a_j$ may only intersect at their boundary and such that the union of all a_i cover $\mathbb R^n$ (i.e. a tiling of $\mathbb R^n$ by a countable set of tiles).

Let A be the statement that for all unions $T_k$ of any collection of $a_i$'s, there is no $a_j$ such that $a_j$'s intersection with $T_k$ is the completey boundary of $T_k$.(i.e. no tile completely enclose another set of tiles).

Is A sufficient for the existence of an ordering (a sequence) of the $a_i$'s such that each $a_i$ in S appear exactly once, and for any two consecutive $a_i$,$a_j$ in the sequence they share a part of their boundary?

Does such a sequence exist if all the tiles are equal upto rotations and translations?

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I am not sure that this question fits [graph-theory], I'm not sure that it doesn't either too. – Asaf Karagila Jun 1 '11 at 5:52

I don't think so. If $a$ has only two neighbors, $x$ and $y$, then the ordering must contain $x,a,y$ or $y,a,x$ or start with $a$. Now it's easy to get three (or more) sets $a,b,c$ such that each has only the two neighbors $x,y$ and now no ordering can have all 5 sets.

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I think you need four such sets? Otherwise you'd have the ordering $axbyc$. (As you say, it's also easy to get four, by considering isolated chunks on the boundary between $x$ and $y$.) – joriki Jun 1 '11 at 6:10
@joriki, sure, you can do $axbyc$, but you haven't done all of ${\bf R}^n$ that way (the sets are all bounded), and you can't leave $c$ to go anywhere else. – Gerry Myerson Jun 1 '11 at 6:15
I see, sorry, I forgot about the condition that they cover $\mathbb R^n$. – joriki Jun 1 '11 at 6:18

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