TITLE: A tree of convex sets?
QUESTION [13 upvotes]: This was suggested by a problem on FreeNode's #math a little while ago...
Construct a directed graph $\Gamma$ with vertex set the set of compact convex sets in $\mathbb R^2$, and an arrow $A\to B$ if $A$ and $B$ are disjoint and there is a point $a$ in $A$ and a $t>0$ such that $a+(t,0)$ is in $B$.
I guess there are no oriented cycles in $\Gamma$, so it is a directed acyclic graph. Can you give a non-ugly proof?
Later: Rahul's example below shows that this is too much to hope for. In fact, what I really wanted is to know if the subgraph spanned by every finite set of disjoint convex sets has no cycles.

REPLY [2 votes]: Here is a simple proof that the directed graph induced by disjoint convex sets is acyclic.
Suppose there is a counter-example, then there must be one with the smallest number of convex sets involved. Suppose such a counter example is given by the convex sets $C_1,C_2,\dots,C_n$. Then the directed cycle must contain all the $C_i$'s by the minimality assumption (otherwise just throw the ones not in the cycle away to get a smaller counter-example). WLOG, assume the graph contains $C_1\to C_2\to \cdots \to C_n\to C_1$.
Lemma: There are no other edges in this graph.
Proof: If we had $A_k\to A_r$ with $k\neq r-1 \pmod{n}$, then $A_k\to A_r\to A_{r+1}\to \cdots \to A_k$ would be a shorter cycle which contradicts our minimality assumption.
So far we haven't used that these are convex sets. Now the trick is that if there is no $A_k$ which intersects the convex hull of $A_r \cup A_{r+1}$ then substituting $A_r$ and $A_{r+1}$ with their convex hull gives a smaller collection of convex sets with a cycle. 
Now by applying the lemma it follows that to show that no such $A_k$ can exist, it is enough to prove that if $A_k$ intersects the convex hull of $A_r \cup A_{r+1}$ then either $A_r \to A_k$ or $A_k \to A_{r+1}$, but this is obvious if you draw the picture, so I'll leave it as an exercise :)
Since there are no two convex sets $A\to B \to A$, we are done.