It turns out it is impossible to have uncountably many disjoint T
s in the plane. I found this solution in Mathematical Puzzles: A Connoisseur's Collection by Peter Winkler.
Let $S$ be the set of ordered triples of disjoint disks in the plane whose centers have rational coordinates and whose radii are rational. Note $S$ is countable.
Given any arrangement of disjoint T
s in the plane, define a map from these T
s to $S$ by choosing, for each T
, a triple in $S$ whose disks each contain one of the three endpoints of the T
, so that each disk only touches one of the three line segments comprising the T
.
I claim this map is at most two-to-one. If not, then there would be a triple $(D_1,D_2,D_3)$ which was joined together by three non-overlapping T
s. But this would be a solution to the water gas electricity problem in the plane, which is impossible. Specifically, the disks are the utilities, and the intersection points of the T
s are the houses.
Since there is an at most two-to-one map from the set of T
s to a countable set, the set of T
s must be countable.
Notes:
Perhaps there is a simpler solution; the one I just gave generalizes to prove that you cannot fit uncountably many disjoint sets in the plane which are homeomorphic to the shape T
.
The axiom of choice is not required to construct this map; $S$ can be well-ordered, so give each T
the smallest legal triple in the well-ordering.