T shaped tetris figures on a plane

Question

I am just wondering how many (and by how many I mean countably or uncountably many) T shaped figures can we place on a XY plane. I assume that that T consists of 2 perpendicular lines and has 0 area.

It is easy to understand for L shaped figures or U shaped ones, one can place uncountably many of those. But I am unable to construct a similar bijection for T shaped ones. I believe that it should be countable and some sort of diagonal argument should be made but I don't know how.

A reasonable followup is for the X shaped ones. How many of those can we fit on a plane?

Answer 1

It turns out it is impossible to have uncountably many disjoint Ts 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 Ts in the plane, define a map from these Ts 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 Ts. 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 Ts are the houses.

Since there is an at most two-to-one map from the set of Ts to a countable set, the set of Ts 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.