TITLE: Prove that there can be at most countably many disjoint letter T's in the plane
QUESTION [6 upvotes]: A letter T in the plane is defined as a non-zero length segment with an orthogonal non-zero length segment that has an end-point in the strict interior of the first segment. Prove that there can be at most countably many disjoint letter T's in the plane. I've tried clumsily to prove that I can find balls around each of the endpoints of a 2 segments defining a T, such that if any other letter T has segment endpoints contained in the balls then the two T's must intersect. This would be enough because for each ball we can choose a pure rational ball that has rational center and rational radius that is contained inside the ball. But I've had a hard time making this proof rigorous. Any help?

REPLY [2 votes]: Given any collection of T's define a function that takes each T to three rational coordinates obtained from the superimposed ice cream cone. say (p,q,r) where p and q are on distinct sides of the cone and r is in the ice cream. If two disjoint T's mapped to the same first two coordinates (match in the cone) which can happen then the r components are distinct. To see this is the case note there is a restrictive angle less than $\pi$. Hence the function is injective and therefore the collection of T's is countable.