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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?

user2566092
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  • What about the width of the segments, is it zero or there is some thickness to them? And their lengths anything? – Georgy Apr 22 '14 at 00:07
  • @Georgy The segments have zero width, i.e. they are just intervals of ${\mathbb R}$ that have each been transformed in the plane by rotation and/or translation. The segment lengths can be arbitrarily small but they are always positive (otherwise the countability claim would be false). – user2566092 Apr 22 '14 at 16:43

3 Answers3

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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.

Mars
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Hint: if you have an uncountable collection, it contains a sequence with distinct centers in a compact subset and lengths of horizontal and vertical intervals bounded away from zero. Therefore....

Moishe Kohan
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Let $\mathscr{A}$ be a collection of disjoint letter $T$'s in the plane (I'm going to call the "second segment of $T$" the segment that has an endpoint in the other, called the "first segment"). For every $n$, let $\mathscr{A}_{n}$ be the set consisting of all $T\in\mathscr{A}$ satisfying:

  1. Both segments of $T$ have length $\geq 1/n$ and it divides the horizontal

  2. The second segment divides the first one in subsegments of length $\geq 1/n$

Let's show that each $\mathscr{A}_n$ is at most countable: Let $n$ be fixed, and suppose that $\mathscr{A}_n$ were not countable. for every $T\in\mathscr{A}_n$, let $c_T$ be the point of intersection of the segments of $T$. Then $\left\{c_T:T\in\mathscr{A}_n\right\}$ is uncountable, hence it has some accumulation point. Since $n$ is fixed, we can take 3 elements $T_1,T_2,T_3\in\mathscr{A}_n$ such that $c_{T_1}$, $c_{T_2}$ and $c_{T_3}$ are sufficiently close for the first segments of $T_1$, $T_2$ and $T_3$ to be "almost parallel". Say that the first segment of $T_1$ is between the first segments of $T_2$ and $T_3$. Also, we can take the $c_{T_i}$ sufficiently close (which will depend only on $n$), so that the second segment of $T_1$ will necessarily intersect the first segment of $T_2$ or $T_3$, contradicting the fact that $\mathscr{A}_n$ consist of disjoint elements.

Hence, each $\mathscr{A}_n$ is at most countable, so $\mathscr{A}=\bigcup_{n=1}^\infty\mathscr{A}_n$ is at most countable.

Luiz Cordeiro
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