A topology question?

Question

I have tried to get some help w/ this problem from other people
but no one has helped me out yet.

Suppose that for each irrational p, an equilateral triangle T_p
(including its interior) is constructed in the plane with one vertex
at (p,0) and its opposite side above and parallel to the x-axis.
Prove that Union{T_p:p in P} (P: irrationals) must contain a rectangle of the form [a,b] x (0,1/k] for some a, b in R and
some k in N.
In the hypothesis, we could weaken “equilateral” to read “ ... ” ?
(Hint: Consider N_k = { p in P : T_p has height >= 1/k })


Thank you.

Answer 1

The above hint is not enough - take a monotone sequence of irrationals (p_n) with attached triangles having fixed height, you'll see that above every point (p_n+p_(n+1))/2 there's a point *not* included in the union of triangles.

Honestly I've been trying to prove this, and the more I think of it, the more I'm unsure it's true. The hint you (asker) gave tells you that there's a k such that there exists an uncountable set of irrationals p such that T_p has height >=1/k, right. An uncountable set in R (real numbers) has an accumulation point, but the set of its accumulation points can have empty interior, so I don't see how you'll get your interval [a,b] (think of Cantor sets, the usual counterexamples related with uncountable sets in the topology of R). So I don't see how that should help. I'm even beginning to wonder if counterexamples can be constructed like this ?

Answer 2

Any point, such that the point belongs to the rectangle, must belong to at least one triangle T_p. If so, then it must belong to the Union of all triangles.

Pick the four corners of the rectangle and show that each corner belongs to at least one triangle. Then show that each corner is a distinct point (otherwise you could have a degenerate rectangle = a single point (e.g., (0,0), being the apex of all T_p, belongs to the Union).