A set in Hilbert space where all triangles are isosceles

Question

In one of P. Erdős papers ("My Scottish Book Problems") he mentioned a set of power $\mathfrak c$ in Hilbert space where all triangles are isosceles. What is the definition of this set?

Answer 1

Found the answer in P. Komjath "Set Theory:Geometric and Real":

The situation changes radically if we replace $R^n$ with the Hilbert space $l^\infty$ of infinite real vectors $(x_0, x_l,...)$ with $\Sigma x^2_i$; finite. An observation due to Erdos, Kakutani, Oxtoby, L.M.Kelly, Nordhaus, and possibly many others is that in this case there are continuum many points with pairwise rational distance.

As it is easy not to find a proof I sketch one. Work in the Hilbert space where an orthonormal basis is $\{b_s\}$ where s can be any finite 0-1 sequence. To every infinite 0-1 sequence z associate $a(z) = \Sigma \lambda_n b_{z|n}$ where z|n denotes the string of the first n terms of z and $\lambda_n = \sqrt{3} \times 2^{-(n+l)}$. If $z \neq z'$ first differ at the (n + 1)-st position then $(a(z) - a(z'))^2 = \Sigma\{\lambda_i^2: i > n\} = 4^{-n}$ so the distance between a(z) and a(z') is $2^{-n}$ . It is easy to see that every triangle in this construction is isosceles. I don't know if there is a similar (or any) construction of continuum many points such that all three-element subsets form a triangle with nonzero rational area.