# Finding monochromatic rectangles in a countable coloring of $R^{2}$

Given a countable coloring of the plane, is it always possible to find a monochromatic set of points $\left\{ \left(x,y\right),\left(x+w,y\right),\left(x,y+h\right),\left(x+w,y+h\right)\right\}$ (the corners of a rectangle)?

This is equivalent to CH.

Quoting "Problems and Theorems in Classical Set Theory" by Komjath and Totik, chapter 16, Continuum hypothesis:$$\DeclareMathOperator{\dist}{dist}$$

CH holds if and only if the plane can be decomposed into countably many parts none containing 4 different points $$a$$, $$b$$, $$c$$, $$d$$ such that $$\dist(a,b)=\dist(c,d)$$

This is a stronger requirement than your problem, so assuming CH the answer is no. Their solution, assuming CH is false, proves that there's a monochromatic rectangle.

CH holds if and only if $$\mathbf R$$ can be colored by countably many colors such that the equation $$x+y=u+v$$ has no solution with different $$x$$, $$y$$, $$u$$, $$v$$ of the same color.
This gives a negative answer assuming CH. Explanation: consider $$\mathbf R$$ as a vector space over $$\mathbf Q$$. Let $$A$$ be some basis. Take any bijection $$A \to A + A$$, where $$+$$ is the disjoint sum. It induces a linear isomorphism $$f:\mathbf R \to \mathbf R \times \mathbf R$$. (You can think that there's a linear isomorphism between reals and complexes if that helps.) Then, if you were given a monochromatic rectangle $$a=(x_1, y_1)$$, $$b=(x_1+x_2, y_1)$$, $$c=(x_1, y_1+y_2)$$, $$d=(x_1+x_2, y_1+y_2)$$, certainly $$a+d=b+c$$. Using that isomorphism, $$f(a)+f(d)=f(b)+f(c)$$ gives a monochromatic solution of quoted equation.