# 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)?

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This is equivalent to CH.

Quoting "Problems and Theorems in Classical Set Theory" by Komjath and Totik, chapter 16, Continuum hypothesis:

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.