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)?



Finding monochromatic rectangles in a countable coloring of $R^{2}$
This is equivalent to CH.


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



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> 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 $\dist(a,b)=\dist(c,d)$
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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.




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Previous version, with added explanation about Hamel basis:


Using



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> 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.
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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.


