Maybe you can help me with that. I was asking myself if you take an uncountable set $S$ and let $S^{(2)}$ be 2-coloured, must there exist an uncountable monochromatic set in $S$?
2026-03-26 20:36:27.1774557387
Uncountable monochromatic set
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Not necessarily. A standard counterexample is Sierpiński's coloring: Well-order the reals, and color $\{a,b\}$ blue if the usual ordering and the well-ordering coincide on $\{a,b\}$, and red otherwise. An uncountable monochromatic set would give us a subset of $\mathbb R$ of order type $\omega_1$ (or its reverse) in the usual ordering, and this is impossible. (Note that if $S^{(2)}$ is the collection of $2$-sized tuples, then a counterexample is trivially obtained by coloring the diagonal different from the rest.)
What is true is that if $S$ has size at least $|\mathbb R|^+$, then $2$-colorings of the collection $S^{(2)}$ of subsets of $S$ of size $2$ give us uncountable monochromatic sets, this is a particular case of a result of Erdős and Rado. The study of these extensions of Ramsey's theorem in called the partition calculus.