ramsey theory - monochromatic cycle-free colouring of the complete graph on R?

Let me add to the above answer by noting that a stronger result holds and that, in the absence of the Continuum Hypothesis, the problems with the attempted construction of a 4-cycle-free coloring are unavoidable.

To fix notation, if $X$ is a set, $[X]^2$ is the set of 2-element subsets of $X$. We show that any coloring $c:[omega_2]^2 rightarrow omega$ has a monochromatic 4-cycle. This will also give a shorter proof of the negative direction of the Erdos-Kakutani result mentioned above.

Fix $c:[omega_2]^2 rightarrow omega$. For each $omega_1 leq gamma < omega_2$, find $n_gamma < omega$ such that $A^gamma_{n_gamma} := {alpha < gamma mid c({alpha, gamma}) = n_gamma}$ is uncountable. Find a stationary $S subseteq omega_2$ and an $n^* < omega$ such that, for all $gamma in S$, $n_gamma = n^*$. For each $gamma in S$, let $alpha_gamma = min(A^gamma_{n^*})$, and let $beta_gamma = min(A^gamma_{n^*} setminus (alpha_gamma + 1))$. By Fodor's Lemma, we can find $alpha^* < beta^* < omega_2$ and a stationary $T subseteq S$ such that, for all $gamma in T$, $(alpha_gamma, beta_gamma) = (alpha^*, beta^*)$. Fix $gamma < delta$, both in $T$. Then $c({alpha^*, gamma}) = c({gamma, beta^*}) = c({beta^*, delta}) = c({delta, alpha^*}) = n^*$, giving us a monochromatic 4-cycle.