monochromatic cycle-free colouring of the complete graph on R? Hi 
So there is an edge-colouring of a complete graph on R (the reals), with countably many colours that as no monochromatic triangle. To find it map R to (0,1) write the numbers in binary and if 2 numbers differ 1st in the kth digit use colour k. 
Now this colouring has cycles of length 4. (1/4, 3/4, 1/3, 2/3 for example). You can get rid of cycles of length 4 by considering the 1st 2  binarary digits in which 2 numbers differ (and of course seperate colours if they only differ in 1 digit). Anyway my question is can we avoid cycles completely? i.e. does there exist a colouring of the complete graph on R such that there is no monochromatic cycle. 
 A: 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. 
A: Turns out that the existence of such a coloring is equivalent to the continuum hypothesis. This was proved by Erdos and Kakutani in 1943 in the paper "On non-denumerable graphs". They prove: 

A complete graph of cardinal number $m$ (that is, the cardinal number of the vertices is $m$) can be split up into a countable number of trees if and only if $m\le \aleph_1$.

