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. 



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



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> 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 $m\le \aleph\_1$.
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