Suppose we 2-colour $K_{2s-1}$ such that no vertex has more than one blue edge incident to it, prove that the graph contains a red $K_s$. I've never seen a Ramsey theory question like this and am struggling to understand how to begin although it seems rather intuitive
I can see that we can draw no more than $(s-1)$ blue edges where at least one vertex cannot have a blue edge incident to it We are left with at least $(s-1)(2s-1) - (s-1) = 2(s-1)^2$ red edges
I've formulated a rough answer but not sure if it suffices.
However we colour blue, we are left with at least one vertex with no blue edges incident
Considering the situation with the maximal number of blue edges then if we show we must have a red $K_s$ then it certainly holds for less blue edges.
Suppose without loss of generality we label the vertices $1,2, ...,2s-2,2s-1$ such that the $(s-1)$ blue edges are $(1,2), (3,4), ..., (2s-3, 2s-2)$ so that $2s-1$ has no blue edges then all edges between vertices $1,3,5,...,2s-1$ are red since they are not blue, forming a red $K_s$
Is my 'without loss of generality' justified?