I have exercise to prove $R(\diamondsuit)=4$, I try to solve this exercise by this proof :
$\ge$ In $Q_3$ (Here $Q_3$ denotes a Boolean lattice of dimension $3$) we can colour the upper two layers red and the lower two layers blue to avoid a monochromatic $2-diamond poset \mathcal D_k$ ($\diamondsuit$) .
$\le$ Consider any colouring of $Q_4$. Asume without loss of generality that $\emptyset$ is red. We consider into two cases :
Case 1: 1234 is red. If there are two incomparable red elements, then together with $\emptyset$ and $1234$, they form a red $\diamondsuit$. If, on the other hand, the red elements are subset of a chain, without loss of generality {$\emptyset,1,12,123,1234$}, then everything else is blue and {$2,23,24,234$} forms a blue $\diamondsuit$.
Case 2: 1234 is blue. $Q_4$ contains a monochromatic $\mathcal N$ (poset $A\lt C, B\lt C, B\lt D$). Note that this $\mathcal N$ will not make use of $\emptyset$ or $123$, since $\mathcal N$ has neither maximum nor minimum. If the copy of $\mathcal N$ is blue, we restrict it to a blue $\mathcal V$ and add $1234$ to obtain a blue $\diamondsuit$. If it is red, we restrict it to a red and add $\emptyset$ to obtain a red $\diamondsuit$.
How about $R(\diamondsuit,\diamondsuit)= . . .$
Anyone can help to solve $R(\diamondsuit,\diamondsuit)= . . .$ or reference to solve this problem.
