An upper bound for poset Ramsey number $2n \le R(Q_n,Q_n) \le n^2+2n$

Question

For each integer $n,m \ge 1$

$2n \le R(Q_n,Q_n) \le n^2+2n$.

Here $Q_n$ denotes a Boolean lattice of dimension $n$ and $R(P,P')$ or $R_{\dim_2}(P,P')$ is the smallest $N$ such that any red/blue coloring of $Q_N$ contains either red copy or $P$ or a blue copy of $Q$. (It is called a poset Ramsey number.)

For the lower bound on $R(Q_n,Q_n) $, with $Q_n$ consider $Q_{2n-1}$. Color the sets of size $0, . . .,n-1$ red and all other sets blue. Then there is no monochromatic chain with $n+1$ elements and there is no monochromatic copy of $Q_n$. it's easy to understand the lower bound and I am stuck to understand the upper bound.

How do you find an upper bound for $n^2+2n$? Any help will be appreciated. Thanks.

I found this result in the paper Maria Axenovich, Stefan Walzer: Boolean lattices: Ramsey properties and embeddings, https://arxiv.org/abs/1512.05565, http://dx.doi.org/10.1007/s11083-016-9399-7

The proof given in the paper is:

For the upper bound on $R(Q_n,Q_n)$, consider a red/blue coloring of $Q_{n^2+2n}$. Let the ground set be $X_0\cup X_1 \cup \dots \cup X_{n+1}$, where $X_i$'s s are pairwise disjoint and of size $n$ each. Consider families of sets $\mathcal B_Y$ for each $Y\subseteq X_0$ with $|Y|\ge1$ to be $\mathcal B_Y = \{Y\cup X_1\cup \dots \cup X_{|Y|} \cup X; X\subseteq X_{|Y|+1}\}$, let $\mathcal B_{\emptyset}=2^{X_1}$. We see that each $\mathcal B_Y$ is a copy of $Q_n$. If this copy is blue, then $\mathcal B_Y$ gives a monochromatic copy of $Q_n$. Otherwise, there is a red element in each $\mathcal B_Y$. This element is $Z_Y=Y\cup X_1 \cup \dots X_{|Y|}\cup S_Y$, where $S_Y\subseteq X_{|Y|+1}$. We claim that these elements form a red copy of $Q_n$. Indeed, we see for $Y, Y' \subseteq [n]$ that $Y\subseteq Y'$ iff $Z_Y\subseteq Z_{Y'}$.

Answer 1

I am not sure how much this is going to help - since this is precisely the proof from this version of the paper linked in the question; I have just added comments to some steps. (But if nothing else, this might help to identify what are the steps which the OP does not understand.)

Notice that in the version actually published, the authors omitted this part, because it is a consequence of more general $$n+m \le R(Q_n,Q_m) \le mn+n+m.$$

For the upper bound on $R(Q_n,Q_n)$, consider a red/blue coloring of $Q_{n^2+2n}$.

That's precisely what we want to do - to show that such coloring of $Q_{n^2+2n}$ contains either red $Q_n$ or blue $Q_n$.

Let the ground set be $X_0\cup X_1 \cup \dots \cup X_{n+1}$, where $X_i$'s s are pairwise disjoint and of size $n$ each.

Certainly, we can divide the set with $n^2+n=n(n+1)$ elements into $(n+1)$ sets with $n$ elements each.

Consider families of sets $\mathcal B_Y$ for each $Y\subseteq X_0$ with $|Y|\ge1$ to be $$\mathcal B_Y = \{Y\cup X_1\cup \dots \cup X_{|Y|} \cup X; X\subseteq X_{|Y|+1}\},$$ let $\mathcal B_{\emptyset}=2^{X_1}$.

We have defined $B_Y$ for each $Y\subseteq X_0$.

We see that each $\mathcal B_Y$ is a copy of $Q_n$. If this copy is blue, then $\mathcal B_Y$ gives a monochromatic copy of $Q_n$.

Indeed, if we fix some $Y\subseteq X_0$,then the only part which changes is $X$. So we have monotone bijection $$Y\cup X_1\cup \dots \cup X_{|Y|} \cup X\mapsto X$$ between $B_Y$ and the sets from $\mathcal P(X_{|Y|+1})$.

So this is (order)-isomorphic to $Q_n$.

Otherwise, there is a red element in each $\mathcal B_Y$. This element is $Z_Y=Y\cup X_1 \cup \dots X_{|Y|}\cup S_Y$, where $S_Y\subseteq X_{|Y|+1}$.

If neither of the sets $\mathcal B_Y$ is blue, at least one element in each $\mathcal B_Y$ is colored red. We chose such element from $\mathcal B_Y$ and denoted it $Z_Y$.

We claim that these elements form a red copy of $Q_n$. Indeed, we see for $Y, Y' \subseteq [n]$ that $Y\subseteq Y'$ iff $Z_Y\subseteq Z_{Y'}$.

If the above is true then $$Y\mapsto Z_Y$$ is an order isomorphism between $\mathcal P(X_0)$ and $\{Z_Y; Y\in X_0\}$.

To see that this maps indeed preserves order, it suffices to see that if $Y\subseteq Y'$ then either $|Y|<|Y'|$ in which case $$Z_Y = Y\cup X_1\cup \dots \cup X_{|Y|} \cup S_Y \subseteq Y\cup X_1\cup \dots \cup X_{|Y'|} \subseteq Z_{Y'}$$ or $|Y|=|Y'|$, which clearly implies $Y=Y'$ and $Z_Y=Z_{Y'}$.