Proof. Lemma 7 provides that there is a red copy of $Q_n$ in $Q$ if and only if there exists a partition $[N]={\mathcal{X}}\cup{\mathcal{Y}}$ of the ground set of $Q$ with $|{\mathcal{X}}|=n$ and $|{\mathcal{Y}}|=k$ as well as an $\mathcal{X}$ -good embedding $\phi$ of $Q({\mathcal{X}})$ into $Q$ with a monochromatic red image.

If there is a red copy of $Q_n$ in $Q$ , then for ${\mathcal{X}},{\mathcal{Y}}$ from Lemma 7 there is also an $\mathcal{X}$ -good copy of ${\mathcal{Q}}({\mathcal{X}})$ . Thus by Theorem 12 there is no blue $\mathcal{Y}$ -shrub.

On the other hand, if there is no red copy of $Q_n$ in $Q$ , there is no red $\mathcal{X}$ -good copy of ${\mathcal{Q}}({\mathcal{X}})$ for any ${\mathcal{X}} \in \binom{[N]}{n}$ . Then for an arbitrary $n$ -element subset $\mathcal{X}$ of $[N]$ , let ${\mathcal{Y}}= [N]\setminus{\mathcal{X}}$ . Now Theorem 12 implies that there exists a blue $\mathcal{Y}$ -shrub. In particular, there is a blue $\mathcal{Y}$ -shrub for any $k$ -element subset $\mathcal{Y}$ of $[N]$ .

Observation 14. $(\varnothing,\varnothing )$ is not embeddable if and only if there is no embedding $\phi \,:\,{\mathcal{Q}}({\mathcal{X}})\to Q$ such that for every $X^{\prime}\subseteq{\mathcal{X}}$ , $\phi (X^{\prime})$ red and $\phi (X^{\prime})\cap{\mathcal{X}}=X^{\prime}$ .

Proof. First suppose that $(X,Y)$ is embeddable. Let $\phi$ be an embedding of ${\mathcal{Q}}({\mathcal{X}})\cap \{X^{\prime}\subseteq{\mathcal{X}}\,:\, X^{\prime}\supseteq X\}$ into $Q$ witnessing that $(X,Y)$ is embeddable.

If $(X,Y)$ is blue, then $\phi (X)\supset (X,Y)$ because $\phi$ has a monochromatic red image. Thus there exists $Y^{\prime}\subseteq{\mathcal{Y}}$ with $Y^{\prime}\supset Y$ such that $\phi (X)=(X,Y^{\prime})$ . But then $\phi$ also witnesses that $(X,Y^{\prime})$ is embeddable, so Condition (i) is fulfilled.

If $(X,Y)$ is red, pick some arbitrary $X^*\subseteq{\mathcal{X}}$ such that $X^*\supset X$ . Then the function $\phi ^*\,:\,{\mathcal{Q}}({\mathcal{X}})\cap \{X^{\prime}\subseteq{\mathcal{X}}\,:\, X^{\prime}\supseteq X^*\}\to Q$ , $\phi ^*(X^{\prime})=\phi (X^{\prime})$ is a restriction of $\phi$ and therefore an embedding with a monochromatic red image such that $\phi ^*(X^{\prime})\cap{\mathcal{X}}=X^{\prime}$ for all $X^{\prime}$ and $\phi ^*(X^*)\supseteq (X^*,Y)$ . Thus for every $X^*\subseteq{\mathcal{X}}$ with $X^*\supset X$ , the vertex $(X^*,Y)$ is embeddable, that is Condition (ii) is fulfilled.

Now, suppose that Condition (i) or Condition (ii) hold. If (i) holds, then $(X,Y)$ is blue and there is some $Y^{\prime}\supset Y$ such that $(X,Y^{\prime})$ is embeddable. Then the embedding witnessing that also verifies that $(X,Y)$ is embeddable.

For the rest of the proof we assume that (ii) holds, that is that $(X,Y)$ is red and for any $X^{\prime}\subseteq{\mathcal{X}}$ with $X^{\prime}\supset X$ the vertex $(X^{\prime},Y)$ is embeddable. We define the required embedding $\phi \,:\,{\mathcal{Q}}({\mathcal{X}})\cap \{X^{\prime}\subseteq{\mathcal{X}}\,:\, X^{\prime}\supseteq X\}\to Q$ for every $X^{\prime}$ with $X\subseteq X^{\prime}\subseteq{\mathcal{X}}$ depending on the number of minimal $X^*$ ’s, $X\subseteq X^*\subseteq X^{\prime}$ such that $(X^*,Y)$ is blue as follows. Let $X^{\prime}$ with $X\subseteq X^{\prime}\subseteq{\mathcal{X}}$ be arbitrary.

1. (1) If for all $X^*$ with $X\subseteq X^*\subseteq X^{\prime}$ , the vertex $(X^*,Y)$ is red, let $\phi (X^{\prime})=(X^{\prime},Y)$ . Note that this case includes $X^{\prime}=X$ .

2. (2) If there is a unique minimal $X^*$ such that $X\subseteq X^*\subseteq X^{\prime}$ and $(X^*,Y)$ is blue, then $(X^*,Y)$ is embeddable by Condition (ii). Let $\phi _{X^*}$ be an embedding witnessing that. Then set $\phi (X^{\prime})=\phi _{X^*}(X^{\prime})$ .

3. (3) Otherwise, let $\phi (X^{\prime})=(X^{\prime},{\mathcal{Y}})$ .

Cases (1)-(3) determine a partition of the set $\{X^{\prime}\subseteq{\mathcal{X}}\,:\, X^{\prime}\supseteq X\}$ into three pairwise disjoint parts. Let ${\mathcal{M}}_j$ , $j\in [3]$ , be the set of those vertices $X^{\prime}$ for which $\phi$ was assigned in Case (j). Note that ${\mathcal{M}}_1\cup{\mathcal{M}}_2\cup{\mathcal{M}}_3=\{X^{\prime}\subseteq{\mathcal{X}}\,:\, X^{\prime}\supseteq X\}$ .

Claim 1. The function $\phi$ witnesses that $(X,Y)$ is embeddable.

• Clearly, for every $X^{\prime}\subseteq{\mathcal{X}}$ with $X^{\prime}\supseteq X$ , $\phi (X^{\prime})\cap{\mathcal{X}}=X^{\prime}$ .

• $(X,Y)$ is red, so $X\in{\mathcal{M}}_1$ . Thus $\phi (X)=(X,Y)$ .

• The argument verifying that $\phi (X^{\prime})$ is red for every $X^{\prime}\subseteq{\mathcal{X}}$ with $X^{\prime}\supseteq X$ depends on $i$ such that $X^{\prime}\in{\mathcal{M}}_i$ . If $X^{\prime}\in{\mathcal{M}}_1$ , it is immediate that $\phi (X^{\prime})$ is red. If $X^{\prime}\in{\mathcal{M}}_2$ , $\phi _{X^*}$ has a monochromatic red image, thus $\phi (X^{\prime})=\phi _{X^*}(X^{\prime})$ is also red. Now consider the case that $X^{\prime}\in{\mathcal{M}}_3$ , that is there are $X_1,X_2\subseteq{\mathcal{X}}$ with $X_1\neq X_2$ and $X\subseteq X_i\subseteq X^{\prime}$ , $i\in [2]$ , such that $(X_i,Y)$ are blue and $X_i$ are both minimal with this property. The latter condition implies that $X_1$ and $X_2$ are incomparable, in particular $(X_1,Y)$ and $(X_2,Y)$ are incomparable as well. Moreover, observe that $X_i\neq X^{\prime}$ , $i\in [2]$ , because $X^{\prime}$ is by definition comparable to both $X_1$ and $X_2$ . Now assume for a contradiction that $\phi (X^{\prime})=(X^{\prime},{\mathcal{Y}})$ is blue. Recall that $(X_1,Y)$ and $(X_2,Y)$ are blue. We know that $X_i\subset X^{\prime}$ and $Y\subseteq{\mathcal{Y}}$ , thus $(X_i,Y)\subset (X^{\prime},{\mathcal{Y}})$ . As a consequence, $(X_1,Y)$ , $(X_2,Y)$ , and $(X^{\prime},{\mathcal{Y}})$ induce a blue copy of $\Lambda$ in $Q$ , which is a contradiction. Thus $\phi$ has a monochromatic red image.

• It remains to show that $\phi$ is an embedding. Let $X_1,X_2\subseteq{\mathcal{X}}$ be arbitrary with $X\subseteq X_i\subseteq X^{\prime}$ , $i\in [2]$ . We shall show that $X_1\subseteq X_2$ if and only if $\phi (X_1)\subseteq \phi (X_2)$ . One direction is easy to prove: If $\phi (X_1)\subseteq \phi (X_2)$ , then $X_1=\phi (X_1)\cap{\mathcal{X}}\subseteq \phi (X_2)\cap{\mathcal{X}}=X_2$ . Now suppose that $X_1\subseteq X_2$ . Let $Y_1,Y_2\subseteq{\mathcal{Y}}$ such that $\phi (X_1)=(X_1,Y_1)$ and $\phi (X_2)=(X_2,Y_2)$ . Then we shall show that $Y_1\subseteq Y_2$ . Note that $Y\subseteq Y_i \subseteq{\mathcal{Y}}$ for $i\in [2]$ . Assume that at least one of $X_1$ or $X_2$ is in ${\mathcal{M}}_1\cup{\mathcal{M}}_3$ . If $X_1\in{\mathcal{M}}_1$ , then $Y_1=Y$ and we are done as $Y\subseteq Y_2$ . Furthermore, if $X_2\in{\mathcal{M}}_3$ , then $Y_2={\mathcal{Y}}$ and we are done as well since $Y_1\subseteq{\mathcal{Y}}$ . If $X_1\in{\mathcal{M}}_3$ , then $X_1\subseteq X_2$ implies that $X_2$ is also in ${\mathcal{M}}_3$ . Conversely, if $X_2\in{\mathcal{M}}_1$ , the fact that $X_2\supseteq X_1$ yields that $X_1\in{\mathcal{M}}_1$ and we are done as before.

As a final step, suppose that $X_1,X_2\in{\mathcal{M}}_2$ . This implies that for each $i\in [2]$ , there is a unique minimal vertex $X^*_i$ such that $X\subseteq X^*_i\subseteq X_i$ and $(X^*_i,Y)$ is blue. Applying the initial assumption, $X^*_1\subseteq X_1\subseteq X_2$ . By minimality of $X^*_2$ , we obtain that $X^*_2\subseteq X^*_1$ . Now this provides that $X^*_2\subseteq X^*_1\subseteq X_1$ . Using the minimality of $X^*_1$ , we see that $X^*_1\subseteq X^*_2$ , so $X^*_1=X^*_2$ .

Recall that $(X^*_1,Y)=(X^*_2,Y)$ is embeddable since $X_1,X_2\in{\mathcal{M}}_2$ . Consider the function $\phi _{X^*_1}=\phi _{X^*_2}$ witnessing that. Because $\phi _{X^*_1}$ is an embedding and $X^*_1\subseteq X_1\subseteq X_2$ , we obtain $\phi _{X^*_1}(X_1)\subseteq \phi _{X^*_1}(X_2)$ . Combining the given conditions,

\begin{equation*}\phi (X_1)=\phi _{X^*_1}(X_1)\subseteq \phi _{X^*_1}(X_2)=\phi _{X^*_2}(X_2)=\phi (X_2),\end{equation*}

that implies that $Y_1\subseteq Y_2$ .

This concludes the proof of the Claim and the Lemma.

Proof. If $(X,\underline{S})$ is blue, we are done. Otherwise Lemma 15 yields an $X_1\subseteq{\mathcal{X}}$ , $X_1\supset X$ such that $(X_1,\underline{S})$ is not embeddable. By repeating this argument, we find an $X^{\prime}\subseteq{\mathcal{X}}$ , $X^{\prime}\supseteq X$ , with $(X^{\prime},\underline{S})$ is blue and not embeddable.

Proof. We construct $\tau \,:\,{\mathcal{O}}({\mathcal{Y}})\to Q$ iteratively and increasingly with respect to the order of ${\mathcal{O}}({\mathcal{Y}})$ . Suppose that $(\varnothing,\varnothing )$ is not embeddable. By Corollary 16 there is some $X_\varnothing \subseteq{\mathcal{X}}$ such that $(X_\varnothing,\varnothing )$ is blue and not embeddable. Let $\tau (\varnothing )=(X_\varnothing,\varnothing )$ . From here, we continue iteratively. Suppose that for $S\in{\mathcal{O}}({\mathcal{Y}})$ , $\underline{S}\neq{\mathcal{Y}}$ , we have defined $X_S\subseteq{\mathcal{X}}$ such that

1. (1) $X_S\supseteq X_T$ for every $T\le _{\mathcal{O}} S$ and

2. (2) $\tau (S)=(X_S,\underline{S})$ is blue and not embeddable.

Consider an arbitrary $S^{\prime}\in{\mathcal{O}}({\mathcal{Y}})$ such that $S\lt _{\mathcal{O}} S^{\prime}$ and $|S^{\prime}|=|S|+1$ . By Lemma 15 applied for $X_S$ and $\underline{S}$ , we obtain that $(X_S,\underline{S^{\prime}})$ is not embeddable. Then Corollary 16 yields that there is some $X_{S^{\prime}}\subseteq{\mathcal{X}}$ , $X_{S^{\prime}}\supseteq X_S$ , such that $(X_{S^{\prime}},\underline{S^{\prime}})$ is blue and not embeddable. Observe that for $T\in{\mathcal{O}}({\mathcal{Y}})$ with $T\le _{\mathcal{O}} S^{\prime}$ , either $T=S^{\prime}$ and so $X_T=X_{S^{\prime}}$ , or $T\le _{\mathcal{O}} S$ and so by $(1)$ $X_T\subseteq X_S\subseteq X_{S^{\prime}}$ . Let $\tau (S^{\prime})=(X_{S^{\prime}},\underline{S^{\prime}})$ .

Using this procedure, we define $\tau$ for all $S\in{\mathcal{O}}({\mathcal{Y}})$ . Let $\mathcal{P}$ be the subposet of $Q$ induced by the image of $\tau$ . We shall show that $\mathcal{P}$ is a weak $\mathcal{Y}$ -shrub witnessed by the function $\tau$ . By (2), for every $S\in{\mathcal{O}}({\mathcal{Y}})$ , $\tau (S)$ is blue and $\tau (S)\cap{\mathcal{Y}}=\underline{S}$ .

Let $S,T\in{\mathcal{O}}({\mathcal{Y}})$ with $S\lt _{\mathcal{O}} T$ . Let $X_S,X_T\subseteq{\mathcal{X}}$ such that $\tau (S)=(X_S,\underline{S})$ and $\tau (T)=(X_T,\underline{T})$ . Clearly, $\underline{S}\subset \underline{T}$ . Moreover, item $(1)$ implies that $X_S\subseteq X_T$ . Consequently, $\tau (S)\subset \tau (T)$ .

Proof of Theorem 12. Let $\mathcal{X}$ and $\mathcal{Y}$ be disjoint sets. Let $Q=Q({\mathcal{X}}\cup{\mathcal{Y}})$ be a blue/red coloured Boolean lattice which contains no blue copy of $\Lambda$ .

First suppose that there is no red $\mathcal{X}$ -good copy of ${\mathcal{Q}}({\mathcal{X}})$ . By Observation 14, $(\varnothing,\varnothing )$ is not embeddable and Lemma 17 provides that there is a blue weak $\mathcal{Y}$ -shrub in $Q$ . Using Proposition 10 we obtain a blue $\mathcal{Y}$ -shrub in $Q$ . This shows that there is either a red $\mathcal{X}$ -good copy of ${\mathcal{Q}}({\mathcal{X}})$ or a blue $\mathcal{Y}$ -shrub.

Next we show that both events could not happen simultaneously. Let $n=|{\mathcal{X}}|$ , $k=|{\mathcal{Y}}|$ and $N=n+k$ . Assume that there exist both an $\mathcal{X}$ -good embedding $\phi \,:\,{\mathcal{Q}}({\mathcal{X}}) \to Q$ with monochromatic red image as well as a $\mathcal{Y}$ -good embedding $\tau \,:\,{\mathcal{O}}({\mathcal{Y}}) \to Q$ with a monochromatic blue image.

We apply an iterative argument in order to find a contradiction. Let $Y_0=\varnothing$ and let $S_0=(Y_0,{\le})$ be the empty ordered set. Now let $X_1\subseteq{\mathcal{X}}$ such that $\tau (S_0)=(X_1,\underline{S_0})$ and let $Y_1\in{\mathcal{Y}}$ such that $\phi (X_1)=(X_1,Y_1)$ . Since $\phi (X_1)$ is red but $\tau (S_0)$ is blue, we know that $\phi (X_1)\neq \tau (S_0)$ and thus $Y_1\neq \underline{S_0}=\varnothing$ , so $|Y_1|\ge 1$ .

Now say that we already defined $X_1,\ldots,X_i$ , $Y_0,\ldots,Y_i$ , $S_0,\ldots,S_{i-1}$ for some $i\in [k]$ such that

• $S_{i-1}\in{\mathcal{O}}$ and $\underline{S_{i-1}}=Y_{i-1}$ ,

• $\tau (S_{i-1})=(X_i,\underline{S_{i-1}})$ ,

• $\phi (X_i)=(X_i,Y_i)$ , and

• $Y_{i-1}\subset Y_i\subseteq{\mathcal{Y}}$ and $|Y_i|\ge i$ .

Fix any ordering $S_i$ of $Y_i$ such that $S_{i-1}\lt _{\mathcal{O}} S_i$ . Such $S_i$ exists because $\underline{S_{i-1}}=Y_{i-1}\subset Y_i$ .

Then let $X_{i+1}$ be such that $\tau (S_i)=(X_{i+1},\underline{S_i})$ . Since $S_{i-1}\lt _{\mathcal{O}} S_i$ and $\tau$ is an embedding, $(X_{i},\underline{S_{i-1}})=\tau (S_{i-1})\subseteq \tau (S_i)=(X_{i+1},\underline{S_i})$ , therefore $X_i\subseteq X_{i+1}$ . Note that $\phi (X_i)=(X_i,\underline{S_i})$ is coloured red but $\tau (S_i)=(X_{i+1},\underline{S_i})$ is blue. Therefore $X_{i}\neq X_{i+1}$ , consequently $X_i\subset X_{i+1}$ and in particular $\phi (X_i)\subset \phi (X_{i+1})$ because $\phi$ is an embedding.

Next let $Y_{i+1}\subseteq{\mathcal{Y}}$ such that $\phi (X_{i+1})=(X_{i+1},Y_{i+1})$ . Then $Y_{i+1}\supseteq Y_i$ and furthermore, because $(X_{i+1},Y_i)$ is blue but $\phi (X_{i+1})$ is red, $Y_{i+1}\neq Y_i$ . Consequently $Y_{i+1}\supset Y_i$ , and in particular $|Y_{i+1}|\ge |Y_i|+1\ge i+1$ .

Iteratively, we obtain $Y_{k+1}\subseteq{\mathcal{Y}}$ with $|Y_i|\ge k+1$ , a contradiction to $|{\mathcal{Y}}|=k$ .

Proof of Theorem 18. Let $\alpha = 21.6$ and $\beta =0.134$ . Let $N\in{\mathbb{N}}$ and $k=\frac{1}{\alpha }\frac{N}{\ln (N)}$ , let $Q={\mathcal{Q}}([N])$ .

The idea of the proof is to construct a $\mathcal{Y}$ -shrub, denoted ${\mathcal{P}}_{{\mathcal{Y}}}$ , for every ${\mathcal{Y}}\in \binom{[N]}{k}$ , with an additional property so that the selected shrubs are independent. Since each shrub does not contain a copy of $\Lambda$ , it follows that the independent union of all the ${\mathcal{P}}_{{\mathcal{Y}}}$ ’s also does not contain a copy of $\Lambda$ . We obtain these shrubs by randomly choosing a $\mathcal{Y}$ -framework for every ${\mathcal{Y}}\in \binom{[N]}{k}$ and then constructing a $\mathcal{Y}$ -shrub based on each of them. Afterwards we define a colouring where every vertex in each constructed shrub is coloured blue and the remaining vertices red.

A $\mathcal{Y}$ -framework of ${\mathcal{Y}}\in \binom{[N]}{k}$ is a $4$ -tuple $({\mathcal{Y}},A_{\mathcal{Y}},Z_{\mathcal{Y}}, X_{\mathcal{Y}})$ such that

• $\mathcal{Y}$ , $A_{\mathcal{Y}}$ $Z_{\mathcal{Y}}$ are pairwise disjoint and ${\mathcal{Y}}\cup A_{\mathcal{Y}}\cup Z_{\mathcal{Y}}=[N]$ ,

• $|A_{\mathcal{Y}}|=\tfrac{3}{2}k\ln k-k$ ,

• $X_{\mathcal{Y}}\subseteq Z_{\mathcal{Y}}$ .

A $\mathcal{Y}$ -framework is random if

• $A_{\mathcal{Y}}\in \binom{[N]\backslash{\mathcal{Y}}}{\tfrac{3}{2}k\ln k-k}$ is chosen uniformly at random, and

• each element of $Z_{\mathcal{Y}}=[N]\backslash ({\mathcal{Y}}\cup A_{\mathcal{Y}})$ is included in $X_{\mathcal{Y}}$ independently at random with probability $\frac{1}{2}$ .

Now draw a random $\mathcal{Y}$ -framework for every ${\mathcal{Y}}\in \binom{[N]}{k}$ . Observe that by choice of $k$ , $k\ln k =\frac{N}{\alpha }\cdot \frac{\ln (N)-\ln (\alpha )-\ln \ln (N)}{\ln (N)}$ , so $\frac{20N}{21\alpha } \le k\ln k\le \frac{N}{\alpha }$ . Since $|Z_{\mathcal{Y}}|= N-\tfrac{3}{2}k\ln k$ , we have $\left(1-\frac{3}{2\alpha }\right)N\le |Z_{\mathcal{Y}}|\le \left(1-\frac{10}{7\alpha }\right)N$ .

Claim 2. W.h.p. for every ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ with ${\mathcal{Y}}_1\neq{\mathcal{Y}}_2$ , $|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|\ge \beta N$ .

Consider some arbitrary ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ , ${\mathcal{Y}}_1\neq{\mathcal{Y}}_2$ . Observe that $\left(1-\frac{3}{\alpha }\right)N\le |Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|\le |Z_{\mathcal{Y}}|\le \left(1-\frac{10}{7\alpha }\right)N$ . In a random $\mathcal{Y}$ -framework, each element of $Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}$ is contained in $X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}$ independently with probability $\frac 12$ . Consequently, $|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_1}|\sim \text{Bin}\left(|Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|,\frac 12\right)$ and ${\mathbb{E}}\left(|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_1}|\right)=\frac 12 |Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|$ . We have $1-\frac 3\alpha \ge 2\beta$ . In addition, $\frac{\big (\frac{1}{2}-\frac{3}{2\alpha }-\beta \big )^2}{1-\frac{10}{7\alpha }}\gt \frac{2}{\alpha }$ , thus there exist some $\epsilon \gt 0$ such that $\frac{\big (\frac{1}{2}-\frac{3}{2\alpha }-\beta \big )^2}{1-\frac{10}{7\alpha }}\geq \epsilon + \frac{2}{\alpha }$ . Applying Chernoff’s inequality gives

\begin{eqnarray*}{\mathbb{P}}\left(|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|\le \beta N\right)&=&{\mathbb{P}}\left (|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|\le \frac{|Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|}{2}-\left (\frac{|Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|}{2}-\beta N\right )\right )\\ &\le & \exp \left ({-}\frac{\Big (\frac{|Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|}{2}-\beta N\Big )^2}{|Z_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|}\right )\\ &\le & \exp \left ({-}\frac{\left(\left(\frac{1}{2}-\frac{3}{2\alpha }\right)-\beta \right)^2}{(1-\frac{10}{7\alpha })}\cdot N\right )\\ &\le & \exp \left ({-}\left (\frac{2}{\alpha }+\epsilon \right )\cdot N\right ). \end{eqnarray*}

Let $E_1$ be the event that for some distinct ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ , $|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|\le \beta N$ . Then

\begin{eqnarray*}{\mathbb{P}}(E_1)&= & \binom{N}{k}\left (\binom{N}{k}-1\right ){\mathbb{P}}(|X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}|\le \beta N)\\ &\le & N^{2k} \exp \left ({-}\left (\frac{2}{\alpha }+\epsilon \right )\cdot N\right )\\ &\le & \exp \left (\frac{2N\ln (N)}{\alpha \ln (N)} - \left (\frac{2}{\alpha }+\epsilon \right )\cdot N\right )\\ &=& \exp ({-}\epsilon N)\to 0, \mbox{ as } N\to \infty . \end{eqnarray*}

This proves Claim 1.

Claim 3. W.h.p. for every ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ with ${\mathcal{Y}}_1\neq{\mathcal{Y}}_2$ , $X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}\not \subseteq X_{{\mathcal{Y}}_2}$ .

We can suppose that the collection of random frameworks fulfils the property of Claim 1. Let ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ be such that ${\mathcal{Y}}_1\neq{\mathcal{Y}}_2$ . Note that each element of $X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}$ is contained in $X_{{\mathcal{Y}}_2}$ with probability $\frac{1}{2}$ . Thus,

\begin{equation*}{\mathbb {P}}\left(X_{{\mathcal {Y}}_1}\cap Z_{{\mathcal {Y}}_2}\subseteq X_{{\mathcal {Y}}_2}\right)=\left (\frac {1}{2}\right )^{|X_{{\mathcal {Y}}_1}\cap Z_{{\mathcal {Y}}_2}|}\le 2^{-\beta N}.\end{equation*}

Let $E_2$ be the event that there exist ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ with ${\mathcal{Y}}_1\neq{\mathcal{Y}}_2$ such that $X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}\not \subseteq X_{{\mathcal{Y}}_2}$ . Since $\frac{2}{\alpha }\lt \ln (2)\beta$ , we have

\begin{equation*}{\mathbb {P}}(E_2)\le N^{2k} {\mathbb {P}}\left(X_{{\mathcal {Y}}_1}\cap Z_{{\mathcal {Y}}_2}\subseteq X_{{\mathcal {Y}}_2}\right) \le N^{2k}\cdot 2^{-\beta N}=\exp \left (\frac {2}{\alpha }N-\ln (2)\beta N\right )\to 0, \mbox { as } N\to \infty .\end{equation*}

This proves Claim 2.

In particular, there exists a collection of $\mathcal{Y}$ -frameworks $({\mathcal{Y}},A_{\mathcal{Y}},Z_{\mathcal{Y}}, X_{\mathcal{Y}})$ , ${\mathcal{Y}}\in \binom{[N]}{k}$ , such that for every ${\mathcal{Y}}_1,{\mathcal{Y}}_2\in \binom{[N]}{k}$ with ${\mathcal{Y}}_1\neq{\mathcal{Y}}_2$ , $X_{{\mathcal{Y}}_1}\cap Z_{{\mathcal{Y}}_2}\not \subseteq X_{{\mathcal{Y}}_2}$ .

Note that $|A_{{\mathcal{Y}}}|=\tfrac{3}{2}k\ln k-k\ge k(\log k + \log \log k)$ . Let ${\mathcal{P}}^{\prime}_{{\mathcal{Y}}}$ be a $\mathcal{Y}$ -shrub in ${\mathcal{Q}}(A_{{\mathcal{Y}}}\cup{\mathcal{Y}})$ as guaranteed by Lemma 11. Note that ${\mathcal{P}}_{{\mathcal{Y}}}$ ’s are not necessarily independent. Let ${\mathcal{P}}_{{\mathcal{Y}}}$ be obtained from ${\mathcal{P}}^{\prime}_{{\mathcal{Y}}}$ by replacing each vertex $W$ of ${\mathcal{P}}^{\prime}_{{\mathcal{Y}}}$ with $W\cup X_{{\mathcal{Y}}}$ . Then ${\mathcal{P}}_{{\mathcal{Y}}}$ is a $\mathcal{Y}$ -shrub in $Q$ .

We consider the following colouring $c\,:\, Q\to \{\text{blue, red}\}$ . For $X\subseteq [N]$ , let

\begin{equation*} c(X) = \begin{cases} \text{blue}, \quad \mbox{ if } X\in \bigcup _{{\mathcal{Y}}\in \binom{[N]}{k}}{\mathcal{P}}_{\mathcal{Y}},\\ \text{red}, \quad\ \mbox{ otherwise.} \end{cases} \end{equation*}

Note that for every ${\mathcal{Y}}\in \binom{[N]}{k}$ , ${\mathcal{P}}_{\mathcal{Y}}$ witnesses that there is a blue $\mathcal{Y}$ -shrub in $Q$ . Recall that a $\mathcal{Y}$ -shrub is an up-tree. Applying Claim 3 the blue subposet of $Q$ is a collection of independent up-trees. Then Lemma 5 provides that the colouring $c$ does not contain a blue copy of $\Lambda$ .

Proof of Theorem 3. Upper Bound: Let $k=(1+\epsilon )\frac{n}{\log (n)}$ and consider an arbitrary blue/red coloured Boolean lattice $Q$ on ground set $[n+k]$ with no blue copy of $\Lambda$ . Pick any ${\mathcal{Y}}\in \binom{[n+k]}{k}$ and assume that there is a blue $\mathcal{Y}$ -shrub in $Q$ . Recall that the maximal elements of the $\mathcal{Y}$ -shrub form an antichain of size $k!$ . Sperner’s theorem provides that the largest antichain in $Q$ has size $\binom{n+k}{\lfloor \frac{n+k}{2}\rfloor }$ , so $k!\le \binom{n+k}{\lfloor \frac{n+k}{2}\rfloor }\leq 2^{n+k}$ .

We also have that $k!\gt \left (\tfrac{k}{e}\right )^k=2^{k(\log k - \log e)}.$ By the choice of $k$ , we obtain for sufficiently large $n$ ,

\begin{equation*}k\log k\ge \tfrac {(1+\epsilon )n}{\log n}\big (\log (n)-\log \log (n)\big )\gt \big (1+\tfrac {\epsilon }{2}\big )n.\end{equation*}

In particular for sufficiently large $n$ , $k\log k -k\log e\gt n+k,$ a contradiction. Thus $Q$ does not contain a blue $\mathcal{Y}$ -shrub for this fixed $\mathcal{Y}$ . Then Corollary 13 yields that there is a red copy of $Q_n$ in $Q$ . Consequently, each blue/red coloured Boolean lattice of dimension $n+k$ contains either a blue copy of $\Lambda$ or a red copy of $Q_n$ .

Lower Bound: Let $N$ sufficiently large, let $k=\frac{10}{216}\frac{N}{\ln (N)}$ and $n=N-k$ . Note that $k\le \frac{N}{2}$ , thus $n\le N\le 2n$ . Let $Q={\mathcal{Q}}([N])$ . By Theorem 18 there exists a colouring of $Q$ with no blue copy of $\Lambda$ such that for every ${\mathcal{Y}}\in \binom{[N]}{k}$ , there is a blue $\mathcal{Y}$ -shrub. By Corollary 13, there is no red copy of $Q_n$ in this colouring, thus $R(\Lambda,Q_n)\ge N= n+ k$ . It remains to bound $k$ in terms of $n$ . Indeed,

\begin{equation*}k=\frac {10}{216}\cdot \frac {N}{\ln (N)}\ge \frac {10}{216}\cdot \frac {n}{\ln (2n)} = \frac {10}{216}\cdot \frac {\log (e)n}{\log (2n)}\ge \frac {1}{15}\cdot \frac {n}{\log (n)},\end{equation*}

which concludes the proof.

Proof of Theorem 2. The lower bound on $R(P, Q_n)$ for $P$ containing either $\Lambda$ or $V$ follows from Theorem 3.

Consider now a poset $P$ that contains neither a copy of $\Lambda$ nor a copy of $V$ . By Corollary 6, $P$ is a union of independent chains. Assume that $P$ has $k$ independent chains on at most $\ell$ vertices each. Let $K$ be an even integer such that $\binom{K}{K/2} \geq k$ . Let $\mathcal{Y}$ be a set of size $K$ and let $\mathcal{X}$ be a set, disjoint from $\mathcal{Y}$ of size $n+ \ell$ . Consider an arbitrary colouring of ${\mathcal{Q}}({\mathcal{X}}\cup{\mathcal{Y}})$ . Assume that there is no red copy of $Q_n$ . We shall show that there is a blue copy of $P$ .

Let $Y_1, \ldots, Y_k$ form an antichain in ${\mathcal{Q}}({\mathcal{Y}})$ , its existence is guaranteed by Sperner’s theorem. Let $Q^i$ be a copy of ${\mathcal{Q}}({\mathcal{X}})$ obtained as an image of an embedding $\phi _i\,:\,{\mathcal{Q}}({\mathcal{X}}) \to{\mathcal{Q}}({\mathcal{X}}\cup Y_i)$ , $\phi _i(X)=X\cup Y_i$ for any $X\subseteq{\mathcal{X}}$ . Consider the blue vertices in $Q^i$ . If there is no blue chain on $\ell$ vertices in $Q^i$ , Corollary 9 implies the existence of a red copy of $Q_n$ in $Q^i$ , a contradiction. Thus for every $i\in [k]$ , there is a blue copy $P_i$ of a chain on $\ell$ vertices in $Q^i$ . Note that for any $A\in Q^i, B\in Q^j$ , $i\neq j$ , $A\not \sim B$ , since $A\cap{\mathcal{Y}} =Y_i\not \sim Y_j=B\cap{\mathcal{Y}}$ . Thus the $P_i$ ’s are independent chains on $\ell$ vertices each. Their union contains a copy of $P$ . This shows that $R(P, Q_n) \leq n + K+\ell = n+ f(P)$ .