$$\P[A_1^0]\ge
\P\left[A_1^{f,N}\right]\P\left[Z^{''\lambda_1}[S(0,N)]=\emptyset\right]>0,$$ a contradiction which proves (\[bondconnect2\]).

Now if $A_1^{\infty}(x,y)$ happens, then there are infinitely many boundary connections between $C^{\lambda_1}(x)$ and $C^{\lambda_1}(y)$ and a.s. no bounded region contains all boundary connections. Therefore $C^{\lambda_1}(x)$ and $C^{\lambda_1}(y)$ will almost surely have been merged into one unbounded component at level $\lambda_2$ by balls that appear in the coupling between level $\lambda_1$ and $\lambda_2$. So $\P[A_2(x,y)|A_1^{\infty}(x,y)]=0$. Thus, since $A_2(x,y)\subset A_1(x,y)$ and $A_1(x,y)$ is partitioned by $A_1^f(x,y)$ and $A_1^{\infty}(x,y)$ we conclude $$\P[A_2(x,y)]=\P[A_2(x,y)|A_1^f(x,y)]\P[A_1^f(x,y)]+\P[A_2(x,y)|A_1^{\infty}(x,y)]\P[A_1^{\infty}(x,y)]=0,$$ for all $x$ and $y$ and so $\lambda_u\le \lambda_{BB}$.

Next we show the easier result that $\lambda_u\ge \lambda_{BB}$. Suppose $\lambda>\lambda_u$. By Theorem \[uniqmon1\] there is a.s. a unique unbounded component in $C^{\lambda}$ which we denote by $C_{\infty}^{\lambda}$. By the continuum version of the FKG inequality (see [@meester]) and the fact that there is an isometry mapping $x$ to $y$ it follows that $$\begin{split}\P_{\lambda}[S(x,R)\leftrightarrow
S(y,R)]& \ge\P_{\lambda}[S(x,R)\mbox{ and }S(y,R)\mbox{ intersects
}C_{\infty}^{\lambda}]\\& \ge\P_{\lambda}[S(x,R)\mbox{ intersects
}C_{\infty}^{\lambda}]^2.\end{split}$$ Since $\lim_{R\rightarrow
\infty}\P_{\lambda}[S(x,R)\mbox{ intersects
}C_{\infty}^{\lambda}]=1$ it follows that $\lambda>\lambda_{BB}$ and thus $\lambda_u\ge \lambda_{BB}$.

The situation at $\lambda_u$ on $\hyp \times \R$
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This section is devoted to the proof of Theorem \[htimesrsats\]. We introduce some new notation: if the points $x,y\in \hyp\times \R$ are in the same component of $C^{\lambda}$ then 