needed to merge them into one component.

If $x,y\in C^{\lambda_1}$ and $C^{\lambda_1}(x)\neq C^{\lambda_1}(y)$, let $B(x,y)$ be the number of boundary connections between $C^{\lambda_1}(x)$ and $C^{\lambda_1}(y)$. Let $$A_1^0(x,y):=A_1(x,y)\cap \{B(x,y)=0\},$$ $$A_1^f(x,y):=A_1(x,y)\cap
\{B(x,y)<\infty\},$$ $$A_1^\infty(x,y):=A_1(x,y)\cap
\{B(x,y)=\infty\},$$ and for $t\in\{0,f,\infty\}$ let $A_1^t$ be the event that $A_1^t(x,y)$ happens for some $x$ and $y$. In the same way as before it is seen that $\P[A_1^t(x,y)]=0$ for all $x$ and $y$ implies $\P[A_1^t]=0$.

Next we will argue that $$\label{bondconnect1} \P[A_1^0(x,y)]=0\mbox{ for all }x\mbox{ and }y.$$ Let $Z^{'\lambda_1}$ and $Z^{''\lambda_1}$ be two independent copies of the Poisson Boolean model with intensity $\lambda_1$ and let $X^{'\lambda_1}$ and $X^{''\lambda_1}$ be their underlying Poisson processes. Since $\lambda_1>\lambda_{BB}$ we may for any $a>0$ pick $R=R(a)$ such that $$\inf_{z_1,z_2}\P_{\lambda_1}[S(z_1,R)\leftrightarrow S(z_2,R)]>1-a.$$ Fix $x$ and $y$ and grow the component of $x$ in $Z^{'\lambda_1}$ (as described earlier) but stop if a ball of radius $R$ is found. Do the same for $y$. Let $F_1$ be the event that the processes are stopped when balls of radius $R$ are found, and note that $A_1^0(x,y)$ is up to a set of measure $0$ included in $F_1$. Let $T_x$ and $T_y$ denote the random times at which the processes are stopped. On $F_1$ we pick $X$ and $Y$ 