in some way independent of $Z^{''\lambda_1}$ such that $S(X,R)\subset Z^{'\lambda_1}[L_{T_x}^{'\lambda_1}(x)]$ and $S(Y,R)\subset Z^{'\lambda_1}[L_{T_y}^{'\lambda_1}(y)]$. Let $$X^{\lambda_1}:=(X^{'\lambda_1}\cap(L_{T_x}^{'\lambda_1}(x)\cup L_{T_y}^{'\lambda_1}(y)))\cup
(X^{''\lambda_1}\cap (L_{T_x}^{'\lambda_1}(x)\cup
L_{T_y}^{'\lambda_1}(y))^c)$$ and $Z^{\lambda_1}:=\cup_{x\in
X^{\lambda_1}}S(x,1)$. The distribution of $Z^{\lambda_1}$ is by construction the distribution of the Poisson Boolean model with intensity $\lambda_1$. Put $$F_2:=F_1\cap \{S(X,R)\leftrightarrow S(Y,R) \mbox{ in
}Z^{''\lambda_1}\}.$$ If we are on $F_2$ then either $\{Z^{\lambda_1}(x)=Z^{\lambda_1}(y)\}$ occurs or $\{B(x,y)\ge 1\}$ occurs and in neither case we are on $A_1^0(x,y)$. Since $$\P[F_2|F_1]=\P[S(X,R)\leftrightarrow S(Y,R)\mbox{ in
}Z^{''\lambda_1}]>1-a$$ it therefore follows that $$\P[A_1^0(x,y)]\le\P[F_1\cap F_2^c]\le \P[F_2^c|F_1]<a$$ proving (\[bondconnect1\]).

Next we show that $$\label{bondconnect2} \P[A_1^f]=0.$$

Let $A_1^{f,N}$ be the event there are two distinct unbounded components in $C^{\lambda_1}$ such there are a finite number of boundary connections between them and they are all contained in the ball $S(0,N)$ for some random finite $N$. Suppose $\P[A_1^f]>0$ and pick $N$ so large that $\P[A_1^{f,N}]>0$. Let $Z^{\lambda_1}$ be the union of the balls from $Z^{'\lambda_1}$ centered outside $S(0,N)$ and the balls from $Z^{''\lambda_1}$ centered inside $S(0,N)$. Now suppose that $A_1^{f,N}$ happens for $Z^{'\lambda_1}$ and that $\{Z^{''\lambda_1}[S(0,N)]=\emptyset\}$ happens. Then we can find two points $\tilde{x}$ and $\tilde{y}$ in separate unbounded components of $Z^{\lambda_1}$ such that there are no boundary connections between them. It follows by the independence of $Z^{'}$ and $Z^{''}$ that 