$\lambda_2$ there is a.s. a unique unbounded component. For $i=1,2$ let $$A_i(x,y):=\{\mu(C^{\lambda_1}(x))=\infty,\,\mu(C^{\lambda_1}(y))=\infty,\,C^{\lambda_i}(x)\neq
C^{\lambda_i}(y)\},$$ and let $$A_i:=\bigcup_{x,y}A_i(x,y).$$

Since $\lambda_{BB}\ge \lambda_c$ we have by Theorem \[uniqmon1\] that any unbounded $\lambda_2$ component a.s. intersects some unbounded $\lambda_1$ component. Therefore $$\label{inklusion1}
 \bigcup_{x,y} \{\mu(C^{\lambda_2}(x))=\infty,\,\mu(C^{\lambda_2}(y))=\infty,\,C^{\lambda_2}(x)\neq
C^{\lambda_2}(y)\}\subset A_2\cup N$$ where $N$ is a set of measure $0$. In the same way as in the second proof of Theorem \[uniqmon1\] we have that $\P[A_i(x,y)]=0$ for all $x$ and $y$ implies $\P[A_i]=0$. By \[inklusion1\], $\P[A_2]=0$ implies $\P[\mbox{there is a unique
unbounded component at level }\lambda_2]=1$. Hence it is enough to show that $\P[A_2(x,y)]=0$ for all $x$ and $y$.

Suppose $C_1$ and $C_2$ are two distinct components in the Poisson Boolean model. A pair of Poisson points $x_1\in C_1$ and $x_2\in C_2$ is called a *boundary-connection* between $C_1$ and $C_2$ if $d(x_1,x_2)<4$ (so that the distance between their corresponding balls is $<2$) or there is a sequence of Poisson-points $y_1,\dots, y_n$ such that

-   the ball centered around $y_i$ intersects the ball centered around $y_{i+1}$ for all $i$.

-   $y_i$ is outside $C_1$ and $C_2$ for all $i$.

-   $d(x_1,y_1)<4$ and $d(x_2,y_n)<4$.

Note that if there is a boundary connection between two components, then at most two more balls are 