this the monotone coupling and is obviously the analogy of the discrete coupling described earlier.

Now suppose $\lambda_1<\lambda_2$ and use the monotone coupling. We say that an unbounded component in $C^{\lambda_2}$ is $\lambda_1$-stable if it contains some unbounded component in $C^{\lambda_1}$. We now state a continuum version of Theorem \[hpsats\].

\[uniqmon1\] Consider the Poisson Boolean model on the homogeneous space $M$. Suppose $\lambda_c(M)<\lambda_1<\lambda_2<\infty$. Then a.s. any unbounded $\lambda_2$-component is $\lambda_1$-stable.

From Theorem \[uniqmon1\], the following corollary is immediate.

\[uniqcor1\] Consider the Poisson Boolean model on the homogeneous space M. Suppose $\lambda_u(M)<\lambda$. Then $\P_{\lambda}[N_C=1]=1$.

[**Remark.**]{} Corollary \[uniqcor1\] is known in the cases $M={\mathbb R}^n$ for any $n\ge 2$ (see [@meester]) and $M=\hyp$ (see [@tykesson]).

We will present two proofs of Theorem \[uniqmon1\]. The first is inspired by the proof of Theorem \[hpsats\] and the second is inspired by the proof of Theorem \[schonmannsats\]. To get a continuum analogy to Theorem \[peressats\] we consider the Poisson Boolean model on a product space.

\[htimesrsats\] Consider the Poisson-Boolean model on $\hyp\times \R$. At $\lambda_u$ there is a.s. not a unique unbounded component.

Note that if one instead considers the model on $\hyp$, then Corollary 5.10 in [@tykesson] says that at $\lambda_u$ there is a.s. 