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What is known about continuum percolation in 1d?

By this, I mean, for $d \in \mathbb{N}$, the Poisson-Boolean model of disks of radius $r_0 \in \mathbb{R}$ with centres arranged randomly in $[0,1]^{d}$, with a Poisson number of disks at density $\lambda$ per unit volume, taking the case $d=1$. A pair of disks is connected if they overlap.

In site percolation on the integer lattice in 1d, the critical site occupation probability $p_c=1$, otherwise the chain of sites will break somewhere.

But in 1d continuum percolation, can $r_0(\lambda)$ scale slowly enough to zero as $\lambda \to \infty$ for the probability of a break goes to zero?

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    $\begingroup$ Whatever $r_0$ and $\lambda$ are, there will be infinitely many gaps of size bigger than $2r_0$, so that there is no percolation. $\endgroup$ Commented Jan 22, 2020 at 7:35
  • $\begingroup$ So basically, only if $2r_0$ is the width of the domain do you get percolation. $\endgroup$
    – apg
    Commented Jan 22, 2020 at 7:37
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    $\begingroup$ OK - sorry - I hadn't read the question properly. So you are looking at a finite segment of the line, namely [0,1] rather than all of $\mathbb R$. My answer applies to the infinite case. So in the finite case, there should be some sort of phase transition, I think. I can't tell you exactly what it is immediately though. $\endgroup$ Commented Jan 22, 2020 at 7:41
  • $\begingroup$ You're answer is correct. So, if the ratio $r_0/w$, where $w$ is the domain width, goes to a constant, is that may be sufficient for percolation? $\endgroup$
    – apg
    Commented Jan 22, 2020 at 7:44
  • $\begingroup$ Can you see any literature on this? Or has it not been studied? $\endgroup$
    – apg
    Commented Jan 22, 2020 at 7:46

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See for example this paper.

Covering by random intervals and one-dimensional continuum percolation

C. Domb, J. Stat. Phys. Vol. 55(1-2), 1989

and also

Exact solution of a one-dimensional continuum percolation model

A. Drory, Phys. Rev. E Vol. 55(4), 1997

among other works, including many by the second author above.

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