zero-one-law in percolation theory I am currently working on a paper by Lyons from 1990 which can be found at https://projecteuclid.org/download/pdf_1/euclid.aop/1176990730.
In chapter 6 Percolation (page 951, 21 respectively) the basic setting is given:
We have $\Gamma$ a countable graph and $p\in[0,1]$.
Every edge is removed with probability $1-p$ independently of the other edges. The random graph which is left after removing is denoted by $\Gamma(\omega_p)$, where $\omega_p$ is a point in a probabilty space $\Omega_p$. I think it is something like $\{0,1\}^{(\operatorname{edges} \operatorname{of} \Gamma)}$, e.g. a $0/1$-vector indexed by the edges of the graph $\Gamma$, where $1$ stands for "edge is there" and $0$ stands for "edge has been removed".
Now, for any vertex $\sigma \in \Gamma$, we denote by $\Gamma_\sigma(\omega_p)$ the connected component of $\sigma$ in $\Gamma(\omega_p)$, which is the subgraph containing $\sigma$, where any two vertices are connected to each other by a path.
Now Lyons states:

By the zero-one-law, the probabilty that $\Gamma_\sigma(\omega_p)$ is infinite for some $\sigma \in \Gamma$ is either $0$ or $1$.

My question now is, which zero-one-law is meant here and how does it apply?
I first thought about the Borel-Cantelli Lemma, which gives a statement about the probability of the limes superior of a sequence of events. So let $\sigma \in \Gamma$. Now as a sequence of events I thought about taking the events $A_1,...$ where $A_n$ stands for the event that the connected component of $\sigma$ has cardinality $n$, e.g. $|\Gamma_\sigma(\omega_p)| = n$. In general, the probability of having a component of size $n$ is the probability of having an edge to the power of the probability, that an edge is not removed, namely $p^n$. Is this right so far? If yes, then Borel-Cantelli will give me that the probability that $\Gamma_\sigma(\omega_p)$ is infinite for some $\sigma \in \Gamma$ is $0$, because of the geometric series $\sum_{n=1}^\infty P(A_n) = \sum_{n=1}^\infty p^n < \infty$.
But what about the probability that $\Gamma_\sigma(\omega_p)$ is infinite for some $\sigma \in \Gamma$ being $1$? Therefore I would need an independent sequence of events $A_1,...$  with $\sum_{n=1}^\infty P(A_n) = \infty$, but in my case I wont have that, so I conclude that I have a basic misunderstanding of the scenario or the construction of the events $A_n$.
Another zero-one-law which came in my mind is the one of Kolmogorow, where I need a sequence of sigma-algebras.
Do you have some ideas on which and how a zero-one-law applies in the probability that $\Gamma_\sigma(\omega_p)$ is infinite for some $\sigma \in \Gamma$?
 A: Before we get going, some terminology: I'll call the edge set of $\Gamma$ $E$ and I'll call an edge $e$ open if it belongs to $\Gamma(\omega)$, i.e. $\omega(e)=1$, where $\omega\in \{0,1\}^{E}$. Furthermore, I'll assume that $\Gamma$ is a priori connected (otherwise, apply the below argument to each connected component separately and use that countable intersections of almost sure events are almost sure).
The Kolmogorov $0-1$ law states that if $(X_n)_{n\in \mathbb{N}}$ is a family of independent variables, and $\mathcal{T}=\cap_{N=1}^{\infty} \sigma((X_n)_{n\geq N})$ denotes the tail $\sigma$-algebra and $F\in \mathcal{T}$, then $\mathbb{P}(F)\in \{0,1\}$.
Note that in order to apply it here, we'll need to go into a countable world, but once we're there, it's intuitively clear enough why this is true: The states of edges are indpendent and clearly I never need to check the state of a specific finite set of edges in order to determine whether or not there is an infinite component (and hence, the event that there is an infinite cluster should lie in the tail $\sigma$-algebra).
In order to formalise the above, there are two cases, we'll start with the pathological one:

*

*There exists a vertex $v$ with countably infinitely many neighbours (by the pigeon-hole principle, this, in particular, covers the case where, for some pathological reason, there are uncountably many edges)

In this case, let $(e_n)_{n\in \mathbb{N}}$ be some countable sub-family of the edges adjacent to $v$, and simply apply Borel-Cantelli to see that the probability that infinitely many of the $e_n$ are open is $1$. Hence, $v$ lies in an infinite component with probability $1$. In particular, there exists an infinite component with probability $1$.


*$\Gamma$ is locally finite (i.e., all vertices have only finitely many neighbours)

In this case, we can enumerate the edges of $\Gamma$. Accordingly, let $X_n=\omega(e_n),$ i.e. the state of the $n$'th edge.
We claim that $(\Gamma(\omega) \; \textrm{contains an infinite component})=:\mathcal{C}_{\infty}\in \mathcal{T}$
For any $N\in \mathbb{N}$, let $\Gamma_N$ denote the spanning sub-graph with edge set $E=E\setminus \{e_n|\; n\leq N-1\}$ and note that $\Gamma_N$ only has finitely many components $\Gamma_N^1,...,\Gamma_N^j$. Let $\mathcal{C}^{k,N}_{\infty}$ denote the event that $\Gamma_N^k(\omega)$ contains an infinite connected component. Clearly,
$\cup_{k=1}^j \mathcal{C}^{k,N}_{\infty}\in \sigma((X_n)_{n\geq N})$ and
$\cup_{k=1}^j \mathcal{C}^{k,N}_{\infty}\subseteq \mathcal{C}_{\infty}$. We'll now argue that the complement of $\cup_{k=1}^j \mathcal{C}^{k,N}_{\infty}$ is contained in the complement of $\mathcal{C}_{\infty}$.
Indeed, for any $\omega\in \{0,1\}^{E},$ if $C_{max}$ denotes the number of edges in the largest connected component, then
$$C_{max}(\Gamma(\omega))\leq N(1+2\max_{1\leq k\leq j}C_{max}(\Gamma_N^k(\omega))),
$$
since every open edge of $\omega$ amongst the $(e_n)_{1\leq n\leq N-1}$ can at best glue the two largest components from the outside together.
Thus, if $\omega\not \in \cup_{k=1}^j \mathcal{C}^{k,N}_{\infty}$, then $C_{max}(\Gamma(\omega))<\infty$ and we get that $\omega\not \in \mathcal{C}_{\infty}$.
In conclusion $\cup_{k=1}^j \mathcal{C}^{k,N}_{\infty}=\mathcal{C}_{\infty}$ for every $N$. Since $\cup_{k=1}^j \mathcal{C}^{k,N}_{\infty}\in \sigma((X_n)_{n\geq N})$, we get that $\mathcal{C}_{\infty}\in \mathcal{T}$. Since the $X_n$ are independent, we get that $\mathbb{P}(\mathcal{C}_{\infty})\in \{0,1\}$ by the Kolmogorov $0-1$ law.
