Not understanding a part of Hugo and Vincent's proof on Percolation

Question

Kindly refer to this paper: https://arxiv.org/abs/1502.03050

In this paper, Hugo Duminil-Copin and Vincent Tassion have given an alternative proof of the well known results. I was reading this paper and could not figure out an argument they have used in the proof of item 1 in theorem 1.1, page 6.

I am mentioning the relevant details in short.

Consider a transitive grah $G=(V,E)$ on which we want to study bond percolation. Each edge $e_i$ is open with probability $p_i$ and closed with probability $1-p_i$ independent of all other edges.

For an event $A$, an edge $e$ is said to be pivotal if changing the status (open or closed) of the edge $e$ causes $E$ to occur and not occur alternatively.

Define for a finite set $S$ of vertices, an edge on the "external boundary" of $S$ to be an edge of the form $\{x,y\}$ where $x\in S,y\notin S$.

Suppose we have a finite set of vertices $\Lambda$ with $0\in\Lambda$. Then define the random subset $\mathcal S$ of $\Lambda$ as: $\mathcal S=\{v\in \Lambda:v$ is not connected to $\Lambda^c$ by an open path$\}$.

Consider now the increasing event $A:=\{0$ is connected to $\Lambda^c$ by an open path$\}$.

They have argued briefly that the event $[xy$ is pivotal for $A$ and $\mathcal S=S]$ is equivalent to $xy$ edge on the outer boundary of $S$ (with $x\in S,y\notin S$) and $0$ is connected to $x$ through vertices in $S$ by an open path, and $\mathcal S=S$.

I do not understand this argument. Why should my pivotal edges be only on the outer boundary of $S$?

If such a pivotal edge exists in $S^c$, then I think changing the status of that edge will not affect the connectivity of $0$ by an open path to $\Lambda^c$. But is it because there are other edges available that can make a path from $0$ to $\Lambda^c$? Why also can the pivotal edge not be in $S$?

Answer 1

I think the detail you missed was that they assumed that $0 \in S$

If $\{x,y\}$ is a pivotal edge, we need that at least one of then connects to $0$, and one of then needs to connect to $\Lambda^c$. Otherwise, opening this edge wouldn't make $0 \longleftrightarrow \Lambda^c$. Suppose then that $0\longleftrightarrow x$ and $ y \longleftrightarrow \Lambda^c$. If $x \in S^c$, then we would have that $0 \in S^c$ as $0 \longleftrightarrow x \longleftrightarrow \Lambda^c$, which is a contradiction and therefore $x \in S$. On the other hand, by definition of $S$, we have that $y \in S^c$. Thus $\{x,y\}$ is in the external boundary of $S$.