Chapter 7 Infinite product spaces

7.5. TWO ZERO-ONE LAWS 17
However P(An) apple P(A)+P(A4An) and so (7.52) gives
This is only possible if P(A) =0 or P(A) =1.
P(A) apple P(A) 2 . (7.57)
Going back to percolation, the above zero-one law tells us:
Corollary 7.27 One of the events {N = 0}, {N = 1}, and {N = •} has probability
one.
Proof. All of these events are translation invariant (or tail) and so they have probability
zero or one. We have to show that P(N = k) =0 for all k 2{2, 3, . . . }. Again,
if this is not the case then P(N = k) =1 because {N = k} is translation invariant;
for simplicity let us focus on k = 2. If P(N = 2) =1 then there exists n 1 such
that the event
( )
d
box [ n, n] is connected to infinity by two disjoint paths that are
Cn =
not connected to each other in the complement of the box [ n, n] d
(7.58)
occurs with probability at least 1/ 2. Indeed, {N = 2} ⇢ S
n 1 Cn and Cn is increasing.
However, the event Cn is independent of the edges inside the box [ n, n] d and
so with positive probability, the two disjoint paths get connected inside the box.
This means that N = 1 with positive probability on the event {N = 2}\Cn, i.e.,
P(N = 1) > 0. This contradicts P(N = 2) =1.
Let us conclude by stating what really happens: we have P(N = •) =0 so we
either have no infinite cluster or one infinite cluster a.s. The most beautiful version
of this argument comes in a theorem of Burton and Keane which combines rather
soft arguments not dissimilar to what we have seen throughout this section.