Translation invariant event in percolation theory in $\mathbb{Z}^d$ At my probability theory proseminar I had a speech about the percolation theory and one of the topics I presented was the uniqueness of an infinite open cluster in Bernoulli bond percolation in $\mathbb{Z}^d$. However, I had real problems with proving that N (the number of infinite open clusters) is constant a.s. I tried using Kolmogorov's 0-1 law, but it only tells us that $\mathbb{P}_p(N \geq 1) \in \{0, 1\}$.
My speech was mainly based on Grimmett's Probability on Graphs (http://www.statslab.cam.ac.uk/~grg/books/USpgs-rev3.pdf, theorem 5.22). The proof says that $\{N = c\}$ is a translation-invariant event, therefore $\mathbb{P}(N = c) \in \{0, 1\}$ for any $c$, but I don't understand why is the sample space $\Omega$ in a product form. In fact, the codomain of a stochastic process representing the percolation model is $\{0, 1\}^{\mathbb{E}^d}$ and this is in a product form, $\textbf{not}$ the $\Omega$ itself. This makes me even more confused, since I'm a complete greenhorn in ergodic theory. I will highly appreciate any kind of explanation or reference to some more elaborate work in this context.
 A: p. 39 states that the sample space $\Omega$ is $\{0, 1\}^{E^d}$ so I would say it is by definition.
You can understand why the event $\{N = c\}$ is translation invariant as follows.  Take any realization of the sample space that has $c$ clusters.  Then if you translate that realization you will find another realization still in the set.  In other words, translation is a bijection on $\{N = c\}$.
To prove that set has measure $0$ or $1$ you will need to use the Kolmogorov $0-1$ law.  See this thread for suggestion on how to do it.
A: You use the fact that the product measure on $\{0,1\}^{\mathbb{E}^d}$ is ergodic with respect to the shifts.
The ergodicity of the product measure w.r.t. the shifts means (by definition) that $\mathbb{P}_p(E)\in\{0,1\}$ for any shift-invariant event $E$.  To show that the product measure is ergodic, you approximate $E$ with a cylinder set $C$.  Now, if you shift $C$ sufficiently far, you get another cylinder set that is independent of $C$.  This will show that the approximated set $E$ (which is shift-invariant) is independent of itself.
