Critical Value of Bond Percolation being 0 My question is summed up by my title. What kind of graph would you have to have such that the bond percolation on that graph had a critical value equal to 0. This would mean that the probability of an edge being open would be 0, and yet it would still be possible for an infinite cluster to exist in your graph. Can you have the probability of an edge being open be dependant on another parameter that doesn't concern the critical value?
Is this possible? Intuitively I think not, but I just wanted to check to see if I misunderstood some of the construction. I also realise that this might be a trivial question, but I just wanted to test my understanding.
 A: The definition of critical value doesn't say anything about what happens at the critical value. If the critical value is $p_c$, then all we claim is that

*

*for $p < p_c$, an infinite cluster exists with probability $0$.

*for $p > p_c$, an infinite cluster exists with probability $1$.

To get $p_c = 0$, it would have to be the case that for all $p>0$, an infinite cluster exists with probability $1$. We're not claiming that an infinite cluster exists when $p=0$: that's impossible, because no open edges exist in that case.
One graph for which bond percolation has a critical value $p_c = 0$ is the graph $G$ with vertex set $\mathbb N = \{0,1,2,3,\dots\}$ and an edge between $0$ and $k$ for all $k>0$. In bond percolation on $G$, there is an infinite cluster containing the vertex $0$; each other vertex has a probability of $p$ of being in this cluster, which is why this cluster is infinite.
For another interesting idea of a construction, take a disjoint union of graphs $G_1, G_2, G_3, \dots$ where $G_n$ has critical value at most $\frac1n$. For any $p>0$, every $G_n$ with $n > \frac1p$ would have an infinite cluster.
