Let $G$ be a connected, undirected graph, with countably infinite set of vertices and countably infinite set of edges. Assume that the degree of each vertex is finite, and moreover, the degrees of all vertices are uniformly bounded.

Let each vertex carry one of two values: $1$ or $-1$.

Now, equip each vertex with a Poisson clock ($\lambda=1$), all clocks are independent. At each time the clock of a vertex ticks, the vertex updates its value to be the value of most of its neighbors (in case of a draw $-$ the value of the vertex remains unchanged).

Does there exist such a graph as described, with certain initial values at vertices, such that with $\mathbf{positive}$ probability, there will be a vertex where the value is $not$ eventually constant?

Thank you.

$\mathbf{EDIT:}$ If you wish, for a beginning, analyze the example given by domotorp in his comment (which could be a solution): take the $3$-regular tree with initial values as follows: pick one vertex, it will be $1$. The vertices around it will be $-1$. The vertices at distance $2$ from the initial vertex are again $1$. And so on, changing the value layerwise. In this graph and initial values, will there be a vertex, that with positive probability, will not converge? (Even if the answer for this example is NO, the fact that every vertex in the example almost surely converges is also nontrivial, and a proof of this will also be upvoted.)

voter model. I know that the answer is yes for the synchronous voter model (where all clocks tick once per second). Indeed: take $\mathbb Z$ with a uniform i.i.d. distribution of $\pm 1$ at the vertices. Then the set of configurations that converge to all 1's is invariant (so of measure 0 or 1); ditto for the configurations that converge to all $-1$'s. By symmetry, they are both of measure 0. From this, it follows that almost surely, each vertex oscillates infinitely often. $\endgroup$ – Anthony Quas Dec 16 '16 at 8:51