Marc LeBrun <mlb@fxpt.com> wrote:
Only if we insist that everything be FINITELY connected are these infinite hairpins a problem. But then it kind of begs the question whether the infinite graph itself is "connected" in the first place.
The only kind of graph connectedness I know of is finite. If you take the adjacency graph of (-1/2)^n, n=0,1,..., then you get two infinite paths with a limit point of zero, but I can't see any graphlike property that can make them connect. Thus any infinite path is either isomorphic to the adjacency graph of N or the adjacency graph of Z, and infinite cycles don't exist. Dan Asimove got what I meant. In general, I'm trying to figure out if we can make conditions on finite subgraphs of an infinite graph that will enforce a Hamiltonian path of either sort on the whole. Dan