[Ore's theorem](http://en.wikipedia.org/wiki/Ore%27s_theorem) states that in a finite graph $G$ with $|V(G)|=n$, there is a Hamiltonian path, provided that the sums of the degrees of 2 distinct, non-adjacent vertices is $\geq n$.


For countable graphs, such a statement cannot hold: consider the disjoint union of two copies of $K\_\omega$.


How about if we restrict ourselves to *connected* countable graphs? Is then the following statement true?



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> > If $G$ is a connected countable graph such that for distinct, non-adjacent vertices $v,w$ we have $\text{max}\{\text{deg}(v), \text{deg}(w)\} = \aleph\_0$, there is an countable Hamiltonian path (extending in two ways, or beginning at one vertex.)
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Ore's theorem for countable graphs
And then here is an obvious counterexample: take a star-like tree with one central vertex of countable degree (and, if you still insist on two non-adjacent vertices, make each ray of length $2$).


