Edwin Clark asked:

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Let G be a graph whose vertices are N, the natural numbers. Suppose for every n the subgraph G_n of G induced on {1,2,...,n} has a Hamiltonian path starting with 1.  Does G necessarily have a Hamiltonian path?
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MIke Reid wrote:

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no.  a simple counterexample is the graph

4 --- 3 --- 6 --- 7 --- 8 --- 9 -- ...
      | \_  |  _/
      |   \ | /  ...
1 --- 2 --- 5

where  5  is connected to  n  for all  n > 5  (as well as to  2  and  3 ). Since  1  and  4  each have degree  1 , they would both have to be endpoints of any infinite hamiltonian path, a contradiction.
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Nice example!  Now suppose every vertex is required to have finite valence.  Then I think this slight modification of Mike's example will still serve as a graph where every {1,...,n} has a Hamiltonian path, but not {1,2,3,...}.

4 --- 3 --- 6 --- 7 --- 9 --- 11 ...
       |\_  |   _/ |   _/ |   _/ |
       |   \ |  /    | /    |  /    |
1 --- 2 --- 5 --- 8 --- 10 ---12 ...

--Dan