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Is there a countable connected graph $G=(\omega, E)$ such that $\text{deg}(v)=\omega$ for all $v\in\omega$, but there is no Hamiltonian path in $G$?

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closed as off-topic by Franz Lemmermeyer, Alexey Ustinov, Wolfgang, Boris Bukh, Myshkin May 11 '16 at 0:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Franz Lemmermeyer, Alexey Ustinov, Wolfgang, Boris Bukh, Myshkin
If this question can be reworded to fit the rules in the help center, please edit the question.

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Take the natural numbers as vertices and connect a pair of numbers with an edge if their difference is divisible by three. Also connect 1 with 2 and 2 with 3.

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