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I recall having heard somewhere that the only 1-manifolds (second countable, Hausdorff, connected spaces locally homeomorphic to $\mathbb R$) are $\mathbb R$ and $S^1$. Is this true? If so, is there a reasonably elementary proof?

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    You will probably want to add a Hausdorffness and connectedness condition.2012-02-26
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    There is the proof at the end of Milnor's "Topology from the differentiable viewpoint" which is "elementary" if my memory isn't playing tricks on me.2012-02-26
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    Make sure you have paracompact in your definition of "manifold", too. Otherwise you can have the "long line" for example.2012-02-26
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    Added Hausdorff and connected to the question.2012-02-26
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    @GEdgar: Second countable is stronger than paracompact for manifolds (equivalent for connected manifolds).2012-02-26
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    If you don't mind only considering submanifolds of Euclidean space (a la Whitney embedding theorem), *Differential Topology* by Guillemin and Pollack has a proof in the appendix.2012-02-26
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    What about links/knots? They are also 1-manifolds. Which condition do they lack?2012-02-26
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    @draks: knots are homeomorphic to $S^1$.2012-02-26
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    @draks: and a link which is not a knot is not connected.2012-02-28

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There's a proof outlined in Problems 17-5 and 17-7 of John Lee's "Introduction to Smooth Manifolds" that uses a basic classification of integral curves of vector fields, specifically that a nonconstant maximally defined integral curve is either injective or periodic, which implies (after a small amount of work) that the image of any nonconstant integral curve is diffeomorphic to either $\mathbb{R}$ or $\mathbb{S}^1$. The problem is finished by showing that any 1-manifold is orientable, and thus admits a nonvanishing global vector field, of which you consider a maximally defined integral curve.

I don't think this is the same proof as given in Guillemin and Pollack or in Milnor, and for my money it's quite a bit simpler than both.

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    Agree. I kind of didn't understand Milnor's proof. It sounded fishy to me. Of course, not casting aspersions on Milnor's mastery of exposition. In fact, I loved every bit of his TFDV book, the first of his classics that I have been planning to read. However, this last proof in the appendix did not speak to me!2016-08-13
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    And when trying to ask if there shouldn't be a simpler, more intuitive proof, I did come up with idea of "choosing a global direction on our manifold (that will be our vector field) and then gluing together integral curves as long as we can. Finishing it as @youler outlined.2016-08-13