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I'm taking a course on Algebraic Topology and I had a question while studying. I know the answer is affirmative but I don't know why. What I want to prove is

There exist a triangulation for every 1-manifold

I tried to use that the manifold is II Axiom of countability, so there exist a countable base for the manifold. Now I would like to prove that I can take intervals to go into these open sets of the manifold in such a way that the triangulation consist on these intervals. And also I have to prove that the triangulation is locally finite, which I don't know how should I try to prove.

Am I on the right track? How can I continue? Any help is welcome.

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  • $\begingroup$ I think the easiest way to prove this is just to directly classify all $1$-manifolds. $\endgroup$ Commented Nov 2, 2015 at 18:03
  • $\begingroup$ But the proof I have for that uses that every 1-manifold has a triangulation. Is there a proof that doesn't use that? $\endgroup$
    – Javier
    Commented Nov 2, 2015 at 18:10
  • $\begingroup$ I think Milnor "topology from the differentiable viewpoint" classifies 1 manifolds without assuming a triangulation. $\endgroup$
    – Thomas Rot
    Commented Nov 2, 2015 at 21:06
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    $\begingroup$ The basic idea is to start at a point and just inductively build larger and larger neighborhoods of it that are homeomorphic to $\mathbb{R}$. If the left and right ends of your neighborhoods run into each other, you get $S^1$; if not, then you get $\mathbb{R}$. The trick is to make sure that after running the induction for long enough, you actually end up exhausting the entire manifold. You can do this by either using paracompactness to start with a locally finite cover by charts which all your neighborhoods are unions of, or by using transfinite induction. $\endgroup$ Commented Nov 2, 2015 at 21:27

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Do you know the classification of 1-manifolds? Every compact one-manifold is a finite union of circles. Now you should be able to explicitly construct triangulations.

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  • $\begingroup$ It is not required to be compact. I know every connected 1-manifold is a circumference or a real line, but I use the fact that it admits triangulation to prove that. $\endgroup$
    – Javier
    Commented Nov 2, 2015 at 17:45

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