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i read that the circle $S^1$ is the only connected compact 1-manifold but don't we have that the interval $I=[0,1]$ is a connected compact 1-manifold and that is not homeomorphic to $S^1$? May be they mean $S^1$ is the only compact not simply connected 1-manifold?

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    I suspect they mean compact and connected without boundary.2012-08-29
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    that is a closed manifold. But a compact manifold may or may not have a boundary2012-08-29
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    The context of the statement you quote probably decided to talk only about manifolds without boundary. Unless you tell us where you read it, it is impossible for us to know what they meant. In any case, it is also true that $S^1$ is the only compact non-simply connected connected manifold...2012-08-29
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    @palio Yes, that's right. But $S^1$ is a manifold without boundary : )2012-08-29
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    here http://en.wikipedia.org/wiki/Classification_of_manifolds#Dimensions_0_and_1:_trivial they say A connected 1-dimensional manifold is either the circle (if compact) or the real line (if not).2012-08-29
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    i think they forgot to mention that it does not have a boundary2012-08-29
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    As I noted above, it is perfectly possible that they consider manifolds to be those without boundary. What the term means is just a convention, it is not carved in stone, and very, very often it is useful to make conventions to make one's like easier.2012-08-29

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Here is the classification of $1$-manifolds (connected):

  1. $[0,1]/(0 \equiv 1)$, the unit circle: compact, without boundary.
  2. $[0,1]$: compact with a non-connected boundary.
  3. $(0,1)$: non-compact, without boundary.
  4. $[0,1)$: non-compact, with connected boundary.