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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?

palio
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    I suspect they mean compact and connected without boundary. – Rudy the Reindeer Aug 29 '12 at 16:59
  • that is a closed manifold. But a compact manifold may or may not have a boundary – palio Aug 29 '12 at 17:00
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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... – Mariano Suárez-Álvarez Aug 29 '12 at 17:01
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    @palio Yes, that's right. But $S^1$ is a manifold without boundary : ) – Rudy the Reindeer Aug 29 '12 at 17:03
  • 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). – palio Aug 29 '12 at 17:06
  • i think they forgot to mention that it does not have a boundary – palio Aug 29 '12 at 17:08
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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. – Mariano Suárez-Álvarez Aug 29 '12 at 17:10

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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.
Ilies Zidane
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