Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
I suspect they mean compact and connected without boundary. –  Matt N. 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
2  
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-Alvarez Aug 29 '12 at 17:01
1  
@palio Yes, that's right. But $S^1$ is a manifold without boundary : ) –  Matt N. Aug 29 '12 at 17:03
2  
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-Alvarez Aug 29 '12 at 17:10

1 Answer 1

up vote 3 down vote accepted

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.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.