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Let $C$ be a nonempty connected subset of $S^2$ and $\partial C$ be the boundary of $C$.

If $\bar C$ (closure) does not separate $S^2$, is $\partial C\cong S^1$?

This seems very true, but I think proof won't be easy. Is this statement true? How do I prove this?

muaddib
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Rubertos
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  • This is false, see here for an example. Also, consider a single point. – Dejan Govc Aug 06 '15 at 13:57
  • @DejanGovc Thank you. However, say it is infinite, then what additional condition would it be to make this statement holds? – Rubertos Aug 06 '15 at 14:21
  • in those cases also you might find some counter example, like if youy consider a line joining north and south poles, or if you consider $S^2$\ {a point} and alll...so I dont think it is an interesting problem but ofcourse you can try to make it interesting by adding some strong conditions like $C^c$ is simple connected and has non-empty interior... – Anubhav Mukherjee Aug 06 '15 at 14:25

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