1
$\begingroup$

The following extension of the Jordan Curve Theorem is well known: every closed connected hypersurface of the sphere $\mathbb S^N$ separates $S^N$ into exactly two connected components. As a consequence, every compact connected hypersurface of the euclidean space $\mathbb R^N$ separates $\mathbb R^N$ into exactly two connected components.

Does this result keep its validity if the word `compact' is replaced by closed? With other words, is it true that every closed connected hypersurface of the euclidean space $\mathbb R^N$ separates $\mathbb R^N$ into exactly two connected components?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ You have to define more precisely that is a "closed connected hypersurface" to make this meaningful. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 29, 2019 at 21:01
  • $\begingroup$ Usually "closed" in this context means compact and without boundary. Your question appears to be about the case where you drop compactness, i.e. your manifold simply has no boundary, like $\mathbb R^{N-1}$. In that case the answer is no, as Euclidean spaces are diffeomorphic to open balls. $\endgroup$
    – Ryan Budney
    Commented Apr 29, 2019 at 21:13
  • 1
    $\begingroup$ @AlexandreEremenko It's possible that the that this is asking for the hypersurface to be proper (i.e. so it is closed as a subset). $\endgroup$
    – RBega2
    Commented Apr 30, 2019 at 0:26
  • 1
    $\begingroup$ In the case of the point-set topology "closed" then the answer is yes, those do separate. The proof is basically the same as when dealing with compact manifolds. $\endgroup$
    – Ryan Budney
    Commented Apr 30, 2019 at 2:25

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .