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Michael Albanese
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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 {\em closed}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?

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 {\em 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?

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?

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Do closed hypersurfaces separate the euclidean space?

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 {\em 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?