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I can prove that there is no $C^2$ embedding of the real projective plane into $\mathbb{R}^3$. In fact, every closed $C^2$ surface of $\mathbb{R}^3$ is orientable (probably this is true for the $C^1$ category as well).

Is there a topological embedding of $\mathbb{R} \mathbb{P}^2$ into $\mathbb{R}^3$?

An answer in this question tells to use the Alexander Duality Theorem, but I don't know how.

Eduardo Longa
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  • I am asking for topological $\mathbb{R}\mathbb{P}^2$, not smooth. – Eduardo Longa Oct 15 '17 at 01:51
  • @EduardoLonga Mariano's answer in the linked question fits the topological case. See page 353 of Bredon's "Topology and Geometry" - specifically, the "Generalized Jordan Curve Theorem" and its corollary. – Aloizio Macedo Oct 15 '17 at 03:46

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