# Proof you can't embed $\mathbb{P}^2$ into $\mathbb{R}^3$

I know you need to assume for a contradiction. The hard bit is the thing I can't do and that is show that if you do put $\mathbb{P}^2$ into $\mathbb{R}^3$ the path components of the complement are two pieces. I know from this it's an easy contradiction as then you would be able to orientate $\mathbb{P}^2$, which is impossible as it contains a Möbius strip.

Note, if you don't know what $\mathbb{P}^2$ is then it the sphere with the antipodal points equal i.e. same conguranchy

Also, is there a general pattern. Like can $\mathbb{P}^3$ be only embedded into $\mathbb{R}^5$

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"Complement" with an "e" and "compliment" with an "i" are two different words that mean two different things. I've changed the spelling in the posting. –  Michael Hardy Nov 8 '11 at 18:20
And here is a complementary comment: Bravo, Michael –  Georges Elencwajg Nov 9 '11 at 20:02
And here is a complimentary comment: Bravo, Michael –  Georges Elencwajg Nov 9 '11 at 20:02
If you know homology, then use Alexander duality with $\mathbb Z_2$ coefficients. The reduced $0$th homology of the complement of $\mathbb{RP}^2$ is isomorphic to $H^2(\mathbb{RP}^2;\mathbb Z_2)\cong\mathbb Z_2$. That means that the complement has $2$ connected components.