Give an explicit embedding from $\mathbb{R}P_2$ to $\mathbb{R}^4$ I have heard that the least dimension $m$ required for $\mathbb{R}P_2$ to be embedded in the Euclidean space is 4, thus I wanted to find an explicit formulae for it. I found two possible strategies, but is not sure that they'll work. 


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*Define $\phi([x_1,x_2,x_3])=(|x_1|,|x_2|,|x_3|,x_1x_2+x_2x_3+x_3x_1)$, where $[x_1,x_2,x_3]$ is the eq.class under quotient from $S^2$. I hope that would be an embedding, since the last is a symmetric polynomial which is equal for $(x_1,x_2,x_3)$ and $-(x_1,x_2,x_3)$. 

*My second stragy is more geometric approach. Note that the projective plane deleted a circle is a Möbius band $M$. Thus if I could "paste" the boundary of a closed disk (via the fourth dimension) onto the boundary circle of a Möbius Band, then I'm done. But since I'm no good at "visualizing" four dimensions, I don't know exactly how to proceed. 
My question is:
1) is 1. an embedding or not? (Or give another more elegant imbedding)
2) is there a way to realize my second approach? or is it hopeless?
3) Is there a more systematic way of doing such embeddings? (Frankly, If our world is in $\mathbb{R}_2$, I would probably not even be able to imagine how to imbed the torus into $\mathbb{R}_3$)

You can do #2 by what are called "Movie moves". Think of $\mathbb{R}^4 = \mathbb{R}^3 \times \mathbb{R}$. Then in level $\mathbb{R}^3\times 0$ embed the Möbius band. Smoothly propagate the boundary of the Möbius band to $\mathbb{R}^3\times \epsilon$. Now, the boundary of the Möbius band is a smooth unknot so we can insert the isotopy to the standard unknot in $\mathbb{R}^3\times [\epsilon,1-\epsilon]$. Now cap off the unknot inside of $\mathbb{R}^3\times [1-\epsilon,1]$.
