Why is it not possible to embed $\mathbb{R}P^2$ in $\mathbb{R}^3$? the title is more or less self-explanatory. 
Why is it not possible to embed the Real Projective Plane $\mathbb{R}P^2$ in to the euclidean space $\mathbb{R}^3$?
I stumbled upon the following example: Let $\mathbb{Z}_2$ act on $\mathbb{R}^3$ by reflections (f(x)=x,f(x)=-x). Then the quotient space $Q$ is a cone over $\mathbb{R}P^2$.
$Q$ is not a manifold. How to prove this statement?
I tried the following: Suppose $0$ would have a neighbourhood which is homeomorphic to $\mathbb{R}^3$. Then by restricting this homeomorphism one could Embed the projective plane into $\mathbb{R}^3$. Which is not possible (by google). However I did not find out why. 
Is there an algebraic invariant, which prohibits it and is easy to visualize? 
 A: Every codimension one closed submanifold of $\mathbb R^3$ is orientable but $\mathbb{R}P^2$ is not orientable .
A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$You have two questions:


*

*Why does $\Reals\Proj^{2}$ not embed in $\Reals^{3}$? (Your title question, which Georges has answered.)

*Why is $Q = \Reals^{3}/{\pm}$ not a manifold? (The motivating question in your post, which 1. answers indirectly.)
Here's an alternative take on 2.: The "link" of the vertex of $Q$ (i.e., the boundary of the image of a ball centered at the origin of $\Reals^{3}$) is $\Reals\Proj^{2}$ instead of $S^{2}$. If some neighborhood $U$ of the vertex of $Q$ were homeomorphic to $\Reals^{3}$, there would exist a closed ball $B$ centered at the origin of $\Reals^{3}$ whose image in $Q$ is (contractible and) contained in $U$. Following with a homeomorphism $\phi:U \to \Reals^{3}$ would give a contractible subset of $\Reals^{3}$ whose boundary is $\Reals\Proj^{2}$.
The same idea implies, for example, that a cone on a $2$-torus is not a manifold, even though a $2$-torus does embed in $\Reals^{3}$.
