# compactification of $\Bbb R^n$

I want to show that $n$ dimensional real projection space is compactification of $\Bbb R^n$, how can I do that?

I can show PR^n is compact, and R^n is localy compact, but I have problem to show R^n is dense in RP^n. Is it done by RP^n=R^n+ RP^n-1 ? Or could you suggest me the better way to show, it is dense?

Is there any direct way?

In some note I read this is not true as 1 point compactification, it's true just for $n=1$. here I saw this:https://en.wikipedia.org/wiki/Compactification_%28mathematics%29

• One-point compactifications are unique up to homeomorphism, so this works only when $S^n$ is homeomorphic to $\mathbb{R}P^n$, which happens only for n = 1. Apr 21, 2016 at 9:02
• Adding to Pedro's comment: for $n>1$ we have $\pi_1RP^n=\mathbb Z/2 \neq 0 = \pi_1S^n$, hence there cannot even be a homotopy equivalence. Apr 21, 2016 at 9:10
• @Pedro, @ Daniel. in which condition we have this uniqueness up to homeomorphism? always in 1 point compactification?
– Tom
Apr 21, 2016 at 9:27
• In the real projective space you add an infinite in each direction, so obviously more than a point for $n>1$. Apr 21, 2016 at 10:00
• @Martin, Could you say a little explicitly why is it? for example for dim 1, why PR is R + {infinity}?
– Tom
Apr 21, 2016 at 10:55

Any closed connected manifold is a compactification of $\mathbb{R}^n$.
To show this, we have to find, for each closed manifold $M$, an embedding $\mathbb{R}^n\rightarrow M$. One way to proceed is to use Morse theory. Any closed manifold admits a Morse function with One minimum. The stable manifold of the minimum is an embedded submanifold diffeomorphic to $\mathbb{R}^n$. As there are no other local minima, this must be dense in the manifold.
Of course this uses some theory. I think you are on the right track with your observation. What is an obvious map $\mathbb{R}^n\rightarrow \mathbb{R}P^n\setminus \mathbb{R}P^{n-1}$?