S^n\{x} isomorphic to R^n

• Nov 21st 2009, 04:32 AM
georgel
S^n\{x} isomorphic to R^n
Could someone please tell me why is $\displaystyle S^n-{x}$ isomorphic to $\displaystyle \mathbb{R}^n$?

Thank you so much.
• Nov 21st 2009, 05:14 AM
Shanks
$\displaystyle S^{n+1}-x$ is isomorphic to $\displaystyle R^n$, not $\displaystyle S^n-x$.
• Nov 21st 2009, 05:18 AM
georgel
Quote:

Originally Posted by Shanks
$\displaystyle S^{n+1}-x$ is isomorphic to $\displaystyle R^n$, not $\displaystyle S^n-x$.

Oh, thanks!
And could you please tell me why is that?
• Nov 21st 2009, 08:32 AM
Jose27
Quote:

Originally Posted by Shanks
$\displaystyle S^{n+1}-x$ is isomorphic to $\displaystyle R^n$, not $\displaystyle S^n-x$.

Well, that depends on your notation. The one I have seen the most is $\displaystyle \mathbb{S} ^n = \{ x\in \mathbb{R} ^{n+1} : \Vert x \Vert =1 \}$ in which case it is indeed true that $\displaystyle \mathbb{S} ^n \setminus {e_n} \cong \mathbb{R} ^n$

To prove it, try to generalize the proof of the stereographic projection used to build the Riemann sphere which can be found in any (good) book on basic complex analysis