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

**georgel** · November 21st 2009, 04:32 AM

Could someone please tell me why is $S^n-{x}$ isomorphic to $\mathbb{R}^n$ ?

Thank you so much.

**Shanks** · November 21st 2009, 05:14 AM

$S^{n+1}-x$ is isomorphic to $R^n$ , not $S^n-x$ .

**georgel** · November 21st 2009, 05:18 AM

Originally Posted by Shanks

$S^{n+1}-x$ is isomorphic to $R^n$ , not $S^n-x$ .

Oh, thanks!
And could you please tell me why is that?

**Jose27** · November 21st 2009, 08:32 AM

Originally Posted by Shanks

$S^{n+1}-x$ is isomorphic to $R^n$ , not $S^n-x$ .

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

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

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