Prove that the n-sphere equals Euclidean n space plus a point at infinity I would like to know if anyone has a proof for n>2 of the fact that the n-sphere equals Euclidean n-space plus a point at infinity.
For n<3, the proof is rather straightforward using stereographic projection.  For higher dimensions, I would prefer to avoid the "That too is just sterographic projection" proof, unless there is a proof that such projection works in higher dimensions.
 A: Stereographic projection works in every dimension. Consider the $n$-sphere canonically embedded in $n+1$-space as the set of points with norm $1$, and consider the copy of $n$-space in $n+1$-space as the set of points $E$ with zero last coordinate, $(y_1,\ldots,y_n,0)$. Given a point in the sphere $x$ different from the north pole $N=(0,\ldots,0,1)$, consider the line $L:\lambda (x-N)+N$ that passes through $N$ and $x$. If you consider the intersection of $L$ with the copy of $n$-space, that is solve the equation $$\lambda (x-N)+N=(y_1,\ldots,y_n,0)$$
you obtain $\lambda (x_n-1)+1=0$ or what is the same, $\lambda =\frac{1}{1-x_n}$. Thus the map sends
$$(x_1,\ldots,x_n, x_{n+1})\in S^n\longrightarrow \frac{1}{1-x_{n+1}} (x_1,\ldots,x_n,0)$$
This is the stereographic projection. The inverse is obtained analogously, since $x_i/(1-x_{n+1})=y_i$ for $i=1,\ldots,n$, squaring this and summing gives $$\frac{1}{(1-x_n)^2}(x_1^2+\cdots+x_n^2)=\frac{1-x_{n+1}^2}{(1-x_{n+1})^2}=y_1^2+\cdots+y_n^2$$
You can obtain then that $x_{n+1}=\dfrac{y_1^2+\cdots+y_n^2-1}{y_1^2+\cdots+y_n^2+1}$ and then $x_i=(1-x_{n+1})y_i$ gives the other coordinates, and the inverse for the projection.
