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From this answer, I can see that to show that for the $2$-dimensional case, I can use

$$ t\mapsto\begin{cases} (x,y)=\left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t\ne\infty, \\[10pt] (x,y)=\lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{otherwise}. \end{cases} $$

But what about the $3$-dimensional case? I feel like it should look something like this:

$$ t\mapsto\begin{cases} (x,y,z)=\left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2}, t \right) & \text{if }t\ne\infty, \\[10pt] (x,y,z)=\lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2},t \right) & \text{otherwise}. \end{cases} $$

as that's kind of what you do when you parametrize a sphere, but it doesn't work. We have that $t=\dfrac{y}{x+1}$, so at $z=0$ we would have $y=0$, which wouldn't describe a circle at all.

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man_in_green_shirt
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  • I guess you have to use stereographic projection here. –  Sep 21 '15 at 17:06
  • Ok, I think I've got it. As $\mathbb{R}^2\cup {\infty}$ is compact (as it is the one-point compactification of $\mathbb{R}^2$), the $2$-sphere is Hausdorff, and the stereographic projection is continuous, it follows that $\mathbb{R}^2\cup {\infty}$ is homeomorphic to the $2$-sphere. Does this look ok? Thanks – man_in_green_shirt Sep 21 '15 at 18:57
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    Looks fine but it is better if you explicitly write the stereographic projection –  Sep 22 '15 at 02:07

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