Stereographic projection continuous at $\infty$ I have trouble with showing that the stereographic projection is continuous on $\infty$. I was given those two functions:
$$\pi:S^2\to\mathbb{C}\cup \{\infty\}$$
$$\pi(x_1,x_2,x_3) =\begin{cases}
\frac{x_1+ix_2}{1-x_3} & \text{if } (x_1,x_2,x_3) \neq (0,0,1) \\
   \infty       & \text{if } (x_1,x_2,x_3) = (0,0,1)
  \end{cases}$$
$$\psi : \mathbb{C}\cup \{\infty\}\to S^2$$
$$\psi(z=x_1+ix_2) =\begin{cases}
\frac{1}{1+x_1^2+x_2^2}(2x_1,2x_2,x_1^2+x_2^2-1) & \text{if } z \neq \infty \\
   (0,0,1)      & \text{if } z = \infty
  \end{cases}$$
And I was told to show that $\pi$ is an homeomorphism. To show that I need to show that $\pi$ is continuous, bijective and that its inverse is also continuous. 
I see that the denominator of the two functions is never equal to zero, because for $\pi$ the only possible case would be (0,0,1) but this has another value and for $\psi$ is always greater than 1 and for this reason we can say that $\psi$ is continuous since composition of continuous functions. I've also shown that $\pi(\psi(z))$ = z and that $\psi(\pi(x_1,x_2,x_3))$ = $(x_1,x_2,x_3)$ (This tells us that $\pi$ is the inverse function of $\psi$ and viceversa.)
My problem tells me to show that $\pi$ and $\psi$ are continuous in infinity, but I don't know how to show this. I wanted to use the limit criterion or the delta-epsilon criterion, but I'm not sure that I can use those for vectors (in case of $\pi$). Do you have any hint that would help me solving this problem?
I've also searched on the internet, but every source that I found said that this problem has an obvious solution, however for me it's not.
 A: An $\epsilon-\delta$ proof would be difficult since you don't have any definition of distance from infinity. So limits would be your best bet.
For a sequence $\mathbf x_n$ on the sphere converging to the north pole, show that $\liminf_{n \to \infty}|\pi(\mathbf x_n)| = \infty$.
Going the other way, if you have a sequence $\mathbf y_n \in \mathbb C\cup \infty$ so that $\liminf_{n \to \infty} |\mathbf y_n| = \infty$, then show that $\psi(\mathbf y_n)$ converges to the north pole.
For a sequence $\mathbf z_n \in \mathbb C$, saying that $\liminf_{n \to \infty} |z_n| = \infty$ means that for any positive real number, only finitely many of the $\mathbf z_n$ are within that distance from the origin. It is, in this setting, the most sensible way to define "converging to infinity".
A: I think there is something wrong in your problem. They're asking you to prove that the Alexandroff's compactification of $\mathbb{C}$ is $S^2$, right?
Well, for this you have indeed the stereographic projection and its inverse that are usual, continuous maps, whit no magic in them, if you just think about them as maps
$$
\pi : S^2 \setminus \left\{ (0,0,1)\right\} \longrightarrow \mathbb{C} \qquad \text{and}\qquad \psi: \mathbb{C} \longrightarrow S^2 \setminus \left\{ (0,0,1)\right\} \ .
$$
This way they're just plain continuous, with the usual topologies on both sides, bijective maps, inverse one to each other. Right?
Particularly, $\psi$ is an immersion into $S^2$ (continuous injective map such that $\mathbb{C} \cong \psi (\mathbb{C})\subset S^2 $ ) that leaves out of its range just one point (namely, $(0,0,1)$).
Ok, now use the following easy lemma.
Lemma. Let $X$ a locally compact and Hausdorff space and $\psi : X \longrightarrow Y$ an immersion into a compact Hausdorff space $Y$ such that $Y\setminus \psi (X)$ reduces to one point. Then $Y$ is the (homeomorphic to an) Alexandroff's compactification of $X$.
