TITLE: Is the riemann sphere compact even though the complex plane isn't?
QUESTION [6 upvotes]: The Complex plane, set of all $z=x+iy$ where $x$ and $y$ are real, surface area equals cross product of $x$ and $y$ equals aleph-something (that's not the question). Projecting the plane onto a sphere, and adding the complex infinity to the set (without a value for $\arg(z)$), gives the Riemann sphere, which, being a sphere, is compact. 
Does the addition of infinity, as some sort of bounder, make the complex plane compact? is the Riemann sphere without infinity compact, since it has an open boundary? also, if the hyperbolic plane can be shown in a disk, what does that mean? The answer to this question might primarily clarify the definitions in my new studies of topology of manifolds, etc. 
thanks.

REPLY [3 votes]: The complex plane $\mathbb{C}$ is not compact. For instance, here is an open cover that does not admit any finite subcover:
$$
{\mathcal U} = \left\{ U_n\right\} \qquad \text{with} \qquad U_n = \left\{ z \in \mathbb{C} \ \vert \ \vert z \vert < n \right\} \ .
$$
The reason why ${\mathcal U}$ does not admit any finite subcover is easy: if you pick just a finite number of $U_n$'s, then their union would be equal to the biggest one of them, which is not the whole $\mathbb{C}$.
Now, thanks to the stereographic projection, think of these $U_n$'s inside the Riemann sphere $S^2$. They do not form a cover of $S^2$, since $\infty$ (= the North Pole) is not included in any of them. So, we need to add an open neighborhood of the infinity. For instance, the image of 
$$
V = \left\{ z \in \mathbb{C} \ \vert \ \vert z \vert > 394 \right\}
$$
by the stereographic projection. Then, we would have an open cover of $S^2$:
$$
{\mathcal U}' = \left\{ U_n \ \vert \ n \in \mathbb{N}\right\} \cup \left\{ V\right\} \ ,
$$
which admits and open subcover. For instance,
$$
U_{395} \ , \quad V \ .
$$