Is the riemann sphere compact even though the complex plane isn't? 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.
 A: Riemann sphere without infinity is just the complex plane again, which is not compact.
The reason that $\mathbb C$ is not compact is because a sequence like $\{1,2,3,\cdots\}$ which tends to infinity does not have a convergent subsequence, whereas any sequence on the Riemann sphere does.
The addition of infinity of called the one point compactification.
(N.B. what you said about surface area equals cross product of x and y equals aleph-something is probably wrong)
A: 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,
$$
U_{395} \ , \quad V \ .
$$
A: For the Riemann sphere, it is the one point compactification of the plane. So yes, rather by definition it is compact. 
Generally, a non-compact topological space can be compactified by suitable addition of "points at infinity". Particularly useful ones (besides the one-point compactification above) include the Stone-Cech compactification, which in some sense adds the "most points", and whose existence for any topological space follows from axiom of choice. 
Another compactification that often arises is the Penrose/conformal compactification of space-time in general relativity. 
A propos your question on hyperbolic plane. Topologically the hyperbolic plane is the same as the normal plane. By considering the Poincare disk model and "adding" the boundary of the disk, you compactify the hyperbolic plane. As mentioned above, any topological space admits some compactification. What's more useful is to consider special compactifications: for example the one-point compactification of the complex plane into the Riemann sphere is conformal relative to the additional geometrical structure on the two surfaces. Similarly the Penrose compactifications respect the geometrical structure on space-times. 
