$\infty\pm\infty$ on a Riemann sphere Here I use "$\infty$" to represent the infinite point on a Riemann sphere. I read that 
(1) $z\cdot\infty = \infty$ for any complex non-zero $z$.
(2) $\infty+\infty=\infty$.
(3) $\infty-\infty$ is indeterminate.
(1) makes sense to me. I would expect that (2) and (3) are indeterminate. Although if  (2) is Inf, I expect that would mean that (3) is also Inf due to (1) and (2). If (1), (2), and (3) are correct as written above, why does it work that way? 
----EDIT----
I take back (3) above, since I can't find a source that says that. 
Howecer, (1) and (2) can be found in Mathematical Analysis, 2nd edition, by Tom M. Apostol. He includes it in his section on the extended complex plane. One response to my initial question and Wikipedia said $\infty+\infty$ is undefined. Does that mean not all mathematicians agree on this?
 A: To say that $\infty-\infty$ is "indeterminate" means that if for every point $c$ in the Riemann sphere, there exist functions $f$ and $g$ and a point $a$ such that
\begin{align}
\lim_{z\to a} f(z) & =\infty, \\[6pt]
\lim_{z\to a} g(z) & = \infty, \\[6pt]
\lim_{z\to a} f(z)-g(z) & = c.
\end{align}
You can choose point $a$ can be chosen to be anything you like and make trivial modifications in $f$ and $g$ so that this all works for that value of $a$.
When working with real numbers rather than on the Riemann sphere, this does not work for $\infty+\infty$ nor for $c\cdot\infty$.  If $f(z)\to\infty$ and $g(z)\to\infty$ then $f(z)+g(z)\to\infty$.  You cannot choose $f$ and $g$ both approaching $\infty$ in such a way that $f(z)+g(z)$ approaches $19$, the way you can with $f(z)-g(z)$.  However, with the Riemann sphere, as some have noted in comments below, similar things happen.
A: The expressions $z\pm\infty$, $z\cdot \infty$ $(z\in{\mathbb C})$, $\infty\cdot\infty$, and $\infty\pm\infty$ are  undefined in the framework of field axioms for ${\mathbb C}$. 
Since for $z\in{\mathbb C}$ (resp. $\in{\mathbb C}^*$) one has $z+ w_n\to\infty$  and $z\cdot w_n\to\infty$ as $w_n\to\infty$ ($n\to\infty$), it makes sense to define "a posteriori" $z+\infty:=\infty$ for all $z\in{\mathbb C}$ and $z\cdot\infty:=\infty$ when $z\ne0$. In this way the functions $$\phi_z: \bar{\mathbb C}\to\bar {\mathbb C}, \quad w\mapsto w+z\ ,\qquad\psi_z: \bar{\mathbb C}\to\bar {\mathbb C}, \quad w\mapsto z\cdot w\ ,$$
where $z\in{\mathbb C}$ (resp. $\in{\mathbb C}^*$) is fixed, are continuous on all of $\bar{\mathbb C}$.
In a real environment (e.g., in measure theory), it makes sense to enlarge ${\mathbb R}$ by  the "ideal point" $+\infty$. As $x_n\to\infty$ and $y_n\to\infty$ imply $x_n+y_n\to\infty$ the convention $x+\infty:=\infty$, $\infty+\infty:=\infty$ makes addition continuous at $(\infty,\infty)$.
A: You are correct, although you should use "undefined" rather than "indeterminate" -- indeterminate is used not when you are doing arithmetic, but rather when you are trying to do other things, such as using limit forms to quickly compute limits.
The arithmetic of the Riemann sphere is the continuous extension of the arithmetic on the plane; if we did not remember that $\infty + \infty$ is undefined, we could use the previous fact to determine that ourselves, since we would have
$$ \lim_{x \to \infty} x + x = \infty + \infty = \lim_{x \to \infty} x + (-1) x $$
If we were working with the extended real numbers rather than the projective complex numbers, we would use two points at infinity $+\infty$ and $-\infty$. It is true that, in the extended real numbers, $(+\infty) + (+\infty) = +\infty$ but $(+\infty) - (+\infty)$ is undefined.
Unfortunately, people often use the same symbol for the projective infinity and the positive extended real infinity, which sometimes leads to confusion: the first equation in the previous paragraph would be written as $\infty + \infty = \infty$
A: This is not a mathematical issue, it's a memory trick or something when you want to find limits for instance.
