$\infty\pm\infty$ 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?
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) &amp; =\infty, \\[6pt]
\lim_{z\to a} g(z) &amp; = \infty, \\[6pt]
\lim_{z\to a} f(z)-g(z) &amp; = 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.
This is not a mathematical issue, it's a memory trick or something when you want to find limits for instance.