Show the countability of the non-zero elements when sum of elements is less than $\infty$ on the extended reals. Show that for $(x_{\alpha})_{\alpha \in A}, x_{\alpha} \in [0, +\infty]$ where $\sum_{\alpha \in A}x_{\alpha} < \infty$ the number of non-zero elements is at most countable. Note $A$ can be uncountable. Here's my attempt:
Since the sum is less than $\infty$ we must have a largest element $x_1$ a second largest element $x_2 < x_1$ and so the sequence:
$$x_1 \geq x_2 \geq x_3 \geq \dots$$ 
We can proceed down like this, counting as we go, until we reach some arbitrary $x_n$. Thus we have a finite number of elements. We can proceed further and reach another finite number of elements. We continue this process until we reach zero but have found a way to count all the non-zero elements.
 A: I think you proof is correct, but I think you have to write details more about existence of largest element. (If $\{x_{\alpha}\}$ is unbounded, then sum cannot be finite. If $\{x_{\alpha}\}$ is bounded, let $M=\sup x_{\alpha}>0$. If $x_{\alpha}<M$ for all $\alpha$, there exists infinitely many $\alpha$ with $x_{\alpha}>\frac{M}{2}$. So there exists a largest element, and continue this process.) 
There is an another approach : Let $[0,\infty)=[1,\infty)\cup[\frac{1}{2}, 1)\cup\cdots [\frac{1}{2^{n}}, \frac{1}{2^{n-1}})\cup\cdots\cup\{0\}$. If there exists $I_{n}=[\frac{1}{2^{n}}, \frac{1}{2^{n-1}})$ s.t. $I_{n}$ contains infinitely many elements of $\{x_{\alpha}\}$, then $\sum x_{\alpha}=\infty$. So each $I_{n}$ must contain finitely many elements of $\{x_{\alpha}\}$ and we are done.
A: You have a good idea, which is to put the list in order so that the positive numbers all come before the negative numbers and every finite partial sum is finite. Whenever you reach zero for the first time, you'll be done adding them up.
There are a few edge cases—what if there isn't a biggest element to start with, as with $\{1,2,3,4,\ldots\}$? Can you show that this problem can't happen when the sum is finite?—but the basic idea is good.
The real missing piece is to explain (to prove) how long it will take to reach zero once you've put the numbers in order. Can it take an infinite number of steps? If your numbers have a finite sum, then the list of positive terms might be finite or even countably infinite (as in $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots + \frac{1}{2^n} + \ldots=1$). But the list of positive terms cannot be un-countably infinite. If your numbers have a finite sum, then the positive terms must stop after a countable number of steps. That's what you must prove.
Sort the numbers into countable bins
Here's one way to do it. You divide up the positive numbers into a countable number of different intervals, as shown in the picture. The first interval $U_1$ consists of 1 and everything bigger than it. The next interval $U_2$ consists of everything between 1/2 and 1. The third interval $U_3$ consists of everything between 1/3 and 1/2. And so on.

These intervals don't overlap. Every number in your list belongs to exactly one of them. (Except the zeros, which don't belong.) Even an uncountably long list of numbers can be sorted into this countable collection of intervals.  Just order the list from biggest to smallest, per your idea, and you'll know where each number belongs.  This is the first key idea, that no matter how big your original list is, you can sort all the numbers in it into a countable number of bins.
Here's the next key idea. Look at an interval like $U_1$ and ask how many terms from your list have been put into it. If it's zero, or a finite number, that's fine. Any more, and you find a contradiction: if $U_n$ contains an infinite number of terms, and each number in $U_n$ is bigger than $\frac{1}{n+1}$, then their sum is at least $\sum_{i=1}^\infty \frac{1}{n+1} = \infty$. Your sum can't be finite the way you assumed.
So when the sum is finite, each interval $U_n$ has a finite number of terms from your list (possibly none). Because there are countably many intervals, each with a finite number of terms, the total number of terms is finite or countably infinite.
Hence when your list has a finite sum, the total number of positive terms must be finite or countably infinite. The rest of the terms, if any, must all be zero. And that's what we set out to prove!
A: Let:
$$B_n = \{ \alpha \in A \mid x_{\alpha} \ge 1/n \}, B = \bigcup_n B_n$$
We have:
$$\# B_n = \sum_{\alpha \in B_n} 1 \le \sum_{\alpha \in B_n} n x_{\alpha} \le n \sum_{\alpha \in A} x_{\alpha} < \infty$$
Hence, each $B_n$ is finite, and $B$ is at most countable. But if $x_{\alpha} \neq 0$, then $x_{\alpha} \in B$, so the set of non-zero $x_{\alpha}$'s is at most countable.
