I wish to prove that there exist uncountably many non-zero real numbers, such that their sum is convergent. I came up with an example that I thought sums to 0. But I know it cannot work, because of this question and similar questions here and here.
Question: The below claim/example doesn't work. Why?
Let $S = \{x| x\in[-1,1]\}$. Note that S has uncountably many elements. Let $f(x)$ be some odd function. In other words, $f(-x) = -f(x)$ for all $x\in \mathbb{R}$.
Let us define $r := \sum\limits_{x \in S} f(x)$
Claim: $r = 0$
Wrong Proof: For each element $y\in S\setminus\{0\}\enspace$, $\exists (-y) \in S\setminus\{0\}$ such that $f(y) + f(-y) = 0$. Therefore the total sum of elements in $S\setminus\{0\}$ is $0$. Further, since the function is odd, $f(0) = 0$. Therefore r, the sum of uncountably many terms, is $0$
Can someone please help me with why the above proof could be wrong? Does it have anything to do with multiple cancellation mappings of which I have given only one?
Best Answer
You say
but then simply continue with a claim about $r$. Where is your definition of what $r$$ is equal to? How do you define
$$\sum_{x\in S} f(x)?$$
For a finite set $S$, the value $$\sum_{x\in S}f(x)$$ can easily be defined as a finite sum (the actual strict definition would of course be recursive).
For any contably infinite set $S$, we already have problems. For a countable sequence $x_1, x_2,\dots$, the typical definition of $$\sum_{i=1}^\infty f(x_i)$$ is the definition $$\sum_{i=1}^\infty f(x_i):=\lim_{N\to\infty}\sum_{i=1}^N f(x_i).$$
but you can already have problems for countably infinite sets. You could just say that if $S$ is countably infinite, then we have $S=\{x_1,x_2,\dots\}$ and we can define $$\sum_{x\in S}f(x) = \sum_{i=1}^\infty f(x_i)$$ but this is actually not a good definition because the definition is dependent on how precisely we order the values of $S$. For example, if $S=\mathbb N$ and $f(n)=\frac{(-1)^n}{n}$, then depending on how you order $\mathbb N = \{n_1, n_2, n_3,\dots\}$, the sum of $$\sum_{i=1}^\infty \frac{(-1)^{n_i}}{n_i}$$ can actually be any real number.
In uncountable sets, the trouble is thus two fold: