Can the sum of uncountably many real numbers be $0$?

Question

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 $0$

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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?

Answer 1

You say

Let us define $r := \sum\limits_{x \in S} f(x)$

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.

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In uncountable sets, the trouble is thus two fold:

1. The ordering problem from countable sets persists in uncountable sets.
2. The sequence definition has no obvious generalization to uncountable "sequences".

Answer 2

Before you can make any claim about the value of $\ \sum_\limits{s\in S}g(s)\ $, you first need to give a precise definition of what it means.

If $\ S\ $ is countably infinite, and there is a bound $\ M\ $ such that $\ \sum_\limits{s\in F}|g(s)|<M\ $ for all finite subsets $\ F\ $ of $\ S\ $ then $\ L=$$\lim_\limits{n\rightarrow\infty}\sum_\limits{j=1}^ng(e(j))\ $ exists and has the same value for any bijective function $\ e:\mathbb{N}\setminus\{0\}\rightarrow S\ $, so you can reasonably define the value of $\ \sum_\limits{s\in S}g(s)\ $ to be $\ L\ $.

If $\ S\ $ is uncountable and there is a bound M such that $\ \sum_\limits{s\in D}|g(s)|<M\ $ for all countable subsets $\ D\ $ of $\ S\ $, then you can (with a little bit more work) similarly define a reasonable value for $\ \sum_\limits{s\in S}g(s)\ $. In this case, however, it's not difficult to show that the set $\ \{s\,|\,g(s)\ne0\}\ $ must be countable, and that the "reasonable value" is just $\ \sum_\limits{\{s\,|\,g(s)\ne0\}}g(s)\ $.

For your function $\ f\ $, however, there exists no such bound $\ M\ $, since, for any that you might propose, there will be a finite subset $ F\ $ of $\ [-1,1]\ $ such that $\ \sum_\limits{s\in F}|f(s)|>M\ $. The definition referred to above is therefore of no use to you. For your statement $\ \sum_\limits{x\in[-1,1]}f(x)=0\ $ to make any sense, you need to first provide a generalisation of the definition which enables you to calculate reasonable values for $\ \sum_\limits{x\in [-1,1]}g(s)\ $ for a broader range of functions $\ g\ $ which includes your $\ f\ $. As far as I know, noone had ever done this, and I doubt if any such generalisation would turn out to particularly useful.

Answer 3

There is a perfectly good and standard way to define an unordered sum. Take any nonempty set $S$ and any real function $f:S\to \mathbb R$. Define the expression

$\sum_{s\in S} f(s)$

as follows... well I have written this up so many times that I prefer to give you a link. See Section 3.3. Infinite Unordered Sums in this free textbook:

Elementary Real Analysis-PDF

In that text there is also a proof of this theorem:

Theorem 3.5: Suppose that $\sum_{i\in I} a_i$ converges. Then $a_i = 0$ for all $i ∈ I$ except for a countable subset of $I$.

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If this proves to be of some interest you will want to know this theorem too, from the same elementary textbook, covering the case when the index set is the set of all positive integers $\mathbb N$:

Theorem 3.50: A necessary and sufficient condition for $\sum_{i\in \mathbb N} a_i$ to converge is that the series $\sum_{i=1}^\infty a_i$ is absolutely convergent and in this case the two values are equal.

Answer 4

If you assume the numerocity of real numbers on an interval $[0,1)$ as $N$, then you can reasonably define for a continuous function $f(x)$

$\sum_{[a,b]} f(x)=N\int_a^b f(x)dx +f(a)/2+f(b)/2$

and

$\sum_{(a,b)} f(x)=N\int_a^b f(x)dx -f(a)/2-f(b)/2$

The question remains, how to define $N$. This question on Mathoverflow proposes a fuzzy idea on how to define $N$ in terms of divergent series. In this approach, $N=\left(\sum_{k=0}^\infty 1\right)^{\sum_{n=1}^\infty 1}$ This is not a theory, just an idea based on decimal representation of real numbers.