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Let $\{ x_i : i \in I \}$ be a family of numbers $x_i \in \mathbb R$ with $I$ an arbitrary index set. We say that this family is summable with value $s$ (and write $s = \sum_{i \in I} x_i$ then) if for every $\varepsilon > 0$ there exists some finite set $I_{\varepsilon}$ such that for every finite superset $J \subseteq I$, i.e. such that $I_{\varepsilon} \subseteq J$, we have $$ \left| \sum_{i \in J} x_i - s \right| < \varepsilon. $$

Does there exists a family of numbers $\{x_i : i \in I\}$ with uncountable $I$ such that $\sum_{i \in I} x_i = 1$ and such that for every countable $J \subseteq I$ we have $$ \sum_{j \in J} x_j < 1 $$ i.e. the countable "sub"-sums have a strictly smaller value?

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  • $\begingroup$ This reminds me of the counterexamples used to clarify the distinction between $\sigma$-additivity and finite additivity in probability theory. $\endgroup$
    – mlc
    Apr 4, 2017 at 11:05
  • $\begingroup$ @mlc What you mean? It is quite easy to give a countable family, like $x_i = 1/2^i$ whose sum is $1$, but every finite subsum is strictly smaller than $1$; when you mean that I ask for a similar example for uncountable summation. $\endgroup$
    – StefanH
    Apr 4, 2017 at 11:20
  • $\begingroup$ See f.i. math.stackexchange.com/questions/256211/… but I am not sure that it directly addresses your question. $\endgroup$
    – mlc
    Apr 4, 2017 at 12:00

3 Answers 3

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You cannot find such a sequence. Take a look at my answer in on-a-necessary-and-sufficient-condition-for-sum-k-in-mathbbza-k-l-a-k-i

If $\sum_{i\in I}x_i=1$, then $\sum_{i\in I}|x_i|\le C$. This implies that $x_i=0$ for all but countably many $i$. To see this let $I_k=\{i\in I:\, |x_i|\ge \frac1k\}$. Then for every finite set $F$ of $I_k$, $$\frac1k \text{card} F\le \sum_{i\in F}|x_i|\le C,$$ which shows that $\text{card} F\le k C$. In turn, $I_k$ must have only finitely many elements. Hence, ${i\in I: x_i\ne 0}=\cup_k I_k$ is countable. So the infinite sum is just a series.

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  • $\begingroup$ Why does convergence imply absolute convergence? Did i miss a prerequisite? If i take a_n=(-1)^n * 1/n, the sum converges by leibniz criterion. However it does not Converge absolutely. It is also easy to modify the a_n, making them convergent towards 1. $\endgroup$
    – axioman
    Apr 15, 2017 at 8:17
  • $\begingroup$ for a series it does not, but for an infinite sum it does. The definitions are different. The proof that a convergent infinite sum is absolutely convergent is in the link I gave. $\endgroup$
    – Gio67
    Apr 15, 2017 at 11:42
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If $\sum_{i\in I} x_i$ converges as a sum in $\Bbb R$ then $I(x_i{\neq0}):=\{i\in I\mid x_i\neq0\}$ has to be countable. This is because $$I_{\neq0}=\bigcup_{n\in\Bbb N} I(x_i{>\frac1n})\cup I(x_i{<\frac{-1}n})$$ and if this were uncountable, there would have to be one term in the union that is uncountable (since a countable union of countable things is countable).

This means you've got an $n$ so that infinitely many elements are larger than $1/n$ (or infinitely many are smaller than $-1/n$). Well for any finite $I_\epsilon$ you can consider $J_k=I_\epsilon\cup \{k\cdot n\text{ elements of }I(>\frac1n)\}$ and then $\sum_{i\in J_k}x_i>\sum_{i\in I_\epsilon} x_i +k$, where you can make $k$ as big as you like.

But you can have uncountable sums in other spaces.

For example if you consider the space of functions $\Bbb R\to\Bbb R$, this is a topological vector space equipped with a family of semi-norms $\{\|\cdot\|_r\mid r\in\Bbb R\}$ where $\|f\|_r:=|f(r)|$. If you let $\delta_{r}$ be the function that is one when $x=r$ and $0$ otherwise you have that $$\sum_{r\in \Bbb R} f(r)\delta_r$$ converges to $f$ for any function in this topology. For example $$\sum_{r\in\Bbb R}\delta_r$$ converges to the constant function $1$ and none of the uncountably many summands are zero elements.

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Assume that $$ \sum_{i\in I} |x_i| < \infty, $$ and define $J_n = \{i \in I: |x_i| \geq 1/n$}, for $n > 0$, and $J_{\infty} = \{i\in I: x_i \neq 0\}$. By assumption, $J_n$ must be a finite set for all $n$, so $J_\infty = \bigcup_{n = 1}^{\infty}J_n$ is countable.

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