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Let $\{x_i : i \in I\}$ be an indexed set in a Banach space $X$ . We say $\sum_{i \in I} x_i$ converges to $x \in X$ , if for every $\epsilon >0$ , there is a finite set $J_{\epsilon} \subseteq I$ such that for every finite set $J \subseteq I$ with $J_{\epsilon} \subseteq J$ , $||\sum_{i\in J} x_i-x||<\epsilon$ .

So if $\sum_{i \in I} x_i$ converges to $x \in X$ , then is it true that $x_i$ is non-zero for at most countably many $i \in I $ ?

I think $x_i=0 , \forall i \notin \cup_{n \in \mathbb N}J_{1/n} $ ( where each $J_{1/n}$ is a finite set such that for every finite set $J \subseteq I$

with $J_{1/n} \subseteq J$ , $||\sum_{i\in J} x_i-x||<1/n$) , but I am not sure as I am unable to prove it . Please help . Thanks in advance

  • Maybe I'm wrong, but consider $X=\mathbb R$, $I=\mathbb N\setminus {0}$ and $x_i=\frac 1{i^2}$. For each $\epsilon$, there is some $N$ such that $\sum_{i=N}^\infty \frac 1{i^2} <\epsilon$. Let $J_\epsilon = {1,\ldots,N}$. If $J_\epsilon \subset J$, $$\sum_{i=1}^\infty \frac 1{i^2} - \sum_{i\in J} \frac 1{i^2} \leq \sum_{i=N}^\infty \frac 1{i^2} \leq \epsilon$$ But all the $x_i$ are non-zero. – Gabriel Romon Mar 19 '17 at 09:44
  • @LeGrandDODOM : your $I$ is countable , so at most countably many $x_i$ are non-zero , that doesn't contradict my claim .. so what's your point ? –  Mar 19 '17 at 12:06
  • sorry, my mistake ! – Gabriel Romon Mar 19 '17 at 12:12

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Assume that uncountably many elements do not vanish. Then (why exactly?), there is some $n \in \mathbb{N}$ such that the set $$ I_n := \{ i \in I \,:\, \| x_i \| \geq 1/n \} $$ is infinite (even uncountable).

But your assumption yields a finite set $J' \subset I$ such that every finite set $J' \subset J \subset I$ satisfies $$ \|x - \sum_{i \in J} x_i \| < \frac{1}{2n}. $$

By the triangle inequality, this implies that if $J' \subset K,L \subset I$ are finite, then $$ \| \sum_{i \in K} x_i - \sum_{i \in L} x_i \| < \frac{1}{n}. $$

But since $I_n$ is infinite, there is some $i_0 \in I_n \setminus J'$. By applying the previous estimate with $K = I_n$ and $L = I_n \cup \{i_0\}$, we finally get $$ \| x_{i_0} \| = \| \sum_{i \in K} x_i - \sum_{i \in L} x_i \| < \frac{1}{n}, $$ in contradiction to $i_0 \in I_n$. This is the desired contradiction.

PhoemueX
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