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I'm working through Dr. Pete Clark's convergence notes here:

and I've been thinking about Exercise 3.2.2 (a) and I am completely stumped.

The exercise says to show that a series converges only if it has at most countably many non-zero terms. I would like to start to argue by contradiction, and suppose that a net $(x_{\alpha})$ has uncountably non-zero terms. I'm drawing a blank on how I can build divergence of the series, based on the definition:

A series $\sum_{\alpha\in I}x_{i}$ is said to unconditionally converge to $x$ if for all $\epsilon > 0$, there exists a finite $J\subset I$ such that whenever $K$ is a finite subset of $I$ such that $I\supset K\supset J$, we have $|\sum_{\alpha\in K}x_{\alpha} - x| < \epsilon$.

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If this is the exercise on p. 10, you don't actually have a net at that point, since $I$ has no structure, and your definition of unconditional convergence isn't quite right. – Brian M. Scott May 8 '12 at 1:52
OK. I see what you mean, when he said $S$ is a naked set, then we cannot assume it is directed. But I double checked my definition of convergence with the one given at the top of page 11, and I don't see the error. – roo May 8 '12 at 2:03
The sets $J$ and $K$ are required to be finite. (I'm writing up an answer, by the way.) – Brian M. Scott May 8 '12 at 2:07
Aha! Thanks! You're right! :) – roo May 8 '12 at 2:07
up vote 2 down vote accepted

Your definition of unconditional convergence to $x$ should read as follows:

$\sum_{\alpha\in I}x_\alpha$ converges unconditionally to $x$ if for each $\epsilon>0$ there is a finite $J(\epsilon)\subseteq I$ such that whenever $I\supseteq K\subseteq J(\epsilon)$ and $K$ is finite, we have $\left|x-\sum_{\alpha\in K}x_\alpha\right|<\epsilon$.

You said that you want to show that $\sum_{\alpha\in I}x_\alpha$ converges unconditionally iff $S=\{\alpha\in I:x_\alpha\ne 0\}$ is at most countable; the problem actually asks only for the implication $(\Rightarrow)$. There's a good reason for this: the other implication is false, as you may see by taking $I=\Bbb N$ and $x_i=1$ for each $i\in I$.

Your idea of assuming $S$ is uncountable to prove the true $(\Rightarrow)$ implication is fine. Suppose that $S$ is uncountable, and for each $n\in\Bbb N$ let $S_n=\{\alpha\in S:|x_\alpha|\ge 2^{-n}\}$. Since $S$ is uncountable, there must be some $n\in\Bbb N$ such that $S_n$ is uncountable. Now split $S_n$ into $S_n^+=S_n\cap(0,\to)$ and $S_n^-=S_n\cap(\leftarrow,0)$, the positive and negative halves of $S_n$; at least one of these must be uncountable, say $S_n^\sigma$. Now let $J$ be any finite subset of $I$. For any $k\in\Bbb N$ you can expand $J$ to a finite sets $K_k\supseteq J$ by adding $k$ elements of $S_n^\sigma$ to to $J$; what happens to the sequence of sums

$$\left\langle\sum_{\alpha\in K_k}x_\alpha:k\in\Bbb N\right\rangle$$

when you do this?

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awesome! thanks so much! – roo May 8 '12 at 2:37
My pleasure!$\;$ – Brian M. Scott May 8 '12 at 2:37

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