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In Elementary Real Analysis of Thomson and Bruckner, I'm stuck in exercise 3.3.4 page 88. Could you please help me giving me hints?

Let the infinite index set $I$. Show that if $\sum\limits_{i\in I}a_i$ converges so too does $\sum\limits_{i\in J}a_i$ for every subset $J\subset I$.

My attempt:

I should use either the definition of convergence:

enter image description here

or Cauchy's criterion: enter image description here

Cauchy's criterion seems to be the one that should be used since having $\sum\limits_{i\in I}a_i=c$ doesn't always mean that $\sum\limits_{i\in J}a_i=c$ (assuming the convergence).

I also tried using $\sum\limits_{i\in J}a_i=\sum\limits_{i\in I}a_i-\sum\limits_{i\in I\backslash J}a_i$ but I couldn't go furthur.

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    $\begingroup$ Use Cauchy's criterion. Given $\varepsilon > 0$, what does the convergence of $\sum\limits_{i\in I} a_i$ give you? How can you use that for $J$? $\endgroup$ – Daniel Fischer Apr 4 '15 at 14:00
  • $\begingroup$ @DanielFischer :o Hi again Daniel Fischer! Remember me? It's a honor to get helped by you again! Given $\varepsilon$ there exist $I_0\subset$ finite set so that $\left|\sum\limits_{i\in J}a_i\right|<\varepsilon$ for every finite subset $J\subset I\backslash I_0$. Given a set $J\subset I$, I really can't see how to use that for $J$ since $J$ is infinite (if $J$ is finite the sum trivially converges). $\endgroup$ – Scientifica Apr 4 '15 at 14:05
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    $\begingroup$ You use $J$ for two different things there, that may confuse you. Call one of the two sets by another name. You have a finite $I_0 \subset I$ such that $\left\lvert \sum_{i\in K} a_i\right\rvert < \varepsilon$ for all finite $K\subset I\setminus I_0$. Now, can you from $I_0$ construct a finite $J_0\subset J$ such that $\left\lvert \sum_{i\in M} a_i\right\rvert < \varepsilon$ for all finite $M\subset J\setminus J_0$? $\endgroup$ – Daniel Fischer Apr 4 '15 at 14:11
  • $\begingroup$ @DanielFischer I see. I'm going to try! $\endgroup$ – Scientifica Apr 4 '15 at 14:18
  • $\begingroup$ @DanielFischer We take $J_0=J\cap I_0$. We have $J\backslash J_0\subset I\backslash I_0$ which makes the proof trivial ^^. Without using a Venn diagram, I really wouldn't be able to solve this. I've always underestimated this diagram always trying to perceive thinks with my mind, but it is really helpful. Thank you very much for your help! $\endgroup$ – Scientifica Apr 5 '15 at 19:43
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Hints

Prove it assuming $a_i\ge 0$. This is much easier since you can apply the Monotone Convergence Theorem.

Then define $$I^+=\{i\in I:a_i\ge 0\}$$ and $$I^-=\{i\in I:a_i< 0\}$$ and the same for $J$.

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  • $\begingroup$ Nice idea! It's done for the case $a_i\ge 0$. Now for the general case, I'm really close. If I prove that the sum on $I^+$ converges if the sum on $I$ converges, then I'll be done. I can't see it now but I'm gonna try. Thank you! $\endgroup$ – Scientifica Apr 4 '15 at 14:15
  • $\begingroup$ Now the proof is complete. Thank you very much! $\endgroup$ – Scientifica Apr 5 '15 at 19:40

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