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Why is a summable family at most countable ?

Let $(a_i)_{n\in I}$($a_i\in [0,\infty],\forall i$) is summable then, $\{i\in I:a_i≠0\}$ is at most countable.

a family is said to be summable if $\sum_{i\in I}a_i<\infty$

Can we always choose a countable set s.t., every member of this set is bigger then some $\epsilon$, but how ?

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derivative
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2 Answers2

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For any positive $\varepsilon$ denote by $I_\varepsilon$ the set of all indices for which $a_i$ is greater than $\varepsilon$: $I_\varepsilon=\{i \in I \mid a_i > \varepsilon\}$.

Now it boils down to two things:

  1. $I_\varepsilon$ is finite for every $\varepsilon > 0$. This is the part where it is necessary to use that the sum $\sum_{i \in I} a_i$ is finite.
  2. Realize that your set $\{i \in I \mid a_i > 0\}$ can be represented as a union: $$ \{i \in I \mid a_i > 0\} = \bigcup_{n \in \mathbb{N}} I_{1/n} $$ Any union of a countable family of finite sets is at most countable, qed.

PS: There was no definition given for the sum $\sum_{i \in I}a_i$ in the question. But, since all $a_i$ are nonnegative, I just assumed that you mean $\sum_{i \in I}a_i = \sup_J \sum_{j \in J} a_j$, where $J$ iterates over all finite subsets of $I$.

Dan Shved
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  • Yes that's the definiton, but does it change anything ? – derivative Mar 24 '14 at 12:26
  • Change compared to what? I find it hard to imagine any other definition for an infinite sum of nonnegative numbers that is essentially different. But the actual execution of step 1 in the plan above does depend on the precise definition that you use. – Dan Shved Mar 24 '14 at 12:30
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Note that $$ \{i :\ a_i\ne0\}=\bigcup_n\{i: \ |a_i|> 1/n\}. $$

If the set on the left is uncountable, then at least one set on the right is infinite.

Now, when considering a series indexed by a net, one has an issue with the definition. The usual way to define the symbol $\sum_{i\in I}a_i$ is by taking the limit along the net of finite subsets of $I$, ordered by inclusion. This implies that absolute convergence is required.

In the case above, if $|a_i|>1/n$ for infinitely many $i$, then $\sum_i|a_i|=\infty$, and so the series cannot be convergent.

Martin Argerami
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