Prove that $\sum_{x\in X}f(x)$ is absolutely convergent iff $\sup\{\sum_{x\in A}|f(x)|:A\subseteq X, A$ finite $\}<\infty$ I’m trying to prove the following result:
“Let $X$ be a countable set , and let $f:X\to\mathbb{R}$ be a function.
Then the series $\sum_{x\in X}f(x)$ is absolutely convergent iff $\sup\{\sum_{x\in A}|f(x)|:A\subseteq X, A$ finite $\}<\infty$.”
I've managed to come up with a proof of the leftward implication but not of the rightward one, so I would appreciate any hints about how to carry out this part of the proof.
Best regards,
lorenzo
--
Note: The book says that the following result might be useful:
"if $f:\{i\in \mathbb{N}:1\leq i \leq n\}\to \mathbb{N}$ is a function then there exists $M \in \mathbb{N}$ such that $f(i)\leq M$ for all $1\leq i \leq n$"
--
The sum on finite and infinite sets are defined as follows:
DEF. (Summations over finite sets)
Let $X$ be a finite set with $n \in \mathbb{N}$ elements, and let $f:X\to \mathbb{R}$ be a function.
Then $\sum_{x\in X}f(x):= \sum_{i=1}^n f(g(i))$, where $g:\{i\in\mathbb{N}: 1 \leq i \leq n\}\to X$ is any bijection from
$\{i\in\mathbb{N}: 1 \leq i \leq n\}$ to $X$; such a bijection exists since $X$ is assumed to have $n$ elements.
DEF. (Series on countable sets)
Let $X$ be a countable sets, and let $f:X\to\mathbb{R}$ be a function.
We say that the series $\sum_{x\in X}f(x)$ is $absolutely$ $convergent$ iff the sum $\sum_{n=0}^\infty f(g(n))$ is absolutely convergent for some bijection $g:\mathbb{N}\to X$. We then define $\sum_{x\in X}f(x):=\sum_{n=0}^\infty f(g(n))$.
It can be shown that these two definitions do not depend on the choice of $g$, and so are well defined.
 A: What is your definition of convergence of $\sum_{x\in A}f(x)$?
Here is a solution in the language of convergence of nets, which seems the natural way to deal with theses kinds of problems.
Some generalities:

*

*A directed set $(D,\leq)$ is a partially order set such that if $c,d\in D$, then there is $g\in D$ such that $c\leq g$ and $d\leq g$.


*A (numeric) net is a function $F:(D,\leq)\rightarrow\mathbb{C}$ . The net $F$ converges to some number $\ell\in\mathbb{C}$ if for any $\varepsilon>0$, there is $d_\varepsilon\in D$ such that $|F(d)-\ell|<\varepsilon$ whenever $ d_\varepsilon\leq d$.

In your setting, consider the collection $\mathcal{D}$ of all finite subsets of $X$ and define $A\leq B$ iff $A\subset B$. Then $F(C)=\sum_{x\in C}f(x)$   defines a net. Your question may be reformulated as showing that the net $\tilde{F}(C)=\sum_{x\in C}|f(x)|$ converges iff the set $V=\{\tilde{F}(C):C\in\mathcal{D}\}$ is bounded.

*

*Suppose $V$ is bounded and let $\alpha=\sup V$. Then $\alpha>\infty$. Given $\varepsilon>0$, there is $C\in\mathcal{D}$ such that $\alpha-\varepsilon < \tilde{F}(C)$.
Notice that for any set $D\in\mathcal{D}$ such that $C\leq D$, $\tilde{F}(C)\leq \tilde{F}(D)\leq\alpha$. Putting things together gives
$$ |\tilde{F}(D)-\alpha |<\varepsilon \qquad\text{whenever} \qquad C\leq D$$
This shows that the net $\tilde{F}(C)$ converges to $\alpha$.


*Conversely, suppose $\lim_{C}\tilde{F}(C)$ exists and equal $\ell$. Then, there is $C_0\in\mathcal{D}$ such that
$$ |\tilde{F}(D)-\ell|<1\qquad\text{whenever}\qquad C\leq D$$
This means that for all finite sets $D\supset C$,
$$\tilde{F}(D)<\ell+1$$
Since for any $0\leq\tilde{F}(E)\leq \tilde{F}(C)$ for all $E\subset C$, we conclude that the set $V=\{\tilde{F}(D): D\in\mathcal{D}\}$ is bounded. The first part will show that in fact $\ell=\sup V$.
