Let $M$ be a non-empty set and let $\mathfrak{X}$ be the set of all finite subsets of $M$. Let $\mathfrak{X}$ be a directed set by the relation $$\forall X,Y \in \mathfrak{X}, \quad X\le Y \leftrightarrow X\subset Y $$ For an arbitrary function $f: M \to \Bbb R$ we define the net $(s_X)_{X \in \mathfrak{X}}$ with $$s_X :=\sum_{x \in X}f(x) \textrm{ for } X \in \mathfrak{X}.$$
Now I have to show that $\lim\limits_{X \in \mathfrak{X}} (s_X)$ exists in $(\Bbb R, | \cdot |)$ if, and only if $\{x \in M : f(x) \not = 0 \}$ is finite OR when there exists a function $\varphi : \Bbb N \to M$ with $$\varphi (\Bbb N) = \{ x \in M:f(x) \not = 0\} \textrm { and } \sum_{n=1}^{\infty}|f(\varphi (n))| \lt \infty,$$ and I also have to specifically determine $\lim\limits_{X \in \mathfrak{X}} (s_X)$, it it exists.
I thinks i have proved the $\leqslant$ direction for the finite set condition but I cannot get the $\varphi $ condition because the $|\cdot |$ from the $|f(\varphi (n))|$ causes me troubles. This is also the problem I have with the $\leqslant$ direction.
Any ideas on how to prove this quickly? (or tipps on how to deal with the $|\cdot |$)
Thanks in advance!
Definition: $\displaystyle\lim_{X\in\mathfrak{X}}s_X=c\Leftrightarrow\forall \varepsilon>0,\exists X_0\in\mathfrak{X}\mbox{ finite},\forall X\in\mathfrak{X}$, $X_0\subset X\Rightarrow |s_X-c|<\varepsilon.$
Hint: Suppose that $\displaystyle\lim_{X\in\mathfrak{X}}s_X=c$ and that $\{x \in M : f(x) \not = 0 \}$ is infinite. Then the net is Cauchy, i.e. for every $\varepsilon>0$, there exists $X_0\in\mathfrak{X}$ finite such that $|s_x-s_y|<\varepsilon$, whenever $X_0\subset X,Y$ for $X,Y\in\mathfrak{X}$.
Thus, for any given $\varepsilon>0$ the set $\{x \in M : |f(x)|\geq\varepsilon\}$ is finite (this is similar to the proof of the implication: $\sum_{n=1}^\infty x_n$ is convergent $\Rightarrow\lim_{n\to\infty}x_n=0$. In that case, we write $|x_n|$ as $|\sum_{i=1}^{n}x_i-\sum_{i=1}^{n-1}x_i|$).
Hence $\{x \in M : f(x) \not = 0 \}$ can be written as the countable union of finite sets. Therefore $\{x \in M : f(x) \not = 0 \}$ is countable. In other words, there exists a function $\varphi:\mathbb{N}\to\mathbb{R}$ with $\varphi(\mathbb{N})= \{x \in M : f(x) \not = 0 \}$. Therefore $\displaystyle\lim_{X\in\mathfrak{X}}s_X=\sum_{n=1}^\infty f(\varphi(n))$.
Remark: Now, not always $\sum_{n=1}^\infty |f(\varphi(n))|$ is convergent. For a counterexample, consider $M=\mathbb{N}$ and $f:M\to\mathbb{R}$ defined by $f(n)=\frac{(-1)^n}{n}$. Here $\varphi(n)=n$. The series $\sum_{n=1}^\infty f(\varphi(n))$ is convergent but $\sum_{n=1}^\infty |f(\varphi(n))|$ is divergent.