Definition
Let $S$ be a set. Let $f: S \to K$ be a function where $K$ is either the field of the real or the complex numbers.
Then, we say that the series $\sum_{s \in S}f(s)$ converges to $F$, and we write $\sum_{s \in S}f(s) = F$ iff
for every $\epsilon > 0$, there exists a finite set $T_{\epsilon} \subseteq T$ such that for all finite sets $T \subseteq S$ with $T_\epsilon \subseteq T$, the inequality $\left|\sum_{s \in T}f(s) -F\right| < \epsilon$ holds
Theorem:
Prove that $\left|\sum_{s \in S}f(s)\right| \leq \sum_{s \in S}|f(s)|$, provided that $\sum_s f(s)$ exists
My attempt:
Write $L = | \sum f(s)|$ and $K = \sum |f(s)|$
Assume to the contrary that $L > K$, then $L-K > 0$
Now, choose a finite set $T \subseteq S$ such that $||\sum_Tf(s)|-L| < L-K$
Then, $K-L < |\sum_T f(s)|-L$ and hence $K < \sum_T |f(s)|$
But then, $\sum_T|f(s)| \leq \sup_{T \subseteq S finite}\sum_T|f(s)| = K < \sum_T|f(s)|$. This is absurd.
Is this correct?