Generalized Sum is Linear

Theorem
Let $\left({z_i}\right)_{i \in I}, \left({w_i}\right)_{i \in I}$ be $I$-indexed families of complex numbers.

That is, let $z_i, w_i \in \C$ for all $i \in I$.

Suppose that $\displaystyle \sum \left\{{ z_i: i \in I }\right\}, \sum \left\{{ w_i: i \in I }\right\}$ converge to $z, w \in \C$, respectively.

Then:


 * $(1): \qquad \displaystyle \sum \left\{{ z_i + w_i: i \in I }\right\}$ converges to $z+w$
 * $(2): \qquad \forall \lambda \in \C: \displaystyle \sum \left\{{ \lambda z_i: i \in I }\right\}$ converges to $\lambda z$

Proof of $(1)$
Let $\epsilon > 0$.

To verify the convergence, it is necessary to find a finite $F \subseteq I$ such that:


 * $\displaystyle d \left({\sum_{i \in G} z_i + w_i, z + w}\right) < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$

Now let $F_z, F_w \subseteq I$ be finite subsets of $I$ such that:


 * $\displaystyle d \left({\sum_{i \in G} z_i, z}\right) < \frac \epsilon 2$ for all finite $G$ with $F_z \subseteq G \subseteq I$
 * $\displaystyle d \left({\sum_{i \in G} w_i, w}\right) < \frac \epsilon 2$ for all finite $G$ with $F_w \subseteq G \subseteq I$

The set $F_z \cup F_w$ will be the sought $F$. Let $G$ be finite such that $F_z \cup F_w \subseteq G \subseteq I$.

It follows that:

From the definition of convergence, conclude that $\displaystyle \sum \left\{{ z_i + w_i: i \in I }\right\} = z+w$.

Proof of $(2)$
Let $\epsilon > 0$.

To verify the convergence, it is necessary to find a finite $F \subseteq I$ such that:


 * $\displaystyle d \left({\sum_{i \in G} \lambda z_i, \lambda z}\right) < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$

Now let $F_z \subseteq I$ be a finite subset of $I$ such that:


 * $\displaystyle d \left({\sum_{i \in G} z_i, z}\right) < \frac \epsilon {\left\vert{\lambda}\right\vert}$ for all finite $G$ with $F_z \subseteq G \subseteq I$

The set $F_z$ will be the sought $F$. Let $G$ be finite such that $F_z \subseteq G \subseteq I$.

It follows that:

From the definition of convergence, conclude that $\displaystyle \sum \left\{{ \lambda z_i: i \in I }\right\} = \lambda z$.