Generalized Sum is Linear

Theorem
Let $\family {z_i}_{i \mathop \in I}$ and $\family {w_i}_{i \mathop \in I}$ be $I$-indexed families of complex numbers.

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

Let $\ds \sum \set {z_i: i \in I}$ and $\sum \set {w_i: i \mathop \in I}$ converge to $z, w \in \C$, respectively.

Then:


 * $(1): \quad \ds \sum \set {z_i + w_i: i \in I}$ converges to $z + w$
 * $(2): \quad \forall \lambda \in \C: \ds \sum \set {\lambda z_i: i \in I}$ 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:


 * $\ds \map d {\sum_{i \mathop\in G} z_i + w_i, z + w} < \epsilon$ 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:


 * $\ds \map d {\sum_{i \mathop \in G} z_i, z} < \frac \epsilon 2$ for all finite $G$ with $F_z \subseteq G \subseteq I$
 * $\ds \map d {\sum_{i \mathop \in G} w_i, w} < \frac \epsilon 2$ 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, it is concluded that:
 * $\ds \sum \set {z_i + w_i: i \in I} = z + w$

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

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


 * $\ds \map d {\sum_{i \mathop\in G} \lambda z_i, \lambda z} < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$

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


 * $\ds \map d {\sum_{i \mathop \in G} z_i, z} < \frac \epsilon {\size \lambda}$

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:
 * $\ds \sum \set {\lambda z_i: i \in I} = \lambda z$

Also see

 * Convergence of Generalized Sum of Complex Numbers establishes a partial converse.