Generalized Sum Restricted to Non-zero Summands/Necessary Condition

Theorem
Let $G$ be a commutative topological semigroup with identity $0_G$.

Let $\family{g }_{i \in I}$ be an indexed subset of $G$.

Let $J = \set{i \in I : g_i \ne 0_G}$

Let $g'= g \restriction_J$ be the restriction of $g$ to $J$.

Let $h \in G$.

Let the generalized sum $\ds \sum_{i \mathop \in I} g_i$ converge to $h$.

Then:
 * the generalized sum $\ds \sum_{j \mathop \in J} g'_j$ converges to $h$.

Proof
Let $U \subseteq G$ be an open subset of $G$ such that $h \in U$.

By definition of convergent net:
 * $(1) \quad \exists F \subseteq I : F \ne \O : \forall E \subseteq I : E \supseteq F : \ds \sum_{i \mathop \in E} g_i \in U$

where $\ds \sum_{i \mathop \in E} g_i$ is the summation over $E$.

Let:
 * $F'= F \cap J$

From Set Difference and Intersection form Partition:
 * $F = F' \cup F \setminus J$

Let $E' \subseteq J$:
 * $E' \supseteq F'$

We have:
 * $E' \cap F \setminus J = \O$

Let:
 * $E = E' \cup F \setminus J$

From Set Union Preserves Subsets:
 * $E \supseteq F$

From $(1)$:
 * $\ds \sum_{i \mathop \in E} g_i \in U$

Case 1 : $F \setminus J = \O$
Let:
 * $F \setminus J = \O$

From Union with Empty Set:
 * $E = E'$

Hence:
 * $\ds \sum_{i \mathop \in E'} g_i \in U$

Case 2 : $F \setminus J \ne \O$
Let:
 * $F \setminus J \ne \O$

We have:

Hence:
 * $\ds \sum_{i \mathop \in E'} g_i \in U$

In either case:
 * $\ds \sum_{i \mathop \in E'} g_i \in U$

By definition of restricted mapping:
 * $\ds \sum_{j \mathop \in E'} g'_j = \sum_{i \mathop \in E'} g_i \in U$

Since $U$ was arbitrary, it follows that $\ds \sum_{j \mathop \in J} g'_j$ converges to $h$ by definition.

Hence:
 * $\ds \sum_{i \mathop \in I} g_i = \sum_{j \mathop \in J} g'_j$