Sum of All Ring Products is Closed under Addition

Theorem
Let $\struct {R, +, \circ}$ be a ring.

Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.

Let $S T$ be defined as:
 * $\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$

Then $\struct {S T, +}$ is a closed subset of $\struct {R, +}$.

Proof
Let $x_1, x_2 \in S T$.

Then:
 * $\ds x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$

for some $s_i, t_i, j, k$, etc.

By renaming the indices, we can express $x_2$ as:
 * $\ds x_2 = \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i$

and hence:
 * $\ds x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$

So $x_1 + x_2 \in S T$ and $\struct {S T, +}$ is shown to be closed.