Sum of All Ring Products is Closed under Addition

Theorem
Let $$\left({R, +, \circ}\right)$$ be a ring.

Let $$\left({S, +}\right)$$ and $$\left({T, +}\right)$$ be additive subgroups of $$\left({R, +, \circ}\right)$$.

Let $$S T$$ be defined as:
 * $$S T = \left\{{\sum_{i=1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1\,.\,.\ n}\right]}\right\}$$

Then $$\left({S T, +}\right)$$ is a closed subset of $$\left({R, +}\right)$$.

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

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

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

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

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

So $$x_1 + x_2 \in S T$$ and $$\left({S T, +}\right)$$ is shown to be closed.