Increasing Union of Subrings is Subring

Theorem
Let $$R$$ be a ring.

Let $$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$$ be subrings of $$R$$.

Then the increasing union $$S$$:
 * $$S = \bigcup_{i \in \N} S_i$$

is a subring of $$R$$.

Proof
Let $$S = \bigcup_{i \in \N} S_i$$.

Clearly $$0_R \in S$$.

Let $$a, b \in S$$.

Then $$\exists i, j \in \N: a \in S_i, b \in S_j$$.

From the construction, we have that either of $$S_i$$ and $$S_j$$ contains the other.

Let $$l = \max \left\{{i, j}\right\}$$ so $$a, b \in S_l$$.

Then $$a + \left({-b}\right) \in S_l$$ and $$a b \in S_l$$, as $$S_l$$ is a subring.

Thus $$a + \left({-b}\right) \in S$$ and $$a b \in S$$.

By the Subring Test, $$S$$ is a subring of $$R$$.