Increasing Union of Ideals is Ideal

Theorem
Let $$R$$ be a ring.

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

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

is an ideal of $$R$$.

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

From Increasing Union of Subrings is Subring, we have that $$S$$ is a subring of $$R$$.

Now we need to show that it is an ideal of $$R$$.

Let $$a \in S$$.

Then $$\exists i \in \N: a \in S_i$$.

Let $$b \in R$$.

Then $$a b \in S_i$$ and $$b a \in S_i$$, as $$S_i$$ is an ideal of $$R$$.

Thus $$a b \in S$$ and $$b a \in S$$.

So $$S$$ is an ideal of $$R$$.