Sum of Ideals is Ideal/General Result

Theorem
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.

Then:
 * $J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.

Proof
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.

Proof by induction:

For all $n \in \N^*$, let $\map P n$ be the proposition:
 * $J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

$\map P 1$ is true, as this just says $J_1$ is an ideal of $R$.

Basis for the Induction
$\map P 2$ is the case:
 * $J_1 + J_2$ is an ideal of $R$

which is proved in Sum of Ideals is Ideal.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $J_1 + J_2 + \cdots + J_k$ is an ideal of $R$.

Then we need to show:
 * $J_1 + J_2 + \cdots + J_k + J_{k + 1}$ is an ideal of $R$.

Induction Step
This is our induction step:

Let $J = J_1 + J_2 + \cdots + J_k$.

From the induction hypothesis, $J$ is an ideal.

From the base case, $J + J_{k + 1}$ is an ideal.

That is:
 * $J_1 + J_2 + \cdots + J_k + J_{k+1}$ is an ideal of $R$.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.