Ring of Sets is Closed under Finite Intersection

Theorem
Let $\RR$ be a ring of sets.

Let $A_1, A_2, \ldots, A_n \in \RR$.

Then:


 * $\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$

$\map P 1$ is true, as this just says $A_1 \in \RR$.

Basis for the Induction
$\map P 2$ is the case:
 * $A_1 \cap A_2 \in \RR$

which is immediate, from definition 1 of ring of sets.

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:
 * $\ds \bigcap_{j \mathop = 1}^k A_j \in \RR$

Then we need to show:
 * $\ds \bigcap_{j \mathop = 1}^{k + 1} A_j \in \RR$

Induction Step
This is our induction step:

We have that:
 * $\ds \bigcap_{j \mathop = 1}^{k + 1} A_j = \bigcap_{j \mathop = 1}^k A_j \cap A_{k + 1}$

But from the induction hypothesis we have that:
 * $\ds \bigcap_{j \mathop = 1}^k A_j \in \RR$

Hence from the basis for the induction, it follows that:
 * $\ds \bigcap_{j \mathop = 1}^{k + 1} A_j \in \RR$

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

Therefore:
 * $\s \bigcap_{j \mathop = 1}^n A_j \in \RR$