Ring of Sets is Closed under Finite Intersection

Theorem
Let $\mathcal R$ be a ring of sets.

Let $A_1, A_2, \ldots, A_n \in \mathcal R$.

Then:


 * $\displaystyle \bigcap_{j \mathop = 1}^n A_j \in \mathcal R$

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \bigcap_{j \mathop = 1}^n A_j \in \mathcal R$

$P(1)$ is true, as this just says $A_1 \in \mathcal R$.

Basis for the Induction
$P(2)$ is the case:
 * $A_1 \cap A_2 \in \mathcal R$

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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \bigcap_{j \mathop = 1}^k A_j \in \mathcal R$

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

Induction Step
This is our induction step:

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

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

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

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

Therefore:
 * $\displaystyle \bigcap_{j \mathop = 1}^n A_j \in \mathcal R$