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 the following results hold:

Ordinary proofs
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Proof
Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition:
 * $$\bigcap_{j=1}^n A_j \in \mathcal R$$ and $$\bigcup_{j=1}^n A_j \in \mathcal R$$.

$$P(1)$$ 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 the definition of ring of sets;
 * $$A_1 \cup A_2 \in \mathcal R$$, which has been proved in Ring of Sets Closed under Various Operations.

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:
 * $$\bigcap_{j=1}^k A_j \in \mathcal R$$ and $$\bigcup_{j=1}^k A_j \in \mathcal R$$.

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

Induction Step
This is our induction step:

We have that:

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

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

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

Therefore:
 * $$\bigcap_{j=1}^n A_j \in \mathcal R$$ and $$\bigcup_{j=1}^n A_j \in \mathcal R$$.