Intersection is Subset

Theorem
$$S \cap T \subseteq S$$

Generalized Result
Let $$S_i \subseteq S: i \in \mathbb{N}^*_n$$.

Then $$\forall i \in \mathbb{N}^*_n: \bigcap_{i = 1}^n S_i \subseteq S_i$$.

Generalized Result
Proof by induction:

For all $$n \in \mathbb{N}^*$$, let $$P \left({n}\right)$$ be the proposition $$\forall i \in \mathbb{N}^*_n: \bigcap_{i = 1}^n S_i \subseteq S_i$$.


 * $$P(1)$$ is true, as this just says $$S_1 \subseteq S_1$$.

Basis for the Induction

 * $$P(2)$$ is the case $$S_1 \cap S_2 \subseteq S_1$$ and $$S_1 \cap S_2 \subseteq S_2$$, which has been proved above. 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:

$$\forall i \in \mathbb{N}^*_k: \bigcap_{i = 1}^k S_i \subseteq S_i$$.

Then we need to show:

$$\forall i \in \mathbb{N}^*_{k+1}: \bigcap_{i = 1}^{k+1} S_i \subseteq S_i$$.

Induction Step
This is our induction step:

We have $$\bigcap_{i = 1}^{k+1} S_i = \bigcap_{i = 1}^k S_i \cap S_{k+1}$$.

From the basis for the induction, we have:


 * $$\bigcap_{i = 1}^k S_i \cap S_{k+1} \subseteq S_i$$ for all $$i \in \mathbb{N}^*_k$$;
 * $$\bigcap_{i = 1}^k S_i \cap S_{k+1} \subseteq S_{k+1}$$ for all $$i \in \mathbb{N}^*_k$$.

Thus $$\bigcap_{i = 1}^k S_i \cap S_{k+1} \subseteq S_i$$ for all $$i \in \mathbb{N}^*_{k+1}$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall i \in \mathbb{N}^*_n: \bigcap_{i = 1}^n S_i \subseteq S_i$$.