Intersection is Largest Subset

Theorem
Let $$T_1$$ and $$T_2$$ be sets.

Then $$T_1 \cap T_2$$ is the largest set contained in both $$T_1$$ and $$T_2$$.

That is:
 * $$S \subseteq T_1 \and S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$$

Generalized Result
Let $$T_i \subseteq T: i \in \N^*_n$$.

Then:
 * $$\left({\forall i \in \N^*_n: S \subseteq T_i}\right) \iff S \subseteq \bigcap_{i = 1}^n T_i$$.

Proof

 * Let $$S \subseteq T_1 \and S \subseteq T_2$$.

Then:

$$ $$ $$

Alternatively:

$$ $$ $$

So:
 * $$S \subseteq T_1 \and S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$$.


 * Now let $$S \subseteq T_1 \cap T_2$$.

From Intersection Subset we have $$T_1 \cap T_2 \subseteq T_1$$ and $$T_1 \cap T_2\subseteq T_2$$.

From Subsets Transitive, it follows directly that $$S \subseteq T_1$$ and $$S \subseteq T_2$$.

So $$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \and S \subseteq T_2$$.


 * From the above, we have:


 * 1) $$S \subseteq T_1 \and S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$$;
 * 2) $$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \and S \subseteq T_2$$.

Thus $$S \subseteq T_1 \and S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$$ from the definition of equivalence.

Generalized Proof
Proof by induction:

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

$$P(1)$$ is trivially true, as this just says $$S \subseteq T_1 \iff S \subseteq T_1$$.

Basis for the Induction
$$P(2)$$ is the case:
 * $$S \subseteq T_1 \and S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$$

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:


 * $$\left({\forall i \in \N^*_k: S \subseteq T_i}\right) \iff S \subseteq \bigcap_{i = 1}^k T_i$$.

Then we need to show:


 * $$\left({\forall i \in \N^*_{k+1}: S \subseteq T_i}\right) \iff S \subseteq \bigcap_{i = 1}^{k+1} T_i$$.

Induction Step
This is our induction step:

$$ $$ $$ $$

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

Therefore $$\left({\forall i \in \N^*_n: S \subseteq T_i}\right) \iff S \subseteq \bigcap_{i = 1}^n T_i$$.