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 $$I$$ be an indexing set.

Then:
 * $$\left({\forall i \in I: S \subseteq T_i}\right) \iff S \subseteq \bigcap_{i \in I} T_i$$

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
Suppose that $$\forall i \in I: S \subseteq T_i$$.

Consider any $$x \in S$$.

By definition of subset, it follows that:
 * $$\forall i \in I: x \in T_i$$

Thus it follows from definition of set intersection that:
 * $$x \in \bigcap_{i \in I} T_i$$

Thus by definition of subset, it follows that:
 * $$S \subseteq \bigcap_{i \in I} T_i$$

So:
 * $$\left({\forall i \in I: S \subseteq T_i}\right) \implies S \subseteq \bigcap_{i \in I} T_i$$

Now suppose that $$S \subseteq \bigcap_{i \in I} T_i$$.

From Intersection Subset we have:
 * $$\forall j \in I: \bigcap_{i \in I} T_i \subseteq T_j$$

So from Subsets Transitive, it follows that:
 * $$\forall j \in I: S \subseteq \bigcap_{i \in I} T_i \subseteq T_j$$

So it follows that $$\forall i \in I: S \subseteq T_i$$.

So:
 * $$S \subseteq \bigcap_{i \in I} T_i \implies \left({\forall i \in I: S \subseteq T_i}\right)$$

Hence:
 * $$\left({\forall i \in I: S \subseteq T_i}\right) \iff S \subseteq \bigcap_{i \in I} T_i$$