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:
 * $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$

General Result
Let $T$ be a set.

Let $\mathcal P \left({T}\right)$ be the power set of $T$.

Let $\mathbb T$ be a subset of $\mathcal P \left({T}\right)$.

Then:
 * $\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$

Proof

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

Then:

Alternatively:

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


 * 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 $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.


 * From the above, we have:


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

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

Proof of General Result
Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.

Suppose that $\forall X \in \mathbb T: S \subseteq X$.

Consider any $x \in S$.

By definition of subset, it follows that:
 * $\forall X \in \mathbb T: x \in X$

Thus it follows from definition of set intersection that:
 * $x \in \bigcap \mathbb T$

Thus by definition of subset, it follows that:
 * $S \subseteq \bigcap \mathbb T$

So:
 * $\left({\forall X \in \mathbb T: S \subseteq X}\right) \implies S \subseteq \bigcap \mathbb T$

Now suppose that $S \subseteq \bigcap \mathbb T$.

From Intersection Subset: General Result we have:
 * $\forall X \in \mathbb T: \bigcap \mathbb T \subseteq X$

So from Subsets Transitive, it follows that:
 * $\forall X \in \mathbb T: S \subseteq \bigcap \mathbb T \subseteq X$

So it follows that $\forall X \in \mathbb T: S \subseteq X$.

So:
 * $S \subseteq \bigcap \mathbb T \implies \left({\forall X \in \mathbb T: S \subseteq X}\right)$

Hence:
 * $\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$

Caution
Be careful of the way the general result is stated:
 * $\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$

Without the brackets, this would read:
 * $\forall X \in \mathbb T: S \subseteq X \iff S \subseteq \bigcap \mathbb T$

That is:
 * For all $X \in \mathbb T$, it is the case that $S \subseteq X$ if and only if $S \subseteq \bigcap \mathbb T$

This is not the same thing at all.

For example, if $\mathbb T = \mathcal P \left({T}\right)$ (as it well might), then $T \in \mathbb T$, and $\bigcap \mathbb T = \varnothing$.

This would imply that:
 * $T \subseteq \bigcap \mathbb T$

That is:
 * $T \subseteq \varnothing$

which is obviously rubbish.

Also see

 * Union Smallest