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

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

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.

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