Union is Smallest Superset

Theorem
Let $$S_1$$ and $$S_2$$ be sets.

Then $$S_1 \cup S_2$$ is the smallest set containing both $$S_1$$ and $$S_2$$.

That is:
 * $$\left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right) \iff \left({S_1 \cup S_2}\right) \subseteq T$$

General Result
Let $$S$$ be a set.

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

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

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

Proof

 * Let $$\left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right)$$.

Then:

$$ $$ $$

So:
 * $$\left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right) \implies \left({S_1 \cup S_2}\right) \subseteq T$$.


 * Next we show $$\left({S_1 \cup S_2}\right) \subseteq T \implies \left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right)$$:

$$ $$ $$

Similarly for $$S$$:

$$ $$ $$


 * So, from the above, we have:


 * 1) $$\left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right) \implies \left({S_1 \cup S_2}\right) \subseteq T$$;
 * 2) $$\left({S_1 \cup S_2}\right) \subseteq T \implies \left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right)$$.

Thus $$\left({S_1 \subseteq T}\right) \and \left({S_2 \subseteq T}\right) \iff \left({S_1 \cup S_2}\right) \subseteq T$$ from the definition of equivalence.

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

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

Consider any $$x \in \bigcup \mathbb S$$.

By definition of set union, it follows that:
 * $$\exists X \in \mathbb S: x \in X$$

But as $$X \subseteq T$$ it follows that $$x \in T$$.

Thus it follows that:
 * $$\bigcup \mathbb S \subseteq T$$

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

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

Consider any $$X \in \mathbb S$$ and take any $$x \in X$$.

From Subset of Union: General Result we have that $$X \subseteq \bigcup \mathbb S$$.

Thus $$x \in \bigcup \mathbb S$$.

But $$\bigcup \mathbb S \subseteq T$$.

So it follows that $$X \subseteq T$$.

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

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