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

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 $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right)$.

Then:

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


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

Similarly for $S$:


 * So, from the above, we have:


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

Thus $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \iff \left({S_1 \cup S_2}\right) \subseteq T$ 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:
 * $\left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$

Also see

 * Intersection Largest