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

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.

Also see

 * Intersection Largest