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

Generalized Result
Let $$I$$ be an indexing set.

Then:
 * $$\left({\forall i \in I: S_i \subseteq T}\right) \iff \bigcup_{i \in I} S_i \subseteq T$$

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.

Generalized Proof
Suppose that $$\forall i \in I: S_i \subseteq T$$.

Consider any $$x \in \bigcup_{i \in I} S_i$$.

By definition of set union, it follows that:
 * $$\exists i \in I: x \in S_i$$

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

Thus it follows that:
 * $$\bigcup_{i \in I} S_i \subseteq T$$

So:
 * $$\left({\forall i \in I: S_i \subseteq T}\right) \implies \bigcup_{i \in I} S_i \subseteq T$$

Now suppose that $$\bigcup_{i \in I} S_i \subseteq T$$.

Consider any $$i \in I$$ and take any $$x \in S_i$$.

From Subset of Union we have that $$S_i \subseteq \bigcup_{i \in I} S_i$$.

Thus $$x \in \bigcup_{i \in I} S_i$$.

But $$\bigcup_{i \in I} S_i \subseteq T$$.

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

So:
 * $$\bigcup_{i \in I} S_i \subseteq T \implies \left({\forall i \in I: S_i \subseteq T}\right)$$

Hence:
 * $$\left({\forall i \in I: S_i \subseteq T}\right) \iff \bigcup_{i \in I} S_i \subseteq T$$