Union of Subsets is Subset/Subset of Power Set

Theorem
Let $S$ and $T$ be sets.

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

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

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

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

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:
 * $\displaystyle \bigcup \mathbb S \subseteq T$

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

On the other direction, suppose $\bigcup \mathbb S \subseteq T$.

Let $X \in \mathbb S$ and $x \in X$. Then, by definition of the set union $\bigcup \mathbb S$, $x \in \bigcup \mathbb S$.

So, $X \subseteq \bigcup \mathbb S$.

By hypothesis and Subset Relation is Transitive, we get $X \subseteq T$, hence $\forall X \in \mathbb S: X \subseteq T$.

Therefore,


 * $\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \Leftarrow \bigcup \mathbb S \subseteq T$.

So, :$\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$ as required.