Union of Elements of Power Set

Theorem
Let $S$ be a set.

Then:
 * $\displaystyle S = \bigcup_{X \in \mathcal P \left({S}\right)} X$

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

Proof
By Subset of Union:
 * $\displaystyle \forall X \in P \left({S}\right): X \subseteq \bigcup_{X \in \mathcal P \left({S}\right)} X$

From Subset of Itself, $S \subseteq S$ and so $S \in \mathcal P \left({S}\right)$.

So:
 * $\displaystyle S \subseteq \bigcup_{X \in \mathcal P \left({S}\right)} X$

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

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

But as $\mathcal P \left({S}\right) \subseteq \mathcal P \left({S}\right)$ from Subset of Itself:
 * $\displaystyle \left({\forall X \in \mathcal P \left({S}\right): X \subseteq S}\right) \iff \bigcup \mathcal P \left({S}\right) \subseteq S$

The LHS is no more than the definition of the power set, making it a tautological statement, and so:
 * $\displaystyle \bigcup \mathcal P \left({S}\right) \subseteq S$

The result follows from Equality of Sets.