Set is Equivalent to Proper Subset of Power Set

Theorem
Every set is equivalent to a proper subset of its power set:


 * $\forall S: \exists T \subset \powerset S: S \sim T$

Proof
To show equivalence between two sets, we need to demonstrate that a bijection exists between them.

We will now define such a bijection.

Let $T = \set {\set x: x \in S}$.

Now we define the mapping $\phi: S \to T$:


 * $\phi: S \to T: \forall x \in S: \map \phi x = \set x$

$\phi$ is an injection:


 * $\forall x, y \in S: \set x = \set Y \implies x = y$ by definition of set equality

$\phi$ is a surjection:


 * $\forall \set x \in T: \exists x \in S: \map \phi x = \set x$

So $\phi$, being both an injection and a surjection, is a bijection by definition.

To show that $T \subset \powerset S$, that is, is a proper subset of $\powerset S$, we merely note that $\O \in \powerset S$ by Empty Set is Element of Power Set, but $\O \notin T$.

Thus $T \subseteq \powerset S$ but $\powerset S \nsubseteq T$.

Hence the result.