Set is Equivalent to Proper Subset of Power Set

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

$$\forall S: \exists T \subset \mathcal{P} \left({S}\right): S \sim T$$

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 $$T = \left\{{\left\{{x}\right\}: x \in S}\right\}$$.

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

$$\phi: S \to T: \forall x \in S: \phi \left({x}\right) = \left\{{x}\right\}$$


 * $$\phi$$ is an injection:

$$\forall x, y \in S: \left\{{x}\right\} = \left\{{y}\right\} \Longrightarrow x = y$$ by definition of set equality


 * $$\phi$$ is a surjection:

$$\forall \left\{{x}\right\} \in T: \exists x \in S: \phi \left({x}\right) = \left\{{x}\right\}$$


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


 * To show that $$T \subset \mathcal{P} \left({S}\right)$$, that is, is a proper subset of $$\mathcal{P} \left({S}\right)$$, we merely note that $$\varnothing \in \mathcal{P} \left({S}\right)$$ by Empty Set Element of Power Set, but $$\varnothing \notin T$$.

Thus $$T \subseteq \mathcal{P} \left({S}\right)$$ but $$\mathcal{P} \left({S}\right) \not \subseteq T$$.

Hence the result.