Set is Equivalent to Proper Subset of Power Set

From ProofWiki
Jump to navigation Jump to search

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

\(\displaystyle \forall x \in S: \ \ \) \(\displaystyle \left\{ {x}\right\}\) \(\subseteq\) \(\displaystyle S\) Definition of Subset
\(\displaystyle \implies \ \ \) \(\displaystyle \left\{ {x}\right\}\) \(\in\) \(\displaystyle \mathcal P \left({S}\right)\) Definition of Power Set
\(\displaystyle \implies \ \ \) \(\displaystyle T\) \(\subseteq\) \(\displaystyle \mathcal P \left({S}\right)\) Definition of Subset


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:

$\forall x, y \in S: \left\{{x}\right\} = \left\{{y}\right\} \implies x = y$ 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 is 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.

$\blacksquare$