Cantor's Theorem (Strong Version)/Proof 2

Theorem
Let $S$ be a set.

Let $\mathcal P^n \left({S}\right)$ be defined recursively by:
 * $\mathcal P^n \left({S}\right) = \begin{cases}

S & : n = 0 \\ \mathcal P \left({\mathcal P^{n-1} \left({S}\right)}\right) & : n > 0 \end{cases}$ where $\mathcal P \left({S}\right)$ denotes the power set of $S$.

Then there is no surjection from $S$ onto $\mathcal P^n \left({S}\right)$ for any $n > 0$.

Proof
The proof proceeds by induction.

The case for $n = 1$ is Cantor's Theorem.

Suppose that the theorem holds for $n$.

Suppose for the sake of contradiction that $f: S \to \mathcal P^{n+1}(S)$ is a surjection.

Define a mapping $g: S \to \mathcal P^n(S)$ by letting $g(x) = \bigcup f(x)$.

This is actually a mapping into $\mathcal P^n(S)$:

$f(x) \in \mathcal P^{n+1}(S) = \mathcal P \left({ \mathcal P^n(S) }\right)$

By the definition of power set, $f(x) \subseteq \mathcal P^n(S)$.

Thus each element of $f(x)$ is a subset of $P^{n-1}(S)$.

Thus by Union of Subsets is Subset:


 * $\bigcup f(x) \subseteq \mathcal P^{n-1}(S)$.

Therefore $\bigcup f(x) \in \mathcal P^n(S)$.

$g$ is surjective:

For each $y \in \mathcal P^n(S)$, $\{y\} \in \mathcal P^{n+1}(S)$.

Since $f$ is surjective, there is an $x \in S$ such that $f(x) = \{ y \}$

Then $g(x) = \bigcup \{ y \} = y$.

As this holds for all such $y$, $g$ is surjective.

But this contradicts the inductive hypothesis.

Thus we conclude that the theorem holds for all $n$.