Cantor's Theorem (Strong Version)/Proof 2/Induction Step/Proof 2

Induction Step
This is our induction hypothesis:
 * There is no surjection from $S$ onto $\map {\PP^k} S$.

We are to show:
 * There is no surjection from $S$ onto $\map {\PP^{k + 1} } S$.

Proof
By Powerset is Nonempty, $\powerset {\map {\PP^k} S}$ is non-empty.

By definition:
 * $\map {\PP^{k + 1} } S = \powerset {\map {\PP^k} S}$

Then:
 * $\map {\PP^{k + 1} } S \ne \O$

By Law of Excluded Middle, there are two choices:
 * $S = \O$

or:
 * $S \ne \O$

Suppose that $S = \O$.

By Image of Empty Set is Empty Set: Corollary 2, there is no surjection from $S$ onto $\map {\PP^{k + 1} } S$.

Suppose that $S \ne \O$.

By the induction hypothesis $\map P k$ is true.

that $f: S \to \map {\PP^{k + 1} } S$ is a surjection.

By Injection from Set to Power Set, there is an injection:
 * $g: \map {\PP^k} S \to \powerset {\map {\PP^k} S}$

By Injection has Surjective Left Inverse Mapping, there exists a surjection:
 * $h: \powerset {\map {\PP^k} S} \to \map {\PP^k} S$

By definition of $\powerset {\map {\PP^k} S}$, this is a surjection:
 * $h: \map {\PP^{k + 1} } S \to \map {\PP^k} S$

By Composite of Surjections is Surjection, we have a surjection:
 * $h \circ f: S \to \map {\PP^k} S$

But this contradicts the induction hypothesis.

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