# Cantor's Theorem (Strong Version)/Proof 2

## Theorem

Let $S$ be a set.

Let $\map {\PP^n} S$ be defined recursively by:

$\map {\PP^n} S = \begin{cases} S & : n = 0 \\ \powerset {\map {\PP^{n - 1} } S} & : n > 0 \end{cases}$

where $\powerset S$ denotes the power set of $S$.

Then $S$ is not equivalent to $\map {\PP^n} S$ for any $n > 0$.

## Proof

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

There is no surjection from $S$ onto $\map {\PP^n} S$.

### Basis for the Induction

$\map P 1$ is Cantor's Theorem.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

There is no surjection from $S$ onto $\map {\PP^k} S$.

Then we need to show:

There is no surjection from $S$ onto $\map {\PP^{k + 1} } S$.

### Induction Step

This is our induction step:

Suppose that $\map P k$ is true.

Aiming for a contradiction, suppose that $f: S \to \map {\PP^{k + 1} } S$ is a surjection.

Define the mapping $g: S \to \map {\PP^k} S$ as:

$\displaystyle \map g x = \bigcup \map f x$

This is actually a mapping into $\map {\PP^k} S$, as follows:

$\map f x \in \map {\PP^{k + 1} } S = \powerset {\map {\mathcal P^k} S}$

By the definition of power set:

$\map f x \subseteq \map {\PP^k} S$

Thus each element of $\map f x$ is a subset of $\map {\PP^{k - 1} } S$.

Thus by Union of Subsets is Subset:

$\displaystyle \bigcup \map f x \subseteq \map {\PP^{k - 1} } S$

Therefore:

$\displaystyle \bigcup \map f x \in \map {\PP^k} S$

That is, $\displaystyle \map g x$ is a mapping into $\map {\PP^k} S$.

Next we have that:

$\forall y \in \map {\PP^k} S: \set y \in \map {\PP^{n + 1} } S$

But by hypothesis $f$ is surjective, and so:

$\exists x \in S: \map f x = \set y$

Then:

$\displaystyle \map g x = \bigcup \set y = y$

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

But this contradicts the induction hypothesis.

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

$\blacksquare$