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 $n$.
$\blacksquare$