Cardinality of Power Set of Finite Set/Proof 3

Proof
Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:
 * $\card S = n \implies \card {\powerset S} = 2^n$

Do not confuse $\map P n$, which is a propositional function on $\N$, with $\powerset S$, the power set of $S$.

Basis for the Induction
From Cardinality of Empty Set:
 * $S = \O \iff \card S = 0$

Then:
 * $\powerset \O = \set \O$

has one element, that is, $\O$.

So:
 * $\card {\powerset \O} = \card {\set \O} = 1 = 2^0$

thus confirming that $\map P 0$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\card S = k \implies \card {\powerset S} = 2^k$

Then we need to show:
 * $\card S = k + 1 \implies \card {\powerset S} = 2^{k + 1}$

Induction Step
This is our induction step:

Let $\card S = k + 1$.

Let $x \in S$.

Consider the set $S' = S \setminus \set x$ where $x$ is any element of $S$.

We see that $\card {S'} = k$.

Now adjoin $x$ to all the subsets of $S'$.

Counting the subsets of $S$, we have:
 * all the subsets of $S'$

and:
 * all the subsets of $S'$ with $x$ adjoined to them.

From the induction hypothesis, there are $2^k$ subsets of $S'$.

Adding $x$ to each of these does not change their number, so there are another $2^k$ subsets of $S$ consisting of all the subsets of $S'$ with $x$ adjoined to them.

In total that makes $2^k + 2^k = 2 \times 2^k = 2^{k + 1}$ subsets of $S$.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: \card S = n \implies \card {\powerset S} = 2^n$