Count of Subsets with Even Cardinality/Proof 2

Theorem
Let $S$ be a set whose cardinality is $n$ where $n \ge 1$.

Then the number of subsets of $S$ whose cardinality is even is $2^{n-1}$.

Proof
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * The number of subsets of $S$ whose cardinality is even is $2^{n-1}$, where $\left|{S}\right| = n$.

Basis for the Induction
When $n = 1$ we have from Cardinality of Power Set that $S$ has $2^1 = 1$ subsets: $\varnothing$ and $S$ itself.

We have that $\left|{S}\right| = 1$ and $\left|{\varnothing}\right| = 0$.

So there is indeed $2^{1-1} = 2^0 = 1$ subset of $S$ whose cardinality is even, that is $\varnothing$.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * The number of subsets of $S$ whose cardinality is even is $2^{k-1}$, where $\left|{S}\right| = k$.

Then we need to show:
 * The number of subsets of $S$ whose cardinality is even is $2^k$, where $\left|{S}\right| = k+1$.

Induction Step
This is our induction step:

Let $\left|{S}\right| = k+1$.

Let $x \in S$.

Consider the set $S\,' = S \setminus \left\{{x}\right\}$.

We see that $\left|{S\,'}\right| = k$.

Let $A \subseteq S$.

If $x \notin A$ then $A \subseteq S\,'$.

If $x \in A$ then there exists a unique $A' \subseteq S\,'$ such that $A' \cup \left\{{x}\right\} = A$.

In fact, $A' = A \setminus \left\{{x}\right\}$.

Let $K$ be the number of subsets of $S\,'$ with an even number of elements.

From the induction hypothesis we have that $K = 2^{k-1}$.

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

Suppose $E\,' \subseteq S\,'$, where $E\,'$ has an even number of elements.

Then $E\,' \cup \left\{{x}\right\}$ has an odd number of elements.

Similarly, $O\,' \subseteq S\,'$, where $O\,'$ has an odd number of elements.

Then $O\,' \cup \left\{{x}\right\}$ has an even number of elements.

So the number of subsets of $S$ with an even number of elements is:
 * $K$ subsets of $S$ that have an even number of elements and are subsets of $S\,'$

and:
 * $K$ subsets of $S$ of the form $O\,' \cup \left\{{x}\right\}$

We have that subset $A$ of $S$ has a unique expression of the form $A'$ or $A' \cup \left\{{x}\right\}$ with $A' \subseteq S\,'$.

Thus there are exactly $2 K$ subsets of $S$ with an even number of elements.

But as $K = 2^{k-1}$ it follows that $2 K = 2^k$.

That is, there are $2^k$ subsets of $S$ with an even number of elements.

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

Therefore:
 * The number of subsets of $S$ whose cardinality is even is $2^{n-1}$, where $\left|{S}\right| = n$.