Count of Subsets with Odd Cardinality/Proof 1

Theorem
Let $S$ be a set whose cardinality is $n$.

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

Proof
Let $F$ be the total number of subsets of $S$ whose cardinality is odd.

From Cardinality of Set of Subsets, the number of subsets of $S$ with $m$ elements is $\displaystyle \binom n m$:
 * $\displaystyle \binom n m = \frac{n!}{m! \left({n-m}\right)}$

where $\displaystyle \binom n m$ is a binomial coefficient.

Thus the total number of subsets of $S$ whose cardinality is Odd is given by:
 * $\displaystyle E = \binom n 1 + \binom n 3 + \binom n 5 + \cdots = \sum_{j \mathop \in \Z} \binom n {2j+1}$

Note the loose limits of the summation sign: the expression truly ranges over all $\Z$.

This is because, when $2 j + 1 < 0$ and $2 j + 1> n$, $\displaystyle \binom n {2j+1} = 0$ by definition of binomial coefficient.

The result then follows from Sum of Odd Index Binomial Coefficients:


 * $\displaystyle \sum_{j \mathop \ge 0} \binom n {2 j + 1} = 2^{n-1}$