Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized

Let $A$ be a non-empty set and $n$ be the number of elements in $A$, i.e. $n:=|A|$.

I know that the number of elements of the power set of $A$ is $2^n$, i.e. $|\mathcal{P}(A)|=2^n$.

I came across the fact that exactly half of the elements of $\mathcal{P}(A)$ contain an odd number of elements, and half of them an even number of elements.

Can someone prove this? Or hint at a proof?

Solutions Collecting From Web of "Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized"

Fix an element $a\in A$ (this is the point where $A\ne\emptyset$ is needed).
Then $$S\mapsto S\operatorname{\Delta}\{a\}$$ (symmetric difference) is a bijection from the set of odd subsets to the set of even subsets.

Hint: One can prove this by induction on the size of $A$. Assume it was true for sets of size $n$ and let $A=\{a_1,\ldots,a_{n+1}\}$. Then every subset of $A$ is either a subset of $\{a_1,\ldots,a_n\}$ or it is a copy of such subset with the addition of $\{a_{n+1}\}$. Use the induction hypothesis to conclude that the sets which do not contain $a_{n+1}$ have this property (with respect to $\{a_1,\ldots,a_n\}$, by adding $a_{n+1}$ you send exactly the same number of odd sets to even size sets, and vice versa; therefore the ratio remains true for $A$.

For $n\in\Bbb Z^+$ let $[n]=\{1,2,\dots,n\}$. Clearly $[1]$ has one even subset and one odd subset. Suppose that $[n]$ has equal numbers of odd and even subsets for some $n\in\Bbb Z^+$. The even subsets of $[n+1]$ are of two types:

  • the even subsets of $[n]$; and
  • the sets of the form $A\cup\{n+1\}$, where $A$ is an odd subset of $[n]$.

By the induction hypotheses there are the same number of sets of the second type as there are of the first, so $[n+1]$ has twice as many even subsets as $[n]$. But $[n+1]$ also has twice as many subsets altogether as $[n]$, so it must have twice as many odd subsets as well, which clearly implies that it has equal numbers of odd and even subsets.

Suppose $n=|A|$. Then there are $$\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}=2^{n-1}$$ sets with even cardinality. Thus, there are exactly half of the sets with an even number of elements.

When $n$ is odd, look at each set and its complement: one will have even number of elements and the other, odd (because odd number can only be written as a sum of an odd and an even number).

When $n$ is even, remove an element to obtain a set with odd number of elements. By the first part, half of its subsets have even cardinality and half odd. Now to form the full $\mathcal P(A)$, we need to join the remainin element to each of the previous subsets: those with odd cardinality with become even, and viceversa.

You can use proprieties of the binomial coefficients.

Denote $\mathcal{P}_k(A)$ the family of subsets of $A$ containing $k$ elements and observe
$|\mathcal{P}_k(A)| = {n \choose k}$ .

Now by a propriety of binomial coefficients $\sum_{k=0}^{n/2} {n \choose 2k} =\sum_{k=0}^{n/2} ({n-1 \choose 2k-1} + {n-1 \choose 2k}) = \sum_{k=0}^{n-1} {n-1 \choose k}$ and similarly $\sum_{k=0}^{n/2-1} {n \choose 2k+1} =\sum_{k=0}^{n/2-1} ({n-1 \choose 2k} + {n-1 \choose 2k+1}) = \sum_{k=0}^{n-1} {n-1 \choose k}$ .

This shows that $\sum_{k=0}^{n/2}|\mathcal{P}_{2k}(A)| = \sum_{k=0}^{n/2-1}|\mathcal{P}_{2k+1}(A)|$ .

Number the members of the set $1,2,3,4,\ldots,n$.

For every subset with an even number of elements, there is a corresponding set with an odd number of elements, that corresponds in this way:

  • If $1$ is a member of the set with an even number of elements, then delete $1$ from the set to get a set with an odd number of elements.
  • If $1$ is not a member of the set with an even number of elements, then add $1$ to the set to get a set with an odd number of elements.

For example, suppose the set is $\{1,2,3,4\}$. Then we have this correspondence between sets with an even number of elements and sets with an odd number of elements:
$$
\begin{array}{rcl}
\text{even} & & \text{odd} \\
\hline
\varnothing & \leftrightarrow & \{1\} \\
\{1,2\} & \leftrightarrow & \{2\} \\
\{1,3\} & \leftrightarrow & \{3\} \\
\{1,4\} & \leftrightarrow & \{4\} \\
\{2,3\} & \leftrightarrow & \{1,2,3\} \\
\{2,4\} & \leftrightarrow & \{1,2,4\} \\
\{3,4\} & \leftrightarrow & \{1,3,4\} \\
\{1,2,3,4\} & \leftrightarrow & \{2,3,4\}
\end{array}
$$

This won’t work with the empty set because we don’t have an element to which we can assign the number $1$.