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?

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 $1$.
