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?

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.
