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?

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. 
