# Prove that any finite non-empty set X has the same number of subsets of even size as it has subsets of odd size?

I'm trying to prove that any finite non-empty set X has the same number of subsets of even size as it has subsets of odd size but finding it quite difficult to form a rigorous argument.

I have come up with an argument that is like this:

To create any subset of X we can go through each element of X and decide whether it is in the subset or it is not in the subset, hence for each element of X we will have that it is either in the subset or is not in the subset. For any even sized subset of X there must exist an odd subset (simply include or remove an element in X) and similarly for all odd sets there must be an even set. Thus the their are an equal number of odd/even subsets of X.

I fear this isn't a good argument and would like some advice.

Thanks.

• I think this has come up before. – Mike F Mar 26 '15 at 21:19

As given, the argument is not complete. Consider the following variation: "For any subset of $X$ with size divisible by $3$, there must exist a subset with size not divisible by $3$ (simply include or remove an element in X) and similarly for all sets with size not divisible by $3$ there exists a subset with size divisible by $3$ (again include or remove an element). Thus their are an equal number of subsets of X that have size divisible by $3$ and size not divisible by $3$."
It seems to me that you are trying to prove the number of odd sets to be equal to the number of even sets by constructing a bijection between the two types of sets. This is to say that we make pairs $(A,B)$ of an even set $A$ and an odd set $B$ where each set occurs in exactly one pair. Then the number of odd sets equals the number of pairs equals the number of even sets. To construct the pairs, one can for instance make all pairs $(A,B)$ where $A$ and $B$ only differ by a fixed element $x$. That is, if $x \in A$ then $B = A \backslash x$ and if $x \not\in A$ then $B = A \backslash x$.
Consider a fixed element $a$. Some subsets contain $a$...
Let's denote (wich is allowed because $X$ is finite): $$X = \{ a_{1},a_{2},...,a_{n}\}$$ Consiser the function $$f:P(X) \rightarrow \{0,1 \}^{\#X}: A \rightarrow (r_{1} ,r_{2}...,r_{n})$$ Where each $r_{i}$ is $1$ if $a_{i}$ is contained in $A$ and $0$ if not. $f$ is a bijection. So we can make the conclusion: $$\#P(X)={2}^{\#X}$$ Because we can make $\#P(X)={2}^{\#X}$ combinations of chosing $\#X$ zeros and ones.