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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.
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:
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.
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.
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:
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 $1$.
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.
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 $\sum_{k=0}^{n/2}|\mathcal{P}_{2k}(A)| = \sum_{k=0}^{n/2-1}|\mathcal{P}_{2k+1}(A)|$ .
Denote $X$ the number of sets with even cardinality, $Y$ the number of sets with odd cardinality, $n$ the cardinality of the power set. $$ X=\sum_{i=0,2,4,...}{n\choose i}, Y=\sum_{i=1,3,5,7,...} {n\choose i}\\\Rightarrow X-Y=\sum_{i=1}^{n}{n\choose i}(-1)^i(+1)^{n-i}=(-1+1)^n=0\\ \Rightarrow X=Y$$
Because $X+Y=n$, it is obvious that $X=n/2$, that is the half of the power set.
