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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?

  • 0
    I'm amazed that so many people bothered to write answers, and that there are some good answers, but there is only one upvote to the question.2012-11-30

8 Answers 8

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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.

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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 $A$.

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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:

  • the even subsets of $[n]$; and
  • the sets of the form $A\cup\{n+1\}$, where $A$ is an odd subset of $[n]$.

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.

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    +1 but you do not even need the induction hypothesis, as you have shown that the number of even subsets of $[n+1]$ is equal to the number of subsets of $[n]$ and a very similar argument would show the number of odd subsets of $[n+1]$ is the same.2012-12-01
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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.

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    It's easiest to prove by induction. If n=1, then there is exactly 1 set with an even number of elements: the set with no elements (the empty set). Suppose it's true for some positive integer k. Then k+1 is either odd or even. If it's odd, then the floor of (k+1)/2=k/2, in which case we're done. If it is even, then break up the sum and binomial coefficient, combine, and you get the desired result.2012-11-30
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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.

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    So easy and natural when $n$ is odd. So futile trying to define a 'semi-complement' function. sigh...2017-12-10
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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 $1$.

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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 $\sum_{k=0}^{n/2}|\mathcal{P}_{2k}(A)| = \sum_{k=0}^{n/2-1}|\mathcal{P}_{2k+1}(A)|$ .