Power set

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In mathematics, given a set S, the power set (or powerset) of S, written $\mathcal{P}(S)$ , P(S), or 2^S, is the set of all subsets of S. In axiomatic set theory (as developed e.g. in the ZFC axioms), the existence of the power set of any set is postulated by the axiom of power set.

Any subset F of $\mathcal{P}(S)$ is called a family of sets over S.

The elements of the power set of the set {x, y, z} ordered in respect to inclusion.

1 An example
2 Properties
3 The notations 2^S and {0,1}^S
4 Relation to binomial theorem
5 Algorithms
6 External links

If S is the set {x, y, z}, then the complete list of subsets of S is as follows:

{ } (also denoted ∅, the empty set)
{x}
{y}
{z}
{x, y}
{x, z}
{y, z}
{x, y, z}

and hence the power set of S is

$\mathcal{P}(S) = \left\{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\right\}\,\!.$

If S is a finite set with |S| = n elements, then the power set of S contains $|\mathcal{P}(S)| = 2^n$ elements. (One can—and computers sometimes do—represent the elements of $\mathcal{P}(S)$ as n-bit numbers; the n-th bit refers to presence or absence of the n-th element of S. There are 2ⁿ such numbers.)

Cantor's diagonal argument shows that the power set of a set (infinite or not) always has strictly higher cardinality than the set itself (informally the power set must be 'greater' than the original set). The power set of the set of natural numbers for instance can be put in a one-to-one correspondence with the set of real numbers (see cardinality of the continuum).

The power set of a set S, together with the operations of union, intersection and complement can be viewed as the prototypical example of a Boolean algebra. In fact, one can show that any finite Boolean algebra is isomorphic to the Boolean algebra of the power set of a finite set. For infinite Boolean algebras this is no longer true, but every infinite Boolean algebra is a subalgebra of a power set Boolean algebra (though this is not always a particularly illuminating representation of an infinite Boolean algebra).

The power set of a set S forms an Abelian group when considered with the operation of symmetric difference (with the empty set as its unit and each set being its own inverse) and a commutative semigroup when considered with the operation of intersection. It can hence be shown (by proving the distributive laws) that the power set considered together with both of these operations forms a commutative ring.

In set theory, X^Y is the set of all functions from Y to X. As 2 can be defined as {0,1} (see natural number), 2^S (i.e., {0,1}^S) is the set of all functions from S to {0,1}. By identifying a function in 2^S with the corresponding preimage of 1, we see that there is a bijection between 2^S and $\mathcal{P}(S)$ , where each function is the characteristic function of the subset in $\mathcal{P}(S)$ with which it is identified. Hence 2^S and $\mathcal{P}(S)$ could be considered identical set-theoretically.

These are special cases of the convention from enumerative combinatorics that provides notations for sets based on their cardinalities.

We can apply this notion to the example above to see the isomorphism with the binary numbers from 0 to 2ⁿ-1 with n being the number of elements in the set. In S a 1 in the position corresponding to the location in the set indicates the presence of the element. So {x, y} = 110

For the whole power set of S we get:

{ } = 000(Binary) = 0 (Decimal)
{x} = 100 = 4
{y} = 010 = 2
{z} = 001 = 1
{x, y} = 110 = 6
{x, z} = 101 = 5
{y, z} = 011 = 3
{x, y, z} = 111 = 7

The power set is closely related to the binomial theorem. The number of sets with $k$ elements in the power set of a set with $n$ elements will be a combination $C (n, k),$ also called a binomial coefficient.

For example the power set of a set with three elements, has:

$C (3,0) = 1$ set with 0 elements
$C (3,1) = 3$ sets with 1 element
$C (3,2) = 3$ sets with 2 elements
$C (3,3) = 1$ set with 3 elements

If S is a finite set, there is a recursive algorithm to calculate $\mathcal{P}(S)$ .

If S = {}, $\mathcal{P}(S) = \{ \{ \} \}$ .
Otherwise $S = T \cup \{ e \} \!$ for some element e and set T.
In which case, $\mathcal{P}(S) = \mathcal{P}(T) \cup \{ X \cup \{ e \} | X \in \mathcal{P}(T) \}$ .