- Power set
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 allsubset s of "S". Inaxiomatic set theory (as developed e.g. in theZFC axioms), the existence of the power set of any set is postulated by theaxiom of power set .Any

subset "F" of $mathcal\{P\}(S)$ is called a "family of sets " over "S".**Example**If "S" is the set {"x", "y", "z"}, then the complete list of subsets of "S" is as follows:

* { } (also denoted Ø, the

empty set )

* {"x"}

* {"y"}

* {"z"}

* {"x", "y"}

* {"x", "z"}

* {"y", "z"}

* {"x", "y", "z"}and hence the power set of "S" is

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

**Properties**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 "m"-th bit refers to the presence or absence of the "m"-th element of "S". There are 2^{"n"}such numbers.)Cantor's diagonal argument shows that the power set of a set (whether infinite or not) always has strictly highercardinality than the set itself (informally the power set must be larger than the original set). In particular,Cantor's theorem shows that thepower set of a countably infinite set is uncountably infinite. For example, thepower set of the set ofnatural number s can be put in a one-to-one correspondence with the set ofreal number s (seecardinality 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 ofsymmetric difference (with the empty set as its unit and each set being its own inverse) and acommutative 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.**Representing subsets as functions**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 correspondingpreimage of 1, we see that there is abijection 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. (Thus there are two distinct notational motivations for denoting the power set by 2^{"S"}: the fact that this function-representation of subsets makes it a special case of the "X"^{"Y"}notation and the property, mentioned above, that |2^{S}| = 2^{|S|}.)We can apply this notion to the example above to see the isomorphism with the binary numbers from 0 to 2

^{n}-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"} = 110For 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**Relation to binomial theorem**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 acombination $C(n,k),$ also called abinomial 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**Algorithms**If $S\; !$ is a

finite set , there is a recursive algorithm to calculate $mathcal\{P\}(S)$.Define the operation $mathcal\{F\}(e,T)\; =\; \{\; X\; cup\; \{\; e\; \}\; |\; X\; in\; T\; \}$

In English, return the set with the element $e\; !$ added to each set $X\; !$ in $T\; !$.

*If $S\; =\; \{\}\; !$,then $mathcal\{P\}(S)\; =\; \{\; \{\; \}\; \}$ is returned.

*Otherwise::*Let $e\; !$ be any single element of $S\; !$.:*Let $T\; =\; S\; setminus\; \{\; e\; \}\; !$, where '$setminus\; !$' denotes the relative complement of $\{\; e\; \}\; !$ in $S\; !$.:*And the result: $mathcal\{P\}(S)\; =\; mathcal\{P\}(T)\; cup\; mathcal\{F\}(e,mathcal\{P\}(T))$ is returned.In other words, the power set of the empty set is the set containing the empty set and the power set of any other set is all the subsets of the set containing some specific element and all the subsets of the set not containing that specific element.

An inductive example:

The basis step: by definition, $mathcal\{P\}(\{\})\; =\; \{\{\}\}$

An inductive step for: $mathcal\{P\}(\{1,2,3\})$

We know $mathcal\{P\}(\{1,2\})=\{\{\},\{1\},\{2\},\{1,2\}\}$

And $mathcal\{F\}(3,mathcal\{P\}(\{1,2\})\; =\; \{\; X\; cup\; \{\; 3\; \}\; |\; X\; in\; \{\{\},\{1\},\{2\},\{1,2\}\}\}\; =\; \{\{3\},\{1,3\},\{2,3\},\{1,2,3\}\}$

Thus $mathcal\{P\}(\{1,2\})\; cup\; mathcal\{F\}(3,mathcal\{P\}(\{1,2\}))\; =\; \{\{\},\{1\},\{2\},\{1,2\},\{3\},\{1,3\},\{2,3\},\{1,2,3\}\}\; =\; mathcal\{P\}(\{1,2,3\})$

There are other more efficient ways to calculate the power set. For example, the algorithm above can be

memoize d usingdynamic programming techniques. Other more complex techniques, such as those usingcombinadic s are also available.**Topologization of power set**Since any family of functions "X"

^{"Y"}from "Y" to "X" might be topologized establishing the so-calledfunction space , the same can be done with the power set 2^{"S"}identified as {0,1}^{"S"}. This particular type of function space is often calledand the topology on the power set is referred to ashyperspace **hypertopology**.**External links***mathworld|urlname=PowerSet|title=Power Set

* [*http://www.mathsisfun.com/sets/power-set.html Power Set*] fromMath Is Fun

*planetmath reference|id=136|title=Power set

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