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Difference between revisions of "Power set"

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The '''power set''' of a given [[set]] <math>S</math> is the set <math>\mathcal{P}(S)</math> of all [[subset]]s of that set.
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The '''power set''' of a given [[set]] <math>S</math> is the set <math>\mathcal{P}(S)</math> of all [[subset]]s of that set. It is also sometimes denoted by <math>2^S</math>.
  
 
==Examples==
 
==Examples==
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Similarly, for any [[finite]] set with <math>n</math> elements, the power set has <math>2^n</math> elements.
 
Similarly, for any [[finite]] set with <math>n</math> elements, the power set has <math>2^n</math> elements.
  
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==Size comparison==
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==Size Comparison==
 
Note that for any [[nonnegative]] [[integer]] <math>n</math>, <math>2^n > n</math> and so for any finite set <math>S</math>, <math>|\mathcal P (S)| > |S|</math> (where [[absolute value]] signs here denote the [[cardinality]] of a set).  The analogous result is also true for [[infinite]] sets (and thus for all sets): for any set <math>S</math>, the cardinality <math>|\mathcal P (S)|</math> of the power set is strictly larger than the cardinality <math>|S|</math> of the set itself.
 
Note that for any [[nonnegative]] [[integer]] <math>n</math>, <math>2^n > n</math> and so for any finite set <math>S</math>, <math>|\mathcal P (S)| > |S|</math> (where [[absolute value]] signs here denote the [[cardinality]] of a set).  The analogous result is also true for [[infinite]] sets (and thus for all sets): for any set <math>S</math>, the cardinality <math>|\mathcal P (S)|</math> of the power set is strictly larger than the cardinality <math>|S|</math> of the set itself.
  
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Note that this proof does not rely upon either the [[continuum hypothesis]] or the [[axiom of choice]].  It is a good example of a [[diagonal argument]], a method pioneered by the mathematician [[Georg Cantor]].
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Note that this proof does not rely upon either the [[Continuum Hypothesis]] or the [[Axiom of Choice]].  It is a good example of a [[diagonal argument]], a method pioneered by the mathematician [[Georg Cantor]].
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==Size for Finite Sets==
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The number of [[element|elements]] in a [[power set]] of a set with n elements is <math>2^n</math> for all finite sets. This can be proven in a number of ways:
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===Method 1===
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Either an element in the power set can have 0 elements, one element, ... , or n elements. There are <math>\binom{n}{0}</math> ways to have no elements, <math>\binom{n}{1}</math> ways to have one element, ... , and <math>\binom{n}{n}</math> ways to have n elements. We add:
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<math>\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n</math>
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as desired.
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===Method 2===
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We proceed with [[induction]].
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Let <math>S</math> be the set with <math>n</math> elements. If <math>n=0</math>, then <math>S</math> is the empty set. Then
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<math>P(S)=\{\emptyset \}</math>
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and has <math>2^0=1</math> element.
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Now let's say that the theorem stated above is true or n=k. We shall prove it for k+1.
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Let's say that Q has k+1 elements.
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In set Q, if we leave element x out, there will be <math>2^k</math> elements in the power set. Now we include the sets that do include x. But that's just <math>2^k</math>, since we are choosing either 0 1, ... or k elements to go with x. Therefore, if there are <math>2^k</math> elements in the power set of a set that has k elements, then there are <math>2^{k+1}</math> elements in the power set of a set that has k+1 elements.
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Therefore, the number of elements in a power set of a set with n elements is <math>2^n</math>.
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===Method 3===
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We demonstrated in Method 2 that if S is the empty set, it works.
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Now let's say that S has at least one element.
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For an element x, it can be either in or out of a subset. Since there are n elements, and each different choice of in/out leads to a different subset, there are <math>2^n</math> elements in the power sum.
  
 
==See Also==
 
==See Also==
 
* [[Set theory]]
 
* [[Set theory]]
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* [[Set]]
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==External Links==
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*[http://mathworld.wolfram.com/PowerSet.html Power Set] at Wolfram MathWorld.
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{{stub}}
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[[Category:Set theory]]

Latest revision as of 10:44, 8 March 2018

The power set of a given set $S$ is the set $\mathcal{P}(S)$ of all subsets of that set. It is also sometimes denoted by $2^S$.

Examples

The empty set has only one subset, itself. Thus $\mathcal{P}(\emptyset) = \{\emptyset\}$.

A set $\{a\}$ with a single element has two subsets, the empty set and the entire set. Thus $\mathcal{P}(\{a\}) = \{\emptyset, \{a\}\}$.

A set $\{a, b\}$ with two elements has four subsets, and $\mathcal{P}(\{a, b\}) = \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$.

Similarly, for any finite set with $n$ elements, the power set has $2^n$ elements.

Size Comparison

Note that for any nonnegative integer $n$, $2^n > n$ and so for any finite set $S$, $|\mathcal P (S)| > |S|$ (where absolute value signs here denote the cardinality of a set). The analogous result is also true for infinite sets (and thus for all sets): for any set $S$, the cardinality $|\mathcal P (S)|$ of the power set is strictly larger than the cardinality $|S|$ of the set itself.

Proof

There is a natural injection $S \hookrightarrow \mathcal P (S)$ taking $x \mapsto \{x\}$, so $|S| \leq |\mathcal P(S)|$. Suppose for the sake of contradiction that $|S| = |\mathcal P(S)|$. Then there is a bijection $f: \mathcal P(S) \to S$. Let $T \subset S$ be defined by $T = \{x \in S \;|\; x \not\in f(x) \}$. Then $T \in \mathcal P(S)$ and since $f$ is a bijection, $\exists y\in S \;|\; T = f(y)$.

Now, note that $y \in T$ by definition if and only if $y \not\in f(y)$, so $y \in T$ if and only if $y \not \in T$. This is a clear contradiction. Thus the bijection $f$ cannot really exist and $|\mathcal P (S)| \neq |S|$ so $|\mathcal P(S)| > |S|$, as desired.


Note that this proof does not rely upon either the Continuum Hypothesis or the Axiom of Choice. It is a good example of a diagonal argument, a method pioneered by the mathematician Georg Cantor.

Size for Finite Sets

The number of elements in a power set of a set with n elements is $2^n$ for all finite sets. This can be proven in a number of ways:

Method 1

Either an element in the power set can have 0 elements, one element, ... , or n elements. There are $\binom{n}{0}$ ways to have no elements, $\binom{n}{1}$ ways to have one element, ... , and $\binom{n}{n}$ ways to have n elements. We add:

$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$

as desired.

Method 2

We proceed with induction.

Let $S$ be the set with $n$ elements. If $n=0$, then $S$ is the empty set. Then

$P(S)=\{\emptyset \}$

and has $2^0=1$ element.


Now let's say that the theorem stated above is true or n=k. We shall prove it for k+1.

Let's say that Q has k+1 elements.

In set Q, if we leave element x out, there will be $2^k$ elements in the power set. Now we include the sets that do include x. But that's just $2^k$, since we are choosing either 0 1, ... or k elements to go with x. Therefore, if there are $2^k$ elements in the power set of a set that has k elements, then there are $2^{k+1}$ elements in the power set of a set that has k+1 elements.

Therefore, the number of elements in a power set of a set with n elements is $2^n$.

Method 3

We demonstrated in Method 2 that if S is the empty set, it works.

Now let's say that S has at least one element.

For an element x, it can be either in or out of a subset. Since there are n elements, and each different choice of in/out leads to a different subset, there are $2^n$ elements in the power sum.

See Also

External Links

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