Difference between revisions of "Power set"

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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.
  
Note that for [[nonnegative]] [[integer]]s, <math>2^n > n</math> so the power set of any set has a [[cardinality]] at least as large as the set itself.  An analogous result holds for [[infinite]] sets: for any set <math>S</math>, there is no [[bijection]] between <math>S</math> and <math>\mathcal{P}(S)</math> (or equivalently, there is no [[injection]] from <math>\mathcal{P}(S)</math> to <math>S</math>).
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Note that for any set <math>\displaystyle S</math> such that <math>\displaystyle a \in S</math>, <math>\displaystyle \{ a \} \subseteq S </math>, so the power set of any set <math>\displaystyle S</math> has a [[cardinality]] at least as large as that of <math>\displaystyle S</math> itself.
  
 
===Proof===
 
===Proof===

Revision as of 21:10, 26 August 2006

The power set of a given set $S$ is the set $\mathcal{P}(S)$ of subsets of that set.

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.

Note that for any set $\displaystyle S$ such that $\displaystyle a \in S$, $\displaystyle \{ a \} \subseteq S$, so the power set of any set $\displaystyle S$ has a cardinality at least as large as that of $\displaystyle S$ itself.

Proof

See Also

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