Power set

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

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.

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Power set

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Size comparison

Proof

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