If a set \(A\) has two elements, show that \(\mathcal{P}(A)\) has four elements. How many elements does \(\mathcal{P}(A)\) have if \(A\) has one element? Three elements? No elements? Why is \(\mathcal{P}(A)\) called the power set of \(A\) ?
Solution
If \(A=\{a,b\}\), then its subsets are exactly \[ \varnothing,\qquad \{a\},\qquad \{b\},\qquad \{a,b\}, \] so \(|\mathcal P(A)|=4\).
In general, if \(|A|=n\), then each element of \(A\) has two independent choices when specifying a subset: include it or omit it. Hence \[ |\mathcal P(A)|=2^n. \] Therefore: \[ \begin{array}{c|c} |A|&|\mathcal P(A)|\\ \hline 0&1\\ 1&2\\ 2&4\\ 3&8 \end{array} \] The name power set refers to this formula: its cardinality is the power \(2^{|A|}\) when \(A\) is finite.