"Now what’s the cardinality of P(X) for some set X? That’s an interesting question, and one well worth pondering. The answer ripples through the heart of a lot of combinatorics and the binary number system, topics we’ll cover later. And the answer is right at our fingertips, if we just extrapolate from the previous example. To form a subset of X, we have a choice to either include, or else exclude, each of its elements. So there’s two choices for the first element, and then whether we choose to include or exclude that first element, there are two choices for the second. Regardless of what we choose for those first two, there are two choices for the third, etc. So if |X| = 2, then its power set has 2 × 2 members. If |X| = 3, then its power set has 2 × 2 × 2 members. In general: |P(X)| = 2^|X|. As a limiting case (and a brain-bender) notice that if X is the empty set, then P(X) has one (not zero) members, because there is in fact one subset of the empty set: namely, the empty set itself. So |X| = 0, and |P(X)| = 1. And that jives with the above formula. I know there’s really no “first” element, but work with me here."