Cardinality of Infinite Sets Definition. Let A and B be two sets. . We say that A and B have the same cardinality, and write | A| = |B|, if there is a bijective map f : A -> B. · We say that |A| ? |B| if there is an injective map f : A -> B. · We say that |A| < |B| if | A| ? |B| but not | A| = | B|. · We sometimes write | A| > |B| to mean | B| < | A|. Example. The open interval (0, 1) has the same cardinality as R, since the map f : (0, 1) -> R defined by f(x) = tan(T(x - 1)) is a bijection (see the graph below). 15 10 5 0 0.2 0.4 0.6 0.8 1.0 -5 -10 -15 Definition. Let N = {1,2,3, ... }. We write No (pronounced 'aleph zero') for the cardinality of N. (So |N| = No.) A set A is countable if |A| < No. A set is countably infinite if | A| = No. A set is uncountably infinite if |A| > No. (We sometimes just say the set is uncountable.) Theorem. Q is countable. Proof. Put the rational number a/b at coordinates (a, b) (just put ? if divid- ing by b = 0), and spiral outwards from the origin: 1
1 ? 23 ? 1 3 0 1 2 3 3 1 ? 3 2 -1 ? 1 2 0 1 2 1 3 2 -3 -2 -1 0 1 2 3 ? ? ? ? ? ? ? 3 2 1 0 -1 -2 -3 3 2 1 ? 1 2 0 ? 12 -1 ? 3 2 3 1 413 0 ? - 13 213 -1 Removing any duplicate entries, we have written the elements of Q in a list: 1, 0, -1, -2, 2, 1, -1, -2, -3, 3, .. .. So Q is countable. Theorem. |A| # |P(A)|. Proof. Let g : A > P(A) be a mapping from A to P(A). (Note that g(a) is always a subset of A.) We need to show that g cannot be a bijection. Define a subset B C A by a E B if and only if a ¢ g(a). But B is not in the image of g. To see why, assume for a contradiction that B is in the image of g. So there is x E A such that g(x) = B. If x E B, then (by definition of B) x $ g(x) = B: this is a contradiction. If x ¢ B = g(x), then (by definition of B), x E B: this is a contradiction also. Since either possibility leads to a contradiction, our original assumption that B is not in the image of g cannot hold. So g is not a bijection, as it is not surjective. 2