Topology (2nd Edition) by James R. Munkres

By James R. Munkres

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Example text

A set that is not countable is said to be uncountable. There is a very useful criterion for showing that a set is countable. 1. Let B be a nonempty set. Then the following are equivalent: (1) B is countable. (2) There is a surjective function f : Z+ → B . (3) There is an injective function g : B → Z+ . Proof. (1) ⇒ (2). Suppose that B is countable. If B is countably infinite, there is a bijection f : Z+ → B by definition, and we are through. If B is finite, there is a 43 46 Set Theory and Logic Ch.

2) ⇒ (3). Let f : Z+ → B be a surjection. Define g : B → Z+ by the equation g(b) = smallest element of f −1 ({b}). Because f is surjective, f −1 ({b}) is nonempty; thus g is well defined. The map g is injective, for if b = b , the sets f −1 ({b}) and f −1 ({b }) are disjoint, so their smallest elements are different. (3) ⇒ (1). Let g : B → Z+ be an injection; we wish to prove B is countable. By changing the range of g, we can obtain a bijection of B with a subset of Z+ . Thus to prove our result, it suffices to show that every subset of Z+ is countable.

If there are injections f : A → C and g : C → A, then A and C have the same cardinality. 7. Show that the sets D and E of Exercise 5 have the same cardinality. 8. Let X denote the two-element set {0, 1}; let B be the set of countable subsets of X ω . Show that X ω and B have the same cardinality. 9. (a) The formula (∗) h(1) = 1, h(2) = 2, h(n) = [h(n + 1)]2 − [h(n − 1)]2 for n ≥ 2 is not one to which the principle of recursive definition applies. Show that nevertheless there does exist a function h : Z+ → R satisfying this formula.

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