现代数学引论.docxVIP

前两章画波浪线的定理证明 Example 1.3 Let be a sequence of sets, then (1) for any , there must be a subsequence of such that for all k, . (2) for any , there must be a positive integer N such that for all , . Proof: (1) Since , then for all , . ? Set , from , it can be concluded that there is at least a positive number such that . ? Set , from , it can be concluded that there is at least a positive number such that . In this way, a subsequence of can be obtained such that for all k, . # Example 2.1 Let A be a set, and the power set of A, i.e., the set/collection of all subsets of A, including the empty set, then . Proof: (1) We first prove that . We define a map such that , It is clear that the map f as described is an injection and therefore . (2) We then prove that . If were not true, i.e., if we assumed that , there would exist a bijection . Based on this bijection f, the set A could be expressed as the union of two disjoint subsets: , where , Note that the subset D is not empty because and f is surjective, which makes sure the existence of an element such that , i.e., . Similarly, there must exist an element such that . But which subset, G or D, does this element belong to? ? If , then , which contradicts the fact that ; ? If , then , i.e., , which contradicts the fact that The awkward position of shows that the assumption is not true. # Example 2.4 Prove that the set is not countable. Proof: If the set were countable, its elements would be all enumerated as follows: where for all k and n. Now let where and for all k. It is clear that the element x as described belongs to , i.e., . But, for all n, which leads to because the elements of have been all enumerated as above. The dilemma of x shows that the set is not countable. # Example 2.8 Prove that . Hint: Let be a map such that for all , Cantor’s Contiuum Hypothesis There is no such a set A that . Theorem 3.1 Let A be a set and ～ be an equivalent relation on A. For all , let , then (1) f