Arithmetik Abelscher Varietaeten mit komplexer by C.-G. Schmidt

By C.-G. Schmidt

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But, since A ∩ B ⊆ A for all A and B, it is sufficient to prove that A ⊆ A ∩ B. So, if x ∈ A, it follows from (i) that x ∈ B and therefore x ∈ A ∩ B. Hence, A ⊆ A ∩ B. 38 SET THEORY To prove that (ii) implies (iii), let’s assume that A ∩ B = A holds. Then, A ∪ B = (A ∩ B) ∪ B = (A ∪ B) ∩ (B ∪ B) = (A ∪ B) ∩ B = B Finally, to prove that (iii) implies (i), we assume that A ∪ B = B holds. Then, since A ⊆ A ∪ B for all A and B, it follows that A ⊆ B. 29 Determine (A ∩ B) ∪ (Ac ∩ Cc ) Solution If A = [0, 1), then Ac = (−∞, 0) ∪ [1, ∞).

19 A set of all natural numbers N is infinite. 20 A set of all integers Z is infinite. ◾ We will discuss the intricacies of infinite sets in a little while. 10 We say that two sets A and B are equivalent (or equinumerous) or that they have the same cardinality, and we write A∼B iff |A| = |B| Following Cantor, we say that cardinal number of a set A is what A has in common with all sets equivalent to A. 21 Given sets A = {1, 2, 3}, B = {a, b, c}, and C = {b, c, a}, we say that A ∼ B, and A ∼ C, but only B = C.

3 (ii). 3 (i). 8 (i) We need to prove that for every x if x ∈ (A ∩ B)c then x ∈ Ac ∪ Bc Suppose x ∈ (A ∩ B)c . By the definition of complement, x ∉ A ∩ B. But this implies that x ∉ A or x ∉ B. Saying that x ∉ A means that x ∈ Ac . Similarly, if x ∉ B, then x ∈ Bc . Hence, x ∈ Ac or x ∈ Bc and by the definition of union this implies that x ∈ Ac ∪ Bc So, we have proved that (A ∩ B)c ⊆ Ac ∪ Bc (*) Let’s now consider the converse, that is, let’s show that for every x if x ∈ Ac ∪ Bc then x ∈ (A ∩ B)c Suppose that ∈ Ac ∪ Bc .

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