Problem 43 gemignani section 62 4 suppose s is a set

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Unformatted text preview: r j ∈ N is in S if and only if kj = 1. This set is ω(k). To check that this is really a one ­to ­one correspondence, we note that ψ and ω are inverses of each other. Start with a set S in P. Apply ψ and then ω; from the definitions you get the same set S. Likewise it is true that if you start with a number r in T and apply ω and ψ, the number resulting is the original number r. Problem 4.3: Gemignani, Section 6.2 # 4 Suppose S is a set with n elements and T is a set with m elements. a) What is the smallest number of elements that S ∪ T can contain? Answer: Since S ⊂ S ∪ T and also T ⊂ S ∪ T, then n ≤ #(S ∪ T) and m ≤ #(S ∪ T), because the cardinal number of a subset is less than or equal to the cardinal number of the containing set. Thus max(m, n) ≤ #(S ∪ T). In fact max(m, n) is the smallest possible value of #(S ∪ T). To show this, it suffices to produce one example with this value, for a given choice of m and n. As the example, let S = (1, … , n} and T = (1, … , m}. Then S ∪ T = S if m ≤ n and S ∪ T = T if n ≤ m. In the first case, #(S ∪ T) = n = max(m, n) and in the second case #(S ∪ T) = m = max(m, n). So max(m, n) is the smallest number possible. b) What is the largest number of elements that S ∪ T can contain? Answer: If S and T are disjoint, then S ∪ T has m+n elements. c) What is the largest number of elements that S ∩ T can contain? Answer: Since S ⊃ S ∩ T and also T ⊃ S ∩ T, then n ≥ #(S ∩ T) and m ≥ #(S ∩ T), because the cardinal number of a subset is less than or equal to the cardinal number of the containing set. Thus...
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This note was uploaded on 01/29/2014 for the course MATH 300A taught by Professor Jamesking during the Winter '11 term at University of Washington.

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