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Question: 169 show all work clearly thanks...

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igtPucted in Exercise 1.6.6 Exerciset the subset B that results using the preceding rule. In each case, alit ater ing that B is not in the range of the function used. We now fo ntradiction arises when we consider whether or not a is an element of B note on the general argument. Because we have assumed that our the t B = f(a) for some a E A. The nd function f : A → P(A) is onto, it must be tha Exercise 1.6.8. (a) First, show that the case a E B leads to a contradiction. B is equally It ts (b) Now, finish the argument by showing that the case a, unacceptable. To get an initial sense of its broad significance, lets apply this result to the set of natural numbers. Cantors Theorem states that there is no onto function from N to P(N); in other words, the power set of the natural numbers is uncountable. How does the cardinality of this newly discovered uncountable set compare to the uncountable set of real numbers? men sections (including the exercises from Section 1.5), give a compelling argu ment showing that P(N)~ R Exercise 1.6.9. Using the various tools and techniques developed in the last sf tho following by establish- L

1.6.9 show all work clearly, thanks.

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