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Question: 163 please show all steps clearly thanks...

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Exerciee 1.6.3, supply rebuttals to the following complaints about the proof of Theorem 1.6.1. a) Every rational number has a decimaLexpansian, so we could apply this same argument to show that the set of rational numbers between 0 and 1 is uncountable. However, because we know that any subset of Q must be countable, the proof of Theorem 1.6.1 must be flawed. (b) Some numbers have tuo different decimal representations. Specifically any decimal expansion that terminates can also be written with repeating 9s. For instance, 1/2 can be written as .5 or as .4999.... Doesnt this cause some problems?

(c) Let A = Uに,an ar (d) Let A = X\A and B-Y\ B. Show g maps B onto 1.6 Cantors Theorem Cantors work into the theory of infinite sets extends far bey of Theorem 1.5.6. Although initially resisted, his creative and re in this area eventually produced a revolution in set theory and a paradigm shi in the way mathematicians came to understand the infinite. xtends far beyond the conclusio ive and relentless ass Cantors Diagonalization Method Cantor published his discovery that R is uncountable in 1874. Although i is some modern polish on it, the argument presented in Theorem 1.5.6 is actually quite similar to the one Cantor originally found. In 1891, Cantor offered another proof of this same fact that is startling in its simplicity. relies on decimal representations for real numbers, which we will accept and use without any formal definitions. Theorem 1.6.1. The open interval (0,1) uncountable. {x E R : 0 < < 1}is Exercise 1.6.1. Show that (0, 1) is uncountable if and only if R is uncountable. This shows that Theorem 1.6.1 is equivalent to Theorem 1.5.6. Proof. As with Theorem 1.5.6, we proceed by contradiction and assume that there does exist a function f : N → (0, 1) that is 1-1 and onto. For each m E N f (m) is a real number between 0 and 1, and we represent it using the decimal notation f(m) = .ama1@m20m30m4amS·

1.6. Cantors Theorem What is meant here is that for each m,n E N, amn is the digit from the set f0,1,2,...,9) that represents the nth digit in the decimal expansion of f (m). The 1-1 correspondence between N and (0, 1) can be summarized in the doubly indexed array I ㈠ f(1) = .a11 a12 a13 a14 a15 a16 2 ㈠ f(2) a21 a22 a23 a24 a25 a26 3f(3) 4f(4) a41 a42 a43 a44 a45 a46 5f(5) a51 a52 a53 a54 a55 a56 6 f(6)a61 a62 a63 a64 a65 a66 a31 a32 a33 a34 a35 a36.. The key assumption about this correspondence is that every real number in Now for the pearl of the argument. Define a real number z e (0,1) with the (0, 1) is assumed to appear somewhere on the list. decimal expansion -.bibabsb4... using the rule 2 if ann 2 3 if ann2 rn Lets be clear about this. To compute the digit b1, we look at the digit a1i in the upper left-hand corner of the array. If a11 2, then we choose bi3 otherwise, we set b 2. Exercise 1.6.2. (a) Explain why the real number .bibabsb4. cannot be f(1). whT f(n) for any n e N

1.6.3 please show all steps clearly thanks

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