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sols22 - 22 NOVEMBER 22ND CARDINALITY IV 51 22 November...

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22. NOVEMBER 22ND: CARDINALITY IV 51 22. November 22nd: Cardinality IV 22.1 . Prove that if S is uncountable, and C S is countable, then | S C | = | S | . Answer: Let N ( S C ) be an infinite countable set, and let C = C N . The union of two countable sets is countable (Example 21.7), so C is infinite countable: C = N . Since we have S = ( S C ) C and S C = ( S C ) N it follows that | S | = | S C | , by Lemma 20.1. 22.2 . Prove that the set of transcendental number has the cardinality of the con- tinuum. Answer: This is an immediate application of Exercise 22.1: Take S = R , and C = A , the set of algebraic numbers; then since A is countable, the result of Exercise 22.1 shows that the set R A of transcendental numbers has the same cardinality as R . 22.3 . Prove that the set C of complex numbers has the cardinality of the contin- uum. Answer: Complex numbers are numbers of the form a + bi , where a and b are real numbers, and i is a square root of 1. Thus, C = R × R : the knowledge of a complex number is equivalent to the knowledge of its real part a and its imaginary part b . It follows that | C | = | R × R

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sols22 - 22 NOVEMBER 22ND CARDINALITY IV 51 22 November...

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