HW - 1 - HOMEWORK - 01 OF MATH 517 CHAPTER 2 BASIC TOPOLOGY...

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Unformatted text preview: HOMEWORK - 01 OF MATH 517 CHAPTER 2 BASIC TOPOLOGY (1) Show that a function f : X Y between two sets X and Y is onto if and only if f ( f- 1 ( B )) = B for all B Y . Proof: ( ) Suppose f : X Y is onto, i.e., f ( X ) = Y . Then B Y , there exists a subset of X , say A , such that f ( A ) = B , denote A = f- 1 ( B ). Thus, f ( f- 1 ( B )) = B for all B Y . ( ) Suppose f ( f- 1 ( B )) = B for all B Y . Denote the subsets of Y by B with Y = S B , then for each , A X , such that A = f- 1 ( B ), i.e., f ( A ) = B for all . Since Y = S B , we have Y = S f ( A ). Since S A X , we have f ( X ) Y . Since f : X Y is a function defined over X , f ( X ) Y . So it follows that f ( X ) = Y . Altogether, we have that a function f : X Y between two sets X and Y is onto if and only if f ( f- 1 ( B )) = B for all B Y . This completes the proof. (2) Let E be the set of all real numbers in (0 , 1) that have a decimal expansion consisting only of the digits 3 and 9. Is E countable or uncountable? Answer: E is uncountable. Proof: ( Cantors diagonal proccess ) Let A be a countable subset of E , and let A consist of the sequences s 1 , s 2 , s 3 , ... . We construct a sequence as follows. If the n th digit in s n is 3, we let the n th digit of s be 9, and vice versa. Then the sequence s differs from every member of E in at least one place; hence s 6 A . But clearly, s E , so that A is a proper subset of E . In other words, every countable subset of E is a proper subset of E . It naturally follows that E is uncountable. This completes the proof. (3) Let E R be a set and f : R R be an increasing function, i.e., if x < y then f ( x ) < f ( y ). (a) Show that E is countable if and only if the image set f ( E ) is countable. Proof: ( ) Suppose E is countable, i.e., E Z + . Since f : R R is strictly increasing, i.e., if x < y then f ( x ) < f ( y ), we have that whenever x 6 = y , f ( x ) 6 = f ( y ). Thus, E f ( E ). So by the transitive property of the equivalence relation, we conclude that if E is countable, so be the image set f ( E ). ( ) Suppose f ( E ) is countable, i.e. f ( E ) Z + . Since f is strictly increasing, i.e., if x < y then f ( x ) < f ( y ), we have whenever x 6 = y , f ( x ) 6 = f ( y ), that is, f : R R is a 1-1 mapping of R to R , i.e. E f ( E ). According to the transitive property of the equivalance relation, we have E Z . So E countable is naturally follows. 1 2 HOMEWORK - 01 OF MATH 517 CHAPTER 2 BASIC TOPOLOGY Altogether, E is countable if and only if the image set f ( E ) is countable. This completes the proof....
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HW - 1 - HOMEWORK - 01 OF MATH 517 CHAPTER 2 BASIC TOPOLOGY...

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