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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 α with Y = S α B α , then for each α , ∃ A α ⊆ X , such that A α = f 1 ( B α ), i.e., f ( A α ) = B α for all α . Since Y = S α B α , we have Y = S α f ( A α ). Since S α A α ⊆ 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: ( Cantor’s 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 11 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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 Fall '08
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 Math, Topology, Sets, Metric space, Topological space, Closed set, y Nr

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