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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 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....
View Full Document

{[ snackBarMessage ]}

Page1 / 7

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online