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Unformatted text preview: Partial Solutions and Comments, Math 74 HW2 and HW3 General Comments: Please split up your proofs into paragraphs when appropriate, such as in longer proofs like HW3, Problem 9. Also, you should define most objects that you’re using, though you don’t need to state what Z , N , ∅ , etc. are. 2.1a. ∀ ε > ∃ δ > ∀ x ∈ R (  x a  < δ = ⇒  f ( x ) f ( a )  < ε ). 2.1b. ∃ ε > ∀ δ > ∃ x ∈ R (  x a  > δ ∧ f ( x ) f ( a )  ≥ ε ). 2.2. Comment: The problem asked to translate the given statements into other statements , not sets . Some possible solutions: a. ( ∀ x )( x ∈ A = ⇒ x ∈ B ) ⇔ A ⊆ B . b. ( ∀ x )( x ∈ A ⇔ x ∈ B ) ⇔ A = B . c. ( ∀ x )( x / ∈ A ) ⇔ A = ∅ . d. ( ∃ x )( x / ∈ A ∧ x ∈ B ) ⇔ A B , or A C ∩ B = ∅ . 2.7. Proposition. Let g ◦ f be injective. Then f is injective. Comment: g ◦ f being injective does not imply that g is injective. Let f : N → R be f ( n ) = n and g : R → R be g ( x ) = x 2 . Proof. Assume g ◦ f is injective and suppose f ( x ) = f ( y ). Then g ( f ( x )) = g ( f ( y )), which is equivalent to g ◦ f ( x ) = g ◦ f ( y ). Since g ◦ f is injective,...
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 Spring '09
 DANBERWICK
 Math, Division, Empty set, Topological space, Finite set, Prop osition

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