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Unformatted text preview: Introduction to Mathematical Analysis SOLUTION MANUAL Wieslaw Krawcewicz October 5, 2011 1 1 Basic Set Theory, Logic and Introduction to Proofs 1.1 Problems 1. Use the Truth table to show that the following statements are tautologies: (a) [( p ⇒ q ) ⇒ p ] ⇒ p (Pierce’s Law); (b) ( ∼ p ⇒ p ) ⇒ p (Clavius Law); (c) [ p ⇒ ( q ⇒ r )] ⇒ [( p ⇒ q ) ⇒ ( p ⇒ r )] (Law of Conditional Syllogism); (d) ( p ∧ q ⇒ r ) ⇔ [ p ⇒ ( q ⇒ r )]. 2. Check if the following statements are tautologies: (a) ( p ⇒ q ) ⇒ [ p ⇒ ( q ∨ r )]; (b) p ∨ [( ∼ p ∧ q ) ∨ ( ∼ p ∧ ∼ q )]; (c) [( p ⇒ q ) ∧ ( q ⇒ p )] ⇒ ( p ∨ q ). 3. Check whether the following statements are true or false: (a) “If a is a multiple of 2 and is also a multiple of 7, then if a is not a multiple of 7 implies that a is a multiple of 3 ;” (b) “If it is not true that the line l is parallel to the line m or the line p is not parallel to the line m , then the line l is not parallel to the line m or the line p is parallel to the line m ;” (c) “If James doesn’t know analysis, then if James knows analysis implies that James was born in the 2nd century B.C..” 4. Check if the following quantified statements are true or false: (a) ∃ x ( p ( x ) ⇒ q ( x ) ) ⇒ [ ∃ x p ( x ) ⇒ ∃ x q ( x ) ] ; (b) ∃ x p ( x ) ∧ ∃ x q ( x ) ⇒ ∃ x ( p ( x ) ∧ q ( x ) ) . 5. Prove the following identities for the sets: (a) ∪ t ∈ T ( A t ∪ B t ) = ∪ t ∈ T A t ∪ ∪ t ∈ T B t ; (b) ∩ t ∈ T ( A t ∩ B t ) = ∩ t ∈ T A t ∩ ∩ t ∈ T B t ; 2 (c) ∪ t ∈ T ( A t ∩ B t ) ⊂ ∪ t ∈ T A t ∩ ∪ t ∈ T B t ; 6. Let A , B , C be sets. Check if the following equalities are true: (a) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ); (b) ( A \ B ) ∪ C = [( A ∪ C ) \ B ] ∪ ( B ∩ C ); (c) A \ [ B \ C ] = ( A \ B ) ∪ ( A ∩ C ) 7. Let f : X → Y be a function from X into Y . Show that if A and B are subsets of X , then (a) f ( A ∩ B ) ⊂ f ( A ) ∩ f ( B ): (b) ( A ′ ⊂ B ′ ) ⇒ ( f − 1 ( A ′ ) ⊂ f − 1 ( B ′ )). (c) f − 1 ( A ′ ∩ B ′ ) = f − 1 ( A ′ ) ∩ f − 1 ( B ′ ). 8. Show that the function f : X → Y is (a) injective if and only if ∀ A ⊂ X f − 1 ( f ( A )) ⊂ A. (b) surjective if and only if ∀ B ′ ⊂ Y f ( f − 1 ( B ′ )) = B ′ . (c) (b) Show that f is bijective if and only if ( f − 1 ( f ( A )) = A ) ∧ ( f ( f − 1 ( B ′ )) = B ′ ) for all A ⊂ X and B ′ ⊂ Y . 9. Let f : X → Y be a function. Write the logic negation to each of the following statements: (a) f is surjective; (b) f is injective; (c) f is bijective. 3 1.2 Solutions of Problems 1. Use the Truth table to show that the following statements are tautologies: (a) [( p ⇒ q ) ⇒ p ] ⇒ p (Pierce’s Law); (b) ( ∼ p ⇒ p ) ⇒ p (Clavius Law); (c) [ p ⇒ ( q ⇒ r )] ⇒ [( p ⇒ q ) ⇒ ( p ⇒ r )] (Law of Conditional Syllogism); (d) ( p ∧ q ⇒ r ) ⇔ [ p ⇒ ( q ⇒...
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 Fall '11
 krwaw
 Set Theory

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