Math 523H#2 - Math 523H Homework#2 Sections 1.2 1.3 1.2#1 Let S be the open interval(1 2 and let T be the closed interval[2 2 Describe the

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Math 523H Homework #2 Sections 1.2, 1.3 1.2 #1. Let S be the open interval (1 , 2) and let T be the closed interval [ - 2 , 2] . Describe the following sets: (a) S T = [ - 2 , 2] = T (b) S T = (1 , 2) = S (c) R \ S = ( -∞ , 1] [2 , ) (d) T \ S = [ - 2 , 1] ∪ { 2 } (e) R \ ( T \ S ) = ( -∞ , - 2) (1 , 2) (2 , ) 1.2 #2. Describe the following sets of real numbers: (a) { x R x 2 > 2 } = { x R x < - 2 or x > 2 } (b) { x R | x | ≤ 3 } = { x R - 3 x 3 } (c) { x R | x - 2 | ≤ 3 } = { x R - 1 x 5 } (d) { x Q | x | ≤ 1 } = { x Q - 1 x 1 } 1.2 #3. If A , B , and C are sets, prove that A ( B C ) = ( A B ) ( A C ) . Proof. Let x A ( B C ). Then x A and x B C . So x is always in A and it may be in B or C or both. Case 1: x B but x / C . Since x A , x A B . Case 2: x C but x / B . Since x A , x A C . Case 3: x B and x C . Then x A B and x A C . In all cases, x A B or x A C . So x ( A B ) ( A C ). Therefore, A ( B C ) ( A B ) ( A C ). Now let x ( A B ) ( A C ). Then x A and x B and x / C , x A and x C and x / B , or x A and x B and x C . Case 1: x A and x B and x / C . Since x B , x B C . So x A ( B C ). Case 2: x A and x C and x / B . Since x C , x B C . So x A ( B C ). Case 3: x A and x B and x C . It is clearly the case then that x A ( B C ). In all cases, x A and x B C . So x A ( B C ). Therefore, A ( B C ) ( A B ) ( A C ). Since A ( B C ) ( A B ) ( A C ) and A ( B C ) ( A B ) ( A C ), A ( B C ) = ( A B ) ( A C ) ± 1.2 #5a. Describe the Cartesian product S × T where S = [0 , 1] and T = { x x 0 } S × T = { ( s,t ) 0 s 1 and t 0 } 1.2 #5d. Describe the Cartesian product S × T where S = N and T = R . S
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This note was uploaded on 12/07/2010 for the course MATH 523 taught by Professor Nahmod during the Fall '10 term at UMass (Amherst).

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Math 523H#2 - Math 523H Homework#2 Sections 1.2 1.3 1.2#1 Let S be the open interval(1 2 and let T be the closed interval[2 2 Describe the

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