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Unformatted text preview: MAT 319/320 Correction of HW2 Exercise 1. Page 38, #2. Proof. The set S 2 is not empty and is bounded below (for example by 0), so it has an infimum. Lets prove that 0 = inf ( S 2 ) : 1. For any x in S 2 , one has lessorequalslant x ; 2. For any > , one can find x in S 2 such that lessorequalslant x < 0 + (take /2 for example). Therefore 0 = inf ( S 2 ) . Now S 2 is not bounded above, so it doesnt have upper bounds (and therefore doesnt have a sup). square Exercise 2. Page 38, #3. Proof. Supremum: Since n N 1 n lessorequalslant 1 , one knows that S 3 is bounded above, since it is also a non empty subset of R , we know that it has a supremum. Lets prove that 1 = sup ( S 3 ) : 1. For any x in S 3 , one has x lessorequalslant 1 ; 2. For any > , one can find an x in S 3 such that 1 <x lessorequalslant 1 (just take x = 1 !) Infimum: Lets prove that 0 = inf ( S 3 ) : 1. For any x = 1/ n in S 3 , one has x greaterorequalslant ; 2. For any > , one can find an x in S 3 such that lessorequalslant x < (indeed by the archimedean property one knows the existence of an integer n > 1/ , then just take x = 1/ n...
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This note was uploaded on 01/31/2011 for the course AMS 319 taught by Professor Mark during the Fall '10 term at SUNY Stony Brook.
 Fall '10
 Mark

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