319HW2Correction

319HW2Correction - MAT 319/320 Correction of HW2 Exercise...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 2

319HW2Correction - MAT 319/320 Correction of HW2 Exercise...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online