hw3solutions23b

# Hw3solutions23b - Anna Medvedovsky [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ Math 23b Spring 2009 HW#3 solutions 1 Let A,B ⊂ R P = R> f R → R Write sentences

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Unformatted text preview: Anna Medvedovsky [email protected] Math 23b / Spring 2009 HW #3 solutions 1. Let A,B ⊂ R , P = R > , f : R → R . Write sentences negating the statements below. a) Negation: There exists an x ∈ A such that for all b ∈ B , we have b ≤ x . b) Negation: For all x ∈ A there is a b ∈ B such that b ≤ x . c) Negation: There exist x and y in R such that f ( x ) = f ( y ) but x 6 = y . (In other words, f is not injective—there’s no inverse map from the image of f back to the source space.) d) Negation: There exists a b ∈ R such that for all x ∈ R , we have f ( x ) 6 = b . (In other words, f is not surjective—it doesn’t hit everything in the target space.) e) Formal negation: There exist x,y ∈ R , and ε ∈ P such that for all δ ∈ P we have | x- y | < δ and | f ( x )- f ( y ) | ≥ ε . However, if you think about this statement carefully, you’ll see that’s it’s false for every function f : for x and y to satisfy | x- y | < δ for all positive δ we must have x = y , but then there’s no way to get | f ( x )- f ( y ) | ≥ ε for any positive ε . So an equally good negation for the original statement is ”0 6 = 0” or any other vacuously false statement. (Undoubtedly, the original statement in the book should have read “For all x ∈ R and all ε ∈ P , there exists a δ ∈ P such that for all y ∈ R , if | x- y | < δ then | f ( x )- f ( y ) | < ε .” This is the condition that f is a continuous function.) This part was not graded. f) Negation: There exists an ε ∈ P such that for all δ ∈ P there exist x and y in R with | x- y | < δ but | f ( x )- f ( y ) | ≥ ε . (The original statement is the condition for f to be what’s called uniformly continuous : you can choose one δ for each ε that works for all x and y .) 2. Negating Lincoln’s “You can fool all of the people some of the time, and you can fool some of the people all of the time, but you cannot fool all of the people all of the time.” The dirty little secret of this exercise is that there are several reasonable ways to translate Lincoln’s poetic quip to a mathematical statement. Let P be the set of all people, T be the space of time, and F ( p,t ) be the function that returns true if person p is fooled at time t , and false otherwise. 1. The statement “You can fool all of the people some of the time” can be interpreted as either (a) “There are times when you can fool everybody”: ( ∃ t ∈ T )( ∀ p ∈ P ) F ( p,t ) , (b) or as “Every person can be fooled sometimes”: ( ∀ p ∈ P )(...
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## This note was uploaded on 10/15/2011 for the course MATH 23b taught by Professor Bonglian during the Spring '09 term at Brandeis.

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Hw3solutions23b - Anna Medvedovsky [email protected] Math 23b Spring 2009 HW#3 solutions 1 Let A,B ⊂ R P = R> f R → R Write sentences

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