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Unformatted text preview: MAT201A  HOMEWORK 1 SOLUTIONS MIHAELA IFRIM 1. Let ( X,d ) be a metric space. Suppose that { x n } X is a sequence and set n := d ( x n ,x n +1 ). Show that for m > n d ( x n ,x m ) m 1 X k = n k X k = n k . Conclude from this that if X k =1 k = X n =1 d ( x n ,x n +1 ) < then { x n } is Cauchy. Moreover, show that if { x n } is a convergent sequence and x = lim n { x n } then d ( x,x n ) X k = n k . Proof. We are working on a metric space ( X,d ) and we define the sequence n = d ( x n ,x n +1 ). The first thing we are asked to prove is that for m > n d ( x m ,x n ) m 1 X k = n k X k = n k . Using triangle inequality , we get very easily our desired result, i.e. d ( x m ,x n ) d ( x n ,x n +1 ) + d ( x n +1 ,x n +2 ) + + d ( x m 2 ,x m 1 ) + d ( x m 1 ,x m ) = n + n +1 + n +2 + + m 2 + m 1 = m 1 X k = n k . The last part of the inequality is trivial since we add positive numbers : each k represends a distance, hence it is a positive number. Now we want to prove that if the series k =1 k = n =1 d ( x n ,x n +1 ) has a finite sum then { x n } n is a Cauchy sequence. Here by s n Ive denoted s m = k = m . This is proved in the following : for each m > n > 0, we have (by triangle inequality) d ( x n ,x m ) d ( x m ,x m 1 ) + ... + d ( x n +1 ,x n ) = m 1 X k = n d ( x k ,x k +1 ) . Since the series k =1 k is convergent, then the sequence of partial sum s m = { m } m =1 is conver gent, where m = m k =1 d ( x k ,x k +1 ). Date : October 8, 2010. 1 2 MIHAELA IFRIM Since any convergent sequence is in particular a Cauchy sequence, this implies { m } m =1 is a Cauchy sequence. Therefore for any > 0, we can find N > 0 so that for any m > n N , we have  s n s m  < . As a consequence, we find m 1 X k = n d ( x k ,x k +1 ) m 1 X k = n d ( x k ,x k +1 ) + d ( x m ,x m +1 ) =  s n s m  < The above things we got imply that for > 0, we can find N > 0 so that d ( x n ,x m ) < whenever m > n N . Thus ( x n ) is a Cauchy sequence. Here to be more explicit you need to use induction. The last part of the problem follows right away from tha fact that the distance function is a contin uous function which allows us to move the limit inside the following: d ( x n ,x ) = lim m d ( x n ,x m ) lim m m 1 X k = m k = s m s n X k = n k 2. Show that the limits of Cauchy sequences in a complete metric space are unique. Proof. Given > 0 we will show that d X ( x,y ) < . Since d X ( x,y ) 0, and d X ( x,y ) > 0 would contradict d X ( x,y ) < for = d X ( x,y ), we must have d X ( x,y ) = 0, so x = y . Since x n x there is some integer N so that n N implies d X ( x n ,x ) < 2 ....
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This note was uploaded on 01/29/2012 for the course MATH 34 taught by Professor Wiley during the Spring '11 term at UC Merced.
 Spring '11
 Wiley
 Differential Equations, Equations

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