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201A_F10_hw1_sol

# 201A_F10_hw1_sol - MAT201A HOMEWORK 1 SOLUTIONS MIHAELA...

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MAT201A - HOMEWORK 1 SOLUTIONS MIHAELA IFRIM 1. Let ( X, d ) be a metric space. Suppose that { x n } ⊂ X is a sequence and set ε n := d ( x n , x n +1 ). Show that for m > n d ( x n , x m ) m - 1 X k = n ε k X k = n ε k . Conclude from this that if X k =1 ε 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 ε k . Proof. We are working on a metric space ( X, d ) and we define the sequence ε 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 ε k X k = n ε 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 ) = ε n + ε n +1 + ε n +2 + · · · + ε m - 2 + ε m - 1 = m - 1 X k = n ε k . The last part of the inequality is trivial since we add positive numbers : each ε k represends a distance, hence it is a positive number. Now we want to prove that if the series k =1 ε 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 I’ve 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 ε k is convergent, then the sequence of partial sum s m = { ε m } m =1 is conver- gent, where ε m = m k =1 d ( x k , x k +1 ). Date : October 8, 2010. 1

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2 MIHAELA IFRIM Since any convergent sequence is in particular a Cauchy sequence, this implies { ε 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 ε k = s m - s n X k = n ε k 2. Show that the limits of Cauchy sequences in a complete metric space are unique. Proof. Given ε > 0 we will show that d X ( x, y ) < ε . Since d X ( x, y ) 0, and d X ( x, y ) > 0 would contradict d X ( x, y ) < ε for ε = d X ( x, y ), we must have d X ( x, y ) = 0, so x = y . Since x n x there is some integer N ε so that n N ε implies d X ( x n , x ) < ε 2 . Since x n y there is some integer M ε so that n M ε implies d X ( x n , y ) < ε 2 . Let n max { N ε , M ε } . Then d X ( x, y ) d X ( x, x n ) + d X ( x n , y ) < ε 2 + ε 2 = ε = d X ( x, y ). 3. Let ( X, d ) be a metric space and suppose that { x n } n =1 X converges to x X . Show that { x n } n =1 is a Cauchy sequence. Give an example where the converse is not true.
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