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# MATH 140A (Lecture B; course code 44775) Fall 2011 Instructor: Professor R. Reilly Assignment #7 COMPLETE Reading Assignments (Note: All section or...

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MATH 140A (Lecture B; course code 44775) Fall 2011 Instructor: Professor R. C. Reilly Assignment #7 COMPLETE Reading Assignments (Note: All section or page numbers refer to the course text, “Elementary Analysis: The Theory of Calculus”, by K. Ross.) Monday 11/07/2011: Section 11 Wednesday 11/09/2011: Section 12 Friday 11/11/2011: HOLIDAY Monday 11/14/2011: Section 14 Some Extra Material Here is some more stuﬀ that was mentioned in the lecture but does not appear to be stated explicitly in the text; or if it is, it is somewhat hidden. (1) Alternate Form of the Deﬁnition of ‘Cauchy Sequence’ The following condition, on a se- quence σ = ( s 1 ,s 2 , ...s n , ... ) of real numbers, is equivalent to the ‘Cauchy Condition’ described in Deﬁnition 10.8: For every ε > 0 there exists N such that if n > N then | s n + k - s n | < ε for all k in IN. Outline of the Proof Given in the Lecture In the text’s deﬁnition, one considers the inequality | s n - s m | < ε , where n,m N . However in this inequality there is no loss of generality in assuming that m n . Likewise, one can further reduce to the case in which m > n , since in the case m = n the desired inequality is trivially true. But if m > n , then m = n + k for some k in IN, namely k = m - n ; and, conversely, for each such k one has n + k > n . The desired result follows easily. (2) Theorem : Let A be an inﬁnite subset of IN. Then A can be described, in exactly one way, in the form A = { n 1 ,n 2 , ...n k , ... } , where 1 n 1 < n 2 < ...n k < ... . Moreover, one has n k k for all k . Outline of the Proof Given in the Lecture The number n k is simply the k -th smallest element of the set A . The inequality n k k follows by induction after noting that n 1 1, since 1 is the smallest natural number, together with the fact that consecutive n k ’s diﬀer by at least 1. (3) If A = { n 1 < n 2 < ...n k < ... } is an inﬁnite subset of IN, with the terms written in strictly increasing order, and σ = ( s 1 ,s 2 , ...s n , ... ) is a sequence, then the subsequence of σ associated with A , denoted σ A , is the sequence ( s n 1 ,s n 2 , ...s n k , ... ). Remark The notation σ A used above is NOT standard. Indeed, about the only place you will ﬁnd it with this meaning is in my Math 140A course. (4) Strengthened Version of Theorem 11.2 Let σ = ( s 1 ,s 2 , ...s n , ... ) be a sequence of real numbers. If L is either a real number or one of the quantities + or -∞ , and if lim n →∞ s n = L , then for every subsequence τ = ( t 1 ,t 2 , ...t k , ... ) of σ , one also has lim k →∞ t k = L . Note The text proves this only when L is ﬁnite. The proof of this stronger version which I gave in the lecture used the result of Hand-in Exercise (I) in Assignment (4). (5) Alternate Proof of Theorem 11.5, the Bolzano-Weierstrass Theorem for Sequences The statement of the theorem in question is as in the text: Every bounded sequence of real numbers has a convergent subsequence. Alternate Proof Let σ = ( s 1 ,s 2 , ...s n , ... ) be the sequence in question. Then, by the ‘bound- edness’ hypothesis, there exist numbers m and M such that m < s n < M for all n . Let a 1 = m and b 1 = M , so that a 1 < b 1 , and s n is in the closed interval I 1 = [ a 1 ,b 1 ] for all n . Let c 1 be
the midpoint of the interval I 1 , so c 1 = ( a 1 + b 1 ) / 2. Then for at least one of the half-intervals [ a 1 ,c 1 ] or [ c 1 ,b 1 ], the following condition holds: there are inﬁnitely many indices n such that s n is an element of that half-interval. If this condition holds for the left half-interval [ a 1 ,c 1 ], let a 2 = a 1 and b 2 = c 1 . If, instead, this condition does not hold for the left-half interval [ a 1 ,c 1 ], so that it must hold for the right-half interval [ c 1 ,b 1 ], let a 1 = c 1 and b 2 = b 1 . Either way, one gets a new subinterval, [ a 2 ,b 2 ], half the length of the original interval [ m,M ], such that s n [ a 2 ,b 2 ] for inﬁnitely many n . Continuing this process indeﬁnitely, one obtains an inﬁnite sequence of intervals [ a k ,b k ], with k IN, such that [ a 1 ,b 1 ] = [ m,M ], and for each k IN the interval [ a k +1 ,b k +1 ] is one of the two halves of [ a k ,b k ], and for each k IN there are inﬁnitely many n IN such that s n [ a k ,b k ]. Clearly b k - a k = ( M - m ) / 2 k - 1 for each k , so that lim k →∞ ( b k - a k ) = 0. Thus it follows from Discussion Exercise (4) from Tuesday, 10/27/2011, that the intersection of these intervals consists of a single point; call it d . Note that d = lim k →∞ a k = lim k →∞ b k . Now deﬁne a subsequence ( s n 1 ,s n 2 , ...s n k , ... ) of σ by the following rule: First, set n 1 = 1; note that n 1 is the ﬁrst index such that s n 1 [ a 1 ,b 1 ]. Likewise, deﬁne n 2 to be the ﬁrst index larger that n 1 such that s n 2 [ a 2 ,b 2 ]. Suppose now that indices n 1 , n 2 , . . . n k have already been constructed so that n 1 < n 2 < ... < n k , and s n j [ a j ,b j ] for each j = 1 , 2 , ...k . Then deﬁne the index n k +1 to be the smallest index such that n k < n k +1 and s n k +1 [ a k +1 ,b k +1 ]. (The existence of such an index which is strictly larger than n k follows from the fact that we have constructed these intervals so that for each of them, s n in the interval for inﬁnitely many indices n ; in particular, n k cannot be the largest such index.) It follows that n 1 < n 2 < ... < n k < ... for all k , so that τ = ( s n 1 ,s n 2 , ...s n k ... ) is a subsequence of σ . It is also clear from the construction that s n k [ a k ,b k ] for each k ; that is, a k s n k leq b k for each k . It now follows from the Squeeze Theorem, i.e., the result of Exercise 8.5(a), that lim k →∞ s k = d as well, as required. Remarks (a) The proof outlined above, of the Bolzano-Weierstrass Theorem for Sequences, is pretty much the standard one, found in most analysis texts. (b) I’ve appended the phrase ‘for Sequences’ to the text’s name of this theorem because there are very similar theorems, found in other texts, which are also called ‘the Bolzano-Weierstrass Theorem’, but which apply in other situations. Be careful, when reading other texts or analysis papers, to know to which version their use of the name ‘Bolzano-Weierstrass Theorem’ refers. Things to Prepare for the Discussions (These are not to be handed in.) Note: These are problems/examples/topics which the Teaching Assistant (TA) will cover in the discussion sections. You are expected to work on them before you go to the discussions. (A) (Items for the discussion on Tuesday 11/08/2011. Do not hand in.) (1) Exercise 11.2 (2) Exercise 12.3 (a, b, d, e) Note: You do not need the results of Section 12 to do this exercise. It could have been placed it in Section 10 just as well. (3) Go over Example 4 in Section 11 (B) (Items for the discussion on Thursday 11/10/2011. Do not hand in.) REVISED (1) Let σ = ( s 1 ,s 2 , ... ) be a sequence of real numbers. Let τ = ( t 1 ,t 2 , ... ) be the subsequence of σ given by the odd-order terms of σ , and let ξ = ( x 1 ,x 2 , ... ) be the subsequence of σ given by the even-order terms of σ . That is, τ = ( s 1 ,s 3 ,s 5 , ...s 2 k - 1 , ... ) and ξ = ( s 2 ,s 4 ,s 6 , ...s 2 k , ... ) ,
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