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Unformatted text preview: Math 131A  Section 2 Spring 2010 Homework 5 For this assignment, the following problems were graded: 11.5 (3 points), 11.7 (3 points), 11.9(b) (2 points), 12.1 (2 points), 12.4 (2 points), 12.12 (3 points). 11.5.(a). Let ( q n ) be an enumeration of the rationals in the interval (0 , 1], and let q ∈ [0 , 1]. Let I n = ( q 1 n ,q + 1 n ) ∩ (0 , 1]. Note that there are infinitely many rationals in I n for all n . Let n 1 be chosen such that q n 1 ∈ I 1 . Having chosen n 1 < ··· < n k with q n j ∈ I j for all 1 ≤ j ≤ k , since there are infinitely many l such that q l ∈ I k +1 , so there exists n k +1 > n k such that q n k +1 ∈ I k +1 . This gives a subsequence ( q n k ), and by construction lim k →∞ q n k = q . Thus the set of subsequential limits of ( q n ) contains [0 , 1]. If x / ∈ [0 , 1], then there exists > 0 (for instance, = 1 2 min { x  ,  x 1 } ) such that ( x ,x + ) ∩{ q n  n ∈ N } = ∅ . Thus, x cannot be a subsequential limit of ( q n ), so we conclude that the set of subsequential limits is in fact [0 , 1]. (b). Taking the supremum and infumum of the above set gives limsup q n = 1 and liminf q n = 0. NOTES: Most people did not provide proofs of these results, just stated that answers. I gave it full credit since the problem is kind of ambiguously stated, but it is a good idea to understand the proofs of these results as well. 11.7. Let ( q n ) be an enumeration of Q . Note that for any M , there are infinitely many rationals q with q > M . Choose n 1 such that q n 1 > 1. Then, having chosen n 1 < ··· < n k with q n j > j for 1 ≤ j ≤ k , we may choose n k +1 > n k with q n k +1 > k + 1. This gives a subsequence q n k , and by construction we have that lim...
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 Spring '10
 Kim
 Limits, Tn, Sn, subsequence, lim supn→∞

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