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# 131a5 - Problems graded 11.9(10 points 12.4(10 points...

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Unformatted text preview: Problems graded: 11.9 (10 points), 12.4 (10 points), 12.12 (15 points) The other problems were graded on completion to be worth five points each (12.3, which was longer, was worth 15); as this made the assignment worth 80 points, all scores were multiplied by 1.25 and rounded up to the next integer to make the assignment scores out of 100. 11.6. Referring to our initial sequence as { x n } , we can refer to a subsequence as { x n k } (indexed by k ) where σ : N → N is a selection function and σ ( k ) = n k for each n ∈ N . We can denote y k = x n k for each positive integer k to simplify notation; our subsequence is { y k } . The object we care about is a subsequence of a subsequence, i.e. a subsequence of { y k } . However, it can be written as { y k j } (indexed by j ) where τ : N → N is a selection function and τ ( j ) = k j for each j ∈ N . To show that this new subsequence is a subsequence of the original sequence, we consider the function σ ◦ τ : N → N ; being the composition of two selection functions, it is itself a selection function. We note that x σ ◦ τ ( j ) = x σ ( k j ) = x n k j = y k j which is exactly the j th term of our desired sequence; therefore, a subsequence of a subsequence is indeed a subsequence of the originial sequence. NOTE: The order in which the selection functions are performed is the REVERSE of the order in which the subsequences are applied; when this type of order-reversing behavior occurs in mathematics, it is called ”contravariant”. 11.7. We select our subsequence by the following steps. Step 1: Let n 1 be the first positive integer such that r n 1 > 1 (we can do this because the rationals are unbounded from above). Step k, k > 1 (assuming step k − 1 is complete): Let n k be the first positive integer greater than n k − 1 such that r n k is greater than the maximum of max r i and k (where the maximum is over i ≤ n k − 1 ); once again we can do this as the rationals are unbounded from above. Once we have completed this procedure we have a subsequence { r n k } with r n k > k for each positive integer k and therefore lim k →∞ r n k = + ∞ ....
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131a5 - Problems graded 11.9(10 points 12.4(10 points...

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