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Unformatted text preview: 4130 HOMEWORK 4 Due Tuesday March 2 (1) Let N N denote the set of all sequences of natural numbers. That is, N N = { ( a 1 ,a 2 ,a 3 ,... ) : a i ∈ N } . Show that  N N  = P ( N )  . We use the Schr¨ oderBernstein Theorem. First, there is an injection from P ( N ) to N N , because we may regard a subset of N as a sequence of zeroes and ones, or equivalently as a sequence of 1’s and 2’s, and this gives the desired injection. The hard part is showing that there is an injection N N → P ( N ). To see this, note that a sequence of natural numbers is the same thing as a function N → N . But by definition, a function is a special kind of subset of N × N . In this way, we get an injection N N → P ( N × N ). But it was shown in class that N × N and N have the same cardinality, and hence so do their power sets. In this way, we get the desired injection N N → P ( N ). Working through the above proof, we can explicitly write down an injection if we like. An example is: ( a 1 ,a 2 ,... ) 7→ { 2 · 3 a 1 , 2 2 · 3 a 2 ,... } ⊂ N . (2) Let { x n } be a Cauchy sequence of rational numbers. Regarding { x n } as a sequence of real numbers, show that { x n } converges to the real number x defined as the equivalence class of the sequence { x n } . The hardest part of this is working out what to prove in the first place. Let ε > 0 be a real number. Choose A ∈ N with 3 / 2 A < ε . Since { x n } is a Cauchy sequence of rational numbers, there exists N ∈ N such that if m,n > N then  x n x m  < 1 /A . In other words, if m,n > N then 1 /A < x m x n < 1 /A ....
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This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell.
 '08
 PROTSAK
 Natural Numbers

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