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Unformatted text preview: Problem Set 2 – Solutions Economics 204  August 2006 1. Theorem 4 from the lim inf/lim sup handout: Let { x n } be a sequence of real numbers. Then lim x →∞ x n = x ∈ R ∪ {−∞ , ∞} if and only if lim inf x →∞ x n = lim sup x →∞ x n = x. Prove this theorem for the case that x is finite. Proof: ( ⇒ ) x n → x ∈ R implies that ∀ > 0 there exist N ( ) such that n ≥ N ( ) ⇒  x n − x  < . This means that x + is an upper bound and x − is a lower bound for { x k : k ≥ N ( ) } . Using α n = sup { x k : k ≥ n } and β n = inf { x k : k ≥ n } , we know that β n ≤ α n (because a lower bound can’t be greater than an upper bound) and for n > N ( ), x − ≤ β n ≤ α n ≤ x + . Since this is true for any , it must be true that α n and β n both converge to x . This completes the proof that lim sup x n = lim inf x n = x . ( ⇐ ) We will prove the contrapositive. Suppose that lim n →∞ x n 6 = x . Then there exists an > 0 such that for all N , there is some n ≥ N such that  x n − x  ≥ . This means that there are infinitely many x n outside of B ( x ) and it must be the case that there are infinitely many of these above x + , infinitely many below x − or both. If the former is true, then α n ≥ x + for all n which means that lim sup x n must be greater than or equal to x + . If the latter is true, then β n ≤ x − for all n , so lim inf x n must be less than or equal to x − . In either case, it is not true that lim sup x n = lim inf x n = x , completing the proof. 2. Using the definition of an open set, prove that (0 , 1) is an open subset of R (a) Under the usual absolute value metric. Let x ∈ (0 , 1) and choose < min { x, 1 − x } . Then B ( x ) ⊂ (0 , 1) because  x −  = x > and  1 − x  = 1 − x > . Therefore (0 , 1) is open. (b) Under the discrete metric. Let x ∈ (0 , 1) and choose = 1 / 2. Then B ( x ) = { x } ∈ (0 , 1) so (0 , 1) is open. 3. Construct a sequence of real numbers 1 (a) That is unbounded and has exactly three cluster points. One of many possibilities: { x n } = { 1 , 2 , 3 , 4 , 1 , 2 , 3 , 8 , 1 , 2 , 3 , 12 , 1 , 2 , 3 , 16 , 1 , 2 , 3 , 20 , 1 , 2 , 3 , 24 , . . . } (b) That is bounded and has infinitely many cluster points. One possibility: { x n } = { 1 , 1 , . 1 , 1 , . 1 , . 01 , 1 , . 1 , . 01 , . 001 , 1 , . 1 , . 01 , . 001 , . 0001 , 1 , . . . } 4. Prove that lim n →∞ ( √ n 2 + n − n ) = 1 / 2. (An epsilondelta proof isn’t neccesary.) Proof: Notice that √ n 2 + n − n = √ n 2 + n − n 1 · √ n 2 + n + n √ n 2 + n + n = n √ n 2 + n + n . Furthermore, n √ n 2 + n + n = 1 √ 1+ 1 n +1 → 1 / 2 (as n → ∞ ). 5. Closure, Boundary, Interior, etc....
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This note was uploaded on 03/06/2009 for the course ECON 0204 taught by Professor Staff during the Summer '08 term at Berkeley.
 Summer '08
 Staff
 Economics

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