204ps22006sols

# 204ps22006sols - Problem Set 2 Solutions Economics 204...

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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 limsup x n = liminf x n = x . ( ) We will prove the contrapositive. Suppose that lim n →∞ x n = 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 limsup x n must be greater than or equal to x + . If the latter is true, then β n x for all n , so liminf x n must be less than or equal to x . In either case, it is not true that limsup x n = liminf 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 0 | = 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

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(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 epsilon-delta 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. (a) Let A , B , denote subsets of a space X . Determine whether the following equations hold; if an equality fails, determine whether one of the inclusions or holds. 1. cl( A B ) = cl A cl B . The equality is not true, but it is true that cl( A B ) cl A cl B . To illus- trate, let A = [0 , 1) and B = (1 , 2]. The sets are disjoint so the closure of their intersection is the empty set. However, the intersection of their closure is { 1 } . To prove this, recall that x X is in the losure of A if every -ball around x contains at least one point in A . So x cl( A B ) means that for any > 0, B ( x ) contains at least on point that is in both A and B . Clearly, then this point is in the closure of A and it is in the closure of B .
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