Math 320-2: Midterm 2 Practice Solutions Northwestern University, Winter 2015
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1 Math 30-: Midterm Practice Solutions Northwestern University, Winter Give an example of each of the following. No justification is needed. (a) A metric on R with respect to which R is bounded. (b) A subset of R (standard metric) with empty interior and whose closure is not R. (c) A nonempty subset of Q (standard metric) which is both open and closed. (d) A non-complete subspace of R which is not open. (e) A proper dense subset of C([0, 1]). Solution. (a) R is bounded with respect to the discrete metric since the open ball of radius around any point contains everything else. (b) The set of integers Z has empty interior in R and has closure equal to Z itself. (c) The set of rationals between 3 and 3 is both open and closed in Q. More generally, if a and b are irrational, then (a, b) Q is both open and closed in Q. (d) The unit disk (disk of radius 1 centered at the origin) with only the top half of the boundary circle is not complete and not open. (e) The set of polynomial functions is dense in C([0, 1]).. Show that for any points p, q in a metric space X, there exist disjoint open subsets U and V of X such that p U and q V. (Draw a picture!) Proof. For distinct p, q X, set r = d(p,q) > 0. We claim that open balls U = B r (p) and V = B r (q) are disjoint. Indeed, suppose that x B r (p). Then d(x, p) < r, so d(x, q) d(p, q) d(x, p) > d(p, q) r = d(p, q) where the first inequality follows from d(p, q) d(p, x) + d(x, q). Thus x / B r (q), so no point of B r (p) is in B r (q) and hence these open sets are disjoint. 3. Let X be a metric space and p a point of X. Show that for any r > 0, the closure of the open ball B r (p) of radius r around p is contained in the closed ball M r (p) of radius r around p. Give an example where the closed ball of radius r around p is strictly larger than the closure of the corresponding open ball. In other words, show that B r (p) M r (p) is always true but B r (p) = M r (p) is not, where M r (p) denotes the closed ball of radius r around p. Solution. Suppose that q B r (p). Then there exists a sequence (p n ) in B r (p) which converges to q. Since each p n is in B r (p), we have Then d(p n, p) < r for all n. d(q, p) d(q, p n ) + d(p n, p) < d(q, p n ) + r. Since p n q, d(p n, q) 0 and taking limits in this previous inequality gives d(q, p) r,
2 showing that r M r (p). Hence B r (p) M r (p) as claimed. Consider R with the discrete metric. Then B 1 (0) = {0} since any nonzero point is at a distance 1 away from 0, and hence B 1 (0) = {0} as well. But the closed ball of radius 1 around 0 is M 1 (0) = R and so is strictly larger than the closure of the corresponding open ball. 4. Show that any Cauchy sequence in a metric space X is bounded. Proof. Suppose that (p n ) is Cauchy. Then there exists N such that In particular, d(p n, p m ) < 1 for m, n N. d(p n, p N ) < 1 for n N. Set r = max{d(p 1, p N ), d(p, p N ),..., d(p N 1, p N )} + 1. Then for any 1 n < N we have while for any n N we have d(p n, p N ) max{d(p 1, p N ), d(p, p N ),..., d(p N 1, p N )} < r d(p n, p N ) < 1 r. Thus every term in (p n ) is in B r (p N ), so (p n ) is bounded. 5. Suppose that A and B subsets of R. Show that A B is closed in R with respect to any of the Euclidean, taxicab, or box metrics if and only if A and B are closed in R with respect to the standard metric. Proof. Suppose first that A and B are closed in R and suppose that (a n, b n ) is a sequence in A B which converges to (a, b) R. Then a n a, so a A since A is closed in R, and b n b, so b B since B is closed in R. Thus (a, b) A B, showing that A B is closed in R. Conversely, suppose that A B is closed in R. Let (a n ) be a sequence in A which converges to some x R. Fix p B and consider the sequence (a n, p) in R. Since a n a and the constant sequence p converges to p, we have (a n, p) (a, p). Since A B is closed in R and each (a n, p) is in A B, we must have (a, p) A B. Thus in particular a A, showing that A is closed in R. A similar argument reversing the roles of A and B show that B is closed in R as well. 6. Given a subset E of a metric space X, define an accumulation point of E to be a point x X such that every open ball around x contains an element of E distinct from x. (This is almost but not quite the same as a limit point of E, where in the case of a limit point we do not require that the element of E the open ball contain be different from x.) Determine the accumulation points and limit points of the following subsets of R. (a) { } 1 n n N (b) Z (c) [a, b] (d) Q Based on this, give a guess without proof as to how to characterize closed sets in terms of accumulation points.
3 Solution. (a) The only accumulation point of this set is 0. Indeed, any open interval ( ɛ, ɛ) contains a number of the form 1 n by the Archimedean property, and no such number equals 0. A number of the form 1 N is not an accumulation point since the interval ( 1 N ɛ, 1 N + ɛ) for ɛ = 1 N 1 N+1 contains no other number of the form 1 n. Finally, no other real number is an accumulation point of E since we can find an interval around it which avoids all numbers of the form 1 n. (b) Z has no accumulation points. Indeed, no n N is an accumulation point since (n 1, n+1) contains no other element of Z, and no non-integer number is an accumulation point since we can find an interval around it which avoids all integers. (c) The accumulation points of [a, b] are the elements of [a, b] itself. For any x / [a, b], we can find an interval around it which is disjoint from [a, b] since the complement of [a, b] is open in R, and any interval around an element y of [a, b] indeed contains an element of [a, b] different from y. (d) Every element of R is an accumulation point of Q since for any x R, any interval around x contains infinitely many rationals by the denseness of Q in R, so at least one such rational will be different from x if x happens to be rational. A subset A X is closed in X if and only if A contains all of its accumulation points. Note that this is true in each of (b) and (c) above, each of which are closed in R. 7. Suppose that (X, d) is a metric space. It is then a fact that the function ρ(x, y) = d(x, y) 1 + d(x, y) also defines a metric on X. Show that a sequence in X converges with respect to d if and only if it converges with respect to ρ. Also, show that X is bounded with respect to ρ. Proof. The point of this result is that is shows an arbitrary metric on a space can always be replaced by a bounded metric which gives the same notion of convergence, and hence the same notion of closed and open sets. First suppose that x n x with respect to d. Then for any ɛ > 0 there exists N such that But d(x n, x) < ɛ for n N. ρ(x n, x) = d(x n, x) 1 + d(x n, x) d(x n, x), so we also have ρ(x n, x) < ɛ for n N, showing that x n x with respect to ρ Conversely suppose that x n x with respect to ρ. We first claim that the set {d(x n, x)} must be bounded in R. (This is a bit tricky.) If not, then we can find a subsequence (x nk ) of x n such that d(x nk, x), which implies that ρ(x nk, x) = d(x n k, x) 1 + d(x nk, x) 1. But since x n x with respect to ρ, we also have x nk x with respect to ρ so ρ(x nk, x) 0, contradicting ρ(x nk, x) 1. Thus {d(x n, x)} is bounded, say by M > 0. Let ɛ > 0. Since x n x with respect to ρ there exists N such that ρ(x n, x) < ɛ 1 + M for n N. 3
4 Hence for n N we have d(x n, x) = (1 + M) d(x n, x) 1 + M (1 + M) d(x n, x) 1 + d(x n, x) = (1 + M)ρ(x n, x) < ɛ, so x n x with respect to x as claimed. Pick any p X. Then for any q X we have ρ(p, q) = d(p, q) 1 + d(p, q) < 1, so q B 1 (p) and hence X B 1 (p), so X is bounded with respect to ρ. 8. For a subset A of a metric space X, show that A = X if and only if both A and its complement A c are dense in X. Proof. Suppose that A = X and let p X. Then p is a boundary point of A. Hence for each n N, B 1/n (p) contains an element a n A and element b n A c. These elements thus satisfy d(a n, p) < 1 n and d(b n, p) < 1 n, which implies that a n p and b n p. Hence p is a limit point of both A and A c, showing that A and A c are both dense in X. Conversely suppose that A and A c are both dense in X and let p X. Then there exist sequences (a n ) in A and (b n ) in A c which both converge to p. Any open ball around p then contains an term from (a n ) and a term from (b n ), and hence contains an element of A and an element of A c. Thus p A so A = X. (The containment A X is always true simply because the possible boundary points of A come from X.) 9. Show that the set of integrable functions on [a, b] is closed in the space C b [a, b] of all bounded functions with the sup metric. Is the subset of differentiable functions on [a, b] closed as well? Proof. Suppose that (f n ) is a sequence of integrable functions on [a, b] which converges to some f C b [a, b] with respect to the sup metric. Then f n f uniformly, so f is integrable since the uniform limit of integrable functions is integrable. Thus the set of integrable functions on [a, b] is closed in C b [a, b]. The subset of differentiable functions is not closed. In particular, f n = x + 1 n is a sequence of differentiable functions on [ 1, 1] which converges uniformly to the absolute value function, meaning that f n converges to the absolute value function with respect to the sup metric. But the absolute value function is not differentiable, and so is not in the subset of differentiable functions. This example can be modified to work on any closed interval [a, b]. 10. Define the function f n : [0, 1] R by f n (x) = { x n x R\Q 0 x Q. Determine, with proof, the distance d(f n, 0) between f n and the constant zero function with respect to the sup metric in C b ([0, 1]). 4
5 Solution. We claim that d(f n, 0) = 1 for every n. Indeed, fix n N. Then x n 0 1 for all x [0, 1] (R\Q), so 1 is an upper bound for the set { f n (x) 0 x [0, 1]} whose supremum defines d(f n, 0). To show that this supremum equals 1, let 0 < ɛ < 1. Then 1 ɛ > 0 so n 1 ɛ make sense. By the denseness of the irrationals in R, there exists an irrational y [0, 1] such that Taking n-th powers gives n 1 ɛ < y < 1. 1 ɛ < y n < 1, so f n (y) = y n is a number in { f n (x) 0 x [0, 1]} which is larger than 1 ɛ, showing that nothing smaller than 1 can be an upper bound of { f n (x) 0 x [0, 1]}. Thus sup{ f n (x) 0 x [0, 1]} = 1, so d(f n, 0) = 1 as claimed. 5
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