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Functional Analysis Exercise Class Week 19 October – 23 October: Deadline to hand in the homeworks: your exercise class on week 26 October – 30 October Exercises with solutions (1) Show that any finite set in a topological space is compact. Solution: Let M be a finite set in a topological space X , and let {Ui }i∈I be an open cover. Then for every x ∈ M we can choose an ix such that x ∈ Uix , and hence {Uix }x∈M is a finite subcover of M . (2) (Discrete topology) Let X be a set, and ( 1, x 6= y, d(x, y) := 0, x = y. a) Show that d is a metric on X , and it generates the discrete topology τd = P(X ) := {M ⊂ X }. b) What are the open/closed/compact sets in this topology? c) Given a set M ⊂ X , what is its closure/interior/boundary? d) Describe the open balls Bε (x) := {y ∈ X : d(x, y) < ε} and the closed balls B̄ε (x) := {y ∈ X : d(x, y) ≤ ε} for every ε ≥ 0. Is it true that Bε (x) = B̄ε (x) for all ε > 0, where Bε (x) denotes the closure of Bε (x)? Solution: a) It is clear that d(x, y) = d(y, x), x, y ∈ X , and d(x, y) = 0 ⇐⇒ x = y. The inequality d(x, z) ≤ d(x, y) + d(y, z) is trivial when x 6= y or x 6= z; on the other hand, if x = y and y = z then x = z and thus d(x, z) = 0, in which case the inequality holds again. 1 b) By definition, all subsets of X are open, and hence all subsets are closed as well. The compact sets are exactly the finite sets. Indeed, if M is finite then it is compact by Exercise (1). On the other hand, if M has infinite cardinality then consider the open cover {{x}}x∈M . Obviously, this open cover has no finite subcover. c) Since every set is open and closed, every set is equal to its closure and its interior. As a consequence, the boundary of every set is empty. d) ε = 0, ∅, Bε (x) = {x}, ε ≤ 1, X, ε > 1, and ( {x}, ε < 1, B̄ε (x) = X, ε ≥ 1. 6 X = B̄1 (x) unless |X | = 1. In particular, B1 (x) = {x} = (3) (Indiscrete/trivial topology) Recall that the indiscrete topology on a set X is τid = {∅, X }. a) Is this topology metrizable? Is it Hausdorff? b) What are the open/closed/compact sets in this topology? c) Given a set M ⊆ X , what is its closure/interior/boundary? Solution: a) If |X | > 1 then any way we choose two different points x, y ∈ X , all neighbourhoods of x will contain y and vice versa, since their only neighbourhood is the whole space. Hence, X is not Hausdorff, and therefore it is not metrizable, either. b) By definition, the only open sets are ∅ and X . The closed sets are their complements, i.e., X and ∅. Every set is compact in this topology. Indeed, the empty set is compact in any topological space, and if M ⊆ X is nonempty then its only open cover is {X }, which is already finite. c) For any non-empty set M ⊆ X , its closure is X . The interior of a set M is the empty set if M 6= X , and the interior of X is X. Hence, the boundary of the empty set and the boundary of X are both the empty set; the boundary of any other set is X . (4) (Half-line topology) Let τhl := {(a, +∞) : a ∈ R ∪ {−∞, +∞}}. a) Show that τhl is a topology on R. 2 b) Is this topology metrizable? Is it Hausdorff? c) What are the open/closed/compact sets in this topology? d) Given a set M ⊂ X , what is its closure/interior/boundary? Solution: a) We have ∅ = (+∞, +∞), and R = (−∞, +∞). Given ai , i ∈ I, where I is any index set, we have ∪i∈I (ai , +∞) = (inf i ai , +∞) ∈ τhl . If I is finite then ∩i∈I (ai , +∞) = (maxi ai , +∞) ∈ τhl . Hence, τhl contains the empty set, the whole space, it is closed under taking the arbitrary union and the finite intersection of its elements, and therefore it is a topology. b) Given x, y ∈ R, x < y, we see that all neighbourhoods of x contain y, and hence they can not be separated by open sets. Thus, R with the half-line topology is not Hausdorff, and therefore not metrizable, either. c) The open sets are given by the definition of the topology. The closed sets are the empty set, R, and (−∞, a] for any a ∈ R. Note that any open cover of M is of the form {(ai , +∞)}i∈I . If inf M ∈ M then there must exist an i such that ai < inf M , and hence {(ai , +∞)} is a finite subcover (with a single open set). On the other hand, if inf M ∈ / M then we can choose a strictly decreasing sequence an ∈ R, n ∈ N, that converges to inf M , and define an open cover of M by {(an , +∞)}n∈N ; this open cover has no finite subcover. Hence, the compact sets are exactly those that contain their infimum. d) By definition, the interior of M is (amin , +∞), where amin := inf{b ∈ R : (b, +∞) ⊆ M }. Note that in particular, if M doesn’t contain an interval of the form (b, +∞) with b ∈ R then amin = inf ∅ = +∞, and (+∞, +∞) = ∅ is the interior of M . The closure of M is (−∞, sup M ] if sup M is finite, otherwise it is R. The boundary of M is (−∞, amin ] if amin 6= +∞, otherwise it is equal to the closure of M . 1 Definition. Let (X , τ ) be a topological space, and let xn ∈ X , n ∈ N, be a sequence in it. We say that x ∈ X is a limit of the sequence, or equivalently, that the sequence converges to x, if for every neighbourhood U of x, there exists an nU ∈ N such that for all n ≥ nU , xn ∈ U . We say that an x ∈ X is a cluster point of the sequence, if for every neighbourhood U of x, and every n ∈ N, there exists an m ∈ N such that m ≥ n and xm ∈ U . 2 Remark. Cluster points are also called accumulation points, condensation points or limit points. Be careful, the latter terminology is a bit confusing, and easy to mix with the notion of a limit. From the definition it is clear that every limit of a sequence is also a limit point, but not every limit point has to be a limit. 3 (5) Show that every sequence in a Hausdorff space has at most one limit. How about limit points? Construct a sequence in [0, 1] (equipped with the Euclidean topology) that has continuum many limit points. Solution: Assume that x 6= y are both limits of a sequence (xn )n∈N . By the Hausdorff property, there exist disjoint open sets Ux 3 x and Uy 3 y. Since x = limn xn , there exists an N1 ∈ N such that for all n ≥ N1 , xn ∈ Ux . Likewise, y = limn xn implies the existence of an N2 ∈ N such that for all n ≥ N2 , xn ∈ Uy . Thus, for all n ≥ max{N1 , N2 }, xn ∈ Ux ∩ Uy = ∅, a contradiction. Let {qn }n∈N be an arbitrary ordering of the rational numbers in [0, 1]. Define a sequence by x1 := q1 ; x2 := q1 , x3 := q2 ; x4 := q1 , x5 := q2 , x6 := q3 , etc. For any x ∈ [0, 1], we can choose a sequence of rational numbers (rm )m∈N such that limm rm = x. Then for any ε > 0 and n ∈ N, there exists a k ∈ N such that rk ∈ (x − ε, x + ε) and an m ∈ N such that m > n and xm = rk ∈ (x − ε, x + ε). Hence, all the points in [0, 1] are limit points of the sequence (xn )n∈N . (6) Describe the set of limits of a sequence in a set equipped with the a) discrete b) indiscrete topology, and c) in the half-line topology on R. Solution: Let X be equipped with the discrete topology. Since {x} is a neighbourhood of any x ∈ X , the only sequences that have a limit are those that are constant after some index. If X is equipped with the indiscrete topology then every point of X is a limit of any sequence in X . Finally, in the half-line topology, a point x ∈ R is a limit of a sequence (xn )n∈N if and only if for all ε > 0, {n : xn < x − ε} is finite. Equivalently, the set of limits of a sequence (xn )n∈N is (−∞, lim inf n xn ]. (7) Let X be a topological space and V be a normed space over the field K = R or K = C. Define kf k∞ := sup kf (x)k , f ∈ V X, and x∈X Cb (X , V ) := {f : X → V, f continuous and kf k∞ < +∞}. a) Show that if fn ∈ Cb (X , V ), n ∈ N, and f ∈ V X is such that lim kfn − f k∞ = 0 then f ∈ Cb (X , V ). n→+∞ b) Show that if V is a Banach space then Cb (X , V ) is a Banach space, i.e., it is a complete metric space with respect to the metric d(f, g) := kf − gk∞ . Solution: a) First, we show that f is continuous, which is equivalent to showing that it is continuous at every x ∈ X . To this end, fix an x ∈ X and an ε > 0; 4 we have to show that there exists a neighbourhood U of x such that for all y ∈ U , kf (y) − f (x)k < ε. Since limn→+∞ kfn − f k∞ = 0, we have an n ∈ N such that kfn − f k∞ < ε/3. Since fn is continuous, there exists a neighbourhood U of x such that for all y ∈ U , kfn (y) − fn (x)k < ε/3. Using now the triangle inequality, we get kf (y) − f (x)k = kf (y) − fn (y) + fn (y) − fn (x) + fn (x) − f (x)k ≤ kf (y) − fn (y)k + kfn (y) − fn (x)k + kfn (x) − f (x)k ≤ 2 kfn − f k∞ + kfn (y) − fn (x)k < ε. Next, we show that f is bounded. This is because for every n ∈ N, kf (x)k = kf (x) − fn (x) + fn (x)k ≤ kf (x) − fn (x)k + kfn (x)k ≤ kf − fn k∞ + kfn k∞ , x ∈ X. Hence, f is continuous and bounded, and thus an element of Cb (X , V ). b) Let fn ∈ Cb (X , V ), n ∈ N, be a Cauchy sequence, i.e., for every ε > 0 there exists an Nε ∈ N such that for all n, m ≥ Nε , kfn − fm k∞ < ε. Since for any x ∈ X , kfn (x) − fm (x)k ≤ kfn − fm k∞ , we see that (fn (x))n∈N is a Cauchy sequence in V . Since V is complete, this sequence is convergent, and we denote its limit by f (x) := limn→+∞ fn (x). We have to show that kfn − f k∞ converges to 0. To this end, let ε > 0, and Nε be such that kfn − fm k∞ < ε for all n, m ≥ Nε . Then, for any n ≥ Nε , and any x ∈ X , we have kfn (x) − f (x)k = lim kfn (x) − fm (x)k < ε, m→+∞ and hence kfn − f k∞ = sup kfn (x) − f (x)k < ε. x∈X This shows that limn→+∞ kfn − f k∞ = 0, and hence, by the previous point, f ∈ Cb (X , V ). (8) According to the definition of a topology, the infinite union of open sets and the finite intersection of open sets are open. Give an example when the infinite intersection of open sets is not open. Solution: For n ∈ N, let Un = (− n1 , n1 ) be an open set in R with a usual metric. Then ∩n∈N Un = {0}, a single point. Since no ball with positive radius around 0 is contained in {0}, it is not an open set. 5 (9) In the lecture it was proved that a compact set K in a Hausdorff space (X , τ ) is closed. Show that this result need not be true if the space X is not Hausdorff. Solution 1: As we have seen in Exercise (3), and is indeed trivial to verify, any set in an indiscrete topological space is compact; however, if the space has at least two points then no proper non-empty subset of it is closed. Solution 2: Consider the Sierpinski two-point topological space - a space that consists of two points, X := {a, b}, and the topology is τ := {∅, {a}, {a, b}}. Then the set {a} is compact since it is finite (see Exercise (1)). But it is not closed, since it is not the complement of an open set. (10) Show that a set in a topological space is closed if and only if it contains its boundary, and open if and only if it is disjoint from its boundary. Solution: By the definition of the boundary, ∂A = A \ int(A) and hence A = int(A) ∪ ∂A with int(A) ∩ ∂A = ∅. Note that A is closed if and only if A = A, and since any set contains its interior by definition, we see that indeed A = A if and only if ∂A ⊆ A. Similarly, A is open if and only if A = int(A), which is equivalent to A being disjoint from its bounday. 6