Analysis II: Topology and Differential Calculus of Several Variables

Transcription

1 Analysis II: Topology and Differential Calculus of Several Variables Peter Philip Lecture Notes Created for the Class of Spring Semester 2016 at LMU Munich October 22, 2018 Contents 1 Topology, Metric, Norm Motivation, Definitions, Convergence Open Sets, Closed Sets, and Related Notions Construction of Topological Spaces Bases, Subbases Subspaces Topics Particular to Metric and Normed Spaces Basic Inequalities Completeness Inner Products and Hilbert Space Equivalence of Metrics and Equivalence of Norms Limits and Continuity of Functions Definitions and Properties Banach Fixed Point Theorem Homeomorphisms, Norm-Preserving and Isometric Maps, Embeddings Further Topologic Properties 61 philip@math.lmu.de 1

2 CONTENTS Separation Compactness Connectedness Differential Calculus Partial Derivatives and Gradients The Jacobian Higher Order Partials and the Spaces C k Interlude: Graphical Representation in Two Dimensions The Total Derivative and the Notion of Differentiability Higher Order Total Derivatives as Multilinear Maps The Chain Rule The Mean Value Theorem Directional Derivatives Taylor s Theorem Implicit Function Theorem Extreme Values, Stationary Points, Optimization Definitions of Extreme Values Quadratic Forms Extreme Values and Stationary Points of Differentiable Functions Constrained Optimization, Lagrange Multipliers A Set-Theoretic Rules for Cartesian Products 117 B Box Topology 118 C Uniform Continuity and Lipschitz Continuity 119 D Viewing C n as R 2n 121 E Pseudometrics and Seminorms 123 F Initial and Final Topologies, Quotient Spaces 126 G Separation: More Counterexamples 129

3 CONTENTS 3 H Compactness 130 H.1 Intersections of Compact Sets H.2 Unit Balls in Normed Vector Spaces H.3 Proof of Tychonoff s Theorem I Topological Invariants 133 J Multilinear Maps 135 K Differential Calculus 138 K.1 Bounded Derivatives Imply Lipschitz Continuity K.2 Surjectivity of Directional Derivatives References 141

4 1 TOPOLOGY, METRIC, NORM 4 1 Topology, Metric, Norm 1.1 Motivation, Definitions, Convergence One major goal of this class is to study convergence in more general contexts than the sets R or C. We have already encountered the convergence of K-valued functions in [Phi16, Sec. 8], where we considered the distinct concepts of pointwise and uniform convergence. The notion of topology provides an abstract concept suitable for all the abovementioned convergences as well as for convergence in countless other situations of interest to us (convergence in K n being just one important example). In Analysis I, we said that a sequence (z k ) k N in K converged to z K if, and only if, every neighborhood of z contains almost all z k (cf. [Phi16, Rem. 7.8]). As it turns out, this is the concept that one can still use in the most abstract situation one merely needs a suitable abstract notion of neighborhood. Recall from [Phi16, Def. 7.7(a)] that a neighborhood of a point z K is a set U K, containing an open ǫ-ball with center z. In the situation of an abstract topological space X, to be defined in Def. 1.1 below, one specifies all the open subsets of X, calling U X a neighborhood of x X if, and only if, there is an open set O X such that x O U. Definition 1.1. Let X be a set and T P(X) a set of subsets of X. Then T is called a topology on X if, and only if, the following three conditions are satisfied: (i) T and X T. (ii) T is closed under finite intersections, i.e. the intersection of finitely many sets in T is again in T : If n N and O i T for each i = 1,...,n, then n O i T. i=1 (iii) T is closed under arbitrary unions, i.e. the union of arbitrarily many(i.e. of finitely ofinfinitelymany)setsint isagainint: IfI isanarbitraryindexsetando i T for each i I, then O i T. i I If T constitutes a topology on X, then the pair (X,T ) is called a topological space. Moreover, a set O X is called T -open or open with respect to T if, and only if, O T. One simply calls O open in case the topology is understood. Given x X, a set U X is called a neighborhood of x if, and only if, there is an open set O T such that x O U (note that U does not have to be in T ). The set of all neighborhoods of x is denoted by U(x); it is also called the neighborhood system or the neighborhood filter of x. If T 1 and T 2 are both topologies on X such that T 1 T 2, then we call T 1 smaller or coarser than T 2, and we call T 2 bigger or finer than T 1.

5 1 TOPOLOGY, METRIC, NORM 5 Lemma 1.2. Let (X,T ) be a topological space. Then O X is open if, and only if, for each x O, there exists an open set O x T such that x O x O. Proof. If O T and x O, then one can just choose O x := O. Conversely, if, for each x O, there exists an open set O x T such that x O x O, then O = x OO x, proving O T by Def. 1.1(iii). Example 1.3. (a) For each set X, T := P(X) constitutes a topology on X, called the discrete topology on X. (b) For each set X, T := {,X} constitutes a topology on X, called the indiscrete or trivial topology on X. (c) For each set X, T := {O X : O = or O c is finite} (where O c := X \O denotes the complement) constitutes a topology on X, called the cofinite topology on X (clearly,,x T ; if O c 1 and O c 2 are both finite, then (O 1 O 2 ) c = O c 1 O c 2 is finite; if O c i, i I, are all finite, then ( i I O i) c = i I Oc i is finite). (d) We call O K open if, and only if, z O B ǫ (z) O, ǫ R + where, as in [Phi16, Def. 7.7(a)], B ǫ (z) := {w K : w z < ǫ}. Then T := {O K : O open} constitutes a topology on K (cf. Th below). Proposition 1.4. Arbitrary intersections of topologies yield again topologies: Let X be a set, let I be a nonempty index set, and let (T i ) i I be a family of topologies on X. Then T := i IT i is a again a topology on X. Proof. Since and X are in each T i, they are also in T. If O 1,O 2 T and i I, then O 1,O 2 T i. Thus, O 1 O 2 T i as well, proving O 1 O 2 T. Similarly, if J is an index set and O j T for each j J, then O j T i for each i I and each j J. Thus, O := j J O j T i for each i I, showing O T. As in K, topologies often arise from so-called metrics or norms, which we define next: Definition 1.5. Let X be a set. A function d : X X R + 0 is called a metric on X if, and only if, the following three conditions are satisfied:

6 1 TOPOLOGY, METRIC, NORM 6 (i) d is positive definite, i.e., for each (x,y) X X, d(x,y) = 0 if, and only if, x = y. (ii) d is symmetric, i.e., for each (x,y) X X, d(y,x) = d(x,y). (iii) d satisfies the triangle inequality, i.e., for each (x,y,z) X 3, d(x,z) d(x,y)+ d(y,z). If d constitutes a metric on X, then the pair (X,d) is called a metric space. One then often refers to the elements of X as points and to the number d(x,y) as the d-distance between the points x and y. If the metric d on X is understood, one also refers to X itself as a metric space. Remark 1.6. The requirement that a metric be nonnegative is included in Def. 1.5 merely for emphasis. Nonnegativity actually follows from the remaining properties of a metric: For each x,y X, one computes 0 showing d(x,y) 0. Def. 1.5(i) Def. 1.5(iii) Def. 1.5(ii) = d(x,x) d(x,y)+d(y,x) = 2d(x,y), (1.1) Definition 1.7. Let X be a vector space over the field K. Then a function : X R + 0 is called a norm on X if, and only if, the following three conditions are satisfied: (i) is positive definite, i.e. ( ) x = 0 x = 0 for each x X. (ii) is homogeneous of degree 1, i.e. (iii) satisfies the triangle inequality, i.e. λx = λ x for each λ K, x X. x+y x + y for each x,y X. If constitutes a norm on X, then the pair (X, ) is called a normed vector space or just normed space. If the norm on X is understood, then one also refers to X itself as a normed space. Lemma 1.8. If (X, ) is a normed space, then the function d : X X R + 0, d(x,y) := x y, (1.2) constitutes a metric on X: One also calls d the metric induced by the norm. Thus, the induced metric d makes X into a metric space.

7 1 TOPOLOGY, METRIC, NORM 7 Proof. Consider x,y X. If x = y, then d(x,y) = x y = 0 = 0. Conversely, if 0 = d(x,y) = x y, then x y = 0, i.e. x = y. Symmetry is verified by the computation d(y,x) = y x = 1 x y = d(x,y). Finally, for the triangle inequality, one lets x,y,z X and estimates d(x,y) = x y = x z +z y x z + z y = d(x,z)+d(z,y), which establishes the case. Remark 1.9. Throughout this class, a multitude of notions will be introduced for metric spaces (X, d), including open sets, balls, closed sets, etc. Subsequently, we will then also use these notions in normed spaces (X, ), always implicitly assuming that they are meant with respect to the metric space (X,d), where d is the metric induced by the norm, i.e. where d is given by (1.2). Definition Let (X,d) be a metric space. Given x X and r R +, define B r (x) := {y X : d(x,y) < r}, B r (x) := {y X : d(x,y) r}, S r (x) := {y X : d(x,y) = r}. (1.3a) (1.3b) (1.3c) The set B r (x) is called the open ball with center x and radius r, also known as the r-ball with center x. The set B r (x) is called the closed ball with center x and radius r. The set S r (x) is called the sphere with center x and radius r. A set U X is called a neighborhood of x if, and only if, there is ǫ R + such that B ǫ (x) U. We call O X open if, and only if, x O B ǫ (x) O. ǫ R + Definition For n N, p [1, [, the function ( n ) 1/p p : K n R + 0, z p := z j p, (1.4) is called the p-norm on K n (here, and in the following, we write vectors z K n in the form z = (z 1,...,z n ), where z 1,...,z n K are the coordinates of z). For p = 2 and K = R, one also speaks of the Euclidean norm. We want to show that the p-norms are, indeed, norms in the sense of Def Before we can do that in Cor below, we need to establish some important inequalities: Theorem 1.12 (Hölder inequality). If n N and p,q > 1 such that = 1, then p q n a j bj a p b q for each a,b K n. (1.5) j=1 }{{} =:a b j=1

8 1 TOPOLOGY, METRIC, NORM 8 Proof. If a = 0 or b = 0, then there is nothing to prove. So let a 0 and b 0. For each j {1,...,n}, apply inequality between the weighted arithmetic mean and the weighted geometric mean [Phi16, (9.45)], i.e. x λ 1 1 x λn n λ 1 x 1 + +λ n x n, with λ 1 = 1/p, λ 2 = 1/q, x 1 = a j p / a p p and x 2 = b j q / b q q, to get a j b j 1 a j p a p b q p a p + 1 b j q p q b q. (1.6a) q Summing (1.6a) over j {1,...,n} yields 1 on the right-hand side, and, thus, n n summed (1.6a) a b = a j bj a j b j a p b q, (1.6b) j=1 j=1 proving (1.5). Theorem 1.13 (Minkowski inequality). For each p 1, z,w K n, n N, one has z +w p z p + w p. (1.7) Proof. For p = 1, (1.7) follows directly from the triangle inequality for the absolute value in K. It remains to consider the case p > 1. In that case, define q := p/(p 1), i.e. 1/p + 1/q = 1. Also define a R n by letting a j := z j + w j p 1 R + 0 for each j {1,...,n}, and notice z j +w j p = z j +w j a j z j a j + w j a j. (1.8a) Summing (1.8a) over j {1,...,n} and applying the Hölder inequality (1.5), one obtains z +w p p ( z 1,..., z n ) a+( w 1,..., w n ) a z p a q + w p a q. (1.8b) As q(p 1) = p, it is a q j = z j +w j p, and, thus a q = ( n j=1 )1 p p q z j +w j p = z +w p 1 p, (1.8c) where p/q = p 1 was used in the last step. Finally, combining (1.8b) with (1.8c) yields (1.7). Corollary For each n N, p [1, [, the p-norm on K n constitutes, indeed, a norm on K n. Proof. If z = 0, then z p = 0 follows directly from (1.4). If z 0, then there is j {1,...,n} such that z j > 0. Then (1.4) provides z p z j > 0. If λ K and z K n, then λz p = ( n j=1 λz j p ) 1/p = ( λ p n j=1 z j p ) 1/p = λ z p. The proof is concluded by noticing that the triangle inequality is the same as the Minkowski inequality (1.7).

9 1 TOPOLOGY, METRIC, NORM 9 Example Let S be an otherwise arbitrary set. According to Linear Algebra, the set F(S,K) of all K-valued functions on S is a vector space over K if vector addition and scalar multiplication are defined pointwise via (f +g) : S K, (f +g)(x) := f(x)+g(x), (λ f) : S K, (λ f)(x) := λ f(x) for each λ K. Now consider the subset B(S,K) of F(S,K), consisting of all bounded K-valued functions on S, where we call a K-valued function f bounded if, and only if, the set { f(s) : s S} R + 0 is a bounded subset of R. Define f := f sup := sup{ f(s) : s S} R + 0 for each f B(S,K). (1.9) We will show that B(S,K) constitutes a vector space over K and sup provides a norm on B(S,K) (i.e. ( B(S,K), sup ) is a normed vector space). To verify that B(S,K) constitutes a vector space over K, it suffices to show it is a subspace of the vector space F(S,K), which, isequivalenttoshowingf,g B(S,K)andλ Kimplyf+g B(S,K) and λf B(S,K). If f,g B(S,K), then s S f(s)+g(s) f(s) + g(s) f sup + g sup R + 0, (1.10a) showing f +g B(S,K) and that sup satisfies the triangle inequality f,g B(S,K) f +g sup f sup + g sup. (1.10b) If f B(S,K), λ K, then, s S λf(s) = λ f(s) λ f sup R + 0 (1.11a) implies λf B(S,K), completing the proof that B(S,K) is a subspace of F(S,K). Moreover, λf sup = sup{ λf(s) : s S} = sup{ λ f(s) : s S} [Phi16, (4.6c)] = λ sup{ f(s) : s S} = λ f sup, (1.11b) proving sup is homogeneous of degree 1. To see that sup constitutes a norm on B(S,K), it merely remains to show positive definiteness. To this end, we notice that the zero element f = 0 of the vector space B(S,K) is the function f 0, which vanishes identically. Thus, f = 0 if, and only if, f sup := sup{ f(s) : s S} = 0, showing sup is positive definite, and completing the proof that sup is a norm, making B(S,K) into a normed vector space.

10 1 TOPOLOGY, METRIC, NORM 10 In generalization of Ex. 1.3(d), every metric (in particular, every norm) induces a topology: Theorem Let (X,d) be a metric space. Then T := {O X : O open} constitutes a topology on X: One also calls T the topology induced by the metric d, making each metric space into a topological space. Proof. Clearly, T and X T. Now consider finitely many open sets O 1,...,O N T, N N, and let O := N j=1 O j. We have to prove that O is open. Hence, let x O. Then x O j for each j {1,...,N}. Since each O j is open, for each j {1,...,N}, there is ǫ j > 0 such that B ǫj (x) O j. If we let ǫ := min{ǫ j : j {1,...,N}}, then ǫ > 0 and B ǫ (x) B ǫj (x) O j for each j {1,...,N}, i.e. B ǫ (x) O, showing O is open. Now let I be an arbitrary index set. For each j I, let O j T. We have to verify that O := j I O j is open. Let x O. Then there is j I such that x O j. Since O j is open, there is ǫ > 0 such that B ǫ (x) O j O, showing O to be open. Definition and Remark A topological space (X,T ) is called metrizable if, and only if, there exists a metric d on X such that T is induced by d. Not every topology is metrizable: For example, the indiscrete topology on a set with at least two distinct elements can never be metrizable (see Rem. 1.42(a) below). Other examples of nonmetrizable topologies are the cofinite topology on an uncountable set (see Rem. 1.39(b) below) and the topology of pointwise convergence (see Ex. 1.53(c) below). Example Let X be a set. We will show that the discrete topology on X (cf. Ex. 1.3(a)) is always metrizable: The corresponding discrete metric is defined by { 0 for x = y, d : X X {0,1}, d(x,y) := (1.12) 1 for x y. We verify that d constitutes a metric on X: Since d(x,y) = 0 holds if, and only if, x = y, d is positive definite. Symmetry, i.e. d(x,y) = d(y,x) is immediate from (1.12). We still need to show x,y,z X d(x,y) d(x,z)+d(z,y). For x = y, it is d(x,y) = 0, and we are done. For x y, it is d(x,y) = 1, and we have to show the right-hand cannot be 0. If z = x, then z y and d(y,z) = 1. If z x, then d(x,z) = 1. Now that we have seen that d is a metric on X, we still need to prove it induces the discrete topology, i.e. that every subset of X is open with respect to d. Thus, let O X and x O. Then B 1 (x) = {x} O, which already shows O to be open. We now want to proceed to introduce convergence on topological spaces(in particular on metric spaces, including K n ). In K we were able to characterize continuity of functions aswellastheclosednessofasetintermsofconvergenceofsequences. Whilethiswillstill be possible in metric spaces, in general topological spaces, the convergence of sequences

11 1 TOPOLOGY, METRIC, NORM 11 no longer suffices for such characterizations. For this reason, we will introduce the more general (and more powerful) notion of net convergence 1 (also referred to as Moore-Smith convergence in the literature). Definition (a) A directed set (I, ) consists of a nonempty set I and a relation on I that is reflexive, transitive, 2 and has the additional property that every set consisting of precisely two elements has an upper bound, i.e. i,j I M I ( ) i M j M. (1.13) (b) Let X be a set. A net in X is a family (x i ) i I in X (i.e. a function from I into X, cf. [Phi16, Def. 2.15(a)]) indexed by a directed set I. If A X, then we say that the net (x i ) i I in X is eventually in A if, and only if, i I j i x j A. (1.14) Example (a) If I is any nonempty set and is a total order on I, then (I, ) is a directed set. In particular, N with its usual order is a directed set, and, thus, every sequence is a net. (b) If I is any nonempty set, then (I,I I) is a directed set, but I I is not a partial order if I has at least two distinct elements. On the other hand, {(0,0),(1,1)} is a partial order on {0,1} that does not make {0,1} into a directed set. (c) Let (X,T ) be a topological space, x X. Then the neighborhood system U(x) is made into a directed set by defining U,V U(x) U V : V U : (1.15) Clearly, is reflexive and transitive (it is even a partial order), and (1.13) is satisfied, since U,V U(x) implies U V U(x) (one still obtains a directed set when using U V in (1.15), but this example turns out to be less useful). (d) If [a,b] R is an interval, a,b R, then the set Π of partitions of [a,b] (cf. [Phi16, Def. 10.3]) is turned into a directed set by setting if, and only if, is a refinement of (cf. [Phi16, Def. 10.8(a)]). The reason one does not require the relation in Def. 1.19(a) to be a partial order is that it is not necessary it would clutter the notion of a directed set without any additional gain for the theory. 1 Alternatively, one can also use the similar, but different, concept of filter convergence. In this class, we only consider net convergence, which seems slightly easier to explain as well as more common in introductory Analysis texts. 2 A relation that is reflexive and transitive is sometimes called a preorder.

12 1 TOPOLOGY, METRIC, NORM 12 Definition Let (X,T ) be a topological space. The net (x i ) i I in X is said to be convergent with limit x X if, and only if, for every neighborhood U of x, the net is eventually in U, i.e. if, and only if, U U(x) i I j i x j U. (1.16) If (x i ) i I converges to x, then we write lim i I x i = x. If I = N (i.e. if the net is a sequence), then we also still write lim i x i = x instead of lim i N x i = x. A net is called divergent if, and only if, it is not convergent. Example (a) Let (X,T ) be a topological space, where T is induced by the metric d on X. Let (x i ) i I be a net in X, and x X. Since every ball B ǫ (x), ǫ > 0, is a neighborhood of x and, conversely, every U U(x) contains some ball B ǫ (x) U, ǫ > 0, we have the equivalence limx i = x i I ǫ R + i I d(x i,x) < ǫ. j i (1.17a) In particular, if I = N and the net is a sequence, then (x i ) i N converges to x if, and only if, the real sequence of distances ( d(x i,x) ) converges to 0 (in the sense of i N Analysis I): lim x i = x lim d(x i,x) = 0. (1.17b) i i (b) Let (X,T ) be an arbitrary topological space and x X. Consider U(x) with the partial order of Ex. 1.20(c). Moreover, let (x U ) U U(x) be a net such that x U U for each U U(x). Then, clearly, lim U U(x) x U = x. (c) We consider K n, n N, with the 2-norm. Let (z k ) k I be a net in K n (for example, a sequence). Here, z k = (z k 1,...,z k n), i.e. the z k j are the coordinates of the vector z k. As in the present example, we will subsequently use upper indices for indices in sequences and nets in situations, where we also need to denote coordinates; different coordinates will be referred to via lower indices. For each coordinate j {1,...,n}, we have the coordinate net (z k j) k I in K. We now claim that (z k ) k I converges with respect to (the topology induced by) the 2-norm to a = (a 1,...,a n ) K n if, and only if, each coordinate net (z k j) k N converges to a j in K, j {1,...,n}: limz k = a k I j {1,...,n} limzj k = a j. (1.18) k I In particular, a sequence in K n converges (with respect to the 2-norm) if, and only if, each coordinate sequence converges in K. Remark: In Th below, we will see that every norm on K n induces the same topology, i.e. the validity of (1.18) does actually not depend on the norm. In preparation for the proof of (1.18), we observe that, for each z K n, one has the following estimates: j {1,...,n} z j z }{{} 2 z }{{} 1 z z n 2 z z n. (1.19)

13 1 TOPOLOGY, METRIC, NORM 13 Let denote the relation that makes I into a directed set. If (z k ) k I converges to a, then, according to (1.17a), given ǫ R +, there is N I such that, for each k N, z k a 2 < ǫ. By (1.19), this implies j {1,...,n} z k j a j z k a 2 < ǫ, provinglim k I z k j = a j. Conversely, if(z k j) k I convergestoa j foreachj {1,...,n}, then, given ǫ R +, (1.17a) yields N I such that, for each k N, Since, by (1.19), this implies z k j a j < ǫ n. z k a 2 n zj k a j < n ǫ n = ǫ, j=1 (z k ) k I converges to a. (d) We can interpret the function limit lim z ζ f(z) = η of [Phi16, Def. 8.17] as a net limit in C: Let M C, f : M C, and let ζ C be a cluster point of M. We make the set I := M \{ζ} into a directed set by defining z,w M z w : z ζ w ζ (i.e. bigger elements are closer to ζ). It is then an exercise to show that, for η C, f(z) = η limf(z) = η. z I lim z ζ (e) We can interpret the Riemann integral b f of a bounded function f : [a,b] R, a a < b, as a net limit in R: As in Ex. 1.20(d), we turn the set Π of tagged partitions of [a,b] into a directed set by setting if, and only if, is a refinement of. Then b f exists if, and only if, lim a Πρ(,f) exists (where ρ(,f) denotes the Riemann sum of [Phi16, (10.7c)]), and, in that case, (see [Wal02, Sec. 5.6]). b a f = lim Π ρ(,f) In [Phi16, Sec. 7.1], we studied subsequences and reorderings of sequences (cf. [Phi16, Sec. 7.21]). We found that subsequences and reorderings of convergent sequences in C are also convergent with the same limit ([Phi16, Prop. 7.23]). Moreover, in [Phi16, Prop.

14 1 TOPOLOGY, METRIC, NORM ], we showed that z K is a cluster point of the sequence (z k ) k N in K if, and only if, the sequence has a subsequence converging to z. We will now study related notions on general nets, where we will find that the mentioned results extend only partially to the most general situation. Reorderings turn out to be much less useful on general nets than on sequences and, while subnets generalize subsequences, the concept of a subnet is somewhat more complicated, such that the concept of a subnet turns out to be noticeably more versatile than that of a subsequence. Definition Let (I, ) and (J, ) be directed sets, let φ : J I, let X be an arbitrary nonempty set, and let σ : I X be a net in X. (a) φ is called final if, and only if, i I j 0 J φ(j) i. (1.20) j j 0 (b) The net (σ φ) : J X is called a subnet of σ if, and only if, φ is final. (c) ForI = J (withthesamerelation ), thenet(σ φ) : I X iscalledareordering of σ if, and only if, φ is bijective. Example Using theidentity for φshows that every net is a subnet of itself. Moreover, every subsequence is a subnet, since, clearly, if φ : N N is strictly increasing, then φ is final. However, sequences can have subnets that are no subsequences. One reason is that final maps do not need to be monotone: For example, (2,1,4,3,6,5,...) is a subnet of (1,2,3,...). The other reason is that a final map does not need to be injective. For example, if σ : N X is a sequence in X, then (σ φ) : R X, is a subnet of σ. φ : R N, φ(s) := min{k N : k s}, Proposition Let (X,T ) be a topological space, x X. (a) Let σ : I X be a net in X. If lim i I σ(i) = x, then every subnet of σ is also convergent with limit x. (b) Let σ : N X be a sequence in X. If lim i σ(i) = x, then every reordering of σ is also convergent with limit x. Proof. (a): Let σ φ be a subnet of σ, where φ : J I is a final map. Moreover, let U U(x) be a neighborhood of x. Since lim i I σ(i) = x, there exists i 0 I such that σ(i) U for each i i 0. As φ is final, there exists j 0 J such that Thus, for each j j 0, we have φ(j) i 0. j j 0 (σ φ)(j) = σ(φ(j)) U,

15 1 TOPOLOGY, METRIC, NORM 15 due to φ(j) i 0, proving lim j J (σ φ)(j) = x as desired. (b): The proof is analogous to the corresponding part of the proof of [Phi16, Prop. 7.23]: Let σ φ be a reordering of σ, where φ : I I is bijective. Let U and i 0 be as before (now with I := N). Define M := max{φ 1 (i) : i i 0 }. As φ is bijective, it is φ(i) > i 0 for each i > M. Then, for each i > M, one has (σ φ)(i) = σ(φ(i)) U due to φ(i) > i 0, proving lim i (σ φ)(i) = x as desired. Caveat: Reorderings of general convergent nets do not necessarily converge, see Ex below. Definition (a) Let X be a set. If A X, then we say that the net (x i ) i I in X is frequently or cofinally in A if, and only if, i I j i x j A. (1.21) (b) Let (X,T ) be a topological space, x X. The net (x i ) i I in X is said to have x as a cluster point (or accumulation point) if, and only if, for every neighborhood U of x, the net is frequently in U, i.e. if, and only if, U U(x) i I j i x j U. (1.22) Proposition Let (X,T ) be a topological space, x X. The net σ : I X in X has x as a cluster point if, and only if, it has a subnet that converges to x. Proof. Let σ φ be a subnet of σ, where φ : J I is a final map. Assume that lim j J (σ φ)(j) = x and let U U(x). If i I, then, as φ is final, there exists j 0 J such that j j 0 φ(j) i. On the other hand, there is j 1 J such that σ(φ(j)) U. j j 1 Since J is a directed set, there exists j 2 J such that j 2 j 0 and j 2 j 1, implying φ(j 2 ) i and σ(φ(j 2 )) U. As we have, thus, shown σ to be frequently in U, x is a cluster point of σ. Conversely, let x be a cluster point of σ. We have to construct a subnet that converges to x. To this end, we let J := { (i,u) I U(x) : σ(i) U }

16 1 TOPOLOGY, METRIC, NORM 16 and define (i,u),(j,v) J (i,u) (j,v) : ( i j U V ). Then is clearly reflexive and transitive on J. Let (i,u),(j,v) J and note U V U(x). As I is a directed set, there is M I such that M i and M j. Moreover, the assumption that σ is frequently in U V implies there is k I such that and, thus, k M i k M j σ(k) U V, (k,u V) J (k,u V) (i,u) (k,u V) (j,v), proving (J, ) to be a directed set. The projection map π : J I, π(i,u) := i, is final: Indeed, if i I, then (i,x) J and (j,u) (i,x) J implies π(j,u) = j i. In consequence, σ π is a subnet of σ. It merely remains to show that σ π converges to x. Thus, let U U(x). Since σ is frequently in U, there must be i I with σ(i) U. Then (i,u) J and if (j,v) (i,u), then (σ π)(j,v) = σ(j) V U, proving lim (j,u) J (σ π)(j,u) = x. In [Phi16, Prop. 7.10(b)], we obtained the result that every convergent sequence in C is bounded. We will see in Prop. 1.29(d) below that this result remains true for sequences in metric spaces (but not for nets, see Ex. 1.30). In general topological spaces, however, one no longer has the concept of boundedness. Definition Let (X,d) be a metric space. (a) The set A X is called bounded if, and only if, A = or A and the set {d(x,y) : x,y A} is bounded in R; A X is called unbounded if, and only if, A is not bounded. For each A X, the number 0 for A =, diama := sup { d(x,y) : x,y A } for A bounded, (1.23) for A unbounded, is called the diameter of A. Thus, diama [0, ] := R + 0 { } and A is bounded if, and only if, diama <. (b) The net (x i ) i I in X is called bounded if, and only if, the set {x i : i I} is bounded in the sense of (a). Proposition Let (X,d) be a metric space.

17 1 TOPOLOGY, METRIC, NORM 17 (a) A X is bounded if, and only if, there is r > 0 and x X such that A B r (x) (in particular, Def. 1.28(a) is consistent with [Phi16, Def. 7.42(a)]). (b) Every finite subset of X is bounded. (c) The union of two bounded subsets of X is bounded. (d) If the sequence (x k ) k N in X is convergent, then it is bounded. Proof. (a): If A is bounded, then diama <. Let r be any real number bigger than diama, e.g. 1+diamA. Choose any point x A. Then, by the definition of diama, for each y A, it is d(x,y) diama < r, showing that A B r (x). Conversely, if r > 0 and x X such that A B r (x), then, by the definition of B r (x), one has d(x,y) < r for each y A. Now, if y,z A, then d(z,y) d(z,x) + d(x,y) < 2r, showing diama 2r <, i.e. A is bounded. (b): Let A be a finite subset of X and a A. Set r := 1+max{d(a,x) : x A}. Then 1 r <, since A is finite. Moreover, A B r (a), showing that A is bounded. (c): Let A and B be bounded subsets of X. Then there are x,y X and r > 0 such that A B r (x) and B B r (y). Define α := d(x,y) and ǫ := r + α. Then A B r (x) B ǫ (x). If b B, then d(b,x) d(b,y) + d(y,x) < r + α = ǫ, showing B B ǫ (x), and, thus, A B B ǫ (x), establishing that A B is bounded. (d) (cf. the proof for sequences in K in [Phi16, Prop. 7.10(b)]): If lim k x k = a X, then there is N N such that x k B 1 (a) for each k > N, i.e. {x k : k > N} is bounded. Moreover, the finite set {x k : k N} is bounded. Therefore, {x k : k N} is the union of two bounded sets, and, hence, bounded. Example The net (e k ) k Z in R, which clearly converges to 0, shows that convergent nets do not need to be bounded. Moreover, using the bijective map φ : Z Z, φ(k) := k, we see that the divergent net (e k ) k Z is a reordering of (e k ) k Z. We conclude the present section by extending the 1-dimensional Bolzano-Weierstrass theorem of [Phi16, Th. 7.27] to sequences in K n : Theorem 1.31 (Bolzano-Weierstrass). Let (z k ) k N be a sequence in K n that is bounded with respect to the 2-norm (in fact, we will see as a consequence of Th below that the property of boundedness in K n does not depend on the chosen norm). Then (z k ) k N has a subsequence that converges in K n. Proof. If (z k ) k N is bounded with respect to the 2-norm, then, due to (1.19), each coordinate sequence (zj) k k N, j {1,...,n}, is bounded in K. We prove by induction over {1,...,n} that, for each j {1,...,n}, there is a subsequence (y k,j ) k N of (x k ) k N such that the coordinate sequences (yα k,j ) k N converge for each α {1,...,j}. Base Case (j = 1): Since (z1) k k N is a bounded sequence in K, the Bolzano-Weierstrass theorem

18 1 TOPOLOGY, METRIC, NORM 18 for sequences in K (cf. [Phi16, Prop. 7.26, Th. 7.27]) yields the existence of a convergent subsequence of (z1) k k N. This provides us with the needed subsequence (y k,1 ) k N of (z k ) k N. Now suppose that 1 < j n. By induction, we already have a subsequence (y k,j 1 ) k N of (z k ) k N such that the coordinate sequences (yα k,j 1 ) k N converge for each α {1,...,j 1}. As (yα k,j 1 ) k N is a subsequence of the bounded K-valued sequence (zα) k k N, by the Bolzano-Weierstrass theorem for sequences in K, it has a convergent subsequence. This provides us with the needed subsequence (y k,j ) k N of (y k,j 1 ) k N, which is then also a subsequence of (z k ) k N. Moreover, for each α {1,...,j 1}, (yα k,j ) k N is a subsequence of the convergent sequence (yα k,j 1 ) k N, and, thus, also convergent. In consequence, (yα k,j ) k N converge for each α {1,...,j} as required. Finally, one observes that (y k,n ) k N is a subsequence of (z k ) k N such that all coordinate sequences (yα k,n ) k N, α {1,...,n}, converge. Let a α := lim k yα k,n for each α {1,...,n}. Then, by (1.18), lim k y k,n = a, thereby establishing the case. 1.2 Open Sets, Closed Sets, and Related Notions Definition Let (X,T ) be a topological space, A X, and x X. (a) A is called open if, and only if, A T (of course, we already defined this in Def. 1.1 here, it is only repeated for the sake of completeness). (b) A is called closed if, and only if, A c = X \A is open (where A c = X \A denotes the complement of A, cf. [Phi16, Def. 1.24(c)]). (c) The point x is called an interior point of A if, and only if, A U(x), i.e. if, and only if, there exists O T such that x O A. Note: An interior point of A is always in A. The set of all interior points of A is called the interior of A. It is denoted by A or by inta. (d) The point x is called a boundary point of A if, and only if, each O U(x) contains at least one point from A and at least one point from A c (A O and A c O ). Note: A boundary point of A is not necessarily in A. The set of all boundary points of A is called the boundary of A. It is denoted by A. The set A A is called the closure of A. It is denoted by A or by cla. The set A is called dense in X if, and only if, A = X; (X,T ) is called separable if, and only if, X has a countable dense subset. (e) The point x is called a cluster point or accumulation point of A if, and only if, every neighborhood of x has a nonempty intersection with A\{x}, i.e. if, and only if, U U(x) U (A\{x}) (cf. [Phi16, Def. 7.33(a)] and Rem. 1.42(a) below). Note: A cluster point of A is not necessarily in A.

19 1 TOPOLOGY, METRIC, NORM 19 (f) The point x is called an isolated point of A if, and only if, it is in A and not a cluster point of A, i.e. if, and only if, there exists U U(x) such that U A = {x} (cf. [Phi16, Def. 7.33(b)]). Note: An isolated point of A is always in A. Example Let (X,T ) be a topological space, where T is induced by a metric d on X. Then, given x X and r R +, the open ball B r (x) is an open set and the closed ball B r (x) is a closed set: That B r (x) T is immediate from the definition of T. To see that B r (x) is closed, we need to show X\B r (x) is open. To this end, let y X\B r (x), ǫ := d(x,y) r. If z B ǫ (y), then d(y,z) < d(x,y) r and d(x,y) d(x,z)+d(z,y). Thus, d(x,z) d(x,y) d(z,y) > r, showing B ǫ (y) X\B r (x), i.e. X\B r (x) is open. Example Consider (X,T ), where X = K and T is induced by the metric d : X R + 0, d(z,w) := z w. (a) Let A :=]0,1] and K = R. Then A =]0,1[, A = {0,1}, A = [0,1]. (b) Let A :=]0,1] and K = C. Then A =, A = A = [0,1]. (c) Let A := Q. In this case, there is no difference between K = R and K = C: A =, A = A = R. (d) Let A := {1/n : n N}. Once again, there is no difference between K = R and K = C: Every element of A is an isolated point. In particular A =. The unique cluster point of A is 0, and A = A = A {0}. Proposition Let (X,T ) be a topological space, A X. (a) A is open if, and only if, A c is closed. (b) The empty set and the entire space X are both open and closed. Such sets are sometimes called clopen. (c) Intersections of arbitrarily many closed sets are closed (cf. [Phi16, Prop. 7.44(b)]). The union of finitely many closed sets is closed (cf. [Phi16, Prop. 7.44(a)]). (d) X is the disjoint union of A, A, and (X \A). (e) A is dense in X if, and only if, for every nonempty O T, one has O A. Proof. (a): AccordingtoDef.1.32(b), A c isclosedif, andonlyif, (A c ) c isopen. However, (A c ) c = X \A c = X \(X \A) = A. (b) is immediate, since,x T. (c): Let I be a (finite or infinite) index set. For each j I, let C j X be closed. We have to verify that C := j I C j is closed. According to the set-theoretic law [Phi16, Prop. 1.39(e)] C c = ( j I ) c [Phi16, Prop. 1.39(e)] C j = Cj. c j I

20 1 TOPOLOGY, METRIC, NORM 20 Now, as we know that C j is closed, we know that Cj c is open. According to Def. 1.1(iii), that means that C c is open, showing that C is closed. Similarly, if we consider finitely many closed sets C 1,...,C N, N N, and let C := N j=1 C j, then the set-theoretic law [Phi16, Prop. 1.39(f)] yields C c = ( N j=1 ) c [Phi16, Prop. 1.39(f)] C j = N Cj. c Since C j is closed, C c j is open, and, by Def. 1.1(ii), C c is open, hence C closed. (d): Onehastoshowfourparts: X = A A (X\A), A A =, A (X\A) =, and A (X \ A) =. Suppose x X \ (A A). Since x / A, there exists U U(x) such that U A or U X \ A. As x / A, it must be U X \ A, i.e. x (X \A). A A = : If x A, then there is U U(x) such that U A, thus, x / A. A (X \ A) = : Since A = (X \ A), this follows from A A =. A (X \A) = holds as A A, (X \A) X \A, and A (X \A) =. (e): If A is dense in X, then X = A A. Let x O. If x A, then x O A. If x A, then O A also holds. Conversely, suppose O A for each nonempty O T. Then, for x X \A and x O T, both O A and O A c, showing x A. Example Due to Prop. 1.35(e), Q is dense in R (endowed with the topology given by ). More generally, Q n is dense in R n, n N, and A := { (x 1 +iy 1,...,x n +iy n ) C n : x j,y j Q, j {1,...,n} } is dense in C n (endowed with the topology given by 2 ). As Q n and A are countable, R n and C n are separable in the norm topology. On the otherhand, if X is uncountable and (X,T ) is discrete, then the space can never be separable (since A = A for each A X). As every discrete space is metrizable (by the discrete metric), this shows that not every metric space is separable. Theorem Let (X,T ) be a topological space, A X. (a) The interior A is the union of all open subsets of A. In particular, A is open. In other words, A is the largest open set contained in A. (b) The closure A is the intersection of all closed supersets of A. In particular, A is closed. In other words, A is the smallest closed set containing A. (c) The boundary A is closed. Proof. (a): Let O be the union of all open subsets of A. Then O is open by Def. 1.1(iii). If x A, then x is an interior point of A, i.e. there is U U(x) and O 1 T such that x O 1 U A (cf. Def. 1.1). Since O 1 is open, x O. Conversely, if x O, then, as O is open, O U(x), showing that x is an interior point of A, i.e. x A. j=1

21 1 TOPOLOGY, METRIC, NORM 21 (b): According to (a), (A c ) is the union of all open subsets of A c, i.e. (A c ) = O, then ( (A c ) ) c [Phi16, Prop. 1.39(f)] = O {S A c :S open} O {S A c :S open} O c = C {S A: S closed} is the intersection of all closed supersets of A (note that C is a closed superset of A if, and only if, C c is an open subset of A c ). As, by Prop. 1.35(d), ( (A c ) ) c = (A c ) A = A A = A A = A, C A is the intersection of all closed supersets of A as claimed. (c): According to Prop. 1.35(d), it is A = X \ ( A (X \A) ). Since A and (X \A) are open, A is closed. We will now proceed to study some relations between cluster points of a set A, the closure of a set A, and convergent nets in A. Sequences suffice instead of nets for topological spaces that are first countable, a notion provided by the following definition: Definition Let (X,T ) be a topological space, x X. A set B U(x) is called a local base at x or a neighborhood base at x if, and only if, every U U(x) contains some B B, i.e. if, and only if, U U(x) B B B U. Moreover, X is called first countable (sometimes also called a C 1 -space) if, and only if, at every x X there exists a countable local base. Remark (a) Indiscrete spaces as well as metric spaces, are first countable: If (X,T ) is indiscrete, then U(x) = {X} for every x X. If (X,T ) is metric (i.e. T is induced by a metric d on X), then, clearly, for each x X, constitutes a countable local base at x. B(x) := { B ǫ (x) : ǫ Q +} (b) Let X be a set endowed with the cofinite topology T of Ex. 1.3(c). The space is first countable if, and only if, X is countable: If X is countable, then the set of finite subsets of X is countable, i.e. T is countable and, for each x X, T U(x) is a countable local base at x. Conversely, assume there exists a countable local base B at x X. It holds that B = U = {x} : (1.24a) B B U U(x) The inclusions in (1.24a) are immediate; at the first equality holds, as every U U(x) contains a B from B as a subset; at the second equality holds, since,

22 1 TOPOLOGY, METRIC, NORM 22 for each y x, one has {y} c U(x), i.e. y / U U(x) U. Taking complements in (1.24a), we have X \{x} = B BB c. (1.24b) SinceB iscountableandeachb c isfinite(asb containsanopenset), (1.24b)proves X to be countable. In particular, for uncountable X, (X,T ) is not metrizable. Another example of a topology that is not first countable is given by the topology of pointwise convergence, see Ex. 1.53(c) below. First countable spaces are typically much easier to handle due to the following result: Proposition Let (X, T ) be a topological space, x X. Assume there exists a countable local base at x. Then each net (x i ) i I in X, converging to x, contains a sequence (x ik ) k N (which is not necessarily a subnet!) such that lim k x ik = x (in particular, for each A X, there is a net in A converging to x if, and only if, there is a sequence in A converging to x). Proof. Assume there exists a countable local base B at x. Without loss of generality, we may assume the elements of B to be open. Let B 1,B 2,... be an enumeration of the elements of B (not necessarily injective). Define, for each k N, B 0k := k B j. Then, since x B 0k B k and each B 0k is open, B 0 := {B 0k : k N} is still a countable local base at x, but with the additional property that B 0,k+1 B 0k for each k N. Now suppose (x i ) i I is a net in X, converging to x. We have to that the net contains a sequence that converges to x as well. Since lim i I x i = x, given k N, there exists i k I such that x ik B 0k. We claim lim k x ik = x. Indeed, if U U(x), then there is N N with B 0N U. Thus, j=1 k N x i k B 0k B 0N U, proving lim k x ik = x as desired. In Ex. 1.53(c) below, we will see that, if there does not exist a local base at x, then it can happen that there is a net in some set A, converging to x, but there is no sequence in A converging to x. Lemma Let (X,T ) be a topological space, A X. Then x X is a cluster point of A if, and only if, there is a net (a i ) i I in A\{x} such that lim i I a i = x (by Prop. 1.40, we may replace net by sequence if X is first countable).

23 1 TOPOLOGY, METRIC, NORM 23 Proof. Let (a i ) i I be a net in A \ {x} such that lim i I a i = x. If U U(x), then there must be i I with a i U. Since a i A and a i x, this already shows U (A\{x}), i.e. x is a cluster point of A. Conversely, if x is a cluster point of A, for each U U(x), there exists x U U (A\{x}). Then (x U ) U U(x) is a net in A\{x} such that lim U U(x) x U = x. Remark Let (X,T ) be a topological space, A X, x X. (a) If T is induced by a metric d on X, then x is a cluster point of A if, and only if, every U U(x) contains infinitely many distinct points from A (exercise) (thus, our new definition of cluster point is consistent with the definition in [Phi16, Def. 7.33(a)]). On the other hand, we now consider a set X with at least two distinct elements and with the indiscrete topology. Let x,y X, x y, A := {y}. Then x is a cluster point of A, even though X (the only neighborhood of x) contains only one point of A distinct from x. In particular, we see that X with the indiscrete topology cannot be metrizable. (b) If x is an isolated point of A and (a i ) i I is a net in A converging to x, then the net must be finally constant with value x, i.e. N I a i = x : (1.25) i N Indeed, since there exists some U U(x) with U A = {x} and (a i ) i I converges to x, (1.25) must hold. Theorem Let (X,T ) be a topological space, A X. Let H(A) denote the set of cluster points of A, and let L(A) denote the set of limits of nets in A, i.e. L(A) consists of all x X such that there is a net (x i ) i I in A satisfying lim i I x i = x (by Prop. 1.40, L(A) consists of all limits of sequences in A if X is first countable). It then holds that A = L(A) = A H(A). Proof. It suffices to show that L(A) A A H(A) L(A). L(A) A : Suppose x / A. Since X \A is open, X \A U(x), showing that no net in A can converge to x, i.e. x / L(A). A A H(A) : Let x A\A. We need to show that x H(A). As A = A A and x / A, we have x A. Thus, if U U(x), then there must exist x U A U. Since x / A, we have x U x, showing x to be a cluster point of A. A H(A) L(A) : If a A, then the constant sequence (a,a,...) converges to a, implying a L(A). If a H(A), then a L(A) according to Lem Corollary Let (X,T ) be a topological space, A X. Then the following statements are equivalent: (i) A is closed. (ii) A = A.

24 1 TOPOLOGY, METRIC, NORM 24 (iii) A contains all cluster points of A. (iv) A contains all limits of nets in A that are convergent in X (by Prop. 1.40, we may replace nets by sequences if X is first countable 3, also cf. [Phi16, Def. 7.42(b)]). In particular, if A does not have any cluster points, then A is closed. Proof. The equivalence of (i) and (ii) is due to Th. 1.37(b) (A is the smallest closed set containing A). The equivalences of (ii), (iii), and (iv) are due to Th. 1.43: Using the notation L(A) and H(A) from Th. 1.43, one has that A = A implies A = A H(A), i.e. H(A) A, i.e. (ii) implies (iii). If H(A) A, then L(A) = A H(A) = A, i.e. (iii) implies (iv). If A = L(A), then A = A, i.e. (iv) implies (ii). Example Let p,q N and consider the metric spaces given by K p, K q, K p+q, each endowed with the topology induced by the respective 2-norm (cf. Ex. 1.22(c)). Let A K p, B K q. (a) If A and B are closed, then A B is closed in K p+q = K p K q : Let (c k ) k N be a convergent sequence in A B with lim k c k = c K p+q. Then, for each k N, c k = (a k,b k ) with a k K p, b k K q. Moreover, c = (a,b) with a K p and b K q. According to (1.18), one has a = lim k a k and b = lim k b k. Since A and B are closed, from Cor. 1.44(iv), we know that a A and b B, i.e. c = (a,b) A B, showing that A B is closed. (b) If A and B are open, then A B is open in K p+q = K p K q : It suffices to show that (A B) c = K p+q \(A B) is closed. To that end, note (A B) c = (A c K q ) (K p B c ) : (1.26) For a point (z,w) K p K q = K p+q, one reasons as follows: (z,w) (A B) c (z,w) / A B ( z / A and w K q) or ( z K p and w / B ) (z,w) A c K q or (z,w) K p B c (z,w) (A c K q ) (K p B c ), thereby proving (1.26). One now observes that A c and B c are closed, as A and B are open. As K p and K q are also closed, by (a), A c K q and K p B c are closed, and, thus, by (1.26), so is (A B) c. In consequence, A B is open as claimed. (c) Intervals in R n, n N: A subset I of R n is called an n-dimensional interval if, and only if, I has the form I = I 1 I n, where I 1,...,I n are intervals in R. The lengths I 1,..., I n R + 0 { } are called the edge lengths of I. An interval I is 3 One calls A sequentially closed if, and only if, A contains all limits of sequences in A. Thus, for first countable X, A is closed if, and only if, A is sequentially closed.

25 1 TOPOLOGY, METRIC, NORM 25 called a (hyper)cube if, and only if, all its edge lengths are equal. For x,y R n, we write x < y (resp. x y) if, and only if, x j < y j (resp. x j y j ) for each j {1,...,n} (clearly, is a partial order on R n, but not a total order for n > 1, for example, neither (0,1) (1,0), nor (1,0) (0,1)). If x,y R n, x < y, then we define the following intervals ]x,y[ := {z R n : x < z < y} =]x 1,y 1 [ ]x n,y n [ open interval, [x,y] := {z R n : x z y} = [x 1,y 1 ] [x n,y n ] closed interval, [x,y[ := {z R n : x z < y} = [x 1,y 1 [ [x n,y n [ ]x,y] := {z R n : x < z y} =]x 1,y 1 ] ]x n,y n ] halfopen interval, halfopen interval. Due to (a),(b) open intervals in R n are open sets and closed intervals in R n are closed sets (with respect to the 2-norm). 1.3 Construction of Topological Spaces Bases, Subbases One can often build a topology T on X from smaller sets B T by taking unions (then B is called a base of T ) or from S T by taking unions of finite intersections (then S is called a subbase of T ). This can be useful, since one might be able to prove something about T by merely proving it for the (smaller) base or subbase (cf. Cor below). Definition Let (X,T ) be a topological space. (a) A set B T is called a base of the topology T if, and only if, every O T is a union of sets from B, i.e. if, and only if, for each O T, there exists an index set I (where I can be empty, finite, or infinite) and sets O i B, i I, such that O = i IO i. X is called second countable (sometimes also called a C 2 -space) if, and only if, there exists a countable base of T (recall the definition of first countable from Def. 1.38). (b) A set S T is called a subbase of T if, and only if, {X} united with all finite intersections of sets from S forms a base of T, i.e. if, and only if, { n } B := β(s) := {X} O i : O 1,...,O n S, n N forms a base of T. 4 i=1 4 Some authors require a subbase to have the additional property of being a cover of X (i.e. every x X must be in some element of S), which leads to a similar, but nonequivalent, notion.

26 1 TOPOLOGY, METRIC, NORM 26 Lemma Let (X,T ) be a topological space. Then B T is a base of T if, and only if, for each x X, B(x) := {B B : x B} is a local base at x (cf. Def. 1.38). Proof. Let B T. If B is a base of T, x X, U U(x), then there is O T with x O U. Since B is a base, there must be B B with x B O U, showing B(x) to be a local base at x. Conversely, assume, for each x X, B(x) is a local base at x. If O T, then, for each x O, there is O x B(x) with x O x O. Then O = x O O x, proving B to be a base of T. In general, not every set of subsets of X is the base of some topology on X. One has the following result: Proposition Let X be a set and B P(X). Then B is the base of some topology T on X if, and only if, B satisfies the following two conditions: (i) B is a cover of X, i.e. X = B BB. (ii) For each B 1,B 2 B and each x B 1 B 2, there exists B 3 B such that x B 3 B 1 B 2. (1.27) Moreover, if B satisfies (i) and (ii), then { } T = O i : O i B, I some index set. (1.28) i I In particular, the topology T is uniquely determined by B; it is the coarsest topology on X containing B, i.e. { } T = min M P(X) : B M M is topology on X (1.29) where P(P(X)) is endowed with the partial order given by. Proof. Suppose B satisfies(i) and(ii). We now define T by(1.28) and show it constitutes a topology on X: Coosing I :=, shows T. As an immediate consequence of (i), X T also holds. Now let O 1,O 2 T and assume O 1 O 2. If x O 1 O 2, then, by (1.28), there exist sets B 1,B 2 B such that B 1 O 1, B 2 O 2, and x B 1 B 2. According to (ii), there exists B x B such that x B x B 1 B 2 O 1 O 2. Thus, setting I := O 1 O 2, O 1 O 2 = x IB x T, showing T is closed under finite intersections. To see that T is closed under arbitrary unions, let I be an index set and O i T for each i I. Then O i = i I j J i B j