A SECOND COURSE IN GENERAL TOPOLOGY

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3 Heikki Junnila, /2014 A SECOND COURSE IN GENERAL TOPOLOGY CHAPTER I COMPLETE REGULARITY 1. Definitions and basic properties Some examples Exercises....9 CHAPTER II CONVERGENCE AND COMPACTNESS 1. Filters Compactness and filters Compactifications Exercises CHAPTER III CONTINUOUS PSEUDOMETRICS 1. Construction of pseudometrics Applications Partitions of unity Continuous selections Exercises CHAPTER IV PARACOMPACT SPACES 1. Definition and basic properties Characterizations and further properties Paracompactness and normality in products The Bing-Nagata-Smirnov Metrization Theorem...83 Exercises CHAPTER V APPROXIMATION 1. The Stone-Weierstrass Theorem Applications Exercises

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5 Introduction Metric spaces and compact Hausdorff spaces are the most important and useful of the various kinds of spaces considered in general topology. In these notes, we deal with these two kinds of spaces, and also with paracompact spaces. One of the main themes in these notes is the construction and the use of continuous pseudometrics. This is connected with two central areas of general topology: covering properties and metrizability. The connection to covering properties results from the Stone Coincidence Theorem (here Stone does not refer to M.H. Stone, famous from Stone-Čech compactifications ja Stone-spaces, but rather to A.H. Stone, who is perhaps best known for his flexagons). Stone proved the coincidence theorem in 1948, and it states the equivalence of two covering properties, full normality, defined by J.W. Tukey in 1940, and paracompactness, defined by J. Dieudonne in Full normality was forgotten after Stone s theorem (and Tukey became later famous as a pioneer of data mining), but results of Tukey and Stone implied one of the fundamental results of modern general topology : every (pseudo)metric space is paracompact. This result makes it possible to use pseudometrics in the theory of paracompact spaces and to use paracompactness to study metrizability. One of the techniques used in these studies is that of partitions of unity. Other topics considered in these notes include Stone-Čech compactifications, continuous selections function spaces and approximation of real and complex functions on compact spaces. I used an early version of these lecture notes for a second course in general topology which I gave at the University of Helsinki in In the lectures, I also dealt with the construction of spaces by means of infinitary combinatorics, but the material on that topic is not included here. Work on these notes was partially supported by Natural Science Foundation of China grant

6 Conventions In the following, a space means a topological space (unless specified otherwise). We usually denote a space by a symbol like X instead of writing, say, (X, τ), and we sometimes denote the topology of a space X by τ X. We write A c X to indicate that A is a closed subset of X, and we write A X to indicate that A is an open subset of X. In particular, we have that τ X = {G : G X}. For a set A X, we denote by A or by Cl A the closure of the set A in the space X. Sometimes we use the more precise notation A X or Cl X A. The symbols Int A and Int X A denote the interior of A in X. A neighbourhood (abreviated nbhd ) of a set A X in the space X is a set E X such that A Int E. If A = {x}, we speak of a neighbourhood of the point x in the space X. Note that here neighbourhoods are not necessarily open sets (unlike in some text-books). A neighbourhood base of a set A X (or of a point x X) is a family B of nbhds of A (of x) such that every nbhd of A (of x) contains some set of the family B. A closed (or open, clopen, etc) neighbourhood base is a nbhd base consisting of closed (open, clopen, etc) sets. For a subset A of a space X, we denote by η A, or by η A (X), the neighbourhood filter {E X : A Int (E)} of the set A. If A = {x}, we write η x and η x (X) in room of η A and η A (X), respectively, and we speak of the neighbourhood filter of the point x. For a product space X = i I X i, we denote by p j the projection (x i ) i I x j of the product onto its jth factor X j. The symbols R, Q, P and I denote the sets consisting of all real numbers, all rational numbers, all irrational numbers and {x R : 0 x 1}, respectively. We use the same symbols to denote the corresponding Euclidean spaces. We let N = {1, 2, 3, 4,...} and ω = {0, 1, 2, 3, 4,...}. TFEA is an abbreviation for the phrase the following are equivalent. 2

7 I COMPLETE REGULARITY I.1. Definition and basic properties Recall that a (topological) space X is T 0 if x, y X, if x y, then U η x with y / U or V η y with x / V. T 1 if x, y X, if x y, then U η x with y / U and V η y with x / V. Note that X is T 1 iff {x} c X for every x X. T 2 if x, y X, if x y, then U η x and V η y with U V =. (T 2 -spaces are also called Hausdorff spaces.) regular if F c X and x X F, there U η x and V η F with U V =. Note that a space is regular iff every point has a closed neighbourhood base. T 3 if X is regular and T 1. normal if F, S c X, if F S =, then U η F and V η S with U V =. Note that a space is normal iff every closed subset has a closed neighbourhood base. T 4 if X is normal and T 1. Note that we have T 4 T 3 T 2 T 1 T 0 Definition A topological space X is completely regular if for all F c X and x X F, there exists a continuous f : X I such that f(x) = 0 and f(f) {1}. The space X is a Tihonov space if X is completely regular and T 1. (Tihonov spaces are sometimes called T spaces.) Note that completely regular regular. Moreover, the Urysohn Lemma shows that T 4 Tihonov. It is well known and easy to see that regularity and the properties T 2, T 1 and T 0 are both hereditary and productive. The same is true for complete regularity. 3

8 1 Proposition (i) Every subspace of a completely regular space is completely regular. (ii) Any product of completely regular spaces is completely regular. Proof. Exercise. Remember that normality is not, in general, preserved either with passing to subspaces or in forming products (even in the case of only two factors). Completely regular spaces have enough continuous real-valued functions. Recall that when X is a set, Y is a topological space and F is a set of mappings X Y, then the weak topology on X determined by F is the coarsest topology on X which makes each f F continuous, i.e., the topology on X which has the family {f 1 (O) : f F and O Y } as a subspace. 2 Lemma X is completely regular iff X has the weak topology determined by a set of functions X I. Proof. Exercise. For spaces X and Y, we denote by C(X, Y ) the set of all continuous mappings X Y. We write C(X) for C(X, R). We show next that Tihonov spaces coincide with spaces which can be embedded in cubes. By a cube we mean a product space of the form I A, where A is any set (recall that I A = {f : f is a mapping A I}, in other words, I A = {(r a ) a A : r a I for every a A}, in other words, I A = a A I a, where I a = I for every a A; the topology on I A is the usual product topology). Recall that a mapping φ : X Y is an embedding of the space X in the space Y if φ is a homeomorphism between X and the subspace φ(x) of Y. 3 Theorem Let X be a Tihonov space. Denote the set C(X, I) by F, and define a mapping ϕ : X I F by the condition ϕ(x) f = f(x). A. The mapping ϕ is an embedding of X into I F. B. Every continuous function ϕ(x) I can be extended to a continuous function I F I. Proof. A. Since X is completely regular and T 1, we see that, for all x, y X, if x y, then there exists f F such that f(x) = 0 and f(y) = 1. It follows that the mapping ϕ is one-to-one. The mapping ϕ is continuous, because for every f F, the composition 4

9 p f ϕ is the same as the (continuous) mapping f: by the definition of ϕ, we have, for every x X, that (p f ϕ)(x) = p f (ϕ(x)) = ϕ(x) f = f(x). By Lemma 2, the family E = {f 1 (O) : f F and O I} is a subbase of X. To show that ϕ is an open mapping X ϕ(x), it suffices (since ϕ is 1 1) to show that ϕ(e) ϕ(x) for every E E. Let h F and O I. Then we have that ϕ(h 1 (O)) = {ϕ(x) : x X and h(x) O} = {ϕ(x) : x X and ϕ(x) h O} = {(r f ) f F ϕ(x) : r h O}. = ϕ(x) {(r f ) f F I F : r h O}. We have shown that the set ϕ(h 1 (O)) is the intersection with ϕ(x) of a standard subbasic set of the product space I F ; hence ϕ(h 1 (O)) ϕ(x). We have shown that ϕ is an open mapping. Hence ϕ is an embedding. B. Let g be a continuous function ϕ(x) I. Then the function h = g ϕ is continuous X I. For every x X, we have that g(ϕ(x)) = h(x) = ϕ(x) h = p h (ϕ(x)). As a consequence, we have that g = p h ϕ(x) and we can set ḡ = p h to obtain a continuous extension of g which is defined on I F. Part A of Theorem 3 has the following important consequence. 4 Corollary (i) A space is a Tihonov space iff the space is homeomorphic with a subspace of some cube. (ii) A space is a compact Hausdorff space iff the space is homeomorphic with a closed subspace of some cube. Proof. (i) Necessity follows from Theorem 3. Sufficiency. Since I is a Tihonov space, it follows from Proposition 1 that any subspace of a cube is Tihonov. (ii) Necessity. A compact Hausdorff space K is normal and hence Tihonov. By (i), X is homeomorphic with a subspace of some cube; moreover, the cube is a Hausdorff space, and hence any compact subspace of it is closed. Sufficiency. By the Tihonov Theorem (see Section II.3), every cube is compact; hence every closed subspace of a cube is compact. 5

10 Note that when ϕ : X Y is an embedding, we can construct an extension of X which is homeomorphic with Y, in other words, we can represent X as a subspace of a space Z, which is a homeomorphic copy of Y. To obtain such a space, we write Z = (Y ϕ(x)) X (here we are assuming that Y ϕ(x) and X have no common points; if this is not the case, then we must first make Y ϕ(x) disjoint from X before we form the union). We declare a set G Z to be open if there exists U Y such that G = (U ϕ(x)) ϕ 1 (U). Now the space Z is homeomorphic with Y and the space X is a subspace of Z. The preceding observation and Corollary 4 have the following consequence. 5 Corollary Tihonov spaces coincide with subspaces of compact Hausdorff spaces. A compactification of a space X is a compact space Z such that X is a dense subspace of Z. If K is any compact space having X as a subspace, then the subspace X of K is a compactification of X. We can restate the preceding corollary as follows. 5 Corollary A space X is a Tihonov space iff X has a Hausdorff compactification. This result gives a characterization for Tihonov spaces which is purely topological in the sense that it (unlike the definition) does not involve the set R with its special properties. However, it is an external characterization depending on other spaces. In Theorem II.4.14 below, we shall give a purely topological internal characterization for Tihonov spaces. We shall also derive some consequences of part B of Theorem 3. First we note that the conclusion of part B can be significantly strengthened. Let Z and Y be topological spaces, and let X be a subspace of Z. We say that X is Y -embedded in Z provided that every continuous mapping f : X Y can be extended to a continuous mapping f : Z Y. 6 Lemma Assume that X is Y -embedded in Z. Then, for every set A, X is Y A -embedded in Z. Proof. Let A be a set, and let f be a continuous mapping X Y A. For every a A, the composition p a f is a continuous mapping X Y and it follows, since X is Y -embedded in Z, that there exists a continuous mapping g a : Z Y such that g X = p a f. We define a mapping g : Z Y A by the rule g(z) a = g a (z), and we note that g is continuous. 6

11 Moreover, for each x X, we have that g(x) a = g a (x) = p a (f(x)) = f(x) a for every a A, and hence we have that g(x) = f(x). 7 Lemma Assume that X is Y -embedded in Z and X is dense in Z. Then X is F- embedded in Z for every closed subspace F of Y. Proof. Let F c Y, and let f : X F be continuous. Then f is continuous X Y and it follows, since X is Y -embedded in Z, that there exists a continuous g : Z Y such that g X = f. By continuity of g, we have that g(z) = g(x) g(x) = f(x) F = F. Hence g is a mapping Z F. Now we can show that every Tihonov space has a very special compactification. 8 Theorem Let X be a Tihonov space. Then X has a Hausdorff compactification C such that, for every compact Hausdorff space K, every continuous mapping X K can be extended to a continuous mapping C K. Proof. Let ϕ be as in Theorem 3. Then X has a compactification C homeomorphic with ϕ(x), and Part B of Theorem 3 shows that X is I-embedded in C. By Lemma 6, X is Y -embedded in C, for every cube Y. It follows by Lemma 7 that X is L-embedded in C for every closed subspace L of a cube. As a consequence, X is K-embedded in C for every space K which is homeomorphic with a closed subspace of a cube. By Corollary 4, X is K-embedded in C for every compact Hausdorff space K. Later we shall see that a compactification C of a Tihonov space X as above is unique (up to a homeomorphism). It is called the Čech-Stone compactification of X. I.2. Some examples We have mentioned above that T 4 Tihonov T 3. In this section, we show that neither of the arrows can be reversed. Recall that a subset of a topological space is clopen if the set is both open and closed. Note that A X is clopen iff the characteristic function χ A of A (i.e., the function X I which has constant value 1 in the set A and constant value 0 in the set X A) is continuous. 7

12 A space is zero-dimensional if the clopen subsets of the space form a base for the topology of the space, in other words, if every point of the space has a clopen nbhd base. By the previous characterization of clopenness it is clear that every zero-dimensional space is completely regular. The basic open sets [a, b) (a, b R and a < b) of the Sorgenfrey line S are clopen, and hence S is zero-dimensional. The topology of S is finer than the Euclidean topology of R and hence S is T 1. As a consequence, S is a Tihonov space. The space S is a well-known example of a normal space whose square is not normal. Using this result and Proposition 1.1, we have the following result. 1 Example The space S 2 is Tihonov but not normal. To obtain a regular non-completely regular space, we cannot rely on any previously known results. 2 Example There exists a regular space X which is not non-completely regular. Proof. We topologize the set Z = R I as follows: points of the set R (0, 1] are isolated; a point of the form (r, 0) has a nbhd base by sets of the form V r F, where V r = {(a, b) Z : b = a r } and F is a finite subset of the set V r {(r, 0)}. We topologize the set X = Z {, + } so that Z is an open subspace of X, the point has a nbhd base by the sets O n = { } {(a, b) Z : a n}, for n N, and the point + has a nbhd base by the sets U n = {+ } {(a, b) Z : a n}, for n N. To see that the space X is regular, note first that the basic nbhds given for points of Z are clopen in X. Moreover, it is easy to see that, for every n N, we have Cl X (O n+2 ) O n and Cl X (U n+2 ) U n. It follows that X is regular. It remains to show that X is not completely regular. We start by proving the following Claim. Let F be a closed G δ -subset of X, and let E = {r R : (r, 0) F }. Assume that n Z and the set [n 1, n] E is countable. Then the set [n, n + 1] E is countable. Proof of Claim. Since F is a G δ -set, we can find, for every point (r, 0) F, a countable set L r V r such that V r L r F. Since the set [n 1, n] E is countable, we can choose a countably infinite set A [n 1, n] E. Denote by L the countable set a A L a, and note that the set H = {s [n, n + 1] : V s L } is countable. To complete the proof of the claim, we show that the set J = [n, n+1] H is contained in E. Let r J. Note that 8

13 V r (V a L a ) for every a A. It follows, since A is infinite and since V a L a F for every a A, that we have (r, 0) Cl F = F. As a consequence, r E. Now we use the Claim to show that we have f( ) = f(+ ) for every f C(X). Let f C(X). We show that we have f( ) f(+ ) ǫ for every ǫ > 0. Let ǫ > 0. By continuity of f, the set F = {p X : f(p) f( ) ǫ} is a closed G δ -set, and there exists n N such that O n F. Let E = {r R : (r, 0) F }. Since O n E, we have that {r R : r n} E. Using the Claim and induction, we get that, for every k N, the set [ n 1 + k, n + k] E is countable. It follows that the set R E is countable. In other words, (r, 0) F for all but countably many r R. As a consequence, we have that ClF = F and hence that f( ) f(+ ) ǫ. We have shown that f( ) = f(+ ) for every f C(X). It follows that the space X is not completely regular. I.3. Exercises 1. Prove Proposition Prove Lemma 1.2. The remaining problems deal with perfect mappings. Recall that a mapping f : X Y is compact if f 1 {y} is compact for every y Y. A mapping is perfect if it is continuous, closed and compact. 3. Let f : X Y be a closed mapping, let y Y and let G X be such that f 1 {y} G. Show that there exists O Y such that y O and f 1 (O) G. 4. Let f : X Y be a perfect mapping. (a) Show that if Y is compact, then X is compact. (b) Show that if Y is Lindelöf, then X is Lindelöf. [Hint: Use Problem 1.] 9

14 5. Let f : X Y be a perfect mapping. (a) Show that the restriction f S is perfect X Y for every S c X. (b) Show that the restriction f f 1 (E) is perfect X E for every E Y. 6. Show that the projection map p X : X C X is perfect when the space C is compact. 7. Let Z be a Hausdorff space, X a subspace of Z, and let f : X Y be a perfect mapping. Show that the graph Γ f = {(x, f(x)) : x X} of f is closed in Z Y. 8. Show that there exists a perfect mapping from a Tihonov space X to a space Y iff there exists a Hausdorff compactification C of X such that X is homeomorphic with a closed subspace of C Y. [Hint: Use Problems 3-5.] 10

15 II CONVERGENCE AND COMPACTNESS In a metric space, convergent sequences (of points of the space) are a very convenient and useful tool in topological considerations. Convergent sequences determine the topology of a metric space X: a set S X is closed iff S contains all limit points of convergent sequences of points of S. This property of a metric space can be stated as follows: every sequentially closed subset is closed. Also we have that a mapping f : X Y between metric spaces is continuous iff f(x n ) f(x) in Y whenever x n x in X. Unfortunately, sequences are not equally useful in connection with general topological spaces. The following simple examples show that convergent sequences do not necessarily tell much about the topology of a non-metric space. A sequence (x n ) n=1 is said to be eventually constant if there exists m N such that x k = x m for every k > m. Every eventually constant sequence of points in a topological space is convergent; these are considered to be trivial convergent sequences. If x is an isolated point of a space X, then x is not the limit of any non-trivial sequence in X. In (i) and (ii) below, we see that also a non-isolated point may have the same property. Examples (i) Let π be the topology on R in which every point p 0 is isolated and the nbhds of 0 are complements (in R) of countable subsets of R {0}. Then 0 is non-isolated, but no non-trivial sequence converges to 0 in (R, π). (ii) Denote by τ the usual topology of Q. Let π be the topology on Q in which every point p 0 is isolated and the point 0 has a nbhd base by sets of the form {x Q : x < r} {x n : n N}, where r > 0, x k 0 for each k and x n 0 in (Q, τ). Note that we have τ π. The point 0 is not isolated in (Q, π). However, 0 is not the limit of any non-trivial sequence. To see this, let x n 0 in (Q, π). Since τ π, we have that x n 0 in (Q, τ), and it follows that the set G = (Q {x n : n N}) {0} is a π-nbhd of 0. Since x n 0 in (Q, π), there exists m N such that {x n : n m} G. We have that {x n : n N} G {0}, and it follows that x n = 0 for every n m. (iii) We denote by [0, ω 1 ] the space obtained when the ordinal ω is equipped with its usual order topology (which has a base {{0}} {(β, α] : 0 β < α ω 1 }). The subset ω 1 = [0, ω 1 ) = [0, ω 1 ] {ω 1 } is not closed in [0, ω 1 ], but it is sequentially closed. To see this, let α n < ω 1 for n = 1, 2, 3,..., i.e., let α n be a countable ordinal for each n. Then 11

16 β = sup n N α n is a countable ordinal, and hence β < ω 1. Now (β, ω 1 ] is a nbhd of the point ω 1 and this nbhd contains no point of the sequence (α n ) n=1 ; as a consequence, the sequence does not converge to ω 1. Even for general topological spaces, there are meaningful ways to define convergence. This can be succesfully achieved with nets or with filters. Of these two approaches to general convergence below, we shall consider only filters. For nets, the reader is adviced to consult the chapter on that topic in the e-book written by Aisling McCluskey and Briam McMaster; the book can be found at II.1. Filters Perhaps the technically most convenient generalization for sequences is obtained with so called filters : here we are not in fact generalizing a sequence (x n ) but rather the family {{x k : k n} : n N} consisting of the tails of the sequence. Definition A family L of sets is a filterbase, if L and the following conditions hold: (i) / L. (ii) For all H, L L there exists K L such that K H L. A filterbase L is a filterbase of a set S if L P(S). A filter of S is a filterbase L of S s.t. (iii) L L and L A S = A L. By a filterbase of a space (Z, τ) we mean a filterbase of the set Z. Remarks (1 ) From (i) and (ii) it follows by induction that a filterbase L is a centered family, i.e., that H for every finite H F. (2 ) For a filter F, we can state (ii) in a stronger form : (ii) f For all H, L F, we have that H L F. We say that a filterbase L is free if L =. If L, then we say that L is fixed. 1 Examples (a) If A S and A, then {A} is a filterbase of S. (b) The family { {x k : k n} : n N } of tails of a sequence (x n ) is a filterbase. This filterbase is free provided that x i x j whenever i j. (c) The fixed filter of a set S determined by s S is the family K s = {A S : s A}. (d) The neighbourhood filter of a point x X is the family η x = {N X : x Int N}. 12

17 1 Lemma A family L of subsets of S is a filterbase iff the family F = {A S : there exists L L such that L A} is a filter of S. Proof. Necessity. Assume that L is a filterbase. We have that L F and hence that F. For every F F, the set F contains some set of the filterbase L and it follows that F. Let F, H F. Then there exist L, N L such that L F and N H. Since L is a filterbase, there exists K L such that K L N. We now have that K F H, and hence that F H F. The definition of F shows directly that we have A F whenever F A S. By the foregoing, the family F is a filter of S. Sufficiency. Assume that F is a filter of S. Then F and hence there exists a set F F; since F contains some set from L, we have that L. Since L F, we have that / L. If L, K L, then L, K F and hence L K F; by the definition of F, there exists H L such that H L K. We have shown that L is a filterbase. The family F above is called the filter of S generated by the filterbase L. Examples (a) The family of tails of the sequence ( 1 n ) n N generates the free filter F = {A R : such n N that 1 k A for every k n} of R. (b) Let s S. The family {{s}} generates the fixed filter K s of S. (c) A neighbourhood base of a point x of a space X is a family of subsets of X which generates the neighbourhood filter η x of x. Neighbourhood bases are fixed filterbases. 2 Lemma Let L and K be filterbases. Then the family L K = {L K : L L ja K K} is a filterbase iff L K. Proof. Exercise. We shall now define the notion of convergence for filters of topological spaces. Definition Let L be a filterbase of a space X and let x X. We say that x is a cluster point of L if we have that x L for every L L. We say that L converges to x, and we write L x, provided that for every V η x, there exists L L such that L V ; then we also say that x is a limit point of L. 13

18 Remarks (1 ) If F is the filter on a space X generated by the filterbase L and x X, then F x iff L x. (2 ) For a filter F of a space X, we can characterize convergence F x by a simpler condition: η x F. Examples (a) If (x n ) n N is a sequence of points in a space X and x X, then x n x iff we have L x for the filterbase L = { {x k : k n} : n N }. Moreover, x is a cluster point of the sequence (x n ) n N iff x is a cluster point of the filterbase L. (b) For every x X, we have that K x x and η x x. 3 Lemma A point x of a space X is a cluster point of a filterbase L of X iff there exists a filterbase N of X such that L N and N x. Proof. Necessity. Assume that x is a cluster point of L. For every L L, we have that x L and hence we have that V L for every V η x. By Lemma 2, it follows that the family N = L η x is a filterbase. Moreover, we have that L N and N x. Sufficiency. Assume that we have L N and N x for a filterbase N of X. To show that x is a cluster point of L, assume on the contrary that there exists L L such that x L. Then X L is a nbhd of x and it follows, since N x, that there exists N N such that N X L. This, however, is a contradiction, since N is a filterbase and N, L N. In particular, the lemma shows that a limit point of a filterbase is a cluster point. The next results show that filter convergence determines the topology of a space. 4 Proposition TFAE for a subset A and a point x of a space X: A. x A. B. The point x is a cluster point of some filterbase of A. C. The family L = {V A : V η x } is a filterbase and L x. Proof. A C: Assume that x A. Then V A for each V η x, and Lemma 2C shows that the family L = η x {A} is a filterbase. Clearly L x. C B: By Lemma 3. B A: Assume that x is a cluster point of the filterbase K of the set A. Let K K. Then K A and x K and hence x A. Note that if X is T 1 and x A, then the filterbase L in C above is fixed iff x A. 14

19 By the above result, we see that a subset A of a space X is closed iff A contains all limit points of its filters. This means that the topology of X is uniquely determined by filter convergence in X. By considering indiscrete topologies {X, }, we see that one filterbase can converge to several points. We now characterize those spaces where this cannot happen. 5 Theorem A topological space X is Hausdorff iff every filterbase of X has at most one limit point. Proof. Necessity. Assume that X is Hausdorff. Let L be a filterbase of X such that L x and L y in X. We show that x = y. Assume on the contrary that x y. Sincea X is Hausdorff, there are nbhds U η x and V η y such that U V =. Since L x and L y, there exist H, K L such that H U and K V. Further, there exists L L such that L H K. Now we have that L H K U V =, in other words, that L =. This is a contradiction, since L is a filterbase. Sufficiency. Assume that every filterbase of X has at most one limit point. We show that X is Hausdorff. Let x and y be points of X such that we have U V for all U η x and V η y. We show that x = y. By Lemma 2, the family L = {U V : U η x and V η y } is a filterbase of X. Directly from the definition of L it follows that L x and L y; our assumption now shows that x = y. Next we show that continuity of mappings between spaces can be very naturally characterized in terms of filter convergence. The intuitive idea of continuity of a map f : X Y is that f does not tear the space X which it is mapping, and this is equivalent with the idea that for points which are near to each other in X, the images are near to each other in Y. We can measure nearness for example with nbhds: points in a small nbhd of a point are thought to be close to the point. This idea leads to following definition of continuity: f is continuous at a point x X provided that for every W η f(x) (Y ), there exists V η x (X) such that f(v ) W. This condition can be stated more simply as follows: f is continuous at a point x X provided that f 1 (W) η x (X) for every W η f(x) (Y ). The mapping f is continuous if f is continuous at each point of X. Using the second formulation of continuity at a point, we easily get the following result: f is continuous iff f 1 (G) X for every G Y. In a metric space, the idea of nearness can be captured by convergent sequences: if 15

20 x n x, then the points x n are getting nearer and nearer to the point x. This translates to the following notion of continuity: f : X Y is continuous at x X provided that f(x n ) f(x) in Y whenever x n x. This condition is, in fact, equivalent with the earlier conditions for metric spaces X and Y. For general spaces, we cannot characterize continuity in terms of convergent sequences, but we obtain a similar characterization using convergence of filters. Let f : X Y. For every family H of subsets of X, we set f(h) = {f(h) : H H}. It is easy to see that f(h) is a filterbase if H is a filterbase. 6 Theorem TFAE for a mapping f : X Y and a point x X: (1) f is continuous at x. (2) For every A X, if x A, then f(x) f(a). (3) If x is a cluster point of a filterbase L of X, then f(x) is a cluster point of f(l). (4) For every filterbase L of X, if L x, then f(l) f(x). Proof. (1) (4): Assume that f is continuous at x and the filterbase L converges to x. To show that f(l) f(x), let W η f(x) (Y ). Since f is continuous at x, we have that f 1 (W) η x (X) and it follows, since L x, that there exists L L such that L f 1 (W). Now f(l) f(l) and f(l) W. We have shown that f(l) f(x). (4) (3): This follows by Lemma 3. (3) (2): Assume that (3) holds and A X is such that x A. The family L = {A} is a filterbase of X and x is a cluster point of L. Since (3) holds, we have that f(x) is a cluster point of f(l), in other words, we have that f(x) f(a). (2) (1): Assume that (2) holds. To show that f is continuous at x, assume on the contrary that there exists W η f(x) (Y ) such that f 1 (W) η x (X). Then we have that x X f 1 (W), and it follows by our assumption that f(x) f(x f 1 (W)); this, however, is a contradiction, since f(x f 1 (W)) Y W and W η f(x) (Y ). Next we characterize filter convergence in product spaces; the result explains why the product topology is often called the topology of pointwise convergence. 7 Theorem Let L be a filterbase in a product space i I X i, and let (x i ) i I X. A. If (x i ) i I is a cluster point of L, then x i is a cluster point of p i (L) for each i. B. L (x i ) i I iff p i (L) x i for every i I. 16

21 Proof. A. This follows from Theorem 6 and continuity of projection maps. B. Theorem 6 and continuity of projection maps show that if L (x i ) i I, then p i (L) x i for every i I. To show the converse, assume that p i (L) x i for every i I. We show that L (x i ) i I. Let G η (xi ) i I ( i I X i). Then there exists a finite set J I and for each j J there exists G j η xi (X i ) such that we have j J p 1 j (G j ) G. For each j J, since p j (L) x j, there exists L j L such that p j (L j ) G j. Since L is a filterbase, there exists L L such that L j J Lj. For every j J, we have that p j (L) p j (L j ) G j and hence that L p 1 j (G j ). It follows that L j J p 1 j (G j ) G. 8 Corollary (x n i ) i I (x i ) i I iff x n i x i for every i I. II.2. Compactness and filters We mentioned above that every filterbase L is a centered family of sets, i.e., that L for every finite L L. It is easy to see that a family N of sets is centered and N = iff N is a filtersubbase, in other words, iff the family { L : L L and L is finite} is a filterbase (this family is called the filterbase generated by the filtersubbase L). The following simple observation connects filtersubbases with compactness. 1 Lemma A non-empty family L of subsets of a set S is a filtersubbase iff no finite subfamily of the family {S L : L L} covers S. 2 Theorem A space X is compact iff every filterbase of X has a cluster point. Proof. Necessity. Assume that X is compact. Let L be a filterbase of X. Then also the family F = {L : L L} is a filterbase. By compactness of X and by Lemmma 1, it follows that the open family {X F : F F} is not a cover of X. Hence there exists a point x X {X F : F F}; for such a point, we have that x F = {L : L L} and x is thus a cluster point of L. Sufficiency. Assume that every filterbase of X has a cluster point. We show that X is compact. Assume on the contrary that X has an open cover U with no finite subcover. By Lemma 1, the family F = {X U : U U} is a filtersubbase. Hence the family ˆF = { F : F F and F is finite} is a filterbase of X. By our assumption, ˆF has a cluster point x. Since ˆF is a closed family, we have that x ˆF, but this is a contradiction, because we have that ˆF = F = X U =. 17

22 The above result remains valid if filterbase is changed to filter. Next we show that we do not even have to consider all filters of a space, but only some rather special ones. Definition An ultrafilter of a set E is a maximal filter of E, that is, a filter of E which is not properly contained in any other filter of E. The next result gives some characterizations of ultrafilters. 3 Theorem TFAE for a filtersubbase L of a set E: A. L is an ultrafilter. B. For every A E, we have either A L or E A L. C. For every finite cover U of E, we have that L U. Proof. It is obvious that C B, and hence it suffices to show that A C and B A. A C: Let L be an ultrafilter of E. To show that condition C holds, assume on the contrary that E has a finite cover U such that U F =. Let U U. Since U does not belong to the maximal filter F, the family F {U} is not a filtersubbase (otherwise the filter of E generated by this family would properly contain F). As a consequence, there exists a finite family F U F such that U F U =. Denote by F the finite subfamily U U F U of F. We have that ( U) F =, i.e., that F =, but this is impossible, because F is a filter. B A. Assume, that condition B holds. We show that L is an ultrafilter. First we note that for all L, T L, we have that L T L; this follows, because otherwise we would have that E (L T) L and L would have a finite subfamily L = {L, T, E (L T)} with L =. Moreover, L is a filter, because whenever L L and L A E, we have that L (E A) = and hence that E A / L and, further, that A L. The filter L is obviously maximal: for every A P(E) L, we have that E A L and hence L {A} is not contained in any filter. The following observations enable us to enlarge filtersubbases to ultrafilters. 4 Lemma (a) If L P(E) is a filtersubbase and A E, at least one of the families L {A} and L {E A} is a filtersubbase. (b) If α is an ordinal and (L β ) β<α is an increasing sequence of filtersubbases (i.e., L β L γ for β < γ < α), then the family L = β<α L β is a filtersubbase. 18

23 Proof. (a) If neither family is a filtersubbase, then we can find finite K, N L such that K A = and N A =, but then we have that (K N) =, and this is impossible. (b) Otherwise, we would find a finite L L such that L = ; since the finite family L is contained in the union of the increasing sequence (L β ) β<α, there would exist β < α such that L L β, but this is impossible, because L β is a filtersubbase. Ultrafilter Theorem (A. Tarski) Every filtersubbase of a set E is contained in some ultrafilter of E. Proof. Let L be a filtersubbase of E. Write P(E) = {A α : 0 < α < λ}, where λ is an ordinal. Set L 0 = L, and define recursively filtersubbases L α of E, for α < λ, as follows: set L α = ( β<α L β) {Aα } if this family is a filtersubbase and if it is not, then set L α = ( β<α L β) {E Aα }. Lemma 4 shows that L α is a filtersubbase for every α < λ, and it also shows that the family F = α<λ L α is a filtersubbase. It follows from the recursive construction that, for every A E, we have either A F or E A F. By Theorem 3, F is an ultrafilter. Moreover, we have that L = L 0 F. Next we consider convergence of ultrafilters. 5 Lemma Let D be a set and X a space, and let φ be a mapping D X. Let F be an ultrafilter of D and let p be a cluster point of the filterbase φ(f) of X. Then φ(f) p. Proof. To show that φ(f) p, let U be a nbhd of p. The set φ 1 (X U) is not a member of the filter F, because we have that p / X U and φ ( φ 1 (X U) ) X U. Since F is an ultrafilter, Theorem 3 shows that φ 1 (U) = D φ 1 (X U) F; hence φ ( φ 1 (U) ) φ(f). Furthermore, φ ( φ 1 (U) ) U. 6 Corollary An ultrafilter of a space converges to each of its cluster points. 7 Theorem A space is compact iff every ultrafilter of the space converges. Proof. Necessity follows from Theorem 2 and Corollary 6. Sufficiency. Assume, that every ultrafilter of the space X converges. We use Theorem 2 to show that X is compact. Let L be a filter of X. By the Ultrafilter Theorem, there exists an ultrafilter F of X such that L F. By our assumption, there exists x X such that F x. Now we have that x {H : H F} {H : H L} and hence x is a cluster point of L. 19

24 We now use Theorem 6 to prove perhaps the most important result of general topology. 8 The Tihonov Theorem Any product of compact spaces is compact. Proof. Let X i be a compact space for every i I. We use Theorem 7 to show that the product space i I X i is compact. Let F be an ultrafilter of i I X i. For every i I, it follows from Theorem 2 that the filterbase p i (F) of the compact space X i has a cluster point x i, and it follows further, by Lemma 5, that p i (F) x i. By Theorem 2.7, we have that F (x i ) i I. In particular, we have the following result (which we have already used earlier). 9 Corollary The space I A is compact for every set A. II.3. Compactifications. We have shown above that every Tihonov space has a Hausdorff compactification. A given Tihonov space may have many different kinds of Hausdorff compactifications, but sometimes we can compare them and find out that one is simpler than another. For Hausdorff compactifications C and K of a space X, we write K C if there exists a continuous mapping C K which keeps the points of X fixed. The relation is obviously reflexive and it easy to see that it is also transitive. The next result shows that is essentially antisymmetric. 1 Proposition Let K and C be Hausdorff compactifications of X such that C K and K C. Then there exists a homeomorphism ϕ : C K which keeps the points of X fixed. Proof. There exist continuous mappings ϕ : C K and ψ : K C such that both ϕ and ψ keep the points of X fixed. We show that ϕ and ψ are inverses of each other. The mapping ψ ϕ is continuous C C and it keeps the points of X fixed; since X is dense in C and since the set of fixed points of a continuous mapping from a Hausdorff space into itself is closed, we see that ψ ϕ = id C. Similarly, we see that ϕ ψ = id K. Hence ϕ and ψ are homeomorphisms. If K and C are Hausdorff compactifications of X such that C K and K C, then we say that C and K are equivalent compactifications of X. Proposition 1 shows that 20

25 we may identify equivalent compactifications. After the identification, the relation is a partial order in the set of all Hausdorff compactifications of a Tihonov space X. Let C be a Hausdorff compactification of a space X. The subspace C X of C is called the remainder of X in C. We shall next point out a relation between the remainders of two compactifications of X which are related to each other in the partial order. 2 Lemma Let C and K be Hausdorff compactifications of a space X, and let ϕ : C K be a continuous mapping which keeps the points of X fixed. Then ϕ maps C X onto K X. Proof. To show that ϕ(c X) = K X, note first that the mapping ϕ is onto: this follows, because ϕ(c) is a compact subset of K and it is also dense, since ϕ(x) = X. It follows that K X ϕ(c X). To see that ϕ(c X) K X, note that the mapping ψ = ϕ ϕ 1 (X) is continuous ϕ 1 (X) X; since X ϕ 1 (X), the mapping ψ is continuous ϕ 1 (X) ϕ 1 (X). Let F = {z ϕ 1 (X) : ψ(z) = z}, and note that, since ϕ 1 (X) is Hausdorff, the set F is closed in ϕ 1 (X). It follows, since X F and X is dense in ϕ 1 (X), that F = ϕ 1 (X). Hence ψ is the identity map on ϕ 1 (X), and it follows, since ψ(ϕ 1 (X)) = X, that ϕ 1 (X) = X. As a consequence, we have that ϕ(c X) K X. A mapping f : X Y is perfect if f is continuous and closed, and f 1 {y} is compact for every y Y. Continuous mappings between compact Hausdorff spaces are perfect. 3 Proposition Let C and K be Hausdorff compactifications of a space X such that K C. Then K X is the image of C X under a perfect mapping. Proof. Since K C, there exists a continuous mapping ϕ : C K such that ϕ(x) = x for every x X. As a mapping between compact Hausdorff spaces, ϕ is perfect. By Lemma 2, we have that ϕ(c X) = K K, and it follows that we have C X = ϕ 1 (K X). By Problem 3 of the Exercises for Chapter I, the mapping θ = ϕ ϕ 1 (K X) is perfect. Moreover, we have that θ(c X) = K X. In Theorem I.1.8, we showed that a special embedding of a Tihonov space X into the cube I C(X,I) gives a special Hausdorff compactification for X. We will now observe that this special compactification is maximal with respect to the partial order. 21

26 4 Proposition Let K and C be Hausdorff compactifications of X. If X is I-embedded in C, then K C. Proof. The proof of Theorem I.1.8 shows that X is K-embedded in C. It follows, since the identity mapping id X of X is continuous X K, that id X can be extended to a continuous mapping ϕ : C K. Let X be a Tihonov space. It follows from the above result that if K and C are Hausdorff compactifications of X such that X is I-embedded both in C and in K, then the compactifications C and K of X are equivalent. This means that the compactification of X described in Theorem I.1.8 is (essentially) unique, and for this reason we call it the Čech-Stone compactification of X. We denote the Čech-Stone compactification of X by βx. For a continuous mapping f from X to a compact Hausdorff space K, we sometimes denote by βf the extension of f to a continuous mapping βx K (the extension is unique, since X is dense in βx). Let X be a Tihonov space. We denote the remainder βx X of X in βx by X, and we call X the Čech-Stone remainder of X. It follows from Proposition 3 that, for every Hausdorff compactification K of X, the remainder K X of X in K is the image of X under a perfect mapping. Perfect mappings preserve many topological properties in both directions, and for such a property P, we have that if some remainder of X has property P, then every remainder of X has property P. Among properties preserved in both ways by perfect maps, we have, for example, compactness, local compactness, σ-compactness and the Lindelöf property. Even for some familiar and simple spaces, like R, Q or N, the Čech-Stone compactification is a very complicated object. We will describe βn below in some detail, but first we give some examples of situations where the compactification is particularly simple. 5 Examples (a) Since a space is dense in any compactification, a compact space has no proper Hausdorff compactifications. Hence we have that βk = K for every compact Hausdorff space K. (b) In Example (iii), in the introduction to this chapter, we considered the ordinal spaces [0, ω 1 ] and [0, ω 1 ). We shall now show that β[0, ω 1 ) = [0, ω 1 ]. It is easily seen that the space [0, ω 1 ] is Hausdorff. To show that [0, ω 1 ] is compact, let U be an open cover of [0, ω 1 ]. We show that U has a finite subcover. Denote by γ 22

27 the least ordinal β such that the interval [β, ω 1 ] is covered by a finite subfamily of U. We show that γ = 0. Assume on the contrary that γ > 0. Let V be a finite subfamily of U such that [γ, ω 1 ] V. Then V η γ, and it follows that there exists δ < γ such that (δ, γ] V. Furthermore, there exists U U such that δ U. Now the finite subfamily V {U} of U covers [δ, ω 1 ], but this is in contradiction with minimality of γ. It follows that γ = 0. As a consequence, a finite subfamily of U covers [0, ω 1 ]. We have shown that [0, ω 1 ] is compact. It remains to show that [0, ω 1 ) is I-embedded in [0, ω 1 ]. Let f : [0, ω 1 ) I be continuous. We show that there exists α < ω 1 such that f has a constant value in the set [α, ω 1 ). Assume that no such α exists. Then we can use transfinite induction to find x γ, y γ ω 1, for γ < ω 1, such that we have, for every γ, that f(x γ ) f(y γ ) and sup{y β : β < γ} < x γ < y γ. For every γ < ω 1, let n γ N be such that f(x γ ) f(y γ ) > 1 n γ. Since ω 1 is uncountable, there exists k N such that the set A = {γ < ω 1 : n γ = k} is uncountable. Choose an increasing sequence γ 1 < γ 2 < γ 3 < of elements of A. Let δ = sup{y γn : n N}. Then we have that x γn δ and y γn δ, but this contradicts continuity of f, since we have that f(x γn ) f(y γn ) > 1 k for every n. It follows from the foregoing that there exists α < ω 1 and r I such that f(β) = r for every β > α. We now define the mapping f : [0, ω 1 ] I by setting f(α) = f(α) for α < ω 1 and f(ω 1 ) = r. It is easy to see that f is continuous. We have shown that [0, ω 1 ) is I-embedded in [0, ω 1 ]. As a consequence, β[0, ω 1 ) = [0, ω 1 ]. We shall next give another description of the Čech-Stone compactification of a Tihonov space. First we study a general way of defining Hausdorff compactifications. In the following, N denotes a family of sets such that N and N is closed under finite unions and finite intersections. An N-filter(sub)base is a filter(sub)base L N. An N-filter is an N-filterbase L such that for all N N and L L, if L N, then N L. An N-ultrafilter is a maximal N-filter, i.e., an N-filter which is not properly contained in any N-filter. Note that since N is closed under finite intersections, so is every N-filter. Many of the results on ultrafilters have analogues for N-ultrafilters. 6 Lemma Every N-filtersubbase is contained in an N-ultrafilter. 23

28 Proof. Let L be an N-filtersubbase. By Zorn s Lemma, there exists a maximal (with respect to inclusion) centered family F such that L F N. Since F is centered, also the family G = { H : H F and H is finite} is centered; moreover, we have that F G N, and it follows by maximality of F that G = F. Hence F is closed under finite intersections. We clearly have that, for all F F and N N, if F N, then N F. It follows from the foregoing that F is an N-filter. By maximality, F is an N-ultrafilter. 7 Lemma Let K be an N-filter. Then K is an N-ultrafilter iff for every N N K, there exists K K such that K N =. Proof. Sufficiency of the condition is obvious. Necessity. Assume that K is an N-ultrafilter, and let N N K. To show that there exists K K such that K N =, assume on the contrary that K N for every K K. By Lemma 2.2, the family I = {K N : K K} is a filterbase. Let J = {N N : L N for some L I}. Then J is an N-filter and K J - a contradiction. An argument from the proof of Theorem 2.3 gives the following consequence of the above result. 8 Corollary Let K be an N-ultrafilter, and let H N be a finite family such that H F. Then H L. We denote by U N be the collection of all N-ultrafilters. For every N N, we let ˆN = {F U N : N F}. 9 Proposition The family { ˆN : N N } is a base for a compact T 1 -topology on U N. Proof. Denote the family by B. We have that U N = ˆ B, and hence that B = U N. Let F, H N and L ˆF Ĥ. Then we have that F L and H L, and it follows from Corollary 8 that F H L. Hence we have that L F H. Since we clearly have that F H ˆF Ĥ, we have shown that B is a base for a topology τ of U N. We show that τ is compact and T 1. Let H, L U N and H L. By maximality of H and L, we have that L H and H L. Let L L H and H H L. Now ˆL and Ĥ are τ-nbhds of H and L, respectively, and we have that L ˆL and H / Ĥ. We have shown that τ is T 1. To show that τ is compact, it suffices to show that every cover of U N by members of B has a finite subcover. Let M N be such that the family { ˆM : M M} 24

29 covers U N. For every L U N, there exists M M such that M L. Hence M is not contained in any N-ultrafilter, and it follows by Lemma 6 that M is not centered. Let K M be a finite family such that K =. Then K is not contained in any N-ultrafilter, and this means that the family { ˆM : M K} covers U N. We have shown that the topology τ is compact. We denote by τ N the topology of U N with base { ˆN : N N }. In the following, we consider U N as a topological space, equipped with the topology τ N. In the next result, we give a sufficient condition for U N to be Hausdorff. 10 Proposition The space U N is Hausdorff provided that for all K, L N, if K L =, then there are M, N N such that M L =, N K = and M N = N. Proof. Assume that the condition stated in the proposition holds for N. Let K, L U N and K L. Let K K L. By Lemma 7, there exists L L such that K L =. By our assumption, there are M, N N such that M L =, N K = and M N = N. Note that we have M L and N K. Hence ˆM and ˆN are nbhds of L and K, respectively. Moreover, we have that ˆM ˆN =, because for every H U N, we have that M N H, and it follows by Corollary 8, that we have M H or N H. Next we define a concept which allows us to use the preceding results to obtain Hausdorff compactifications for topological spaces. A closed base of a space X is a family F of closed subsets of X such that every closed subset of X is the intersection of sets belonging to F. Note that F is a closed base of X iff the family {X F : F F} is a base of X. A network for X is a family L of subsets of X such that every open set is the union of sets belonging to L. A base of X is a network consisting of open sets. A normal closed base of X is a closed base F of X satisfying the following conditions: (i) F is a network of X; (ii) F is closed under finite unions and finite intersections; (iii) For all disjoint F, H F, there exist S, T F such that F T =, H S = and S T = X. Note that a T 1 -space is normal iff the family consisting of all closed subsets is a normal closed base. 25

30 11 Proposition Let N be a normal closed base of a T 1 -space X. A. For every x X, the family N x = {N N : x N} is an N-ultrafilter. B. The mapping x N x is an embedding of X into U N. C. The set {N x : x X} is dense in U N. Proof. A. Let x X. By the corresponding property of N, the family N x is closed under finite intersections. Moreover, it is clear that we have E N x if E N and there exists D N x such that D E. As a consequence, N x is an N-filter. To see that N x is maximal, let N N N x. Then X N is a nbhd of x and it follows, since N is a network, that there exists L N x such that L X N. It follows that there is no filter F such that N x {N} F. We have shown that N x is a maximal N-filter. B. We denote the mapping x N x by ϕ. Since X is T 1 and N is a network of X, the mapping ϕ is one-to-one. For every N N, we have that ϕ 1 ( ˆN) = {x X : N N x } = X N. Since the family { ˆN : N N } is a base of U N and {X N : N N } is a base of X, it follows from the foregoing that ϕ is an embedding. C. Let N N, and consider the basic set ˆN = {F U N : N F} in U N. Assume that ˆN. Then we have that N X. Let x X N. Then N x ˆN. It follows from Propositions 9-11 that if a T 1 -space X has a normal closed base, then X has a Hausdorff compactification and hence X is Tihonov. We shall next show that the converse holds: every Tihonov space has a normal closed base. The zero-set of a function f : X R is the set f 1 {0}. A zero-set of a space X is the zero-set of a continuous function X R. We denote by Z X the collection of all zero-sets of X. If f : X R is continuous, then so is the function g, defined by the rule g(x) = min(1, f(x) ). Moreover, g has the same zero-set as f. It follows that all zero-sets of X can be obtained as zero-sets of continuous functions X I. Note that a zero-set is always closed. If Y is a metric space, then every closed subset F of Y is a zero-set: F is the zero-set of the continuous function y d(y, F), where d is a compatible metric for Y. Note also that if f : X Z is continuous, then for every zero-set S of Z, the set f 1 (S) is a zero-set of X. It follows from the preceding observations that for all f, h C(X), the set {x X : f(x) h(x)} is a zero-set. Further zero-sets are indicated in the following result. 26