Axioms of separation

Axioms of separation These notes discuss the same topic as Sections 31, 32, 33, 34, 35, and also 7, 10 of Munkres book. Some notions (hereditarily normal, perfectly normal, collectionwise normal, monotonically normal,...) are not discussed in Munkres book. A topological space (X, T ) is: 1. Axioms T 0 if for every two distinct points x, y X, there is an open set that contains exactly one of x, y. T 1 if for every x X and every y X \ {x} there is an open set that contains x but does not contain y. Equivalent definition: X is T 1 iff all finite subsets of X are closed. T 2 (also called Hausdorff) if for every two distinct points x, y X, there are U, V T such that x U, y V, and U V =. T 3 (also called regular) if X is T 1 and for every x X and every closed set H X such that x H there exist U, V T such that x U, H V and U V =. T 3 1 (also called Tychonoff) if X is T 1 and for every x X and every closed 2 set H X such that x H there exist a continuous function f : X R such that f(x) = 0 and f(h) = {1}. T 4 (also called normal) if X is T 1 and for every every pair of disjoint closed sets H, K X there exist U, V T such that H U, K V and U V =. T 5 (also called hereditarily normal) if every subspace of X is normal. T 6 (also called perfectly normal) if X is normal and every closed set in X is a G δ -set. 1 In (optional) Sections 14 and 15 we briefly mention two more axioms of separation called collectionwise normality and monotone normality. 2. Some criteria and implications Proposition 1. A space X is regular iff for every x X and every neighborhood U x, there is a neighborhood V x such that x V V U. Proposition 2. A space X is normal iff for every closed set H X and every neighborhood U H, there is a neighborhood V H such that H V V U. T 0 T 1 T 2 T 3 T 3 1 T 4 T 2 5 T 6 The imlications T 0 T 1 T 2 T 3 are trivial. To see that T 3 T 3 1, let x X and H be a closed set of X not containing 2 x. If X is T 3 1, then there is a continuous function f : X R with f(x) = 0 and 2 f(h) = {1}. Putting U = f 1 ((, 1 2 )) and V = f 1 (( 1 2, )) proves regularity. T 3 1 T 4 is not trivial at all. This implication uses: 2 1 A set H is called a Gδ -set if H = n N On for some open sets On X. 1

2 Theorem 3. (Urysohn s Lemma) A space X is normal iff for every pair of disjoint closed sets H, K X, there is a continuous function f : X R such that f(h) = {0} and f(k) = {1}. The proof is postponed until Section 11. Modulo Urysohn s Lemma, T 3 1 2 T 4 becomes easy to see. T 4 T 5 is trivial. T 5 T 6 is not, but we skip this, and in general, axioms T 5 and T 6 are out of syllabus. 3. Some Examples Example 4. A T 0 space which is not T 1. X = {1, 2}, T = {{1, 2}, {1}, }. Example 5. A T 1 space which is not T 2. Let X be any infinite set and T fin = { } {X \ F : F X is finite}. Example 6. A T 2 space which is not T 3. Consider (R, T 23 ) where the topology T 23 is defined as follows: the points of Q are isolated while a basic neighborhood of a point r P = R \ Q takes the form O ε (r) = {r} {q Q : q r < ε} where ε > 0. Remark 7. Examples of T 3 which are not T 3 1 are pretty exotic. For example, 2 there is a regular space X such that every continuous function from X to R is constant. We will skip this topic. Remark 8. Examples of T 3 1 spaces which are not normal involve some properties 2 of cardinal and ordinal numbers. We consider such examples in Sections 4. Subspaces, products, images Proposition 9. If X is a T i space for some i {0, 1, 2, 3, 3 1 2 } then so is every subspace of X. A subspace of a normal space does not have to be normal (we will see many examples of this below). This is the reason to introduce axiom T 5. Every subspace of a T 6 space is T 6 (we are skipping the proof of this). Theorem 10. A closed subspace of a normal space is normal. Proof. Very easy. Proposition 11. Let i {0, 1, 2, 3, 3 1 2 }. If for each a A, X a is a T i space then so is the Tychonoff product a A X a. A product of two normal spaces does not have to be normal (we will see examples of this below). Continuous mappings, in general, do not preserve axioms of separation. Just consider the identity mapping from (X, T ) to (X, T trivial ).

3 5. Axioms of separation in some classes of spaces Here we will see results of two sorts: In the presence of some property, say P, an in general weaker axiom of separation may become equivalent to an in general stronger axiom of separation. In the presence of some axiom os separation, an in general weaker topological property may become equivalent to an in general stronger topological property. Definition 12. A topological space X is called zero-dimensional if X has a base consisting of clopen 2 sets. Theorem 13. Every T 1 zero-dimensional space is Tychonoff. Proof. Let x X, H be closed in X and x H. Then U = X \ H is a neighborhood of x. Since X is zero-dimensional, there is a clopen set V such that x V U. Define f : X R by { 0 if x V ; f(x) = 1 if x X \ V. Then f is continuous, f(x) = 0, and f(h) = {1}. Theorem 14. Every regular Lindelöf space is normal. Proof. Let H and K be two disjoint closed sets in a regular Lindelöf space (X, T ). For every h H, fix U h T so that h U h U h X \ K. For every k K, fix V k T so that k V k V k X \ H. Being a closed subspace of a Lindelöf space X, H is Lindelöf. Therefore the cover {U h : h H} of H by open sets of X has a countable subcover, say {U hi : i N}. Likewise, there is a countable cover {V ki : i N} of K. For i N, put Ũi = U hi \ j<i V k j and Ṽi = V ki \ j i U h j. Then U = i N Ũi and V = i N Ṽi are disjoint neighborhoods of H and K. Exercise 15. Give an example of a Hausdorff Lindelöf space which is not regular. Theorem 16. Every Hausdorff compact space is normal. Proof. Let (X, T ) be a Hausdorff compact space, x X and H X be a closed set such that x H. For each y H, fix U y, V y T so that x U y, y V y, and U y V y =. Being a closed subspace of a compact space X, H is compact. Therefore the cover {V y : y H} of H by open sets of X has a finite subcover, say {V y1,..., V yn }. Then U = U y1... U yn and V = V y1... V yn are disjoint neighborhoods of x and H respectively. This proves the regularity of X. Now, since every compact space is Lindelöf, the normality of X follows from Theorem 14. Exercise 17. Give an example of a T 1 compact space which is not Hausdorff. Recall that X is countably compact if every countable open cover of X has a finite subcover (which for T 1 spaces is equivalent to all closed discrete subsets of X be finite); X is pseudocompact if every continuous function from X to R is bounded. A Ψ-space constructed using an infinite m.a.d. family is Hausdorff and zero dimansional, hence Tychonoff and pseudocompact but not countably compact. 2 That is, closed and open at one time

4 Theorem 18. Every normal pseudocompact space is countably compact. Proof. Suppose X is normal and not countably compact. Then there is an infinite closed discrete D X. Select a countably infinite D 0 = {d n : n N} D. Define f 0 : D 0 R by f(d n ) = n for all n. Since D 0 is discrete, f 0 is a continuous function. By Tietze Theorem 53, f 0 extends to a continuous function f : X R. Clearly f is unbounded, so X is not pseudocompact. Corollary 19. A Ψ-space constructed using an infinite m.a.d. family is not normal. Theorem 20. Every metrizable space is perfectly normal. Proof. To see that a metric space is normal, Let (X, d) be metric and H and K two disjoint closed sets in X. Put U = {x X : d(x, H) < d(x, K)} and V = {x X : d(x, K) < d(x, H)}. To see the perfection, note that, given a closed set H, H is of the form H = n N O n where O n = {x X : d(x, H) < 1 n }, so a a G δ-set. Theorem 21. Every LOTS 3 is normal. If time allows, we will discuss the proof in class. 6. Some cardinal arithmetic Here we review some information on cardinal numbers without proofs. Definition 22. Two sets X and Y are called equipotent (or we say that X and Y have the same cardinality and write X = Y ) if there is a bijection from one of these sets onto the other. Thus we assign to every (finite or infinite) set a so called cardinal number (also finite or infinite). We write X = m etc. For finite sets, cardinalities are just natural numbers (including zero for the empty set). The cardinality of N is denoted by ℵ 0, the cardinality of R is denoted by c. Definition 23. Say that X Y if there is a injection from X into Y. Theorem 24. (Bernstein) Let X and Y be two sets. If there is an injecton from X to Y and there is an injection from Y to X then there is a bijection from X to Y. So if X Y and Y X then X = Y. If X Y and it is not true that X = Y then we write X < Y. Every two cardinal numbers are comparable 4 (that is, given cardinals m and n exactly one of the following three cases takes place: (1) m < n; (2) m = n; (1) m > n). Every set of cardinals has minimum. The minimum of cardinalities of uncountable sets is denoted by ℵ 1. We have ℵ 0 < ℵ 1 c. The assumption that ℵ 1 c is called the Continuum hypothesis (abbreviated as CH). It is known that CH is consistent with, and independent from, the standard set of axioms of set theory. 3 = Linearly Ordered Topological Space 4 This fact is non trivial!

5 If X and Y are disjoint sets such that X = m and Y = n, then m + n denotes the cardinality of the set X Y. If X and Y are non empty sets such that X = m and Y = n, then m n denotes the cardinality of the set X Y. Theorem 25. For every infinite cardinal m, m + m = m and m m = m. More generally, for every infinite cardinals m and n, m + n = max{m, n} and m n = max{m, n}. Recall that, given a set X, the power set of X, denoted by P(X), is the set of all subsets of X. 5 If X = m then P(X) is denoted by 2 m. 6 Theorem 26. For every cardinal number m, 2 m > m. It is known (and not difficult to see) that 2 ℵ0 = c. If X and Y are two sets, X = m and Y = n then the cardinality of all functions from X to Y is denoted by n m. It is known that c ℵ0 = c. 7 7. Some examples of non normal Tychonoff spaces Theorem 27. (Johns Lemma) If a topological space X contains a dense subspace D and an infinite closed discrete subspace Z such that 2 D < 2 Z then X is not normal. Proof. Suppose X were normal. It is clear that P(Z) \ {, Z} = P(Z). Let H P(Z) \ {, Z}. Then H and Z \ H are disjoint non empty closed subsets of X. Fix disjoint open sets U H, V H such that H U H and Z \ H V H. Put D H = D U H. We claim that D H D H whenever H H. Indeed, if H H then either (1) H \ H or (2) H \ H. Consider case 1. Pick z H \ H. Then z U H V H and thus U H V H is a non empty open set. Since D is dense we can pick d U H V H D. Then d U H D V H (U H D) (X \ U H ) (U H D) (X \ (U H D)) = D H (X \ D H ) = D H \ D H. Case 2 is similar. So H D H defines an injection of a set of cardinality 2 Z into a set of cardinality 2 D, a contradiction. Examples 28. The following spaces are (Tychonoff but) non normal: The Niemytzky plane; The square of the Sorgenfrey line. Indeed, each of these spaces is separable and contains a closed discrete subspace of cardinality c, and by Theorem 26, 2 c > c = 2 ℵ0. 8. Ordinal numbers, and more examples of non normal Tychonoff spaces Here we review some information on ordinal numbers without proofs. Recall that a linear order on a set X is a reflexive, transitive, and antisymmetric binary relation on X such that every two elements of X are comparable. 5 For example, if X = {a, b} then P(X) = {, {a}, {b}, {a, b}}. 6 The reason for this notation: one can associate with every A X the indicator function of A, denoted by χ A, a function from X to {0, 1} that equals 1 on A and 0 outside A. This provides a bijection between P(X) and the set of all functions from X to {0, 1} (note that the set {0, 1} is often denoted by 2). 7 We will use this in Section 13.

6 Definition 29. A linear order on X is called a well order if every non empty subset of X has minimum. Examples 30. The standard order on N is a well order. The standard order on Z or Q or R or I is not a well order. Exercise 31. (1) Is the lexicographic order on N N a well order? (2) Is the lexicographic order on {0, 1} N a well order? Theorem 32. (Zermelo) On every set, there is a relation that well orders it. Theorem 32 follows from (in fact, is equivalent to) the Axiom of Choice. Definition 33. Two ordered sets (X, <) and (Y, ) are called order-isomorphic if there is a bijection f : X Y such that for every a, b X, a < b iff f(a) f(b). Proposition 34. Let (X, <) and (Y, ) be two well ordered sets. Then exactly one of the following three cases takes place: (X, <) is order isomorphic to some initial interval of (Y, ). (X, <) and (Y, ) are order isomorphic. (Y, ) is order isomorphic to some initial interval of (X, <). Remark 35. Note that the previous proposition establishes a specific property of well ordered sets. For example, neither of the two (just) linearly ordered sets R and N is order isomorphic to an initial interval of the other. If two well ordered sets are order isomorphic, then we say that they are of the same order type. Order types of well ordered sets are called ordinal numbers (or simply ordinals). It follows from Proposition 34 that every two ordinals are comparable. Also it follows from the previous that every non empty set of ordinals has minimum. Examples 36. Let n N. The order type of the set {0,..., n 1} (with the standard order) is denoted by n. 8 The order type of N (with the standard order) is denoted by ω. ω is the smallest infinite ordinal. Consider a well ordered set that consists of N (with the standard order) and one more element which is greater than every element of N. The order type of this well ordered set is denoted by ω + 1. 9 Then follows ω + 2, ω + 3,...ω + ω,... The smallest order type of an uncountable set is denoted by ω 1. Then follows ω 1 + 1.... etc.... Definition 37. A subset A of a linearly ordered set (X, <) is called cofinal if for every x X there is a A with x a. n, considered as an ordinal, has a one-element cofinal sub- Examples 38. set {n 1}. 8 In this sense, n = {0,..., n 1}, and in general, every ordinal is, in fact, the set of all ordinals smaller than itself. 9 Exercise: show that ω + 1, considered with the order topology, is homeomorphic to the space convergent sequence together with its limit S = {1/n : n N} {0} (with the topology inherited as a subspace of R).

7 No finite subset of ω is cofinal but every infinite subset is. ω + 1, as well as ω 1 + 1 has a one-element cofinal subset. Proposition 39. Every cofinal subset of ω 1 is uncountable. Proof. Since ω 1 is the minimum order type of an uncountable set, every initial interval of ω 1 is countable (otherwise we would have a contradiction with Proosition 34). So, if ω 1 would have a countable cofinite subset, say A, then ω 1 will be the union of countably many coutable subsets (initial intervals with the right endpoints at a A) and thus would be countable - a contradiction. Lemma 40. For every point a ω 1, [0, a] is a neighborhood of a. 10 Proof: Indeed, [0, a] = [0, a + 1). Proposition 41. ω 1 is first countable. Proof. Let a ω 1. Then [0, a] is countable. Then {(c, a] : c < a} is a countable base of neighborhoods of a. Proposition 42. ω 1 is sequentially compact (hence countably compact). Proof. Let ξ = {a n : n N} be a sequence of elements of ω 1. Then ξ is not cofinal in ω 1 and thus the set M = {b ω 1 : infinitely many a n are b} is non empty. Since ω 1 is well ordered, M has minimum, call it m. Then some subsequence of ξ converges to m (because every neighborhood of m contains infinitely many elements of ξ and ω 1 is first countable). Lemma 43. Every strictly decreasing sequence of ordinals is finite. Proof. Otherwise we would have a non empty subset in a well ordered set which does not have minimum. Proposition 44. ω 1 + 1 is compact. Proof. If time allows, we will discuss proof in class. Example 45. Note that ω 1 is a proper dense subset (hence not a closed subset) in ω 1 + 1. Therefore ω 1 is another example of a countably compact space which is not compact. Example 46. (The Tychonoff plank) Consider X = (ω 1 +1) (ω+1)\{ ω 1, ω } as a subspace of (ω 1 + 1) (ω + 1) where ω 1 + 1 and ω + 1 bear the order topology and the product bears the Tychonoff product topology. 11 Then X is Tychonoff but non normal. The two bad disjoint closed sets are H = { ω 1, n : n ω} and K = { a, ω : a ω 1 }. In class, we will discuss the proof that H and K can t be separated by open sets. 10 Here and henceforward, when discussing the topological properties of ordinals or their subsets we mean the order topology. 11 The notation here is difficult to grasp for the first time. Note that ω is the greatest element of ω + 1 (just like n is the greatest element of n + 1 = {0,..., n}), and ω 1 is the greatest element of ω 1 + 1.

8 9. The Tychonoff embedding Theorem Definition 47. The weight w(x) of a topological space X is the minimum cardinality of a base of X. Thus X is second countable iff w(x) = ℵ 0. Definition 48. Say that a family F of functions from X to R separates points if for every distinct x, y X there is f F such that f(x) f(y). separates points from closed sets if for every x X and every non empty closed set H X such that x H there is f F such that f(x) = 0 and f(h) = {1}. 12 It is clear that if X is T 1 and F separates points from closed sets, then F separates points. Proposition 49. If X is a Tychonoff space then there is a family F of continuous functions from X to I such that: F separates the points of X from closed sets; F = w(x). Proof. Let w(x) = τ and let B be a base of X such that B = τ. Call an ordered pair U, V good if U V and there is a continuous function f : X I such that f(u) = {0} and f(x \ V ) {1}. Denote the family of all good pairs by G. For every U, V G, fix a continuous function f U,V : X I such that f U,V (U) = {0} and f U,V (X \ V ) {1}. Put F = {f U,V : U, V G}. Then F = G τ. We claim that F separates points from closed sets. Let x X and let H be a closed subset of X such that x H. Since B is a base for X there is V B such that x V X \H. Since X is Tychonoff there is a continuous function g : X I such that g(x) = 0 and g(x \ V ) = {1}. Since g 1 ([0, 1/2)) is a neighborhood of x and B is a base for X, there is U B such that x U g 1 ([0, 1/2)). Define f : X I by { 0 if g(y) 1/2 f(y) = 2(g(y) 1/2) if g(y) 1/2. Then f is continuous, f(u) = {0} and f(x \ V ) = {1}. So U, V G. Then f U,V F, f U,V (x) = 0 (since x U) and f U,V (H) = {1} (since H X \ V ). 13 Recall 14 that, given a family F = {f a : a A} where f a : X Y a, the diagonal product F : X a A Y a = P is defined by π a ( F(x)) = f a (x) for all a A. If all f a are continuous then F is continuous with respect to the Tychonoff product topology on P. Note that if F separates points, then F is an injection. 12 A variation of this definition says that F separates points from closed sets if for every x X and every non empty closed set H X such that x H there is g F such that g(x) g(h). Having such a function g, put h(y) = g(y) f(x). This defines a function h : X R such that h(x) = 0 h(h). Put m = min{ z : z h(h)}. Then m > 0. Define f : x R by f(y) = h(y) /m if h(y) < m and f(y) = 1 if h(y) m. Then f is like in the definition. Note that if g is continuous then so is f. Also note that instead of f : X R one can consider f : X I where I is the unit interval. 13 Formally, we have proved so far only that F w(x). But note that the family of sets {f 1 ([0, 1)) : f F} is a base for X, so the cardinality of F can t be less than w(x). 14 See notes on mappings

9 Theorem 50. (The Tychonoff Embedding Theorem) Every Tychonoff topological space X is homeomorphic to a subspace of I w(x). Proof. Let w(x) = τ. By Proposition 49 there is a family of mappings F = {f a : a A} from X to I such that F = A = τ and F separates the points of X from closed sets. Put f = F : X I A. 15 We claim that f is an homeomorphic embedding. Being a diagonal product of continuous mappings, f is continuous. It is clear that F separates points, so F is an injection. It remains 16 to verify that for every open set O X, the image f(o) is an open subset of f(x) (in the subspace topology inherited from I τ ). Let y f(o). Pick x f 1 (y) O. Since F separates points from closed sets there is f a F such that f a (x) = 0 and f a (X \ O) = {1}. Put W = πa 1 ([0, 1)). Then W is an open set in I A and W f(x) f(o). So, every point y of f(o) is contained in f(o) together with some open neighborhood (in the subspace topology) and thus f(o) is open. The powers I τ are often called Tychonoff cubes. So Tychonoff spaces are, up to a homeomorphism, subspaces of Tychonoff cubes. Remark 51. Being products of compact spaces, Tychonoff cubes are compact hence by Theorem 16 normal. So every Tychonoff space is (homeomorphic to) a subspace of some normal space. This provides a plenty of examples of non normal subspaces of normal spaces. 10. The Urysohn Metrization Theorem Theorem 52. (The Urysohn Metrization Theorem) Every regular second countable space is metrisable. Proof. Let X be a regular second countable space. Being second countable, X is Lindelöf. Being regular and Lindelöf by Theorem 14 X is normal hence Tychonoff. Hence by Theorem 50 X is homeomorphic to a subspace of I ℵ0. Being the product of countably many metrizable spaces, I ℵ0 is metrizable, and thus so is every its subspace. So, being homeomorphic to a metrizable space, X is metrizable. 11. Proof of Urysohn s Lemma We have to prove: X is normal for every pair of disjoint closed sets H, K X, there is a continuous function f : X R such that f(h) = {0} and f(k) = {1}. is easy: having sauch a function f, put U = f 1 ([0, 1/2)) and V = f 1 ((1/2, 1]). Then U and V are disjoint neighborhoods of H and K. To show, let H and K be non empty disjoint closed subsets of a normal space X. Put enumerate r 0 = 0, r 1 = 1, and enumerate all rational numbers in the open interval (0, 1) as r 2, r 3,... Using the normality of X and the criterium of normality from Proposition 2 we can inductively define the open sets U r for r = r 0, r 1, r 2,... so that U r U r whenever r, r {r 0, r 1, r 2,...} and r < r. To start, we find U 0, U 1 so that H U 0 U 0 U 1 = X \ K and then continue using Proposition 2. open 15 I A takes the role of I w(x) 16 Because a homeomorphism is a mapping that has three properties: bijection, continuous,

10 Having all sets U r constructed, define f : X I R by { inf{r : x Ur }, if x U f(x) = 1 ; 1, if x X \ U 1. Then obviously f(h) = {0} and f(k) = {1}. It remains to show that f is continuous. To show this, it suffices to check that (*) f 1 (W ) is open for every interval W = (p, q) R. And to check (*) it suffices to verify that for every p and q, the sets f 1 ((, q)) and f 1 ((p, )) are open (and obviously, only p, q [0, 1] are of interest). Note that f 1 ((, q)) = {x : inf{r : x U r } < q} = r<q U r which is obviously open. On the other hand, let x f 1 ((p, )). Then p < f(x). Pick rational r, r (0, 1) so that p < r < r < f(x). Then x X \ U r X \ U r f 1 ((p, )). Note that X \ U r is open. It follows that f 1 ((p, )) is open. 12. The Tietze Extension Theorem Theorem 53. Let X be a normal space, Z a non empty subspace of X, and f : Z I or f : Z R a continuous function. Then there is a continuous function g from X to I or R respectively such that g Z = f. 17 The proof is a more complicated variation of the proof of Urysohn s Lemma. If time allows, we will discuss it in class (drawing pictures helps.) Remark 54. In some texts, the Tietze Extension Theorem is called the Tietze- Urysohn Extension Theorem. Urysohn s Lemma can be treated as a partial case of the Tietze-Urysohn Extension Theorem. Indeed, consider Z = H K and consider f : Z I given by f(z) = 0 for z H and f(z) = 1 for z K. Then f is continuous since H and K are disjoint closed sets. 13. C-embedded subspaces Definition 55. A subspace Z of a topological space X is called C-embedded in X if for every continuous function f : Z R there is a continuous function g : X R such that g Z = f. So Tietze Theorem 53 says that every closed subspace of a normal space is C- embedded. Exercise 56. Show that Tietze Theorem 53 can be inverted: if a space is not normal, then there is a closed subspace which is not C-embedded. For a topological space X, C(X) denotes the set of all continuous functions from X to R. Proposition 57. If X is a separable topological space, then C(X) c. Proof. Let D be a dense countable subspace of X. If f 1, f 2 C X and f 1 D = f 2 D, then f 1 = f 2. 18 Therefore C(X) = C(D) R D = c ℵ0 = c. Example 58. The bottom line B in the Niemytzky plane N is (a closed subspace but) not C-embedded in N. 17 That is g(z) = f(z) for every z Z. 18 This was one of homework exercises.

11 Proof. B is discrete, so C(B) = R B and thus C(B) = R B = c c = (2 ℵ0 ) c = 2 ℵ0 c = 2 c > c. On the other hand, N is separable, so C(N) c. Since there are more elements in C(B) than in C(N), not every functions extends continuously from B to N. Theorem 59. Let X be a Tychonoff space and K a compact subspace of X. Then K is C-embedded in X. Proof. By Theorem 50 there is a homeomorphic embedding f : X I τ where τ = w(x). Put X = f(x) and K = f(k). It suffices to show that K is C- embedded in X. I τ is compact, hence normal. Being compact, K is closed in the normal (hence Hausdorff) space I τ hence by Theorem 53 C-embedded in I τ, hence C-embedded in X. Exercise 60. Show that for every topological spaces X and Y and for every x X, the subspace {x} Y is C-embedded in X Y. 14. Collectionwise normality This section is optional. We are going to show that metrizable spaces are much better than just normal in another direction than being perfectly normal. Definition 61. A family F of subsets of a space (X, T ) is called discrete if for every x X there is a neighborhood U x such that U x intersects at most one element of the family F. Exercise 62. (1) Note that every discrete family of sets is pairwise disjoint. (2) Show that any finite family of pairwise disjoint closed sets in any space is discrete. (3) Note that any subfamily of a discrete family of sets is discrete. (4) Show that if F is a discrete family of closed sets, then F is closed. Definition 63. A space (X, T ) is called collectionwise normal if for every discrete family F of closed sets in X, one can find open U F F for all F F so that the family {U F : F F} is pairwise disjoint. Obviously, every collectionwise normal space is normal. Proposition 64. A space (X, T ) is collectionwise normal if for every discrete family F of closed sets in X, one can find open V F F for all F F so that the family {V F : F F} is discrete. Proof. Let U F be like in the definition of collectionwise normality. Then H = F is closed in X and U = F F U F is a neighborhood of of H. By normality there is an open set W such that H W W U. Put V F = U F W. Then the family {V F : F F} is discrete. Indeed, for x X there are only two possibilities: x U F for some (only one!) F. Then U F is a neighborhood of X that intersects exactly one element of {V F : F F} (namely, V F ). x X \ F F U F. Then X \ W is a neighborhood of x that does not intersect any element of {V F : F F}.

12 The definition of collectionwise normality can be expressed like this: every discrete family of closed sets can be expanded to a disjoint family of open sets. Proposition 64 can be expressed like this: every discrete family of closed sets can be expanded to a discrete family of open sets. Theorem 65. Every metrizable space is collectionwise normal. Sketch of proof. Let F be a discrete family of non empty closed sets in X. The case when F consists of one element is trivial, so assume F 2. For each F F, put F = (F \ {F }). Then F is a closed set disjoint from F. Put U F = {x X : d(x, F ) < d(x, F )}. 15. Monotone normality This section is optional. There is yet one more direction in which metrizable spaces are better than just normal. In the definition of normality of X, we assign to every pair of 19 non empty disjoint closed sets H, K X disjoint neighborhoods U H and V K. This assignment is being made separately for each pair H, K. Here is a (looking pretty natural) attempt to make such an assignment more systematic: Definition 66. A space X is called monotonically normal if one can assign to every ordered pair H, K of non empty disjoint closed sets a pair of open sets U(H, K), V (H, K) such that: U(H, K) H and V (H, K) K; U(H, K) V (H, K) = ; If H H and K K then U(H, K ) U(H, K) and V (H, K ) V (H, K). Theorem 67. Every metrizable space is monotonically normal. Proof. Put U(H, K) = {x X : d(x, H) < d(x, K)} and v(h, K) = {x X : d(x, H) > d(x, K)} Exercise 68. Show that every T 1 space having at most one non isolated point is monotonically normal. Many nice space are monotonically normal, and yet this property is pretty restrictive. Some information without proofs: Every LOTS is monotonically normal; I ℵ1 is not monotonically normal. 19 Let s restrict to