Lecture notes, lecture 12 PDF

Preview

Summary

Download Lecture notes, lecture 12 PDF

Description

12 | Urysohn Metrization Theorem

Recall that a topological space X is metrizable if there exists a metric � on X such that the topology on X is induced by � . We have seen already that any metrizable space must be normal, but that not every normal space is metrizable. Our goal here is to show that if a normal space space satisfies one extra condition then it is metrizable. Recall that a space X is second countable if it has a countable basis. We have: 12.1 Urysohn Metrization Theorem. Every second countable normal space is metrizable. The main idea of the proof is to show that any space as in the theorem can be identified with a subspace of some metric space. To make this more precise we need the following: 12.2 Definition. A continuous function � : X → Y is an embedding if its restriction � : X → �(X )is a homeomorphism (where �(X ) has the topology of a subspace of Y ). 12.3 Example. The function � : (0� 1) → Rgiven by �(�) = �is an embedding. The function � : (0� 1) → R given by �(�) = 2� is another embedding of the interval (0� 1) into R. 12.4 Note. 1) If � : X → Y is an embedding then � must be 1-1. } the 2) Not every continuous 1-1 function is an embedding. For example, take N = {0� 1� 2� � � �with discrete topology, and let � : N → R be given � 0 if � = 0 � (�) = 1 if � > 0 � 74

12. Urysohn Metrization Theorem

75

The function � is continuous and it is 1-1, but it is not an embedding since � : N → � (N) is not a homeomorphism. 12.5 Lemma. If � : X → Y is an embedding and Y is a metrizable space then X is also metrizable. Proof. Let µ be a metric on Y. Define a metric � on X by �(�1 � �2 ) = µ(�(�1 )� �(�2.))It is easy to check that the topology on X is induced by the metric � (exercise). Let now X be a space as in Theorem 12.1. In order to show that X is metrizable it will be enough to construct an embedding � : X → Y where Y is metrizable. The space Y will be obtained as a product of topological spaces: � 12.6 Definition. Let {X� }�∈I be a family of topological spaces. The product topology on �∈I X� is the topology generated by the basis �� � B= �∈I U� | U� is open in X� and U� �= X� for finitely many indices � only 12.7 Note. 1) If X1 , X2 are topological spaces then the product topology on X1 × X2 is the topology induced by the basis B = {U1 × U2 | U1 is open in X1 , U2 is open in X2 }. X2 U2

X1 × X2 U1 × U2

U1

X1

2) In general if X1 � � � � �� Xare topological spaces then the product topology on X1 × · · · × X� is the topology generated by the basis B = {U1 × · · · × U� | U� is open in X� }. ∞ 3) If {X� }�=1 �is an infinitely countable family of topological spaces then the basis of the product topology on ∞�=1 X� consists of all sets of the form

U1 × · · · × U� × X�+1 × X�+2 × X�+3 × � � �

where � ≥ 0 and U� ⊆ X� is an open set for � = 1� � � � � �. 12.8 Proposition. Let {X� }�∈I be a family of topological spaces and for � ∈ I let � X� → X� �� : �∈I

be the projection onto the �-th factor: �� ((�� )�∈I ) = �� . Then:

12. Urysohn Metrization Theorem

76

1) for any � ∈ I the function �� is continuous. � 2) A function � : Y → �∈I X� is continuous if and only if the composition �� � : Y → X�is continuous for all � ∈ I Proof. Exercise. 12.9 Note. Notice that the basis B given in Definition 12.6 consists of all sets of the form −1 �−1 �1 (U1 ) ∩ · · · ∩ ��� (U�� )

where �1 � � � ���∈ � I and U�1 ⊆ X�1 � � � � ���U⊆ X�� are open sets. ∞ is a countable family of metrizable spaces then 12.10 Proposition. If {X� }�=1 metrizable space.

�∞

�=1

X� is also a

Proof. Let �� be a metric on X�. We can assume that for any �� � � ∈ X� we have �� (�� � � ) ≤ 1. Indeed, if �� does not have this property then we can replace it by the metric �� � given by: � �� (�� � � ) if �� (�� � � ) ≤ 1 � ��� (�� � ) = 1 otherwise The metrics �� and ��� are equivalent (exercise), and so they define the same topology on the space X�. � Given metrics �� on X� satisfying the above condition define a metric �∞ on ∞�=1 X� by: �∞ 1 �� (�� ���)� �∞ ((�� )� (�� )) = 2� �

�=1

The topology induced by the metric �∞ on

�∞

�=1

X� is the product topology (exercise).

12.11 Example. The Hilbert cube is the topological space [0� 1]ℵ0 obtained as the infinite countable product of the closed interval [0� 1] : �∞ [0� 1] [0� 1]ℵ0 = �=1

[0� 1]ℵ0

) �� ∈ [0� 1]for � = 1� 2� � �The � Elements of are infinite sequences (�� ) = (�1 � 2�� � � �where Hilbert cube is a metric space with a metric � given by �∞ 1 |�� − �� | �((�� )� (�� )) = 2� �=1

Theorem 12.1 is a consequence of the following fact:

12. Urysohn Metrization Theorem

77

12.12 Theorem. If X is a second countable normal space then there exists an embedding� : X → [0� 1]ℵ.0 Theorem 12.12 will follow in turn from a more general result on embeddings of topological spaces: 12.13 Definition. Let X be a topological space and let {�� }�∈I be a family of continuous functions . We say that the family {�� }�∈Iseparates points from closed sets if for any point �0 ∈ X �� : X → [0� 1] and any closed set A ⊆ X such that �0 �∈ A there is a function �� ∈ {�� }�∈Isuch that �� (�0 ) > 0and �� |A = 0. 12.14 Embedding Lemma. Let X be a T1-space. If {�� : X → [0� 1]}�∈Iis a family that separates points from closed sets then the map � �∞ : X → [0� 1] �∈I

given by �∞ (�) = (�� (�))�∈I is an embedding. 12.15 Note. If the family {�� }�∈Iin Lemma 12.14 is infinitely countable then �∞ is an embedding of X into the Hilbert cube [0� 1]ℵ0 . We will show first that Theorem 12.12 follows from Lemma 12.14, and then we will prove the lemma. ∞ Proof of Theorem 12.12. Let B = {V� }�=1 be a countable basis of X , and let S the set given by

S := {(�� � ) ∈ Z+ × Z+ | V � ⊆ V� } If (�� � ) ∈ Sthen the sets V � and X r V� are closed and disjoint, so by the Urysohn Lemma 10.1 there is a continuous function ��� : X → [0� 1] such that � 1 if � ∈ V� ��� (�) = 0 if � ∈ X r V� We will show that the family {��� }(���)∈Sseparates points from closed sets. Take �0 ∈ X and let A ⊆ X ∞ be an closed set such that �0 �∈ A. Since B = {V� }�=1 is a basis of X there is V� ∈ B such that �0 ∈ V� and V� ⊆ X r A. Using Lemma 10.3 we also obtain that there exists V� ∈ B such that �0 ∈ V� and V � ⊆ V� . We have ��� (�0 ) = 1. Also, since A ⊆ X r V� we have ��� |A = 0. By the Embedding Lemma 12.14 the family {�� � }(��� )∈S defines an embedding � [0� 1] �∞ : X → (���)∈S

� The set S is countable. If it is infinite then (���)∈S [0� 1]=∼[0� 1]ℵ0. If S is finite then (��� )∈S [0� 1]=∼ [0� 1]N for some N ≥ 0 and [0� 1]N can be identified with a subspace of [0� 1]ℵ0 . �

12. Urysohn Metrization Theorem

78

Proof of Theorem 12.1. Follows from Theorem 12.12, Lemma 12.5, and the fact that the Hilbert cube is a metric space (12.11). It remains to prove Lemma 12.14: Proof of Lemma 12.14. We need to show that the function �∞ satisfies the following conditions: 1) �∞ is continuous; 2) �∞ is 1-1; 3) �∞ : X → �∞ (X ) is a homeomorphism. � 1) Let �� : �∈I [0� 1] → [0� 1]be the projection onto the �-th coordinate. Since �� �∞ = �,� thus �� �∞ is a continuous function for all � ∈ I. Therefore by Proposition 12.8 the function �∞ is continuous. 2) Let �� � ∈ X, � �= �. Since X is a T1-space the set {�} is closed in X . Therefore there is a function �� ∈ {�� }�∈Isuch that �� (�) > 0and �� (�) = 0. In particular �� (�) �= �� (�) . Since �� = �� �∞this gives �� �∞ (�) �= �� �∞ (�). Therefore �∞ (�) �= �∞ (�). 3) Let U ⊆ X be an open set. We need to prove that the set �∞ (U) is open in � (X ) . It will suffice to show that for any �0 ∈ U there is a set V open in �∞ (X ) such that �∞ (�0 ) ∈ V and V ⊆ �∞ (U). � Given �0 ∈ U let �� ∈ {�� }�∈Ibe a function such that �� (�0 ) > 0 and �� |X rU = 0. Let �� : �∈I [0� 1] → [0� 1] be the projection onto the �-th coordinate. Define V := �∞ (X ) ∩ ��−1 ( (0� 1] ) The set V is open in �∞ (X ) since �−1 � ( (0� 1] ) is open in

�

�∈I

[0� 1] . Notice that

V = {�∞ (�) | � ∈ X and �� �∞ (�) > 0} Since �� �∞ (�0 ) = �� (�0 ) > 0we have � (�0 ) ∈ V. Finally, if �∞ (�) ∈ V then �� (�) > 0 which means that � ∈ U, and so �∞ (�) ∈ �∞ (U). This gives V ⊆ �∞ (U). One can show that the following holds: 12.16 Proposition. Every second countable regular space is normal. Proof. Exercise. As a consequence Theorem 12.1 can be reformulated as follows: 12.17 Urysohn Metrization Theorem (v.2). Every second countable regular space is metrizable.

12. Urysohn Metrization Theorem

79

Exercises to Chapter 12 E12.1 Exercise. Show that the product topology on R� = R × · · · × R is the same as the topology induced by the Euclidean metric. � E12.2 Exercise. Let {X� }�∈I be a family of topological spaces. The box topology on �∈I X� is the topology generated by the basis �� � B= �∈I U� | U� is open in X�

Notice that for products of finitely many spaces the box topology is the same as the product topology, but that it differs if we take infinite products. � Let X = ∞ �=1 [0� 1] be the product of countably many copies of the interval [0� 1]. Consider X as a topological space with the box topology. Show that the map � : [0� 1] → X given by � (�) = (�� �� �� � � � ) is not continuous. E12.3 Exercise. Prove Proposition 12.8 E12.4 Exercise. Let X and Y be non-empty topological spaces. Show that the space X × Y is connected if and only if X and Y are connected. ∼ X × Y . Show that either X or Y is E12.5 Exercise. Assume that X � Y are spaces such that R = consists of only one point. E12.6 Exercise. Let X � Y be topological spaces. For a (not necessarily continuous) function � : X → Y the graph of � is the subspace Γ(� ) of X × Y given by Γ(� ) = {(�� � (�)) ∈ X × Y | � ∈ X } Show that if Y is a Hausdorf f space and� : X → Y is a continuous function then Γ(� ) is closed in X × Y. E12.7 Exercise. Let X1 , X2 be topological spaces, and for � = 1� 2let �� : X1 × X2 → X� be the projection map. a) Show that if a set U ⊆ X1 × X2 is open in X1 × X2 then �� (U) is open in X� .

b) Is it true that if A ⊆ X1 × X2 is a closed set then �� (A) must be closed is X� ? Justify your answer. E12.8 Exercise. The goal of this exercise is to complete the proof of Proposition 12.10. For � = 1� 2� � � � �� ) be a metric space such that �� (�� � � ) ≤ 1 for all �� � � ∈ X�. Let �∞ be a metric the Cartesian let (X� � � product ∞�=1 X� given by �∞ 1 � = �∞ ((�� )� (�� )) �� (�� ���)� 2� �=1

Show that the topology defined by �∞ is the same as the product topology.

12. Urysohn Metrization Theorem

80

E12.9 Exercise. The goal of this exercise is to give a proof of Proposition 12.16. Let X be a second countable regular space and let A� B ⊆ X be closed sets such that A ∩ B = ?. a) Show that there exist countable families of open sets {U1 � U 2 � � � � } and {V1 �2 �V� � � } such that �∞ �∞ (i) A ⊆ �=1 V� U� and B ⊆ �=1 (ii) for all � ≥ 1 we have U � ∩ B = ? and V � ∩ A = ? b) For � ≥ 1 define U�� := U� r

� �

V�

and

V�� := V� r

Let and

and ?.

V�

=

�∞

� �=1 V�.

Show that

U�

�=1

�=1

�∞

= �=1 U�� that U � ∩ V � =

U�

��

U�

and

V�

are open sets, that A ⊆ U � and B ⊆ V � ,...