Download 12 | Urysohn Metrization Theorem

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) → R given 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.
2) Not every continuous 1-1 function is an embedding. For example, take N = {0� 1� 2� � � � } with the
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
U1 × U2
X1 × X2
U1
X1
2) In general if X1 � � � � � X� are 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� }∞
is an infinitely countable family of topological spaces then the basis of the product
�=1�
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.
12.10 Proposition. If {X� }∞
�=1 is a countable family of metrizable spaces then
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:
�∞ ((�� )� (��� ))
The topology induced by the metric �∞ on
�∞
∞
�
1
�� (�� � ��� )
=
2�
�=1 X�
�=1
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] :
∞
�
ℵ0
[0� 1] =
[0� 1]
�=1
[0� 1]ℵ0
Elements of
are infinite sequences (�� ) = (�1 � �2 � � � � ) where �� ∈ [0� 1] for � = 1� 2� � � � The
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]ℵ0 .
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
�� : X → [0� 1]. We say that the family {�� }�∈I separates points from closed sets if for any point �0 ∈ X
and any closed set A ⊆ X such that �0 �∈ A there is a function �� ∈ {�� }�∈I such that �� (�0 ) > 0 and
�� |A = 0.
12.14 Embedding Lemma. Let X be a T1 -space. If {�� : X → [0� 1]}�∈I is a family that separates points
from closed sets then the map
�
�∞ : X →
[0� 1]
given by �∞ (�) = (�� (�))�∈I is an embedding.
�∈I
12.15 Note. If the family {�� }�∈I in 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 (�� �) ∈ S then 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 {��� }(���)∈S separates 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
�
�∞ : X →
[0� 1]
(���)∈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] ∼
=
N
N
ℵ
0
[0� 1] for some N ≥ 0 and [0� 1] can be identified with a subspace of [0� 1] .
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
�� ∈ {�� }�∈I such that �� (�) > 0 and �� (�) = 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 �� ∈ {�� }�∈I be 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 ) > 0 we 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
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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.
E12.5 Exercise. Assume that X � Y are spaces such that R ∼
= X × Y . Show that either X or Y is
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 Hausdorff 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� 2 let �� : 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� � � �
�
�
let (X� � ��
� ) be a metric space such that �� (�� � ) ≤ 1 for all �� � ∈ X� . Let �∞ be a metric the Cartesian
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
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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 � U2 � � � � } and {V1 � V2 � � � � } such that
�
�∞
(i) A ⊆ ∞
�=1 U� and B ⊆
�=1 V�
(ii) for all � ≥ 1 we have U � ∩ B = ? and V � ∩ A = ?
b) For � ≥ 1 define
U�� := U� r
�
�
�=1
V�
and
V�� := V� r
�
�
�=1
U�
�
�∞
�
�
�
�
�
�
�
Let U � = ∞
�=1 U� and V =
�=1 V� . Show that U and V are open sets, that A ⊆ U and B ⊆ V ,
�
�
and that U ∩ V = ?.