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14 | Compact Spaces
14 | Compact Spaces
14.1 Definition.
� Let X be a topological space. A cover of X is a collection Y = {Y� }�∈I of subsets of
X such that �∈I Y� = X .
X
Y3
Y1
Y2
Y1
Y2
14.1 Definition.
� Let X be a topological space. A cover of X is a collection Y = {Y� }�∈I of subsets of
X such that �∈I Y� = X .
Y3
X
If the sets Y� are open in X for all � ∈ I then Y is an open cover of X . If Y consists of finitely many
sets then Y is a finite cover of X .
14.2 Definition. Let Y = {Y� }�∈I be a cover of X . A subcover of Y is cover Y� of X such that every
element of Y� is in Y.
If the sets Y� are open in X for all � ∈ I then Y is an open cover of X . If Y consists of finitely many
sets then Y is a finite cover of X .
14.3 Example. Let X = R. The collection
14.2 Definition. Let Y = {Y� }�∈IY be
a cover
ofRX|. ��
A �subcover
of
Y
is cover Y� of X such that every
=
{(��
�)
⊆
∈
Z�
�
<
�}
element of Y� is in Y.
is an open cover of R, and the collection
14.3 Example. Let X = R. The collection
Y� = {(−�� �) ⊆ R | � = 1� 2� � � � }
Y = {(�� �) ⊆ R | �� � ∈ Z� � < �}
is a subcover of Y.
is an open cover of R, and the collection
14.4 Definition. A space X is compact
if every open cover of X contains a finite subcover.
Y� = {(−�� �) ⊆ R | � = 1� 2� � � � }
14.5 Example. A discrete topological space X is compact if and only if X consists of finitely many
is a subcover of Y.
points.
14.4 Definition. A space X is compact if every open cover of X contains a finite subcover.
87
14.5 Example. A discrete topological space X is compact if and only if X consists of finitely many
points.
87
92
Y = {(�� �) ⊆ R | �� � ∈ Z� � < �}
is an open cover of R, and the collection
is a subcover of Y.
Y� = {(−�� �) ⊆ R | � = 1� 2� � � � }
14.4 Definition. A space X is compact if every open cover of X contains a finite subcover.
14.5 Example. A discrete topological space X is compact if and only if X consists of finitely many
points.
87
93
�� . We have:
14.7
Example.
The real line R is not compact since the open cover
� ∈U
1 ∪ ··· ∪ U
X {0}
= U∪
�0 ∪
X=
{U
� �1| � = 1� 2� ��N� � }
Y = {(� − 1� � + 1) ⊆ R | � ∈ Z}
so {U�0 � U�1 � � � � � U�N } is a finite subcover of U.
The space
X is compact. Indeed, let U = {U� }�∈I be any open cover of X and let 0 ∈ U0 . Then there
1
does not
any finite
exists
N have
> 0 such
that subcover.
� ∈ U�0 for all � > N. For � = 1� � � � � N let U�� ∈ U be a set such that
14.7
Example. The real line
R is not compact since the open cover
1
� ∈ U�� . We have:
function.
If X is compact and � is onto then Y is
14.8 Proposition. Let � : X → Y be aXcontinuous
=−U1�
· ·R
· ∪| U
�0 ∪
Y = {(�
�U
+�11)∪⊆
��N∈ Z}
compact.
so {U�0 � U�1 � � � � � U�N } is a finite subcover of U.
does not have any finite subcover.
Proof.Example.
Exercise.The real line R is not compact since the open cover
14.7
14.8 Proposition. Let � : X → Y be a continuous function. If X is compact and � is onto then Y is
compact.
14.9
Corollary. Let � : X → Y be Y
a continuous
function.
⊆∈
X Z}
is compact then �(A) ⊆ Y is compact.
= {(� − 1� �
+ 1) ⊆ If
R A| �
does
have
any finite
subcover.
Proof.not
Exercise.
The
function
�|A : A
→ �(A) is onto, so this follows from Proposition 14.8.
Let
X be
→
Y abecontinuous
a continuous
If X
X is
isand
compact
and
� Y
is ⊆
onto
Y is
14.8
Proposition.
14.9
� :X�X, : Y→
Y topological
be
function.
A⊆
compact
then
�(A)
Ycompact.
isthen
compact.
14.10Corollary.
Corollary.Let
Let
spaces.
Iffunction.
X isIf compact
Y ∼
then
is
=X
compact.
Proof. The
function
is onto, so this follows from Proposition 14.8.
Follows
from �|
Proposition
14.8.
A : A → �(A)
Proof. Exercise.
14.10 Example.
Corollary. For
Let any
X , Y� be
spaces. (��
If X�)is⊆compact
Y ∼
then(��
Y �)
is ∼
compact.
= Xsince
14.11
< �topological
the open interval
R is notand
compact
R.
=
14.9 Corollary. Let � : X → Y be a continuous function. If A ⊆ X is compact then �(A) ⊆ Y is compact.
14.12
any � < 14.8.
� the closed interval [�� �] ⊆ R is compact.
Proof. Proposition.
Follows from For
Proposition
Proof. The function �|A : A → �(A) is onto, so this follows from Proposition 14.8.
14.11
Foropen
any cover
� < �ofthe
(�� �) ⊆ R is not compact
since (�� �) ∼
= R.
Proof. Example.
Let U be an
[��open
�] andinterval
let
∼
14.10 Corollary. Let X , Y be topological spaces. If X is compact and Y = X then Y is compact.
A = {� ∈ [�� �]
[�� �]
can be
covered
aR
finite
number of elements of U}
14.12 Proposition.
For| the
any interval
� < � the
closed
interval
[�� by
�] ⊆
is compact.
Proof. Follows from Proposition 14.8.
Let �0 := sup A.
Proof. Let U be an open cover of [�� �] and let
14.11
For any
� the open interval (�� �) ⊆ R is not compact since (�� �) ∼
= R.
Step 1.Example.
We will show
that� �<
0 > �. Indeed, let U ∈ U be a set such that � ∈ U. Since U is open we
the interval
�] can be
covered
finite
of elements
have [��A�=+{�ε)∈⊆[��
U �]
for| some
ε > 0. [��
It follows
that
� ∈ A by
for aall
� ∈ number
[�� � + ε).
Thereforeof�0U}
≥ � + ε.
14.12 Proposition. For any � < � the closed interval [�� �] ⊆ R is compact.
Let �0 := sup A.
Proof.
be an
open
cover
�] and let
let U ∈ U be a set such that � ∈ U. Since U is open we
Step 1.Let
WeUwill
show
that
�0 >of�.[��Indeed,
have [�� � + ε) ⊆ U for some ε > 0. It follows that � ∈ A for all � ∈ [�� � + ε). Therefore �0 ≥ � + ε.
A = {� ∈ [�� �] | the interval [�� �] can be covered by a finite number of elements of U}
Let �0 := sup A.
Step 1. We will show that �0 > �. Indeed, let U ∈ U be a set such that � ∈ U. Since U is open we
have [�� � + ε) ⊆ U for some ε > 0. It follows that � ∈ A for all � ∈ [�� � + ε). Therefore �0 ≥ � + ε.
94
Proof. Follows from Proposition 14.8.
14.11 Example. For any � < � the open interval (�� �) ⊆ R is not compact since (�� �) ∼
= R.
14.12 Proposition. For any � < � the closed interval [�� �] ⊆ R is compact.
Proof. Let U be an open cover of [�� �] and let
A = {� ∈ [�� �] | the interval [�� �] can be covered by a finite number of elements of U}
Let �0 := sup A.
Step 1. We will show that �0 > �. Indeed, let U ∈ U be a set such that � ∈ U. Since U is open we
have [�� � + ε) ⊆ U for some ε > 0. It follows that � ∈ A for all � ∈ [�� � + ε). Therefore �0 ≥ � + ε.
95
covered
∈
a
��00 ]] can
by
of
elements
of
U,
0
covered
by�0U
U∈
U.If As
As
a consequence
consequence
[��
can
be0 covered
covered
by[�a
a0 �finite
finite
number
elements
of for
U, and
and
such thatby
U.U.
�0 <
� then there [��
exists
ε2 >be
such that
�0 + εnumber
. Notice
that
any
0 ∈
2 ) ⊆ U0of
so
�
∈
A.
0
so
� ∈�0(�∈� A.
� + ε ) the interval [�� �] can be covered by a finite number of elements of U, and thus � ∈ A.
0
0
2
Since
� > �view
the assumption
that ��0 =
supTo
A.
0 this contradicts
Step
Step 3.
3. In
In view of
of Step
Step 2
2 it
it suffices
suffices to
to show
show that
that �00 =
= �.
�. To see
see this
this take
take again
again U
U00 ∈
∈U
U to
to be
be a
a set
set
such that
that ��00 ∈
∈ U.
U. If
If ��00 <
<�
� then
then there
there exists
exists ε
ε22 >
>0
0 such
such that
that [�
[�00 �� ��00 +
+ε
ε22 )) ⊆
⊆U
U00 .. Notice
Notice that
that for
for any
any
such
�� ∈
∈ (�
(�00 �� ��00 +
+ε
ε22 )) the
the interval
interval [��
[�� �]
�] can
can be
be covered
covered by
by a
a finite
finite number
number of
of elements
elements of
of U,
U, and
and thus
thus �� ∈
∈ A.
A.
Since
�
>
�
this
contradicts
the
assumption
that
�
=
sup
A.
0
0
Since
�
>
�
this
contradicts
the
assumption
that
�
=
sup
A.
0
14.13 Proposition.
Let X be a compact space. If Y0 is a closed subspace of X then Y is compact.
Proof.
14.13
14.13
14.14
Exercise.
Proposition.
Proposition. Let
Let X
X be
be a
a compact
compact space.
space. If
If Y
Y is
is a
a closed
closed subspace
subspace of
of X
X then
then Y
Y is
is compact.
compact.
Proposition. Let X be a Hausdorff space and let Y ⊆ X . If Y is compact then it is closed in X .
Proof. Exercise.
Exercise.
Proof.
Proposition 14.14 is a direct consequence of the following fact:
14.14 Proposition.
Proposition. Let
Let X
X be
be a
a Hausdorff
Hausdorff space
space and
and let
let Y
Y ⊆
⊆X
X .. If
Y is
is compact
compact then
then it
it is
is closed
closed in
in X
X ..
14.14
If Y
14.15 Lemma. Let X be a Hausdorff space, let Y ⊆ X be a compact subspace, and let � ∈ X r Y .
There
exists 14.14
open sets
U� V ⊆consequence
X such thatof� the
∈ U,following
Y ⊆ V and
Proposition
is a
a direct
direct
fact:U ∩ V = ?.
Proposition
14.14 is
consequence of
the following
fact:
14.15 Lemma.
Lemma. Let
Let X
X be
be a
a Hausdorff
Hausdorff space,
space, let
let Y
Y ⊆
X be
be a
a compact
compact subspace,
subspace, and
and let
let �� ∈
∈X
Xr
rY
Y ..
14.15
⊆X
X
V
There
exists
open
sets
U�
V
⊆
X
such
that
�
∈
U,
Y
⊆
V
and
U
∩
V
=
?.
There exists open sets U� V ⊆ X such that � ∈ U, Y ⊆ V and U ∩ V = ?.
X
X
U
�
U
U
��
Y
Y
Y
V
V
Proof. Since X is a Hausdorff space for any point ��∈ Y there exist open sets U� and V� such that
� ∈ U� , � ∈ V� and U� ∩ V� = ?. Notice that Y ⊆ �∈Y V� . Since Y is compact we can find a finite
number of points �1 � � � � � �� ∈ Y such that
Proof.
point �
∈Y
there exist
Proof. Since
Since X
X is
is a
a Hausdorff
Hausdorff space
space for
for Yany
any
exist open
open sets
sets U
U�� and
and V
V�� such
such that
that
⊆ point
V�1 ∪ �
·�
·∈
· ∪YV�there
�
�
�� ∈
U
,
�
∈
V
and
U
∩
V
=
?.
Notice
that
Y
⊆
V
.
Since
Y
is
compact
we
can
find
finite
� . Since Y is compact we can find a
∈ U�� , � ∈ V�� and U�� ∩ V�� = ?. Notice that Y ⊆ �∈Y
V
a
finite
�
�∈Y
number
of V
points
�1 �� �� V
�� �� �� �
��� ∈
∈U
Y :=
such
that
number
of
points
Y
such
Take
V =
U�that
�1 ∪ · ·�·1∪
�� and
1 ∩ · · · ∩ U�� .
Y ⊆
⊆V
V��1 ∪
∪ ·· ·· ·· ∪
∪V
V���
Y
�
1
Take
Take V
V =
=V
V��11 ∪
∪ ·· ·· ·· ∪
∪V
V���� and
and U
U :=
:= U
U��11 ∩
∩ ·· ·· ·· ∩
∩U
U���� ..
96
Proof Compact
of Proposition
14.14. By
14.
Spaces
Lemma 14.15 for each point � ∈ X r Y we can find an open set U� ⊆
X
90
such that � ∈ U� and U� ⊆ X r Y . Therefore X r Y is open and so Y is closed.
14.16 Corollary. Let X be a compact Hausdorff space. A subspace Y ⊆ X is compact if and only if Y
is closed in X .
X
U�2
V�1
Y
�1
Proof. Follows from Proposition 14.13 and� Proposition 14.14.
�2
14.17 Proposition. Let � : X → Y be a continuous bijection.
If X is a compact space and Y is a
Hausdorff space then � is a homeomorphism.
U�1
V�2
Proof. By Proposition 6.13 it suffices to show that for any closed set A ⊆ X the set �(A) ⊆ Y is closed.
Let A ⊆ X be a closed set. By Proposition 14.13 A is a compact space and thus by Corollary 14.9 �(A)
is a compact subspace of Y . Since Y is a Hausdorff space using Proposition 14.14 we obtain that �(A)
is
closed
in Y .
Proof
of Proposition
14.14. By Lemma 14.15 for each point � ∈ X r Y we can find an open set U� ⊆ X
such that � ∈ U� and U� ⊆ X r Y . Therefore X r Y is open and so Y is closed.
14.16
a compact
Hausdorff
space.
Y ⊆ X is compact if and only if Y
14.18 Corollary.
Theorem. IfLet
XX
is be
a compact
Hausdorff
space
thenA Xsubspace
is normal.
is closed in X .
Proof. Step 1. We will show first that X is a regular space (9.9). Let A ⊆ X be a closed set and
Proof.
Proposition
and Proposition
let � ∈Follows
X r A. from
We need
to show14.13
that there
exists open14.14.
sets U� V ⊆ X such that � ∈ U, A ⊆ V and
U ∩ V = ?. Notice that by Proposition 14.13 the set A is compact. Since X is Hausdorff existence of
14.17
Let � :from
X →
Y be 14.15.
a continuous bijection. If X is a compact space and Y is a
the setsProposition.
U and V follows
Lemma
Hausdorff space then � is a homeomorphism.
Step 2. Next, we show that X is normal. Let A� B ⊆ X be closed sets such that A ∩ B = ?. By Step 1
for
every
∈ A we can
findit open
sets
V� for
such
� ∈set
U� A
, B⊆⊆X Vthe
V�Y=is?.
The
� and
� and
� ∩⊆
Proof.
By�Proposition
6.13
suffices
to U
show
that
anythat
closed
set U
�(A)
closed.
collection
U
=
{U
}
is
an
open
cover
of
A.
Since
A
is
compact
there
is
a
finite
number
of
points
�
�∈A
Let A ⊆ X be a closed set. By Proposition 14.13 A is a compact space
� and thus by Corollary
�� 14.9 �(A)
�is1 �a� �compact
� � �� ∈ A
such that
U�is
a cover ofspace
A. Take
U Proposition
:= �
and Vwe
:=obtain
�� . Then
� }aisHausdorff
1� � � � � Y
�=1 U�� 14.14
�=1 Vthat
subspace
of {U
Y . �Since
using
�(A)
U
and
V
are
open
sets,
A
⊆
U,
B
⊆
V
and
U
∩
V
=
?.
is closed in Y .
14.18 Theorem. If X is a compact Hausdorff space then X is normal.
Proof. Step 1. We will show first that X is a regular space (9.9). Let A ⊆ X be a closed set and
let � ∈ X r A. We need to show that there exists open sets U� V ⊆ X such that � ∈ U, A ⊆ V and
U ∩ V = ?. Notice that by Proposition 14.13 the set A is compact. Since X is Hausdorff existence of
the sets U and V follows from Lemma 14.15.
Step 2. Next, we show that X is normal. Let A� B ⊆ X be closed sets such that A ∩ B = ?. By Step 1
for every � ∈ A we can find open sets U� and V� such that � ∈ U� , B ⊆ V� and U� ∩ V� = ?. The
collection U = {U� }�∈A is an open cover of A. Since A is compact �
there is a finite number
�� of points
�
�1 � � � � � �� ∈ A such that {U�1 � � � � � U�� } is a cover of A. Take U := �=1 U�� and V := �=1 V�� . Then
U and V are open sets, A ⊆ U, B ⊆ V and U ∩ V = ?.
97
Let A ⊆ X be a closed set. By Proposition 14.13 A is a compact space and thus by Corollary 14.9 �(A)
is a compact subspace of Y . Since Y is a Hausdorff space using Proposition 14.14 we obtain that �(A)
is closed in Y .
14.18 Theorem. If X is a compact Hausdorff space then X is normal.
Proof. Step 1. We will show first that X is a regular space (9.9). Let A ⊆ X be a closed set and
let � ∈ X r A. We need to show that there exists open sets U� V ⊆ X such that � ∈ U, A ⊆ V and
U ∩ V = ?. Notice that by Proposition 14.13 the set A is compact. Since X is Hausdorff existence of
the sets U and V follows from Lemma 14.15.
Step 2. Next, we show that X is normal. Let A� B ⊆ X be closed sets such that A ∩ B = ?. By Step 1
for every � ∈ A we can find open sets U� and V� such that � ∈ U� , B ⊆ V� and U� ∩ V� = ?. The
collection U = {U� }�∈A is an open cover of A. Since A is compact �
there is a finite number
�� of points
�1 � � � � � �� ∈ A such that {U�1 � � � � � U�� } is a cover of A. Take U := �
U
and
V
:=
�=1 ��
�=1 V�� . Then
U and V are open sets, A ⊆ U, B ⊆ V and U ∩ V = ?.
98