ma4266 topology
DESCRIPTION
Lecture 11. MA4266 Topology. Wayne Lawton Department of Mathematics S17-08-17, 65162749 [email protected] http://www.math.nus.edu.sg/~matwml/ http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1. Basics. Theorem 6.11: A subset of. is compact iff it is. closed and bounded. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/1.jpg)
MA4266 Topology
Wayne LawtonDepartment of Mathematics
S17-08-17, 65162749 [email protected]
http://www.math.nus.edu.sg/~matwml/http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1
Lecture 11
![Page 2: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/2.jpg)
Basics
Theorem 6.11: A subset of is compact iff it isnR
Definition: A topological space is countably compact
closed and bounded.
if every countable open cover has a finite subcover.
Definition: A topological space is a Lindelöf space
if every cover has a countable subcover.
Theorem 6.12: If X is a Lindelöf space, then X iscompact iff it is countably compact.
Theorem 6.13: The Lindelöf Theorem Everysecond countable space is Lindelöf.
Proof see page 175
![Page 3: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/3.jpg)
Bolzano-Weierstrass PropertyDefinition: A topological space X has the BW-propertyif every infinite subset of X has a limit point. Theorem 2.14: Every compact space has the BWP.Proof Assume to the contrary that X is a compactspace and that B is an infinite subset of X that has no limit points. Then B is closed (why?) and
B is compact (why?). Since B has no limit points, for every point x in B there exists an open set
XOx such that }.{xBOx Therefore
}:{ BxOx O is an open cover of B. FurthermoreO does not have a finite subcover of B (why?).
Definition p is an isolated point if {p} is open. See Problem 10 on page 186.
![Page 4: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/4.jpg)
ExamplesExample 6.3.1(a) Closed bounded intervals [a,b] have the BWP.
(b) Open intervals do not have the BWP.
(c) Unbounded subsets of R do not have the BWP.
(d) The unit sphere
does not have the BWP (why?).
)(2 N,...}4,3,2,1{N
}||||,:,...),({)(1
2221
2
k kk xxRxxxN}1||||:)({ 2 xNxS
in the Hilbert space S
![Page 5: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/5.jpg)
Definition: Let be a metric space and
such that every subset of
.X A Lebesgue number foran open cover of
),( dX
is a positive number X
Lebesgue Number of an Open Cover
OO
having diameter less than is contained in some
element in .OTheorem 6.16 If ),( dX is a compact metric space
then every open cover of X has a Lebesgue number.
Proof follows from the following Lemma 1 since eachsubset having diameter less than an open ball of radius
is a subset of .
![Page 6: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/6.jpg)
Lemma 1: Let be a metric space that satisfies
of
1}{ nnx
and assume to the
the Bolzano-Weierstrass property. Then every open
),( dX
cover X
BWExistence of Lebesgue Number
O
O
has a Lebesgue number.
OxB nn ),( 1
be an open cover of
contrary that
O
Then there exists a sequence Xdoes not have a Lebesgue number.
in
X
for every
Proof Let
and for everynThen
such that
.OO1}{ nnx is infinite (why?) so the BW property
implies that it has a limit point a so there exists 0and OO with .),( OaB Then ),( 2
aB contains
infinitely many members of .}{ 1nnx
![Page 7: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/7.jpg)
Hence with
Then for
BWExistence of Lebesgue Number
22),(),(),( zxdxadzad nn
so
.21 nnx
This contradicting the initial assumption that for all
),( 1nnxBz
),( 2aB contains some
.),(),( 1 OaBxB nn
O OnOxB nn ,1,),( 1
and completes the proof of Lemma 1.
![Page 8: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/8.jpg)
Definition: Let
An
be a metric space and
Total Boundedness
net for
.0XA
such that
),( dX
is a finite subset
.,),( XxAxd The metric space
XX
is totally bounded if it has an net for every .0Lemma 2: Let ),( dX be a metric space that satisfiesthe Bolzano-Weierstrass property. Then X is TB.
Proof Assume to the contrary that there exists 0such that X does not have an net. Choose
Xa 1 and construct a sequence 1}{ kka with
),(11 k
j jk aBa that has no limit point.
![Page 9: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/9.jpg)
Theorem 6.15: For metric spaces compactness = BWP.
Compactness and the BWP
For the converse let
space
O be an open cover of a metric
Lemma 1 implies that there exists
),( dX having the Bolzano-Weierstrass property.
such that for
Xx is contained in some
Proof Theorem 4.14 implies that compactness BWP.
0the open ball
XxxA n },...,{ 1
every ),( xB
subset
Lemma 2 implies that there exists a finitemember of
an open cover of
.Osuch that }:),({ ,...,1 nkkxB
.X Choose nkkk OxB ,...,1,),( Oand observe that }:{ ,...,1 nkkO covers .X
![Page 10: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/10.jpg)
Theorem 6.17: For a subset
Compactness for Subsets of
conditions are equivalent:
(a)
nR
is compact.
the following
has the BWP.
nRA
A(b) A(c) A is countably compact.
(d) A is closed and bounded.
Question Are these conditions equivalent for ?)(2 NA
![Page 11: MA4266 Topology](https://reader036.vdocument.in/reader036/viewer/2022082422/56812dcc550346895d930acc/html5/thumbnails/11.jpg)
Assignment 11
Read pages 175-180
Exercise 6.3 problems 2, 3, 4, 5, 9, 13, 14, 15