#### Transcript MA4266_Lect11 - Department of Mathematics

```MA4266 Topology
Lecture 11. Friday 5 March 2010
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
R
n
is compact iff it is
closed and bounded.
Definition: A topological space is countably compact
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 is
compact iff it is countably compact.
Theorem 6.13: The Lindelöf Theorem Every
second countable space is Lindelöf.
Proof see page 175
Bolzano-Weierstrass Property
Definition: A topological space X has the BW-property
if 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 compact
space 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
O  X such that O  B  {x}. Therefore
x
x
O  {Ox : x  B} is an open cover of B. Furthermore
O 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.
Examples
Example 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
S
in the Hilbert space
2
 (N )
N  {1,2,3,4,...}
 ( N )  { ( x1 , x2 ,...) : xk  R, || x ||   k 1 x }
2
S  { x   ( N ) : || x ||  1 }
2
does not have the BWP (why?).
2

2
k
Lebesgue Number of an Open Cover
( X , d ) be a metric space and O
an open cover of X . A Lebesgue number for O
Definition: Let
is a positive number

such that every subset of X
having diameter less than
element in

is contained in some
O.
Theorem 6.16 If
(X ,d)
is a compact metric space
then every open cover of X has a Lebesgue number.
Proof follows from the following Lemma 1 since each
subset having diameter less than  is a subset of
an open ball of radius  .
BWExistence of Lebesgue Number
Lemma 1: Let
(X ,d)
be a metric space that satisfies
the Bolzano-Weierstrass property. Then every open
cover
O
Proof Let
of X has a Lebesgue number.
O
contrary that
be an open cover of
X
and assume to the
O
does not have a Lebesgue number.
 in
X such that
Then there exists a sequence {x }
n n 1
B( xn , 1n )  O for every n and for every O  O.

Then {xn }n 1 is infinite (why?) so the BW property
implies that it has a limit point a so there exists   0
and O  O with B ( a,  )  O. Then B(a, 2 ) contains

infinitely many members of {xn }n 1.
BWExistence of Lebesgue Number

xn with
B
(
a
,
Hence
2 ) contains some
Then for z  B( xn , 1n )
1
n
 2 .
d (a, z )  d (a, xn )  d ( xn , z )  2  2  


so
B( xn , 1n )  B(a,  )  O.
This contradicting the initial assumption that for all
B( xn , 1n )  O, n  1, O  O
and completes the proof of Lemma 1.
Total Boundedness
( X , d ) be a metric space and   0.
An   net for X is a finite subset A  X
such that d ( x, A )   , x  X . The metric space X
is totally bounded if it has an   net for every   0.
Definition: Let
Lemma 2: Let ( X , d ) be a metric space that satisfies
the Bolzano-Weierstrass property. Then X is TB.
Proof Assume to the contrary that there exists 
such that X does not have an   net. Choose

a1  X and construct a sequence {ak }k 1 with
ak 1   j 1 B (a j ,  ) that has no limit point.
k
0
Compactness and the BWP
Theorem 6.15: For metric spaces compactness = BWP.
Proof Theorem 4.14 implies that compactness  BWP.
For the converse let
space
O be an open cover of a metric
( X , d ) having the Bolzano-Weierstrass property.
Lemma 1 implies that there exists
  0 such that for
x  X the open ball B( x,  ) is contained in some
member of O. Lemma 2 implies that there exists a finite
every
subset A  {x1 ,..., xn }  X such that {B( xk ,  ) : k
Choose B( xk ,  )  Ok  O, k
and observe that {Ok : k  1,..., n} covers X .
an open cover of
X.
 1,..., n}
 1,..., n
Compactness for Subsets of R
Theorem 6.17: For a subset
A R
n
n
the following
conditions are equivalent:
is compact.
(c)
A
A
A
(d)
A
is closed and bounded.
(a)
(b)
has the BWP.
is countably compact.
Question Are these conditions equivalent for A   2 ( N ) ?
Assignment 11
Prepare to solve on the board Tuesday 9 March
Exercise 6.3 problems 2, 3, 4, 5, 9, 13, 14, 15
Supplementary Materials
Definition: A compact, connected, locally connected
Metric space is called a Peano space (or P. continuum).
Examples: closed balls in
R
n
Theorem (Hahn-Mazurkiewicz): A topological space
X
is a Peano space iff it is Hausdorff and there exists a
continuous surjection
f : [0,1]  X .
http://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn-Mazurkiewicz_theorem
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