Transcript MA4266_Lect10 - Department of Mathematics

MA4266 Topology
Lecture 10
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
Covers
A
X.
A cover of A is a collection C  P(X) of subsets of
X such that A   C   CC C.
Definition: Let
be a subset of a set
CC
Examples Ck  2 (2 Z  1)  Z , k  0
k
A  {18,121, 24}  X  Z , C  {Ck : k  0}
A  {18,121, 24}  X  Z , C  {C0 , C1 , C3}
A  2Z  X  Z , C  {Ck : k  1}
A  R  X , C  {Q  x : x  R}
Question Which covers are finite ? Countable ?
Subcovers
Definition: Let
A
be a subset of a set
X
and
C
be
A. A subcover (of A derived from C ) is a
cover C  of A such that C  C.
a cover of
Lemma If
A
is finite then every cover of
subcover.
Question Give an example of this lemma.
A has a finite
Open Covers
A be a subset of a topological space X .
An open cover of A is a cover of A whose elements are
open. An finite cover of A is a cover of A that is finite.
Definition: Let
Example 6.1.1
A  [0,5]  X  R, O  {( n  1, n  1) : n  Z }
is an open cover and
O  {(1,1), (0,2), (1,3), (2,4), (3,5), (4,6)}
is an subcover (obviously this subcover is open)
Compact Spaces
Definition: A topological space X is compact if every
open cover of X has a finite subcover. A subspace A
of X is compact if it satisfies either of the following
equivalent conditions:
1.
A is a compact topological space (regarded as
a topological space with the subspace topology).
2. Every open cover of
subsets of
X
A that consists of open
has a finite subcover.
Examples
Example 6.1.2 (of compact spaces)
(a) Finite spaces
(b) Closed bounded intervals
(c) Closed bounded subsets of
(d)
R
R
n
with the finite complement topology
Example 6.1.3 (of noncompact spaces)
(a) Any infinite discrete topological space
(b) The open interval
(c)
R
n
(0,1)
Finite Intersection Property
Definition A family of subsets of X has the finite
intersection property if every finite subcollection
has nonempty intersection.
Example 6.1.4
{( 1n ,1) : n  1}  X  R
Question What is the intersection of (all) these sets?
Theorem 6.1: A space
closed sets in
X
is compact iff every family of
X with the finite intersection property
has nonempty intersection.
Proof Follows from De Morgan’s Laws.
Cantor’s Theorem of Deduction
Theorem 6.2: Let
E1  E2  E3  
be a nested
sequence of nonempty, closed, bounded subsets of
Then

k 1
Ek   .
Proof ???
Question Why assume both closed and bounded ?
Theorem 6.3: Closed subset of compact  compact.
Proof ???
R.
Properties of Compact Sets
Theorem 6.4: Compact subset of Hausdorff  closed.
Proof Let A be a compact subset of a Hausdorff space
X and let x X \ A. It suffices to show that there
x V  X \ A.
Since X is Hausdorff, for every y  A there exist
disjoint open subsets U y , V y with y  U y , x  V y .
Clearly { U y : y  A } is an open cover of A.
Since A is compact there exist { y1 , , yn }  A
such that { U y ,  , U y } is a cover of A. Construct
exists an open subset
n
1
VX
n
n
U  k 1 U k ,V  k 1 Vk .
V  X \ A
with
Then A  U , U  V  
and V is open.
Properties of Compact Sets
Theorem 6.5: If A and B are disjoint compact subsets
of a Hausdorff space X then there exist disjoint open
subsets U and V of X with A  U , B  V .
Proof Similar to the proof of theorem 6.4.
Question Construct an example of a compact subset
of a topological space
Hint:
X
X
has two points.
such that
A is not closed.
A
Compactness and Continuity
Theorem 6.6: The continuous image of a compact
space is compact. Proof Easy.
Theorem 6.7: If
f : X Y
is compact and
is continuous then
Theorem 6.8: If
f : X Y
X
X
Y
f
is compact and
is Hausdorff and
is closed.
Y
is Hausdorff and
is a continuous bijection then
f
is a
homeomorphism. Proof Easy.
f :X R
is continuous then there exists c, d  X such that
f (c)  f ( x)  f (d ), x  X . Proof Easy.
Theorem 6.9: If
X
is compact and
Compactness and Continuity
Definition Let ( X , d ), (Y , d ) be metric spaces. A
function f : X  Y is uniformly continuous if for
every   0 there exists   0 such that
d ( x, y )    d ( f ( x), f ( y ))   , x, y  X .
Theorem 6.10: If ( X , d ) is a compact metric space
and (Y , d ) is a metric space and f : X  Y is
continuous then f is uniformly continuous.
Proof see pages 170-171.
Question Show that uniform continuity  continuity.
Question Give an example of a continuous function
that is not uniformly continuous.
Tutorial Assignment 10
Read pages 161-174
Exercise 6.1 problems 3, 6
Exercise 6.2 problems 1, 2, 3, 4
Complete the proof of the Lemma, page 172 for n > 2.