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MA4266 Topology
Lecture 13. Friday 12 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
Finite Products
basis the set
Question Is
2
R with the usual topology has a
B  {O1  O2 : O1 , O2 open  R }
Example 7.1.1
B
a topology for
2
R ?
Question What is the ‘usual’ basis for
2
R ?
Question How can this example be generalized to
define a basis for a topology on the product of spaces
( X 1 , T1 ),..., ( X n , Tn ) ?
Question Why are the projections maps continuous ?
Finite Products
Theorem 7.1 Let
f :Y  X
where
Y
is a space and
X   i 1 X i .
n
f
is
continuous iff the composition
pi  f : Y  X i
is
continuous for each projection
pi .
X
is a product space
Proof First observe that a subbasis for
1
i
Then
X
is given by
S  { p (Oi ) : 1  i  n, Oi open  X i }.
Theorem 4.11 (p 116) implies that f is continuous iff
each f 1 ( pi1 (Oi ))  ( pi  f ) 1 (Oi ) is open.
Finite Products
Theorem 7.2 The product of a finite number of
Hausdorff spaces is a Hausdorff space.
Theorem 7.3 The product of a finite number of
connected spaces is a connected space.
Theorem 7.4 The product of a finite number of
separable spaces is separable.
Theorem 7.5 The product of a finite number of
1st (2nd) countable spaces is 1st (2nd) countable.
Theorem 7.6 If ( X 1 , d1 ),..., ( X n , d n ) then the product
n
topology on  i 1 X i is the topology generated by the
product metric defined on page 83.
Finite Products
X be compact it is
sufficient that there exist a basis B for X such that
every cover B  B of X has a finite subcover.
c
Lemma In order that a space
Proof Let
B
be such a basis and let
open cover of
Ox  O with
be an
X . For each x  X there exists
x  Ox and Bx  Bc with
x  Bx  Ox .
open cover of
O
Then
{Bx : x  X } is an
X so it has a finite subcover
{Bx1 ,..., Bxn } hence {Ox1 ,..., Oxn }
covers
X.
Finite Products
Theorem 7.7 The product of a finite number of
compact spaces is compact.
Proof It suffices to show that if
X1
and
X2
are
X1  X 2 is compact. Let
B  {U V : U open  X1,V open  X 2 }
be the basis for the product topology and O  B
be a cover of X1  X 2 . For x  X1 the subset
m
{x} X 2 is compact so it is a subset of  i 1 U x,i  Vx,i
compact then
x
where each U x ,i  Vx ,i  B. Let U x   i 1 U x ,i . Then
mx
U x  X 2   i 1 U x  Vx,i .
mx
Read rest of proof p. 202-203.
Arbitrary Products
Definition If
A is an index set and { X  :   A}
a family of sets, their Cartesian product is the set:


X   X    x : A   X  : x  x ( ) X  ,   A
 A
 A


For   A, the function p : X  X 
given by p ( x)  x is called the projection map
th
of X onto the 
coordinate set X  .
Definition If each X  is a nonempty topological space,
X is the topology generated
1
by the subbasis S  p (O ) :   A, O open  X  .
the product topology on


Arbitrary Products
Theorem 7.8 Generalization of Theorem 7.1 that
characterizes continuity of maps using projections.
Theorem 7.9 Hausdorff space products are Hausdorff
Theorem 7.10 Connected space products are conn.
Proof Let y  X   X  and let
 A
be the subset of
X
Yk , k  1
of points that differ from
y
in at most k coordinates. Problem: Show that each
Yk
connected. Then
and
X Y
Y   k 1 Yk is connected
therefore
X
is connected.
Arbitrary Products
Lemma: The Alexander Subbasis Theorem A space
X is compact iff there exists a subbasis S for X
such that every cover Sc  S of X has a finite
subcover. Proof See exercise 12 on p. 210-211.
Theorem 7.11: The Tychonoff Theorem The product
X    A X  of compact spaces is compact.
1

 S  { p (O) : O open  X  } cover X .
We show that Sc has a finite subcover. For each   A
let U be the collection of open O  X  such that
1
p (O )  Sc . Problem: prove that U does not cover X .
O
.
Construct x  X so that for every   A, x  
OU
Then x   OS O contradicting Sc covering X .
Proof: Let Sc
c
Examples
Example 7.2.1 The Hilbert Cube
I   [0,1] is the product
of a countably infinite number of closed intervals. It is
H   2 ( N ) by the map
xn
x1 x2
f ( x1 , x2 ,..., xn ,...)  ( 1 , 2 ,..., n ,...)

Infinite Dimensional Euclidean Space R can be
embedded in Hilbert space
embedded in the Hilbert Cube by the map
g ( x1 , x2 ,..., xn ,...)  ( 12  1 arctan x1 ,..., 12  1 arctan xn ,...)
A deeper result: H   2 ( N ) is homeomorphic to R  .
Proof. R. D. Anderson and R. H. Bing, A complete elementary
proof that Hilbert space is homeomorphic to the countable
infinite product of lines, Bull. Amer. Math. Soc. 74 (1968),771-792.
Examples
Example 7.2.2 The Cantor Set is an Infinite Product
The Cantor set, consisting of all numbers in
[0,1]
having the ternary (base 3) expansion
x  .x1 x2 x3 ...   j 1 x j 3 , x j {0,2}

j
is homeomorphic to the infinite product
by the map
F (.x1 x2 x3...)  ( , ,...)
x1
2
Proof See page 209.
x2
2
{0,1}  {0,1}N
Assignment 13
Read pages 195-203, 204-209
Prepare to solve during Tutorial Thursday 18 March
Exercise 7.1 problems 4, 10, 11
Exercise 7.2 problems 5, 11, 12