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MA4266 Topology
Lecture 18. Friday 16 April 2010
Wayne Lawton
Department of Mathematics
S17-08-17, 65162749 matwml@nus.edu.sg
http://www.math.nus.edu.sg/~matwml/
http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1
Bounded Continuous Functions
Definition For a topological space
( X , T ) we denote by
C ( X , R) the vector space of bounded, continuous,
real-valued functions with domain X and topology
determined by the metric
 ( f , g )  lub {| f ( x)  g ( x) | : x  X }.
Exercise 8.6 Problem 4 (proof similar to Example 3.7.6
on pages 93-94) that
(C ( X , R),  ) is complete so it
is a Banach space (a complete normed vector space).
Prove this !
Complete Regularity
( X , T ) is called
completely regular (page 247) if it is T1 and T3 1 .
Definition A topological space
2
The latter condition means that for every closed
C  X and p  X \ C there exists a continuous
f :X R
with
f ( p)  0, f (C )  1.
Prove that then we can choose
f  C ( X , R ).
Definition A topological space ( X , T ) is a Urysohn space
if for every pair of distinct points
a continuous
p, q  X there exists
f : X  R with f ( p)  0, f (q)  1.
Prove that completely regular  Urysohn.
Complete Regularity
( X , T ) is completely regular, then the
weak topology for X generated by C ( X , R) is T .
Theorem 8.20 If
T  be the weak topology generated by C ( X , R).
Clearly T   T Why ?
For every U  T and x  U the set C  X \ U is a
Proof Let
closed set satisfying
of
X
x  C so the complete regularity
implies there exists a continuous
f : X  [0,1]
f ( x)  0, f (C )  1. Therefore
1
x  f ([0, 12 ))  T  so U is a union of sets in T 
hence U T . Therefore T  T .
such that
Stone-Cech Theorem
Theorem 8.21 If X is completely regular then there exists
a compact Hausdorff space  ( X ) which contains X
as a dense subspace and for which every f  C ( X , R)
~
can be extended to f  C (  ( X ), R ).
Proof: For f  C ( X , R) let I f be the smallest closed
interval containing f ( X ) and let Y 
I f and
define
e : X Y

f C ( X , R )
by
p f (e( x))  f ( x), x  X , f  C ( X , R), and  ( X )  e( X )
Clearly e is continuous Why ?
If x  y  X there exists f  C ( X , R ) such that
f ( x)  f ( y ) Why ? Hence
e
is one-to-one Why ?
Stone-Cech Theorem
To show that e : X  Y is an embedding it suffices
to show that it is open. Since it is one-to-one it suffices
To show that there exists a subbasis S for the topology
of X such that e maps every element of S to an open
set in e( X ). Theorem 8.20 implies that
1
S  { f (U ) : f  C ( X , R), U open  R }
is such a subbasis for the topology of
1
X.
1
f
e ( f (U ))  p (U )  e( X )
is an open set in the subspace topology on e( X )  Y .
Furthermore
~
Finally, the extension is given by f ( y)  p f ( y). Why ?
Answers to Homework #3
1. Let
X
be a nonempty set. A subset
F  P( X )
is called ‘frisbee’ (on X) if it satisfies the following:
  F, ( A  F and A  B)  B  F,
( A  F and B  F )  A  B  F.
Frisbee’s are Filters in Mathematical Language
http://en.wikipedia.org/wiki/Filter_(mathematics)
is called a ‘freeflyer’ if

AF
A  ,
Freeflyers are Free Filters,
and is called a ‘highflyer’ it is a frisbee that
is not a proper subset of another frisbee.
Highflyers are Ultrafilters.
Answers to Homework #3
(a) For nonempty
S  X define FS  { A  X : S  A}.
Prove that FS is always a frisbee and that it is a
a highflyer if and only if S is a singleton set.
 F
and
( A  F and A  B)  B  F
( A  F and B  F)  A  B  F
Choose any
Then
Fp
pS
and define
is a filter, FS
This shows that if
FS
and
are obvious.
Fp  { A  X : p  A}.
 Fp , and FS  Fp
iff
S  {p}.
an ultrafilter  S is a singleton.
The converse follows since
Fp
is clearly an ultrafilter.
Answers to Homework #3
(b) For
| X | 
show that
F  { A  X : | X \ A |  }
is a freeflyer.
 F
follows since
| X \ |  | X |  
( A  F and A  B)  X \ B  X \ A
 | X \ B |  | X \ A |    B  F.
( A  F and B  F)  X \ ( A  B)  ( X \ A)  ( X \ B)
 | X \ ( A  B) |     A  B  F.
F is a filter. Furthermore, for every p  X
X \ { p}  F   AF A   so F is a free filter.
therefore
Answers to Homework #3
(c) Use the Hausdorff Maximal Principle to prove
that every frisbee is contained in a highflyer.
X and let P be the set of
filters that contain F. Then P s partially ordered by
inclusion. For any linearly ordered L  P the set
U   L  P and is an upper bound for L .
Let
F
be any filter of a set
LL
Hence by Zorn’s Lemma (equivalent to the HMP) there
M P.
ultrafilter that contains F.
exists a maximal filter
Then
M
is an
Answers to Homework #3
F is a highflyer if and only if
for every A  X , either A  F or X \ A  F.
Only If Part: Assume that F is an ultrafilter and let
A  X and A  F. Construct the set
G  {G : G  F  ( X \ A) for some F  F}.
Then every set in G is nonempty since
F  ( X \ A)    F  A  A  F.
Hence G is clearly a filter and F  G .
Since F is an ultrafilter F  G hence
X \ A  X  ( X \ A)  G  X \ A  F.
(d) Show that a frisbee
Answers to Homework #3
F is a highflyer if and only if
for every A  X , either A  F or X \ A  F.
If Part: Assume that F is a filter satisfying the property
and F  G where G is a filter. If A  G
then X \ A  F else we obtain the contradiction
X \ A  G    A  ( X \ A)  G .
Therefore A  F so F  G and F is an ultrafilter.
(d) Show that a frisbee
Answers to Homework #3
2. Let
X
be a nonempty topological space and
x X .
F (on X) ‘lands on’ x if every open
set that contains x belongs to F.
A frisbee
A filter lands on x if it converges to x.
A filter hovers near x if x is a limit point of the filter.
X is a topological space and x  X the set
Nx  { A  X : x  int A } of neighborhoods of x is
a filter called the neighborhood filter of x.
Result A filter F converges to x (F  x) iff Nx  F.
Definition If
Answers to Homework #3
(a) Let
X
and
a function
Y
be topological spaces. Show that
f : X Y
is continuous at
x X
F that converges to x,
the filter G  { G : G  f ( A) for some A  F }
converges to f (x ).
if and only if for every filter
f ( x)  O open and filter F  x.
since f : X  Y is continuous there exists open
U  x such that f (U )  O. Since F  x
U  F  f (U )  G  O  G so G  f (x).
Only If Part: Let
Answers to Homework #3
(a) Let
X
and
a function
Y
be topological spaces. Show that
f : X Y
is continuous at
x X
F that converges to x,
the filter G  { G : G  f ( F ) for some F  F }
converges to f (x ).
if and only if for every filter
If Part: Let
f ( x)  O open
F be the filter
x. Then F  x
and let
consisting of all neighborhoods of
G  f(x)  O  G. Therefore there exists
a neighborhood A of x with f ( A)  O so
x  int A and f (int A)  O  f is continuous .
hence
Answers to Homework #3
2. Let
X
be a nonempty topological space and
A filter
F
(on X) converges to
if every open
x belongs to
X  A X  be a product space and let
set that contains
(b) Let
x
F.
x X .
F be a filter on X. Prove that F  x if and only if
G  { G : G  p ( A) for some A  F }  p ( x)
for every   A.
Fx
iff every subasic open set
1
x  p (O)  F
so the conclusion follows from part (a).
Answers to Homework #3
(c) Prove that a space X is compact if and only if
every ultrafilter converges to some point in X .
Only If Part. Let F be a filter on
X . First we show that
F hovers near some point. Otherwise for every p  X
there exists open O p  p and Fp  F with O p  Fp   .
Clearly {O p : p  X } is an open cover of X hence since
X is compact there exists a finite subcover O p ,..., O p .
m
Then F   j 1 Fp  F satisfies F  F  X   .
This contradiction shows that there exists p  X such
that F hovers near p. Then the set below is a filter
1
m
j
G  {G : G  A  B, for some A  Np , B  F} containing
F and Np .
So F ultrafilte r  F  G  Np  F  p.
Answers to Homework #3
(c) Prove that a space X is compact if and only if
every ultrafilter converges to some point in X .
X is not compact so
O of X that does not
If Part. Assume to the contrary that
that there exists an open cover
have an open subcover. Define the set
FO  { G : G  X \ O for some O  O },
observe that it is a filter, and let F be an ultrafilter that
contains F so by assumption there exists p  X
O
such that F  p. Choose a O  O with p  O
so X \ O  FO  X \ O  F. Now F  p  O  F
hence   O  ( X \ O)  F - a contradiction.
Answers to Homework #3
(d) Prove The Tychonoff Theorem using ultrafilters.
Theorem (Tychonoff) Let { X  :   A} be a family of
compact topological spaces. Then the product space
X  A X 
F
is compact.
X . By 2 (c) it
suffices to show F  x for some x X . Define
G  { G : G  p ( A) for some A  F },   A
Proof Let
be an ultrafilter on the set
and observe they are ultrafilters so 2(c) implies there
exist x  X  such that G  x ,   A. Define
x  X by p ( x)  x ,   A. Therefore 2 (b) implies
that
F  x.
Exam Suggestions
Knowledge
Definitions: Basis, Hausdorff, Connected, Compact, etc.
Results: Statements of Theorems, Lemmas, Corollaries
Examples: Construction and Properties of Spaces
Skills
Derive: Proofs
Construct: Examples
Explain: Proofs of Major Theorems in Outline Form