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MA4266 Topology
Lecture 15. Tuesday 30 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
Chains and Maximal Elements
Assume that ( X , ) is a partially ordered set.
Definition A subset C  X is called a chain if (C , )
is a linearly (or totally) ordered set.
Definition An element m  X is a maximal element if
n  X , n  m  n  m.
Remark If X has a maximum m  max ( X )
then m
is a maximal element. However a maximal element need
not be a maximum element.
Example X  P({a, b, c}) \ {{a, b, c}},U  V  U  V
{ , {a}, {a , b}} and {{ b}, {b , c}} are chains,
{ ,{a},{b, c}, X } and {{a},{b}} are not chains,
{a , b}, {b , c}, {a , c}} are (all 3) maximal elements.
Zorn’s Lemma
( X , ) is a partially
ordered set. If every chain C  X has an upper bound
then X has a maximal element.
Lemma (Zorn) Assume that
Proof This is equivalent to the axiom of choice, the well
ordering principle, and the Hausdorff maximal principle.
http://en.wikipedia.org/wiki/Zorn%27s_lemma
Lemma Hausdorff Maximal Principle (HMP) Every
chain in X is contained as a subset of a maximal chain.
Proof It suffices by Zorn’s lemma to show that every
chain of chains has a maximal element. This follows
since the union of a chain of chains is itself a chain
which is also an upper bound for the chains in the chain.
Alexander Subbasis Theorem
X is compact if and only if there
exists a subbasis S for X with the following Property F:
Every cover of X by members of S has a finite subcover.
Proof If X is compact then every subbasis has prop. F.
To prove the converse assume that S is a subbasis
with property F and assume that X is not compact.
It suffices to obtain a contradiction. Let O be an open
cover of X that has no finite subcover. Covers are
Theorem A space
partially ordered by inclusion and the union of any chain
of covers, each having no finite subcover, is an upper
bound of the covers in this chain. So the HMP implies
Alexander Subbasis Theorem
that there exists a maximal open cover of X with the
property that it has no finite subcover. So we can and
will choose O to be a maximal open cover with this
property. (remark: this required the Axiom of Choice)
Lemma If U 1 ,..., U n are open sets and there exists
OO
I  U1    U n  O then there
exists i  {1,..., n } such that U i  O. Proof Else
 j  { 1,..., n }, U j  O   O j ,1 ,..., O j ,m j  O
such that X  O j ,1    O j , m j  U j . This gives the
such that
contradiction
n
X  O   j 1 O j ,1    O j ,m j
Alexander Subbasis Theorem
Now for every
U1 ,..., U n  S
and every p  O there exist
such that x  U 1    U n  O.
OO
Question Why does this statement hold ?
Then the preceding lemma implies that there exists
U x ,O  S such that x  U  O. Therefore
{U x ,O : x O  O } is an open cover of X
of
S
by members
and hence admits a finite subcover
{U x1 ,O1 ,..., U xL ,OL }. Since U xi ,Oi  Oi it follows that
{ O1 ,..., OL } is a finite subcover of X by members of O.
This contradiction completes the proof.
The Tychonoff Theorem
Theorem7.11The product of compact spaces is compact
Proof Let { X  :   A } be a collection of compact
be the product space and
X
spaces, let X 

 A
1
let S  { p (O ) :  A, O open  X  } be the
‘standard’ subbasis. It suffices, by the AST to show
that every cover of X by elements of S has a finite
subcover. Let U be such a cover and assume to the
contrary that U did not have a finite subcover. For each
1
  A let U  { O open  X  : p (O )  U }.
No finite subset of U covers X  , so U doesn’t cover
X  , so  x  X  that is not in any member of U .
If x  X , x ( )  x ,   A then x  U U U .


Assignment 15
Read pages 234-237, 237-241, 243-251.
Written Homework #3 Due Friday 9 April
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
is called a ‘freeflyer’ if  AF A   ,
and is called a ‘highflyer’ it is a frisbee that
is not a proper subset of another frisbee.
Assignment 15
(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.
(b) For | X |   show that F  { A  X : | X \ A |  }
is a freeflyer.
(c) Use the Hausdorff Maximal Principle to prove
that every frisbee is contained in a highflyer.
(d) Show that a frisbee F is a highflyer if and only if
for every A  X , either A  F or X \ A  F.
Remark (b) and (c) ensure the existence of freeflying
highflyers, and (d) shows that they are mind boggling !
Assignment 15
2. Let X be a nonempty topological space and x  X .
A frisbee F (on X) ‘lands on’ x if every open
set that contains x belongs to F.
(a) Let X and Y be topological spaces. Show that
a function f : X  Y is continuous at x  X
if and only if for every frisbee F (on X) that lands on x,
f (F ) (clearly a frisbee on Y) lands on f (x).
(b) Let
X   A X 
be a product space and let
F lands on x  X
if and only if p (F ) lands on p ( x )  X  ,   A.
F
be a frisbee on X. Prove that
Assignment 15
(c) Prove that a space
X
is compact if and only if
every highflyer lands on some point in
X.
(d) Prove The Tychonoff Theorem using highflyers.