Transcript MA4266_Lect8 - Department of Mathematics

MA4266 Topology
Lecture 8. Friday 12 Feb 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
Separation
Definition Let (X,T) be a topological space. A
separation of X is are elements A, B in T with
A B 
A  , B  
A B  X
Connected and Disconnected Spaces
Definition A topological space (X,T) is disconnected
if it admits a separation, otherwise it is connected.
Example 5.1.1
A discrete space with > 1 point is disconnected
{ p, q}  X  A  { p}, B  X \ A gives separation
A trivial space is connected
R \ {0} is disconnected
R \ R  {0} is disconnected
2
R {0} {(0,1)},  F {(0,1)} : F  R {0}
is connected
Real Numbers with Usual Topology
Example 5.1.2 It is CONNECTED !
Proof Assume to the contrary that there exists a
separation A, B for R and choose a  A, b  B.
Define
A0  A  [a, b]. Since A0 is nonempty
and bounded above (why?) it has a least upper
bound
c.
Then
c  A0
(why?) so
c  b (why?)
Therefore (c, b]  B (why?).
Since B is closed c B.
This leads to a contradiction c  B  A   .
A Subset Y of Topological Space (X,T)
is called disconnected if the subspace Y is
disconnected, this means that there are
elements A, B in T with
A  B Y  
A Y  , B Y  
Y  A B
Examples
[0,1]  [2,3]  ( R, Tusual ) is disconnected
Proof Choose
A  (1, 32 ), B  ( 32 ,4)
Q  ( R, Tusual ) is disconnected
Proof Choose
A  (, 2 ), B  ( 2 , )
Separated Sets
Definition Let (X,T) be a topological space.
Nonempty
A, B  X
are separated if
A  B  A B  
Theorem 5.1 The following statements are equivalent
1. X is disconnected
2. X is the union of two disjoint, non-empty closed sets
3. X is the union of two separated sets
4. There exists a continuous function f:X { a, b }_disc.
5. X has a proper subset that is both open and closed
6. X has a proper subset A such that


A  X \ A 
Connectedness is a Continuous Invariant
Theorem 5.2 Let X be a connected space and Y be a
space and f : X  Y be continuous. Then Y is
connected.
Proof Assume to the contrary that Y is disconnected.
Then Y has a separation A, B. Then the sets
U  f 1 ( A), V  f 1 ( B) are
(a) open (why?),
(b) disjoint (why?),
(c) nonempty (why?),
(d) have union X (why?).
What does this say about these sets? Are we done?
Closure of Connected is a Connected
Theorem 5.4 If Y is a connected subset of a space X
then Y is connected.
Proof (see page 136 for a different proof).
Assume to the contrary that Y is disconnected.
Then there exists open subsets A, B  X
A  Y , B  Y are a separation of Y .
Then A  Y   , B  Y   (why?).
Therefore A  Y   , B  Y   (why?).
Hence A  Y , B  Y are a separation of Y .
such that
What does this imply about Y ? Are we done ?
Example 5.2.3 Topologists Sine Curve
Unions of Connected Sets
Theorem 5.5 Let X be a space and { A :   I }
a family of subsets with W  I A   . Then
Y  I A is connected.
U ,V be open subsets of X such that
U  Y   , U V  Y   , Y  U V .
It suffices to show that V Y   . Choose b W
and choose   I so that U  A    . Since
Proof Let
A 
is connected
Since each
so
Y U
A
A   U
is connected each
and hence
b U .
A  U
and hence
V  Y  U  V  Y  .
Components
Definition A component of a topological space X
C  X which is not a
proper subset of any connected subset of X .
is a connected subset
Theorem If X is a topological space then the relation
a  b if there exists a connected subset C  X
such that
a, b  C
is an equivalence relation.
Proof Reflexivity and symmetry are obvious, and
transitivity follows theorem 5.5.
Corollary Each equivalence class is a component.
Corollary Each component is closed. If there are a
finite number of components, they are also closed.
Connected Subsets of R
Theorem 5.7 R  C   , C connected  C is an interval.
Proof Clearly a  C , b  C , a  c  [ a, b]  C.
Hence if C is not bounded above then
a  C  [a, )  C and if C is not bounded below
then
a  C  (, a]  C.
So if C is not bounded above or below then C = R.
If R is bounded below but not above then either
C  [a, ) if a  inf( C )  C or C  (a, ) if a  inf( C )  C
and similarly for 6 cases where R is bounded above.
Theorem 5.8 The Intermediate Value Theorem Let
f : [a, b]  R be continuous and y 0 be between f (a )
and f (b ). Then there exists c  [ a, b] with f (c)  y0 .
Assignment 8
Read pages 131-146
Do Exercise 5.1 problem 3
Prove the six properties after the definition on page 139
Do Exercise 5.2 problems 5, 8, 14, 16
Do Exercise 5.3 problem 3
Do Exercise 5.4 problems 1, 2