Download ADVANCE TOPICS IN TOPOLOGY - POINT

ADVANCE TOPICS IN TOPOLOGY - POINT-SET
NOTES COMPILED BY KATO LA
19 January 2012
Background
Intervals:
pa, bq “ tx P R | a ă x ă bu
Ó
Ó
calc. notation
set theory notation
,
/
/
/
/
/
/
/
/
/
/
.
“Open” intervals
/
/
/
/
/
/
/
/
/
/
-
pa, 8q
p´8, bq
ra, bs, ra, 8q: Closed
pa, bs, ra, bq: Half-openzHalf-closed
Open Sets: Includes all open intervals and union of open intervals. i.e., p0, 1q Y p3, 4q.
Definition: A set A of real numbers is open if @ x P A, D an open interval containing x which is a subset of A.
Question: Is Q, the set of all rational numbers, an open set of R?
ˆ
˙
1
1
1
´ ε, ` ε is a subset of Q. We can
- No. Consider . No interval of the form
2
2
2
ask a similar question in R2 .
Is R open in R2 ? - No, because any disk around any point in R will have points above and
below that point of R.
Date: Spring 2012.
1
2
NOTES COMPILED BY KATO LA
Definition: A set is called closed if its complement is open.
In R, p0, 1q is open and p´8, 0s Y r1, 8q is closed. R is open, thus Ø is closed. r0, 1q
is not open or closed. In R, the set t0u is closed: its complement is p´8, 0q Y p0, 8q. In
R2 , is tp0, 0qu closed? - Yes.
Chapter 2 - Topological Spaces & Continuous Functions
Definition: A topology on a set X is a collection T of subsets of X satisfying:
(1) Ø, X P T
(2) The union of any number of sets in T is again, in the collection
(3) The intersection of any finite number of sets in T , is again in T
Alternative Definition: ¨ ¨ ¨ is a collection T of subsets of X such that Ø, X P T and
T is closed under arbitrary unions and finite intersections.
Definition: A set X for which a topology has been defined is called a topological space.
Example: X “ ta, b, cu
T1
T2
T3
T4
“ tØ, X, tau , ta, buu is a topology on X.
“ tØ, Xu is known as the “trivial ” topology.
“ tall subsets of Xu is known as the “discrete” topology.
“ tØ, X, tb, cu , tcuu has complements of sets in T1 .
Notice which Tx are subsets of Ty and vice versa.
Definition: A set which belongs to T is called an open set.
Example: If we let T contain all the sets which, in a calculus sense, we call open - We
have “R with the standard [or usual] topology.”
Example: [Example 3, Page 77 in the text]
X is a set.
Tf contains all sets whose complements is either X or finite OR contains Ø and all sets
whose complement is finite.
Recall: pA X BqA “ AA Y B A and pA Y BqA “ AA X B A
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
3
Checking condition (2): Suppose tUα u is a collection of sets in Tf .
´ď ¯A č ` ˘
ď
Taking the complement of
Uα ñ
UαA ñ which must be finite because
Uα “
each of the UiA is finite.
˜
Checking condition (3):
n
č
i“1
¸A
Ui
“
n
ď
`
UiA
˘
each UiA is finite. ñ
ď`
˘
UiA must
i“1
be finite.
Basis for a Topology
Definition: If X is a set, a basis for a topology T on X is a collection B of subsets of
X [called “basis elements”] such that:
(1) Every x P X is in at least one set in B
(2) If x P X and x P B1 X B2 [where B1 , B2 are basis elements], then there is a basis
element B3 such that x P B3 Ă B1 X B2
Question: How in fact do you know that you get a topology from basis elements?
Examples: [of bases]
(i) Open intervals of the form pa, bq are a basis for the standard topology on R.
(ii) Interior of circle are a basis for the standard topology in R2 .
(iii) All one-point subsets of X are a basis for the discrete topology.
26 January 2012
Examples:
(i) X “ ta, b, c, d, eu, T1 “ tØ, X, tau , ta, buu, T2 “ tØ, X, tau , tcu , ta, bu , ta, b, cu , ta, cuu.
T2 Ą T1 , so T2 is “finer” than T1 . i.e., The set that is “finer” has more open sets.
We say, “tau is open in T1 ” which is equivalent to tau P T1 .
In T2 , tau is open. tau is not closed since tauA “ tb, c, d, eu is not open.
(ii) On the real number line, T is referred to as the “usual” or “standard” topology
on R:
(a) Sets in T are sets that are “open” in the calculus sense - i.e., p1, 3q, p0, 8q,
and p1, 3q Y p4, 6q.
(b) Sets in T are those that are unions of all sets of the from pa, bq where a ă b.
4
NOTES COMPILED BY KATO LA
Recall the definition of a basis: A basis for a topology pX, T q is a collection of subsets B
such that:
(1) @ x P X is in some B P B
(2) If x P X and x P B1 X B2 , then there is B3 such that x P B3 Ă B1 X B2
Note: Every basis “generates” a topology this way: A subset U is “open” if every element
in U satisfies: x P B Ă U for some B P B.
Three Classic Lemmas
Lemma 1: Every element in T is a union of basis elements.
Lemma 2: If pX, T q is a topological space and C is a collection of open sets such
that for every open set A, if x P A, then x P C Ă A for some C P C , then C is a basis.
Lemma 3: Let B and B 1 be bases for T and T 1 on X. Then T 1 is finer than T if
and only if for each x and each B P B containing x, there is B 1 P B 1 such that x P B 1 Ă B.
Further examples:
(i) R` is the “lower limit topology” on R where a basis is the set of all intervals of the
form ra, bq where a ă b.
Is (0, 1) open in R` ? i.e., Can we write (0, 1) as a union of sets of the form
ra, bq?
˙ „
˙
„
˙
„
1
1
1
,1 Y ,1 Y ¨¨¨ Y
,1 Y ¨¨¨
- Yes. (0, 1) =
2
3
n
How do R` and R compare? i.e., Can we write [0, 1) as a union of sets of the
form pa, bq?
- No. Consider using the same method as above. If we attempt to include zero, say
„
˙
1
´
, 1 Y ¨ ¨ ¨ , will include zero, but as well as erroneous values less than zero.
100
Every set open in R is open in R` , but not the converse. Thus, R` is finer than R.
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
5
(ii) Rk is the so-called “k-topology” where the basis elements are intervals of the form
"
*
1 1 1
pa, bq OR pa, bq z k where k “ 1, , , , ¨ ¨ ¨ . How do R and Rk compare?
2 3 4
Well, all sets open in R are open in Rk . Are sets open in Rk open in R as well?
˙ ˆ
˙ ˆ
˙
ˆ
1 1
1 1
1
?
,1 Y
,
Y
,
Y ¨¨¨.
i.e., p´1, 1q z k “ p´1, 0q Y
2
3 2
4 3
This is very close because 0 P left-hand side, but 0 R right-hand side. So Rk is
finer than R.
How do R` and Rk compare?
ˆ
˙ ˆ
˙
1
1
r0, 1q ‰
,1 Y
,1 Y ¨¨¨.
2
3
r0, 1q is open in R` , but not in Rk and p´1, 1q is open in Rk , but not in R` .
So R` and Rk are not comparable. i.e., Just like skew lines!
A topology on Z:
"
Open sets in the basis have two forms:
tnu
if n is odd
tn ´ 1, n, n ` 1u if n is even
This is the “Digital Line Topology .”
Is p3, 6q open? - No. p3, 6q “ t3u Y t3, 4, 5u Y t5u Y t5, 6, 7u. In fact, is any set like
t6u [“even,” in other words] open? - No.
But r3, 7s is open: r3, 7s “ t3, 4, 5u Y t5, 6, 7u
Note: t3u and t5u are open sets in the basis, but not necessary as they are included
in r3, 7s by t3, 4, 5u and t5, 6, 7u respectively in the digital line topology.
6
NOTES COMPILED BY KATO LA
2 February 2012
Basis ñ Topology
If we have a basis, we get a topology this way:
Definition - Basic Property [BP]: A set A is called open if every element in A is
in some basis element which is a subset of A. i.e., Definition of open when given a basis.
Note: If B is a basis element, is B open? - Yes! If x P B, then x P ljhn
B Ă B.
basis element
Do all sets U that satisfy the basic property form a topology?
(1) Does Ø satisfy BP? - Yes, vacuously. Does X satisfy BP? i.e., If x P X, is there a
B so that x P B Ă X? - Yes, by definition of a basis.
ď
(2) If tUα u is a collection of sets satisfying BP, does
Uα ? Since all Uα satisfy BP,
α
ď
for each Uα and x P Uα , there is a B such that x P B Ă Uα . So if x P
Uα , then
α
ď
x P Uα1 for at least one set. So there is a B such that x P B Ă Uα Ă
Uα .
α
(3) If U1 , ¨ ¨ ¨ , Un satisfy BP, does U1 X U2 X ¨ ¨ ¨ X Un satisfy BP?
Proof for Two Sets [Easily extended to n sets]:
U1 , U2 satisfy BP. ñ x P U1 , then x P B1 Ă U1 and similarly, x P U2 , then
x P B2 Ă U2 and Does U1 X U2 satisfy BP? i.e., Is the intersection open?
If x P U1 X U2 , then x P B1 Ă U1 and x P B2 Ă U2 . So x P B3 Ă U1 X U2 by
property (2) by the definition of as basis.
Proof of Lemma 1:
?
T generated by basis “ Unions of basis elements
pðq A set in the right-hand side is a union of open sets. So it is open, and thus, in the
left-hand side.
pñq A set in the left-hand side os open, so it satisfies BP. Form a union of all those
basis elements over all X in the set. Thus, it is a union of basis elements. ♣
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
7
The Order Topology
We digress to discuss order relations [Page 24 in the text]. :
Definition: A relation R on a set A is called an order relation or linear relation or
simple order if and only if
(1) If x, y P A with x ‰ y, then xRy or yRx
(2) xRx is never true
(3) If xRy and yRz, then xRz
Suppose we have two linear orders on ta, b, cu. Are they essentially the same relation or
not? In other words, are the following the same?
*
aRb, bRc, aRc
Same Order Type
cRb, bRa, cRa
One classic linear order on R ˆ R:
"
pa, bq ă pc, dq ðñ
a ă c or
a “ c and b ă d
This is usually referred to as the “Dictionary Order .” Notice there is no “smallest”
element.
An order relation on the unit circle:
"
pa, bq ă pc, dq ðñ
b ă d or
b “ d and a ă c
Is there a “smallest” one? - Yes, p0, ´1q. End of digression. :
Definition: Let X be a set with a linear order. Assume X has ě two elements. Then
the collection B is a basis for the order topology on X if B consists of all sets of the
following type:
(1) All open intervals pa, bq
(2) If X has a smallest element a0 , then all sets ra0 , bq
(3) If X has a largest element b0 , then all sets pa, b0 s
B is a basis. Check for confirmation [Check it out!] Is the order topology on R the usual
topology? - Yes! Parts (b) and (c) do not apply on R, so there is only (a) to follow!
8
NOTES COMPILED BY KATO LA
9 February 2012
Order Topology [Continued...]
Examples:
(1) Order Topology on R
(2) The dictionary order on R ˆ R. There is no first or last element.
(3) ta, bu ˆ Z` “ ta1 , a2 , a3 , ¨ ¨ ¨ ; b1 , b2 , b3 , ¨ ¨ ¨ u
pa1 , a4 q “ ta2 , a3 u
pa6 , b1 q “ ta7 , a8 , ¨ ¨ ¨ u
This topology has a first element, but no last element. We can even look at sets of
the form:
ra1 , a4 q “ ta1 , a2 , a3 u
ra1 , a2 q “ ta1 u
The Product Topology
Definition: If X and Y are topological spaces, the product topology, X ˆ Y , is the
topology having as a basis all sets of the form U ˆ V , where U is open in X and V is open
in Y .
Example: R ˆ R
The product topology on R ˆ R has basis elements of the form U ˆ V which are open
rectangles. The usual topology on R has basis elements which are open disks. It turns out
these two topologies are equivalent. By Lemma 3, to compare topologies, we can compare
their respective basis elements. i.e., Given a point in one basis element of a topology, can I
fit a basis element from a different topology in the original topology basis? In our example,
Given a basis element in R ˆ R [an open rectangle], given any point in our open rectangle,
can I fit an open disk around that point that is completely contained in the open rectangle?
- Yes! And vice versa.
Another example is to ask ourselves,“What is an example of a basis given a topology?”
Question: Is “a basis all sets of the form U ˆ V , where U is open in X and V is open in
Y ” really a basis? That is,
open in X open in Y
hnlj
(1) Is every element x in some set U
xPX ˆY
hnlj
ˆ
V ?
some V -type set
(2) If x P pU1 ˆ V1 q X pU2 ˆ V2 q, then x P pU1 X U2 q ˆ pV1 X V2 q Ă pU1 ˆ V1 q X pU2 ˆ V2 q
some U -type set
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
9
Theorem: If B1 is a basis for X1 , B2 a basis for Y , then a basis for X ˆ Y is all sets of
the form
tB1 ˆ B2 | B1 P B1 , B2 P B2 u
Proof is left to the reader.
Subspaces
Definition: Let pX, T q be a topological space. If Y Ă X, then the subspace topology
on Y consists of open sets of the form U X Y , where U is open in X.
Question: Is the above definition really a topology?
(1) Ø “ ljhn
Ø X Y ; ljhn
Y X Y.
open in X
open in X
(2) If tUα X Y u are all in the topology, is
ď
Uα XY open? ñ
˜
ď
ď
pUα XY q “
¸
Uα
XY.
α
α
open in X
(3) If U1 X Y, U2 X Y, ¨ ¨ ¨ , Un X Y are open, is
˜
¸
n
č
Ui X Y .
n
č
pUi X Y q open? But
č
pUi X Y q “
i“1
i“1
open in X
Examples:
(1) Consider R with the usual topology and subset Z. What topology does Z have as
a subspace of this R?
ˆ
˙
1 1
t3u is open because t3u “ 2 , 3
X Z.
2 2
In fact, every point-set is open. Every set is open. This is the discrete topology.
(
(2) Let T “ R2 with the usual topology. Let S “ px1 , x2 q P R2 | x21 ` x22 “ 1 . Let
A “ tpx1 , x2 q P S | x1 ą 0u.
Is A open in T ? - No, it cannot be written as the union of interior of circles.
Equivalently, a point in A does not lie inside a circle, which is completely in A.
Is S open in T ? - No, because x21 ` x22 “ 1, if the equality was less than or equal,
then yes.
Is A open in S if S has the subspace topology
č in relation to T ? - In subspace S of
T , an open set is any normal open set in
S. Thus, A is open in the subspace
topology.
10
NOTES COMPILED BY KATO LA
16 February 2012
Lemma: Let Y be a subspace of X. If U is open in Y and Y is open in X, then U is open
in X.
Proof : Suppose U is open Y . i.e., Y intersect with something originally open in the
original set. Then U “ Y X U 1 , which is open in X. Remember: Any open set in a
subspace is the intersection of the subspace with an open set in X. So U is open in X
because it is the intersection of two sets, both open in X.
Theorem: If A is a subspace of X, B is a subspace of Y , then A ˆ B has the same
topology as A ˆ B inherits as a subspace of X ˆ Y .
Proof is left to the reader.
Examples:
(1) Z ˆ Z in the order topology on Z ô Z with the discrete topology. p2, 1qpoint “
p1, 3qinterval ˆ p0, 2qinterval .
But consider the set Z ˆ Z in R2 . What topology does it have if we think of it as
a subspace of R2 with the usual topology? - Any point p can be written as the
intersection of an open set in R2 with ZˆZ. So any tpu is open. This is the discrete
topology.
(2) Consider R with the usual topology and Y “ r0, 1s. Y is not open, which may
cause a problem.
(a) Whatˆis Y in
 the order topology? - Open sets are of the form: pa, bq, r0, bq, and pa, 1s.
1 3
, .
Not
2 4
(b) What is Y in the subspace topology that it gets [inherits] from R, the order
topology? - Here, the open sets are normal open sets intersected with Y :
pa, bq, r0, bq, and pa, 1s. We get precisely the same sets! Note this is not
always the case.
(3) Y “ r0, 1s Y t2u. What is Y in the order topology? - Basic open sets are of the
form: pa, bq, r0, bq, and pa, 2s. t2u is not open in the order topology. What is Y in
the topology
it˙inherits as a subspace of R with the usual topology? - t2u is open:
ˆ
1 1
X Y . t2u in both topologies stated above are not the same.
t2u “ 1 , 2
2 2
(4) Do t1, 2u ˆ Z` and Z` ˆ t1, 2u have the same order type in the dictionary order
topology? - No. It is helpful to simply state where the topologies differ. For
example, the first topology has order type ω 2 , intervals where there is an infinite
amount of points between the endpoints, and two points have no immediate predecessors. While the second topology has order type ω, intervals where there is a finite
number of elements between the end points, and every element has an immediate
predecessor except (1, 1).
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
11
(5) Consider I “ r0, 1s and I ˆ I. Is I ˆ I as a subspace of R ˆ R with the dictionary
order topology the same as I ˆ I with its own dictionary order topology? - The
short answer is no. Consider the first topology as T1 and the second topology as
T2 . Everything open in T2 is open in T1 . Though not proven, it appears convincing
that T2 Ă T1 .
Closed Sets
Definition: A set in a topological space is called closed if and only if its complement is
open.
Examples:
(1) In R with the usual topology, all singleton sets are closed: t0u is closed, and its
complement is p´8,(0q Y p0, 8q, which is open.
(2) px, yq | x2 ` y 2 “ 1 in R2 with the usual topology. - Take any point not in the
circle,
we can*make an open set around said point that does not touch the circle.
"
1 1 1
, , , ¨ ¨ ¨ in R with the usual topology. - The complement is not open. So the
(3)
2 3 4
set, call it K, is not closed. Is K open? - No.
(4) r0, 1q is not open or closed.
(5) Y “ r0, 1s Y p2, 3q in the subspace topology of R with the usual topology. What
is p2,ˆ3q, open
˙ or closed? - Open. Y X p2, 3q. r0, 1s is open as well: r0, 1s “
1
Y X ´1, 1 . In Y , the complement of r0, 1s is p2, 3q. So, r0, 1s being open, p2, 3q
2
must be closed. In fact, both r0, 1s and p2, 3q are both open and closed.
Question: Give an example of a topology with a closed set A which becomes open when
one point is removed.
Using R with the usual topology: A “ txu Ñ Ø or A1 “ r1, 8q Ñ p1, 8q or A2 “
R Ñ p´8, xq Y px, 8q.
12
NOTES COMPILED BY KATO LA
23 February 2012
Facts about Closed Sets
Theorem: In a topological space X,
(1) Ø, X are closed
(2) An arbitrary intersection of closed sets, is closed.
(3) The union of a finite number of closed sets, is closed.
Theorem: If Y is a subspace of X, A is closed in Y , Y is closed in X, then A is closed in X.
Theorem: If Y is a subspace of X and A is closed in Y , then A “ B X Y where B
is closed in X [and conversely].
Note: It is noteworthy to compare the above theorems with what we have already learned
about open sets and see how they are alike and differ.
Closure and Interior
Definition: The closure of a set A, denoted A, is the intersection of all the closed sets
that contain A.
Note: The above definition is “nice” because we see at once that A is closed. It is a “poor”
definition because envisioning all the closed sets containing A can be difficult.
Definition: The interior of A is the union of all open subsets of A.
Note: The interior of A is open. A open ô A = interior of A, A closed ô A “ A.
Examples:
"
(1) A “
*
1 1 1
, , , ¨ ¨ ¨ ; Closure: A Y t0u; Interior: Ø.
2 3 4
(2) A “ Q; Closure: R; Interior: Ø.
(3) Weird topological space - X “ ta, b, cu; Open sets: Ø, X, tau, ta, bu; If A “ tau;
Closed sets: X, Ø, tb, cu, tcu; If A “ tau, B “ tbu; A “ X, B “ X X tb, cu “ tb, cu.
Theorem: If Y is a subspace of X, A a subset of Y , A is the closure of A in X, then the
closure of A in Y is A X Y .
ˆ
˙
„

1
1
Question: If A Ă B, is A Ă B? - A “
, 1 , B “ p0, 1q. A “
, 1 Ć B. Now, if
2
2
A Ă B, is A Ă B?
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
13
“Big” Theorem: Let A be a subset of X, a topological space:
(1) x P A ô Every open set containing x intersects A.
(2) x P A ô Every basis element containing x intersects A.
Note: The implication of this writing style is to ensure the intersections are nonempty.
Alternative Wording: x P A ô Every neighborhood of x intersects A.
this the case?
Why is
contrapositive
A Word on Logic: P ñ Q ô
Qñ
P ; P ô Q is P ñ Q and Q ñ P
conditional statement
The contrapositive of the alternative wording is as follows: x R A ô D a neighborhood
of x that does not intersect A.
Proof : Suppose x R A is the intersection of all closed sets containing A. Then x is
not in at least on closed set B that contains A : x R B. So x P B A . Since B is closed,
it contains A. Then B A is open not containing A. ñ A neighborhood of x that does not
intersect A. ♣.
Limit Points
Definition: x is a limit point of A if and only if every neighborhood of x intersects A in
a point other than x itself.
Examples:
A
A
Set of Limit Points pA1 q
r0, 1s
r0, 1q
r0, 1s
t0u
R
t0u
Q
Ø
R
"
*
1 1
0, , , ¨ ¨ ¨
2 3
"
1 1
, ,¨¨¨
2 3
*
t0u
14
NOTES COMPILED BY KATO LA
Main Result: A “ A Y A1
Proof :
pñq x P A ñ Every neighborhood of x intersects A.
If x P A, then x P A Y A1 .
If x R A, then every neighborhood of x intersects A in a point other than x itself So x P A1 .
pðq x P A Y A1 .
If x P A, then x P A because A Ă A.
If x P A1 , then every neighborhood of x intersects A - So x P A. ♣
If R has the discrete topology, then whatever A is, ljhn
A “
č
Bα
All closed sets con-
Will equal A in this topology
taining A.
The minimal separation axiom is T0 .
T0 : Given two distinct points x and y, there is either an open set containing x and not y
or an open set containing y and not x.
T1 : The same condition as T0 , but the “or” condition becomes and .
T2 [Hausdorff Space]: Given two distinct points x and y, there are disjoint open sets A and
B such that x P A and y P B.
What do we mean when we say that a sequence tx1 , x2 , x3 , ¨ ¨ ¨ u in a topological space
converges to a point x? - It means that every neighborhood of x contains all points in
the sequence from some xn on.
Odd behavior in a non-T2 space:
X “ ta, b, cu, Open sets: Ø, X, ta, bu, tb, cu, tbu. What does tb, b, b, ¨ ¨ ¨ u converge to?
Claim: The sequence converges to b.
Neighborhoods of b: X, ta, bu, tb, cu, tbu. i.e., Every term is in this neighborhood of b.
Some for all neighborhoods. So the sequence converges to b.
In fact, tb, b, b, ¨ ¨ ¨ u converges to a, to b, and ti c!
But this is not a Hausdorff space because we can find an open set around b, but any open
set around a will contain b.
Munkres explains in a Hausdorff space, sequences have unique limits and singleton sets
are always closed. Why is, say txu, always closed in a Hausdorff space?
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
15
22 March 2012
Functions
f : ljhn
A ÞÝÑ B. We can speak of f pT q where T is a subset of the domain.
domain
“12 “22
Example: f : R ÞÝÑ R by f pxq “
x2 .
If A “ p1, 2q, then f pAq “ p 1 , 4 q.
ljhn
image set of f on everything in A
We also write f ´1 pAq to represent all elements that f maps to A. Above, for example, f ´1 pp1, 4qq “ p´2, ´1q Y p1, 2q.
Continuous Functions
Calculus: F is continuous at “a” if: @ ε ą 0, D δ ą 0 such that |x´a| ă δ ñ |f pxq´f paq| ă
ε.
“f is continuous.” ðñ “f is continuous for every ‘a’ in the domain.”
Topologically:
$
’
’
’
’
&
f ´1 pAq is open whenever A is open
OR
f is continuous ðñ
’
’
’
’
%
the inverse image of every open set is open
If f : pX, T1 q ÞÝÑ pY, T2 q, then f is continuous if and only if the inverse image of every open set in Y is an open set in X.
Question: What does it mean to claim that a function f : X ÞÝÑ Y is not continuous? - For at least one open set in Y , f ´1 pY q is not open.
16
NOTES COMPILED BY KATO LA
$
1
’
’
& x´ 2
Suppose f pxq “
’ 2
’
% x´ 2
3
3
ˆ
f ´1 pp0, 3qq “
if x ă 2
in R2
if x ě 2
˙
ˆ
˙
˙
„
1
1 1
1
, 3 . Now, let A “ 1 , 3 . Then f ´1 pAq “ 2, 3 .
2 2
2
2
open
not open
"
What about f pxq “
0 if x ă 2
in R2
1 if x ě 2
What set in A provides a counter-example of showing f is not continuous?
˙
ˆ
1 1
, 1 , then f ´1 pAq “ r2, 8s
- Let A “
2 2
not open
open
Next, consider R in the usual topology and X “ ta, b, cu with open sets: tX, Ø, tau, ta, buu.
f : R ÞÝÑ X this way: f p1q “ b, f pxq “ a, if x ‰ 1. Is f ´1 pAq open when A is open?
f ´1 pØq “ Ø
f ´1 pXq “ R
,
/
/
/
/
/
/
/
/
.
all open!
/
/
f ´1 ptauq “ R z t1u “ p´8, 1q Y p1, 8q /
/
/
/
/
/
´1
f pta, buq “ R
Note that a maps to all numbers except 1.
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
17
Suppose we have the same whole space X on the usual topology on R and
$
& a ÞÑ 1
b ÞÑ 2 Is f continuous?
f pxq “
%
c ÞÑ 3
ˆˆ
f ´1
ˆˆ
f
´1
1 1
,1
2 2
˙˙
1 1
2 ,3
2 2
open
ˆˆ
“ tau which is open, so
f ´1
1 1
,1
2 2
˙˙
is open.
˙˙
“ tcu ñ So f is not continuous.
ljhn
not open
Identity Maps
Definition: The so-called identity map is defined as ipxq “ x.
Suppose i : X ÞÝÑ X. Will it always be continuous? Note: If i : R ÞÝÑ R, then i is
continuous. In fact, it is our usual y “ x.
Example: Let X be the domain R in the usual topology. Let Y be the range Rl , the
lower limit topology ra, bq.
r0, 1q ÞÝÑ r0, 1q
“f ´1 pAq
A
Notice f ´1 pAq is not open in R under the usual topology. A is open under the lower limit
topology. Thus, i is not continuous.
Now, suppose i : Rl ÞÝÑ R. Then i is an identity function as it takes pa, bq ÞÝÑ pa, bq.
open
open
If we have one set, two topologies, and i : pX, T1 q ÞÝÑ pX, T2 q and i is continuous [as
in our preceding example], what can we deduce about T1 and T2 ? - T1 Ą T2 . i.e., T1
must be finer than T2 .
Theorem 18.1: Several statements equivalent to “f is continuous.”
We will prove the following: f : X ÞÝÑ Y is continuous ðñ @ x P X and each neighborhood V of f pxq, there is a neighborhood U of f pxq such that f pU q Ă V .
18
NOTES COMPILED BY KATO LA
Proof :
(ñ) Suppose f is continuous. Let x P X and let V be a neighborhood of f pxq. f is
continuous, so f ´1 pV q is open in X. Let U “ f ´1 pV q.
Now f pU q “ f pf ´1 pV qq “ V . So, f pU q Ă V . [In fact, equality holds here.]
f ´1 pf pV qqdoes not always equal V
(ð) Suppose V is open in Y . Why is f ´1 pV q open in X? For each x P f ´1 pV q D an open
neighborhood that maps into V . Take the union of all these open neighborhoods.
The union will be open and equal to f ´1 pV q. So f ´1 pV q is open.
Definition: A homeomorphism is one-to-one and onto. Two “homeomorphic” spaces
are ones that are topologically equivalent. Given f : X ÞÝÑ Y , f continuous, and f ´1
continuous. f is a homeomorphism and X, Y are homeomorphic.
Example: p0, 1q and p10, 20q as interval subspaces of R under the usual topology are homeomorphic. What is a one-to-one and onto map: p0, 1q ÞÝÑ p10, 20q which is continuous and
for which the inverse function is continuous?
29 March 2012
Metric Spaces
Definition: A metric on a set X is a function d : X ˆ X ÞÝÑ R such that
(1) dpx, yq ě 0 and dpx, yq “ 0 ô x “ y
(2) dpx, yq “ dpy, xq
(3) dpx, zq ď dpx, yq ` dpy, zq
Every metric gives rise to a topology on X through the use of “epsilon-balls,” [ε-balls]
denoted Bpx, εq “ ty P X | dpx, yq ă εu which is the set of all points whose distance from
x is less than epsilon.
Once we know that ε-balls form a basis, we can get all the open sets by taking union
of the ε-balls.
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
19
Examples:
(i)
"
If dpx, yq “
ˆ
1
B x,
2
˙
“ txu
0
if
x“y
1 otherwise
ˆ
˙
1
This is open. Thus, B x,
is open. ñ txu is open.
2
This metric generates the discrete topology.
(ii) Let P “ px1 , y1 q, Q “ px2 , y2 q, and dpP, Qq “
ε-balls generate the usual topology on R2 .
a
px1 ´ x2 q2 ` py1 ´ y2 q2 . These
Do these ε-balls indeed for a basis?
(1) If x P X, then x P Bpx, 1q since dpx, xq ă 1.
Before proving the next [and last] basis property, we would like to justify the
following statement:
If y P Bpx, εq, then there is a Bpy, δq such that Bpy, δq Ă Bpx, εq
For δ, choose something that is less than ε ´ δpx, yq. Then, Bpy, δq Ă Bpx, εq.
Why? If z P Bpy, δq, then is z P Bpx, εq?
dpx, zq ď dpx, yq ` dpy, zq
ă dpx, yq ` δ
ă dpx, yq ` pε ´ dpx, yqq
“ε
(2) If x P By py, ε1 q X Bz pz, ε2 q, then x P Bx px, ?q Ă By py, ε1 q X Bz pz, ε2 q. What is “?”.
Look at ε1 ´ dpx, yq and ε2 ´ dpx, zq. Use the smaller of the two for “?”.
Definition: A metric space is a topological space that has a metric whose ε-balls generate the space in question. A space for which no metric can generate the open sets is called
non-metrizable.
20
NOTES COMPILED BY KATO LA
Examples:
˙
ˆ
`
˘
1
“ 12 , 1 21 .
(i) R with dpx, yq “ |x ´ y|. Open balls are open intervals. e.g., B 1,
2
The metric topology is the usual topology on R.
a
(ii) dpP, Qq “ px1 ´ x2 q2 ` py1 ´ y2 q2 on R2 gives the usual topology on R2 . This
metric is sometimes called the 2-metric and can be extended to Rn .
(iii) On R2 : dpP, Qq “ |x1 ´ x2 | ` |y1 ´ y2 |. This is known as the 1-metric or the
“taxicab metric.”
(iv) On R2 : dpP, Qq “ max t|x2 ´ x1 |, |y2 ´ y1 |u. This is the “infinity metric,” denoted d8 .
One way to begin understanding a strange metric is to look at a few specific
ε-balls. For (ii), (iii), and (iv), what are the points which are, say one unit
away from p0, 0q?
It turns out (ii) yields the usual open unit circle, (iii) yields an open rhombus,
and (iv) yields an open square.
If Cr0, 1s is the set of all continuous functions: r0, 1s ÞÝÑ R, then
ż1
|f pxq ´ gpxq| dx
(v) d1 pf, gq “
0
„ż 1
(vi) d2 pf, gq “

pf pxq ´ gpxqq2 dx
0
(vii) d8 pf, gq “ max |f pxq ´ gpxq|
0ďxď1
1
2
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
21
Theorem: Every metric space is Hausdorff.
Proof : Let p, q P X where p ‰ q and X is a metric space. Let ε “
Bpp, εq and Bpq, εq. Is Bpp, εq X Bpq, εq “ Ø?
1
dpp, qq and consider
2
Suppose r P Bpp, εq X Bpq, εq, then
dpp, qq ď dpp, rq ` dpr, qq
ăε`ε
“ 2ε
“ dpp, qq
ñ dpp, qq ă dpp, qq ñð
But this is a contradiction! Thus, there must have been nothing in the intersection from
the beginning. ♣
It is common [or as Smith stated: “It is not uncommon”] to define a new metric in terms
of a familiar metric, particularly if the familiar metric has some undesirable property.
Suppose we have an unbounded metric. We can turn it into a bounded metric that
generates the same topology.
Note: A bounded metric is one which always gives values between M and ´M for some
M P Z.
Example: If d is unbounded, the define d˚ px, yq “ mintdpx, yq, 1u. The above is a metric
[Check it out!], and its values are less than or equal to one. But d˚ will generate the same
topology as d.
22
NOTES COMPILED BY KATO LA
12 April 2012
Chapter 3 - Connectedness & Compactness
Connected Topological Spaces
Consider R under the usual topology. r0, 1s Y r2, 3s is not connected; R is connected;
p´8, 0q Y p0, 8q is not connected; and p´8, 0s Y r0, 8q “ R.
ˇ ˇ
ˇ1ˇ
2
In R : (The y-axis) Y The graph of y “ ˇˇ ˇˇ is not connected. This is the usual asx
1
ymptotic graph of reflected along the y-axis.
x
ˆ ˙
1
Consider (The y-axis) Y The graph of y “ sin
where x ą 0. Is this connected?
x
Definition: A topological space is separated [or not connected or disconnected] if it
can be written as the union to two disjoint, nonempty, open sets:
X “U YV
open and nonempty
A topological space is connected if it cannot be written as a separation.
Facts:
(1) X is connected ô the only sets that are both open and closed are X and Ø.
(1.5) X is not connected ô D proper, nonempty subsets that are both open and closed.
(2) X is not connected ô There are sets U and V which are disjoint, nonempty, and
neither set contains a limit point of the other.
Reiteration: X “ U Y V . V is closed, so it contains all its limit points. So
none of these limit points an be in U since U X V “ Ø.
ˆ ˙
1
, this actually is connected
Note: Addressing (The y-axis) Y The graph of y “ sin
x
by fact (2) [or the failure of fact (2)].
ADVANCE TOPICS IN TOPOLOGY - POINT-SET
23
19 April 2012
Lemma: If U, V are a separation of X and if Y is a connected subspace of X, then either
Y Ă U or Y Ă V .
Proof : Suppose X “ U Y V where U, V are open, non-empty, disjoint sets of X. Consider
Y X U, Y X V . These sets are open in Y and are disjoint. Both these sets cannotboth be
non-empty or Y would not be connected. Thus, Y X U, Y X V is empty - Say Y X U “ Ø.
This means that Y X V “ Y which tells use that Y Ă V . ♣
Theorem: The image of a connected space under a continuous map is connected.
Proof : Let f : X ÞÝÑ Y be continuous. Suppose f pXq “ Z. Then f : X ÞÝÑ Z is
onto and continuous. [Use the face that inverse image of open sets is open].
Compact Spaces
Definition: An open covering of a subspace or space is a collection of open sets whose
union is the
ˆ subspace or˙ space. e.g., Let X “ R, Y “ p0, 1q. If we consider all intervals of
1
1
where 0 ă a ă 1, we have an open covering of Y .
the form a ´ , a `
3
3
Some spaces have the following property: For every open covering, there is a finite subcollection which is still a covering. Such spaces are called compact.
Example: Y “ p0, 1q is not compact subset of R under the usual topology - Although
the covering just given does have a finite subcovering, not every covering of p0, 1q has a
ˆ
˙ ˆ
˙ ˆ
˙
˙
8 ˆ
ď
1
1
1
1
finite subcover. e.g.,
,1 ,
,1 ,
,1 ,¨¨¨ “
, 1 Ă p0, 1q is an open cover2
3
4
n
n“1
ing, but there does not exist a finite number of elements that will still cover p0, 1q.
However, r0, 1s is compact.
Proof : Suppose r0, 1s has an open covering. D an a P r0, 1s such that r0, as is contained in
a finite number of these open sets. What is the largest a such that r0, as is covered by a
finite number of these open sets? Could a ă 1? - No, because a would be “inside” some
open interval that covers a and that means we could extend r0, as even further to the
right. Contradiction. 6 a “ 1. ♣
24
NOTES COMPILED BY KATO LA
Theorem: The image of a compact space under a continuous map is compact.
Proof : Proof is on page 116 in the text.
In R with the usual topology:
(a) A subspace Y is connected ô Y is in an interval
(b) A subspace Y is compact ô Y is closed and bounded
Why is (a) true?
Proof :
pñq Suppose Y Ă R is connected. Is it an interval?
connected ñ interval ô interval ñ connected
A is not an interval ñ There is a, b, t P A and a ă t ă b such that t R A.
e.g., U “ p´8, tq, V “ pt, 8q and Y X U, Y X V is a separation of Y .
pðq Suppose Y is an interval. Is it connected?
Suppose Y “ p´8, aq. If Y is not connected, then Y “ U Y V where U, V are
open, non-empty, and disjoint sets. Let t “ sup V . Can t “ a? Suppose t “ a. Is
t P U or t P V ? If t P U , then D pa ´ ε, as Ă U , etcetera.
Think about this: One can argue the entire interval p´8, as must be in U . So V ‰ Ø.
Then, what if t P V ? What about other interval types?