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General Topology
§1. Topological Space
1.1 Definitions of topology & topological space
Definition 1. Topology on set : T is a collection of subsets
in X, satisfied that a) Ø, X T ;
b) intersection of any
finite number of elements of T belongs to T ;
c) union of elements of any subfamily of T belongs to T .
Remark: T includes Ø, X, and closed under finite
intersection & arbitrary union .
Example 1 Anti-discrete topology : {Ø, X } (trivial topology).
Ex 2 Discrete topology :
2 X (Exp X) , all subsets of X.
Ex 3 Sierpinski topology: X={0,1}, T ={Ø, {0}, X }.
Definition 2. Topology space
are called open sets.
( X ,T );
the elements of T
For any set, normally we can define several different topologies
Definition 3 Let T , T 2 be the topologies on set X, if T 1 T 2 ,
T 1 is weaker or coarser than T 2 . T 2 is stronger or finer than T 1 .
1
Let T : A be a collection of topologies on X.
T = T : A=U X :U T for all A is weaker than
all the T .
1.2 Sub-bases & bases
LetE be a collection of subsets of a set X . LetBE be the
smallest collection of subsets of X which contains all the
elements of E together with the empty set and the whole
set X and which is closed under finite intersection. Let
T E { : BE } , then
T E is a topology on X , the smallest topology containing E .
The collection E is called a subbase of the topology T E and
we say that the topology T E is generated by the subbase E .
Definition 3. A collection B of open subsets of a topology
space X is called a base of the space if each open set in X is a
union of some elements of B .
Since T is closed under arbitrary union . T itself is also
a base. We want to get a base of smaller cardinality.
Theorem B is a base of (X ,T ) if and only if all the
elements of B are open and for any x∈X, open set U containing
x, there exists V ∈B such that x ∈V
U.
Proof : For any x∈X , open set U containing x , U is the
union of some elements B , i.e. B ' B such that
U = V :V B ' , x∈U , then V B ', x ∈V. It is
clear that V U .
For any open set U and any x U ,
xVx U , thus U xU Vx. U.
Vx B
such that
Topology T is uniquely determined by its base B , and
T = : B
As is known, single topology T can have different bases .
Uncountable T can have countable base.
Proposition 1. A collection B of subsets of a set X is a base of
a topology on X if and only if, B X and for any U, V B
and any x U V , there exists W B such that x W U V.
Proof : ( ) X T , hence X B . U ,V B U V . T
Therefore, for x U V , W B , x W U V .
( ) Let T U X : BU B s.t. U = BU .;, X T
and T is closed under arbitrary union. For U , V T .
we prove that U V T . For any x U V , since U BU
V = BV , U 1 ,V1 B such that x U 1 V1 , then Wx B
such that xWx U1 V1 . Thus U V = xU V Wx .
We let BU V Wx , xU V , then U V = BU V. This shows that
T is closed under finite intersection. T is a topology. It is easy
to see that B is a base for T .
Note: (1) A collection B of subsets of a set X is a base of
a topology on X if B X , and B is closed under finite
intersection.
(2) BE in definition of sub-base is a base of T E .
Example 4. Given a set R. B
. ( a,b , a x isba base for
the usual topology T . One can show that U T if and only if
U is a union of pairwise disjoint intervals .
2
Example 5. Given a plane R 2 , B
for natural topology on plane,
where B x {(x,y): x 2 + y 2 } .
B (x):x R
is a base
Example 6 Bs =a,b : a x b . It is clear that Bs is closed
under finite intersection & Bs R , form a base for T S .
The topological space ( R ,Bs ) is called the Sorgenfrey line.
1
(a, b)
a , b , so
T S is stronger than the
n
usual topology T on the line R . T T S and T T S .
T S does not have a countable base.
n =1
B0 B
B0 P
B0
Proof : Let t= U,V , U , V P . Let :P P B , t t
t W such that V W U if such W exists; otherwise (t)= .
Let B0 (t) : (t) . B0 B and B0 c P P c P . B0 is a
base for T .
How to prove that φ is surjective and B is a bsae?
We can prove it as follows: for any open set O and xO , there
exists U P such that xU O . Choose W B with xW U
Choose V P with xV W U .We can do this since both P
and B are bases. By definition of . Let t= U ,V and
W =(t)B0 , then x (t) O .Thus B0 is a base since x , O are
arbitrarily chosen.
Corollary No countable subset of Bs can be a base for X,T s .
Let
A {a : b R , a,b } ,
if it is countable . Choose c,d
cR\ A and c d , then c,d Bs can not be written as union
of elements in .
If there is some a < c, then…. Otherwise, …
1.3 Neighborhood of points. Nearness of a point to a set and
the closure operator
Definition 4 A set O X is called a neighborhood of a point x
in the topological space X if there exists an open subset U of X
such that xO U .
Remark 1. Open set is neighborhood of every point it contains.
Remark 2. Intersection of two nbds is again a neighborhood.
Neighborhood is a “measure” of nearness to the point , also it is an
approximation to the point. There is no distance in topological
spaces.
Definition 5. Let X,T be a topological space and let A X
and x X .We say that the point x is near to(or at distance zero
from) the set A and write x A (at distance 0) if for every open
nbhd U of x , U A .
Otherwise we say x is far from A ,write x A (or x,A 1 ) .
is said to be induced by topology T , to emphasize this we
write δ= δ(T ) .
Properties of : x X ,
1) x ( x, 1 )
3) x A B x A x B
A X
,
2)
B X
x A x A
4) x B yB y A x A (weak form of triangle axiom)
In condition 3, if a point far from A B , then it is also far
from A and B . In condition 4 , the transitivity of the nearness
relation 0+0=0 .
The set of all points near to a subset A X is called the closure
of A in the topological space X. Thus , A x X : x A .The map
A A is called the closure operator in the topological space X.
The closure operator possesses the following properties
1)
2) A A for all A X
3) A B A B
4) A A
Definition : The set which contains all points near to it is said
to be closed. A is closed if and only if A A .
Proposition 3 A set A is closed in a topological space X,T if
and only if X\A is open in X,T
Proof : A= A {x : x A} For y X \ A , there exists a nbhd U of
y,such that U A= , hence y U X \ A
every point y
in X\A is its interior point X\A is open.
By De Morgan law, closedness is closed under finite union &
arbitrary intersection.
Note that: A is a closed set ( A A). and is the smallest
closed set containing A.
Example 7. Let X,T be an anti-discrete space. T =,X .
For any nonempty A, A X . In discrete space, T 2 X , A= A .
In Sorgenfrey line R ,T s , R R .
Proposition 4. If U and V are disjoint open subsets of X , then
U V ( and V U )
Proof: For any xV , V itself is an open neighborhood of x
and V U = . Hence xU by the definition, U V .
1.4 Definition of a Topology using a Nearness relation or a
closure operator
An (abstract) closure operator on X is a rule which assigns to
each subset A of X . Map 2 X 2 X, A A , satisfies
1)
2) A A for all A X
3) A B A B
4) A A
Closure operator can be obtained via (abstract) nearness relation
by defining A x X : x.A
Let F A X A= A ,then T X \ A AF is a topology on X .
Proposition 5. The topology of a space X is uniquely determined
by any one of the following objects: the nearness relation , the
closure operator , or the collection of all closed sets.
Proof: Only for closure operator. Let T X \ A AF
1) X \ = X T ; X X hence X X , X\ X T .
2) A,B T X \ A X \ A X \ B X \ B
X \ A B X \ A X \ B X \ A X \ B X \ A X \ B
= X \ A B . Hence A BT .
3) Suppose that G T . For any AG , X \ A X \ A . On the other
hand for any A1 G , we have X \ AG A AG X \ A X\ A1 .
Hence X \ AG A X\ A1 X\ A1 and
X \
AG
A
It is clear that
X \
AG
A X \
A1G
X \
AG
X \ A1 X \
AG
A
A X \
,
AG
A1G
AG
A
AT
A1 X \
AG
A
, therefore
.
Thus T is a topology on X , and - is the closure operator. uniquely determines the closed subsets in X, hence uniquely
determines its topology .
Definition. For any set A X , the derived set of A is denoted
by d(A) , it consists of all its limit points (accumulation point).
A point x is called to be a limit point of A if for any neighborhood
O of x , O A\x .
We have that A A d A by definition of the derived set.
Four properties of derived set
1) d
3) d A B d A d B
2) A B d A d B
4) d d A A d A A
The above four axioms uniquely determined a topology T such
that in this topological space, the derived set of A is exactly d(A) .
The relation of interior , closure and complement
, -,'
IntA A x A| an open nbhd O of x , O A , which is an open set
Remark: (1) A is open iff A = Int(A); (2) Int(A) is the union of
all open sets contained in A.
1)
3) A
, X X
B A B
2) A A
4) A A
Theorem 1. A A' - ' , A- A' ' .
Proof : For any x A , A is a neighborhood of x. Since A A' ,
then x A' - x A' - ' . Thus A A' - ' .
On the other hand , for any x A' - ' , we have x A' - . Then
there exists a neighborhood V of x , such that V A' V A .
That is A' - ' A .
Theorem 2 (C.Y.Yang) d(A) is closed
d{x} is closed. P 77 6.
Û
Theorem 3 (Kuratowski)
P 81 4.
Definition 2 Boundary set of A, Bd(A):
Bd (A)= ¶ (A)= A
\A .
Bd ( A) = A \ A = A Ç ( X \ A ) = A Ç A'- = Bd ( A¢).
Bd ( A) = A Ç ( X \ A)- .
Definition 3. The neighborhood system U x of point x : the
collection of all x's neighborhoods. It satisfies
1) For any U x , U U . xU
2) Closed under finite intersection: U,V U x , U V U .
3) Closing up : U U x , U V V U .
4) Contains an open neighborhood V (V is a neighborhood of all
its points) U Ux V Ux V U y V V U y
Neighborhood system uniquely determine a topology T and in
(X,T ) , the neighborhood system coincide with pre-assigned one.
1.5
Subspaces of topological space
To each subset Y of a topological space (X, T ) associated a
new topological space Y , T |Y where T | U Y : U T is the
set of all “traces” in Y of the open subsets of X. T |Y is said to
be the topology generated(or induced) by T and Y , T |Y is
called a subspace of (X, T ) .
Y
Properties of subspace can be very different from the whole
space.
Any figure on the plane: disk, circle, disk with a hole.
Q, J rational & irrational number with induced usual topology.
Thm. If B is a base of (X, T ), then B ' =
base of the subspace (Y, T Y ).
{B Ç Y : B Î B }
is a
Proof: (1) Obviously, each U Y B is open in Y.
(2) Suppose x U and U is open in Y, we have U = V Y
where V is open in X. By the definition of bases, B B such
that x B V. It follows that x B Y V Y = U.
Thm. 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.
Relative nature of closed set . A topological space (X, T ) ,
(Y , T Y ) is a subspace. (1) A Ì Y can be closed in Y, but not in X.
(2) clY (A)= clX (A)Ç Y for A Ì Y Ì X .
A = cl X (A)
Proof: (1) Let x cly(A), we prove that x clx(A). For each
nbhd U of x in X, U Y is a nbhd of x in Y. So, (U Y) A =
. That is U A . Therefore, x clx(A) Y.
Let x clx(A) Y. For each nbhd V of x in Y, nbhd U of x in
X s.t. V = U Y. From x clx(A), U A . This implies
that (U Y) A , i.e., V A . Thus, we conclude that
x cly(A).
Thm. Closed family in Y := {F Ç Y : F is closed in X }, using
subspace we can construct many interesting examples.
Proof: (1) Let A be closed in Y, then Y \ A is open in Y. So,
open set U in X s.t. Y \ A = U Y. It is easy to check that A = Y
(X \ U), and X \U is closed in X.
(2) Assume that F is closed in X, then X \ F is open in X. So,
(X \ F) Y is open in Y. Since (X \ F) Y = Y \ (F Y), we
know that F Y is closed in Y.
Example 8. (The Cantor perfect set) We use the word segment
to refer to a set of the form [ a, b] and interval to refer to a set of
the form (a, b) . A1= segment [0,1] , An is a union of a finite
number of pairwise disjoint segment. An + 1denote set obtained by
removing from each segment of An the middle 13 interval.
Then we will get a decreasing sequence An : n ΠN . C = ǥn= 1 An
of the line R is called the Cantor perfect set . C is closed, has no
isolated points, but does not contain any interval , uncountable and
measure 0.
Remark: A point of the set A \ d(A) is called an isolated point
of A. A point x is an isolated point of X iff the one-point set {x}
is open.
1.6 The Product Topology
Def. Let ( X , T X ) and (Y , T Y ) be topological spaces. The product
topology on X ´ Y is the topology T having as base the collection
{U ´ V : U Î T X ,V Î T Y }
Prop: If B is a base of the space ( X , T X ) andC is a base of the
space (Y , T Y ) , then the collection {B ´ C : B Î B , C Î C } is a base of
the product space (X, T ) .
Def: Let p1 : X ´ Y ® X be defined by the equation p1 ( x, y) = x
Let p 2 : X ´ Y ® Y be defined by the equation p 2 ( x, y) = y
The map p1 and p 2 are called the projections of X ´ Y onto its
first and second factors, respectively.
Thm. The collection S { 11 (U ) : U T X } { 2 1 (V ) : V T Y }
is a subbase of the product topology on X Y.
Proof: Let T be the product topology and T the topology
generated by S.
(1) Because S T , we have T T.
(2) On the other hand, every base element U V of the
topology T has the form U V 11 (U ) 2 1 (.V ). Thus, U V
T . So that T T as well.
1.7 The metric topology
Def: A metric on a set is a function d: X Y R having the
following properties:
(1)d(x, y) 0 for all x, y X; equality holds iff x = y.
(2)d(x, y) = d(y, x) for all x, y X.
(3)d(x, y) + d(y, z) d(x, z), for all x, y, z X. (triangle inequality)
Def: Given a metric d on X, the number d(x, y) is called the
distance between x and y. Given > 0, the set {y: d(x, y) < }
is called the –ball centered at x. It is usually denoted by B(x, ).
Def: If d is a metric on the set X, then the collection of all balls B(x, ), for x X and > 0, is a base of a topology on X,
called the metric topology induced by d.
Def: If X is a topological space, X is called to be metrizable if there
exists a metric d on the set X that induces the topology of X.
Remark: A metric space is a metrizable space.
Def: Let X be a metric space with metric d. A subset A of X is
said to be bounded if there is some number M such that d(x, y)
M for every pair x, y of A.
Thm: Let X be a metric space with metric d. Define d: X X
R by the equation d(x, y) = min {d(x, y), 1}. Then d is a
metric that induces the same topology as d.
1.8 Continuous Functions
Def: Let X and Y be topological spaces. A function f: X Y is
said to be continuous if for each open set V of Y, the set f- -1 (V)
is an open set of X.
Prop.: Let X and Y be topological spaces, and B a base of the
topology T of Y. A function f: X Y is continuous iff the
inverse image of every base element is open.
Proof: The arbitrary open set V of Y can be written as a union
of base elements: V = {B : J}, and we have that f- -1 (V)
= {f- -1 (B ): J},
Prop.: Let X and Y be topological spaces, and S a subbase of
the topology T of Y. A function f: X Y is continuous iff the
inverse image of every subbase element is open.
Proof: The arbitrary base element B of Y can be written as a
finite intersection of subbase elements: B = S1 S2 … Sn ,
and we have that f- -1 (B) = f- -1 (S1 ) f- -1 (S2 ) … f- -1 (Sn ).
Ex.: Let T be the usual topology on R, and T l the lower limit
topology on R. Then, the identity function f: x x is not a
continuous function from (R, T ) to (R, T l ). On the other
hand, the identity function g: x x is a continuous function
from (R, T l ). To (R, T )
Thm.: Let X and Y be topological spaces and f: X Y is a
function . Then, the following are equivalent:
(1)
(1)
(2)
(3)
f is continuous.
For every subset A of X, we have f- (cl(A)) cl( f (A)).
For every closed subset B of Y, the set f- -1 (B) is closed in X.
For each x X and each nbhd. V of f (x), there is a nbhd. U
of x such that f (U) V.
Proof: (1) (2) Assume that x cl(A). If V is an open nbhd. of
f (x), then f -1 (V) is an open nbhd. of x. So, it must intersect A in
some point y. It follows that V intersects f (A) in the point f (y).
This implies that f (x) cl( f (A)).
(2) (3) Let B be closed in Y and A = f- -1 (B). If x cl(A),
then f (x) f (cl(A)) cl(f (A)) cl(B) = B. Therefore,
x f- -1 (B) =A.
(3) (1) Let V be an open set in Y and B = Y – V. Then
f- -1 (B) = f- -1 (Y) - f- -1 (V) = X - f- -1 (V) . This means that
f- -1 (V) is open in X.
(1) (4) Let x X and let V be an open nbhd. of f (x). Then U
= f- -1 (V) is an open nbhd. of x, and f (U) V.
(4) (1) Let V be an open set in Y. For each x f- -1 (V), we
have f (x) V. There is an open nbhd. U of x such that f (U) V,
This implies that U f- -1 (V). Therefore, f- -1 (V) can be
written as the union of open sets.
Def: Let X and Y be topological spaces, f: X Y be a bijection .
If both the function f and its inverse f- -1 : Y X are continuous,
then f is called a homeomorphism.
Remark 1: Another way to define a homeomorphism is to say
that f: X Y is a bijective correspondence such that f (U) is
open U is open.
Remark 2: A homeomorphism f: X Y gives us a bijective
correspondence not only between X an Y, but between the
collection open sets of X and Y. So, any property of X that is
entirely expressed in terms of the topology of X yields the
corresponding property of Y.
Ex: A bijective function f: X Y can be continuous without
being a homeomorphism.
Let S1 denote the unit circle, S1 = {(x, y): x2 + y 2 =1},
considered as a subspace of the plane R2 , let f : [0, 1) be the
map defined by f (t) = (cos2t, sin2t).
(1) f is bijective and continuous.
(2) The image of the open set U = [0, 0.2) [0, 1) under f is not
open in S1.
Local formulation of continuity: The map f: X Y is continuous
if X can be written as the union of open sets U such that f U is
continuous for each B .
Proof: Let X be the union of open sets U and each f U is
continuous. For each open set V in Y, it is easy to check that
f- -1 (V) U = (f U) -1 (V). Since f U is continuous, the set
f- -1 (V) U is open in U, and thus open in X. So,
f- -1 (V) = (f- -1 (V) U ) is open in X.
The pasting lemma: Let X = A B, where A and B are closed in
X. Let f: A Y and g: B Y be continuous. If f (x) = g (x) for
every x A B, then f and g combine to give a continuous
function h: X Y, defined by setting h (x) = f (x) if x A, and
h (x) =g (x) if x B.
Proof: Let C be closed subset of Y, then we have
h -1 (C) = f- -1 (C) g -1 (C) by elementary set theory. Since f is
continuous, f- -1 (C) is closed in A, and thus closed in X.
Similarly, we can prove that g -1 (C) is closed in B, and therefore
closed in X. It follows that h -1 (C) is closed in X.
1.8 Sequential Spaces.The Sequential Closure Operator
Definition 9. A topological space X is called sequential if , for
every set A X which is not closed in X , there exists a sequence
xn n 1 of points of A converging to a point of the set A \A .
Sequential closure : Aseq x X : sequencexn n1 in A , xn x; A Aseq is
a map from 2 X 2 X , it is called the sequential closure operator. It
can be defined in any topological spaces.
Properties :
0 seq
1seq
2 seq
Remark:
. seq
. A Aseq A
. A B A B
A A
seq
seq seq
seq
seq
in general.
seq
Example. Let X be the set consisting of 3 different type points:
isolated points : xmn .
limit points : yn of sequence xmn : m N
point z : with most complicated local structure
The collection B = xmn : n N , m N Vk yn : n, k N is a base of
a topology on X , where Vk yn yn xmn : m k .
and W : z W , p N ,Vk yn \ W is finite and yn W for all n p .
1. X is sequential space (Arens Fort space, Modified) Because
only those yn ' s and z could be limit point of sequence with
distinct points . If yn ' s or z in G \ G , there must be a sequence in
G converging to it.
2. Let
A xmn : m N , n N , B yn : n N
B seq B z ,therefore z Aseq seq
A A
seq
seq seq
, then Aseq A
, but z A
. Thus
B
seq
.
Definition 10 . A topological space X is called a Frechet-Uryshon
space if the closure of every subset A X in X coincides with the
sequential closure of A : A Aseq .
It follows that Frechet-Uryshon space is sequential and from
the above example, it shows that not every sequential space is
Frechet-Uryshon space.
Example 14. Consider the subspace Y A z of the space X
in example. It is easy to verify that no sequence of points in A
converges to the point z. But A A z . Consequently, Y is not
a sequential space. Thus a subspace of a sequential space need
not be sequential.
Proposition 6. A topological space X is a Frechet-Uryshon
space if and only if each subspace Y of X is sequential.
Proof: Let Y X with subspace topology. A Y , if A is
not closed in Y.Then clY A \ A , clY A cl X A Y Aseq Y , For any
point x in( Aseq Y ) \ A , there is a sequence in A converging to x.
Therefore Y is sequential.
For any subset
, let y A \ A , consider Y A y , then Y
is sequential. Therefore in A sequence xn n 1 converging to y,
this shows that A Aseq hence A Aseq . X is Frechet-Uryshon
space.
A X
1.9. The First Axiom of Countability and Bases of a Space at a
Point (and at a Set)
Definition11. A collection of open neighborhoods of a point x
in a topological space X is called a base of the space X at the point
x if each neighborhood of x contains an element of .
First Axiom of countability: each point has a countable base. C
Example: Space with countable base, all discrete spaces
satisfies C .
Note that: the discrete space X has countable base X itself is
countable. Hence not every C space is C space.
Proposition 7. If X satisfies C X is a Frechet-Uryshon space.
Proof : For any set A X ,we show A Aseq . Let x A , suppose
is a countable base at x, say On : n , WLOG we may
assume that Om On when m n , otherwise we can simply let
Om mi 1 Oi , then replace Om by Om . It is easy to see that A Om
for any m , pick xm A Om , then xn : n is a sequence in A
converging to x. For any neighborhood U of x, Om such that
x Om U , then xn Om U for all n m .
Proposition 8. A countable space X is
base.
C it
has countable
Proof : Trivial.
For each point x, Let Bx be a countable neighborhood base
S Bx : x X is a countable collection, use it as subbase, the
topology generated by S is the original topology T .
B S1 Sn : Si S is a countable base for T .
Example 15. X in Example 13, Y in Example 14 are countable
, but do not have C .They are not even Frechet-Uryson space.
Thus , not every countable space has a countable base. Note that
all single point sets (singletons) in Y are closed and only one point
(the point z) is not isolated.
1.10 Everywhere Dense Sets and Separable Space.
Definition 13. A subset A of a topological space X is called
everywhere dense in X if its closure is equal to X: A X .
: equivalent to for every nonempty open U , U A
: enough for every nonempty base element U B U A
X is separable if it contains a countable dense subset.
Example 20. Every countable space is separable.
is dense in
,T as well as , T S . ,T has countable base, , T S
does not.
Proposition 10. Not every separable space satisfying the first
axiom of countability possesses a countable base.
Proposition 11. Every space with a countable base is separable.
Proof : Choose a point in each nonempty base element B . A of
all such points is countable and dense in X .
C is hereditary property.
Note 1. Separability is not hereditary.
Example(P149): X , T is a topological space , X ,X * X ,
T * A {}| A T , X * , T * is a space.
1 X * , T * is separable . belongs to any nonempty open set ,
then is a dense set in X * , T * .
2 X , T can have any properties we want . For example choose
any nonseparable space , say uncountable discrete space. X , T
is a subspace of X * , T * .
Note 2. Even every subspace of X is separable(X is hereditarily
separable ). X need not be C .
Sorgenfrey line: not C but hereditarily separable. Think of how
to prove it.
Countable space without a countable base Example 14 or
Example 19 , countable Frechet-Uryshon fan.