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Proposition 7. Metrizable spaces satisfies CI .
Proof.
{B1 ( x) : n  N  } is
a base at x for any x∈X.
□
n
Closure operator in a metrizable space (X, T  ) can be described
in terms of sequences. x  A   sequence{xn }n1 in A converging to x.
lim x
n 
n
x
iff lim  ( x , x)  0
n 
n
( cl(A) = [A]seq )
Any singleton, finite subset of metrizable space is closed, therefore
antidiscrete space is not metrizable.
Proposition 8. Every separable metric space X has a ctble base.
Proof. Let  be a metric on X which generatesT . Let A be a ctble
dense subset. B ={B1 ( x) : x∈A, n∈ N}is ctble and consists of
n
open subsets. We show
B is a base. Let x∈X, U be any open

subset contain x. Then a= (x, X\U)>0. Choose n∈ N so that
1
a
 . Fix y∈A such that y∈ B 1 (.x )For any z∈
B1 (, y )
n
3
n
 ( x, z )   ( x, y )   ( y , z ) 
Therefore,
x  B1 ( y)  U .
1 1 2
  a
n n n
n
□
3
Internal characterization of metrizable spaces. Theorem of
Metrization. Nagata-Smirnov-Bing Theorem 1950.
Needs paracompactness defined by Dieudonne and Stone’s
Theorem.
In class of C II spaces, it is very easy. T3  CII Metrizable
2.6. The Separation Axioms T0 , T1 , T2
T0 : x  y,  open
U containing exactly 1 of these pts. |U∩{x,y}|=1
T1 : x  y,  open
U, x∈U∧y U
 and
T2 : x  y,  open
U, V, x∈U∧y∈V∧U∩V=Φ
T2  T1  T0 anti-discrete space X containing more than 1 point
does not satisfy
T0
Ex 19. X={0,1}, T ={Φ,{0},{0,1}}
T0 not T1
Ex 21. A commutative ring with unit. X set of all prime ideals
of A. For any a∈A, let X denote
the set of all prime ideals
a
which does not contain a. it is clear that X a  X b  X ab for all
a,b∈A, X 0  , X1  X
consequently, B ={ X a: a ∈A} is a base of topology.
This topology is called the spectral or Zariski (X,T ) prime
spectrum of the ring A, denote Spec(A).
Cl{x}={y∈X: y prime ideal & y containing Spec(A)}
(X,T ) satisfies T0 not T1, the only closed pts in X are the
maximal ideals of ring A.
Ex 22. Topology of finite complements (Ex. 2.2.4)
X infinite, T ={X\F: F is finite}
Any two non-empty open sets intersect, not T2
All singletons are closed, hence T1
Proposition 13. X satisfies T1 finite subsets are closed.
Proof: first prove that singleton {p} is closed. For any q≠p,
 open nbd. U q so that p  U q . Then X \ { p}   U is open, hence
{p} is closed. Finite subsets are finite union of singletons, the
finite union of closed sets is closed.
q p
 For any p≠q, since {p}, {q} are closed,
both are open and
p U p , q U p  q U q , p U q
q
U p  X \ {q},U q  X \ { p}
.□
Note. Finite space X satisfied T1 X is discrete.
For T0 , this is not true as Ex 19 shows.
Metric space satisfies T2 . x ≠ y, then  (x, y)>0.
Let   1  ( x, y), B ( x)  B ( y)  .
2
T0 , T1 , T2 are hereditary property.
Proposition 14. T 1 T 2 , (X,T 1 ) satisfies Ti for i=0,1, so does
(X, T2 ).
Stronger topology satisfies stronger separating Axioms.
2.7 Regular ＆ Normal spaces T3 ,T4
Definition 7. X is regular, if for any closed set A and x∈X\A
 open U, V such that x∈V, A U and U∩V=Φ.
It _____
is equivalent to: for any nbhd Ox of x,  a nbhd O1x of x such that
O1x  Ox .
Note: regular need not T1in general, antidiscrete space are
regular,but not T1.
Axiom T3 : regular + T0 ; T3  T2
Subspace of a regular space is regular.
T3 is also a hereditary property.
Metrizable space is regular, for closed A  X and x∈X\A, since
 (x,A)>0, then B ( x)  B ( A)   if   1  ( x, A).
2
Ex 23. A T2 but not regular space.

X=R, A={1/n: n∈ N},
closed sets are of the form B∪C, where B
is closed in usual topology and C  A. This defines a topology T 1
on R, if we let C=Φ, then we get all closed sets in (R,T ), that
means T  T 1 , also T ≠T 1 since R \ A∈T 1 , but R \ A
T .
A is closed in X, 0  A, but 0 and A can’t be separated by open
sets.
Definition 8. Normal: disjoint closed subsets A, B can be
separated by disjoint nbds,  open U, V, such that
A  U , B  V ,U  V  .
Axiom T4 : normal + T1 ; T4  T3
Proposition 15. Any metrizable space X is normal.
Proof: Let  be a metric generating topology T . A, B disjoint
nonempty closed subsets. For x∈A, y∈B, let
 ( x, B )
where


.
Vx  B ( x)  {z  X :  ( x, z )   x },
x
3
x
U y  B y ( y)  {z  X :  ( y, z)   y }, where  y   ( y, A) .
3
Let V  {Vx : x  A},U  {U y : y  B}, then U, V are disjoint open
nbds of A and B respectively. It is clear U, V are open.
U∩V=Φ follows from the triangle axiom.
Note. Normality is not a hereditary property. This causes a lot of
inconvenience.
Example: prove later. [need Stone-Cech compactification or
Tychonoff Theorem]
Proposition 16. Every closed subspace of normal space is
normal.
Proof: Let Y be a closed subspace of normal space X, then disjoint
closed subsets of Y …
Ex 24. (The Nemytskii plane)
2
L={(x, y) ∈ R:y≥0}
upper-half plane. M=x-axis. For z  M ,   0,
Let W(z, ) denote the set of all pts inside the circle in L with
1
radius  and is tangent to M at z. Let Wi ( z )  W ( z, i )  {z} . If
z∈L\M, then take Wi ( z)  B1 ( z)  L, B = {Wi ( z ) : i  N  , z  L}forms
i
a base on L. With this topology, L is called the Nemytskii
plane.
1. It is regular Hausdorff space.
On P=L\M, subspace topology from Nemytskii plane is just the
usual topology on L\M. M is closed and discrete subspace of L.
It follows that Nemytskii plane is a union of two metrizable
subspaces M and P, one is open and the other closed.
2. Proposition 17. It is not metrizable, since it is not normal
Proof: Let C  P  L \ M be set of all pts both of whose coordinates
are rational. C is countable dense in L, since it intersects all base
elements.
Suppose L was normal. An arbitrary subset A of M is closed in L,
so M\A is also closed in L. Let U A ,VA are disjoint open
nbds
of A
_____
_____
A, U A  U B ,since
and M\A respectively.
Let CA  U A  C . If B  M , B _____
_____
U A  VB  A  ( X \ B), A \ B  , while VB  VB  . But
A\B≠Φ,
then
_____
____________
_____
U A  U A  C  C A , since C is everywhere dense and U A is open.
Thus for A, B  M , A  B, CA  CB (other we would have). But in M
there are more than 2 different subsets, while C only have 2
many different subsets. That is a contradiction. This shows that L
is not normal.
□
0
0
Remark 1. Not every regular space in which all finite sets are
closed is normal.
Remark 2. Clearly topology of L is stronger that usual topology on
upper-half plane. Upper-half plane is metric space, hence normal,
but L isn’t.
Proposition 18. If X Hausdorff, all but finitely many pts are
isolated, then X is normal.
Proposition 19. Subspace of linearly ordered space is normal.
Omit the proof.
X is normal  for closed
A and nbd U of A,  open nbd V
____
of A such that A  V  V  U .
Proposition 20. Combinatorial shrinking lemma.
X is normal for any finite open cover   {U1 ,,U k } ,  open
____
shrinking  '  {V1 , ,V2}, i.e. Vi  U i .
Proof. Let A, B be disjoint closed subsets. Then U1  X \ A,U2  X \ B
are open cover of X. F1 , F2 with F1  U1, F2  U2 are closed cover of X
iff X \ F1 , X \ F2 are disjoint open nbds of A and B. If Fi is closed
 U i and if X is normal, then  open Vi such that
with Fi ____
Fi  Vi  Vi  U i . This is the argument of the case k=2. Then by
induction we can prove the statement.
Suppose we proved X normal  for any open____cover   {U1 ,,U k }
 open shrinking  '  {V1,,V2},  ' is a cover and Vi  U i .
k
For open cover   {U1 ,,U k ,U k 1}, consider   {i 1U k ,U k 1} , then we
can get shrinking  '  {V ,Vk 1} , then {U1 ,,U k } is an open cover for
closed subset X \ U k 1 , then use inductive hypothesis. □
“Combinatorial thickening lemma”
Proposition 21. Let X be normal , and {F1 ,, Fk } a finite
collection of closed subsets. Then  open collection {U1 ,,U k }
such that Fi  U i and Fi  Fi    Ui Ui   for any
1
p
1
p
i1 ,  , i p  {1,  , k}.
Urysohn metrization theorem. Every normal T1 space with
countable base is metrizable.
Tychonov: replace normality by regular.
Non metrizable normal spaces: Sorgenfrey and space of all
ordinal smaller than the uncountable ordinal.
This 2 spaces are hereditarily normal.
P167 2, 5, 6 p151-156
§3 Continuous Maps of Topological Space
The Foundations of the theory 54-55 P84-86
3.1 Different Definitions of Continuity of Maps of Top. Space
Definition 1. (X, T 1 ), (X, T ) spaces, f: X→Y
is called
a
___
_______
( x  A  f ( x)  f ( A
continuous map if xδA→f(x)δf(A)
. ))
Map which is not continuous is called discontinuous.
2
It is easy to prove that a map is continuous iff the image of the
closure of any set is contained in the closure of the image of the
set: f ( A)  f ( A)
Proposition 1. TFAE 1) f: X→Y is continuous;
2) f 1 (V ) is open for any open V in Y;
3) f 1 ( P) is closed for any closed P in Y.
___________
1
1
1
f
( P)
f
(
P
)

f
(
P
),
Proof: 1) →3) Let__________
P is closed,__________
we show
hence
_____
is closed. For x  f 1 ( P) , f ( x)  f ( f 1 ( P))  P  P, hence x  f 1 ( P).
3) →2) For any V open in Y, Y\V is closed.
closed in X, hence f 1 (V ) is open.
f 1 (Y \ V )  X \ f 1 (V )
is
2) →1)
It’s enough to show that f(x) ¬δf(A) → x ¬δA.
________
f ( x)  f ( A) ,  open neighborhood V of f (x ) such that V  f ( A)  .
So, we have f ( x) V  X \ f ( A), and
x  f 1 (V )  f 1 (Y \ f ( A))  X \ f 1 ( f ( A))  X \ A.
1
f (V )
is an open nbd of x.
f 1 (V )  A  ,
therefore x  A .
Proposition 2. f: X→Y is continuous iff  x∈X and nbd. V
of f(x),  nbd U of x such that f (U )  V .
Proof: f (x) ∈V, V is a nbd of f (x), open W such that f ( x) W  V
so x  f 1 (W )  f 1 (V ). U  f 1 (W ) is an open nbd of x such that
f (U )  f ( f 1 (W ))  W  V .
For any open V in Y, we show that f 1 (V ) is open in X. For any
x  f 1 (V ) , f (x)∈V. Since V is a nbd for f (x),  open nbd U of x
such that f (U )  V , hence U  f (V ), that is f 1 (V ) is open.
()
1
Definition 2. (X, 1), (Y,  2) metric spaces, f: X→Y is
1 ( x, x) , 
continuous if  x∈X and ε > 0, δ> 0,
2 ( f ( x), f ( x))  
Remark. f: X→Y continuous as metric spaces 
f: X→Y continuous as topological spaces. (homework)
Definition 3. f: X→Y is called sequentially continuous if for
any {xi }i1 converging to x, { f ( xi )}i1converges to f(x).
Continuous →sequentially continuous. Converse is not true.
Remark. f: X→Y is continuous iff
x∈X and nbd. V
of f(x),  nbd U of x such that f (U )  V .
Ex 1. A uncountable，b  A, X=A∪{b}, U open if b  U,
A\U is countable. This defines a topology on X. X is not discrete,
since b is not isolated. a P-space, the intersection of countably
family of open subsets is open. ( G sets are open.) All
countable sets in X are closed. X is not sequential, no nontrivial
convergent sequences. The identity map
f: X→ Y, Y=X with discrete topology is not continuous. bδA
in X, but far from A in Y. But the map is sequentially
continuous, since convergent sequences in X are eventually
constant.
Proposition 3. f: X→Y sequentially continuous, X sequential,
then f is a continuous map.
Proof. Omit.
Definition4. A continuous map f: X→Y is called a condensation
if f is one to one & onto, that is f (X)=Y and x1  x2 , f ( x1 )  f ( x2 ).
If T  T , i: (X, T ) →(X, T ) is an condensation form (X, T )
onto (X, T ).
Definition 5. f: X→Y is homeomorphism if f is 1-1 & onto &
xδA f(x)δf(A).
1
2
1
2
1
Proposition 4. f: X→Y is homeomorphism  f is 1-1 & onto,
both f and f 1 are continuous, that is both are condensation. In
this case f 1 is also a homeomorphism.
Proposition 5. Any topological space Y is the image under a
condensation of a discrete space.
Thus not every condensation is homeomorhpic.
2
Definition 6. X, Y are called homeomorphism if
 homeomorphism f: X→Y.
Homeomorphism is an equivalence relation.
Definition 7. Topological invariant Φ: X has Φ, every space
homeomorphic to X has Φ.
Ex 2. 2 disks are homeomorphic.
disk homeomorphic to the inside of square or triangle
(a, b) ≌R
circle is not homeomorphic to [a, b]
The circle has no fixed point property, but [a, b] has fixed fixed
point property.
fixed point is a nontrivial topological invariant.
R is not homeomorphic to R 2 .
Rm  Rn if m  n
Ex 3. countable metrizable space without isolated points
homeomorphic to space of rational numbers with usual topology.
Space of irrationals homeomorphic to Baire space B
Cantor perfect set homeomorphic to countable product discrete 2
points space D={0,1}.
One can have 2 different topologies T , T on X, (X, T ), (X, T )
homeomorphic
1
2
Storgenfrey line is not homeomorphic to R
not metrizable
metrizable
Metrizability is topological invariant.
Other: separability, all separation Axioms, compactness,
dimension
P58. 11, P78. 6, P100, 5
3.2 General Aspects of the comparison of topological space
Question 1. Determine if a space X can be continuously
mapped onto Y.
Fact 1: segment can be continuously mapped onto a square
(Peano curve P99 this book)
Fact 2: Cantor set can be continuously mapped onto any closed
bounded subset of R n
Question 2. We want to get the “invariant” that completely
determine the “standard” spaces.
Fact 1: Hausdorff spaces + continuous images of Cantor set ≡
compact metric space
Fact 2: Hausdorff spaces + continuous images of segment ≡
locally connected, metrizable comp connected spaces
General question
Question 3. Properties preserved by continuous map
Note. Every space is continuous image of discrete spaces. So
metrizability, normality are not preserved.
Can not solve: continuously map one space onto another, need
more restriction.
Replace a space by a class of spaces
Problem 4. P, Q classes of spaces
When can space in Q be represented as the continuous image of
some spaces in P
Problem 5. when  X∈P such that any Y in Q is a
continuous image of X, X is called universal space.
Replace continuous map by different kind of map get many
interesting problems
For example, replace continuous map by condensation.
Problem 6. a) when any space in Q is the image under
condensation of some space in P
b) When any space in Q can be condensed or some space in P .
1. What kind of space condense onto metrizable space?
Nontrivial answer.
2. What kind of space condense onto complete space?
No complete answer.