EASY NOTES
TAKLEED
WITH SOLU SOLUTION TION
Introduction to Topology Topology Pure and Applied By Collin Adams Willams College Rebert Farzosa University of Maine
EFFORED BY MUHAMMAD TAHIR AZIZ TAKLEED (BHAKKAR) GOVT COLLEGE UNIVERSI UNIVERSITY TY FAISALABAD
Fb: www.fb.com/ www.fb.com/ta tahir hir azizta aziztakleed kleed E -ma -mail: il: ta tahiraziz92@ hiraziz92@ g ma mail.com il.com tahiraziz1042 ta hiraziz1042@ @ g ma mail.com il.com Web: ta tahiraziz92.bl hiraziz92.blog og s pot pot.com .com
Phone: +92 301 78 10 067 +92 333 88 39 067 +92 343 69 74 245 +92 313 69 74 245
PAKISTAN ZINDABAD
TAKLEED
2
Chapter # 1
Topological spaces Definition: A topology on a set X is a collection of subsets of X having the following properties (T1) Ø and X are open open in Τ (T2) The union of any collection of T opens in Τ (T3) The intersection of elements elements of any finite sub collection of T opens in Τ A set X for which a topology has been specified is called topological space. Thus a collection of subset of a set X is a t opology on X if it includes empty „Ø‟ set, X, finite intersection i ntersection and arbitrary union of sets in collection Example 1.1 Let X be three – three – point point set {a,b,c} we consider four different collections s.t ={{a},{a,b},{a,b,c}=X} a C. b
c
a b c ={{a},{b},{a,b},{b,c},{a,b,c}=X} ={{a},{c},{a,c},{b,c},{a,b,c}=X} ={{a},{b},{c},{a,b},{a,b,c}=X} Check which is Topology SOL: we check all axioms of topology
(T1) Ø, X ∈ T1 (T3 ) Intersection of sets in collection ∩∈ T1 (T2) Union of sets in collection ∪∈ T1 ⇒ T1 Is topology Similarly T2 ,T3 &T4 are topologies
Example 1.2: Let X be nonempty set and define T= ∅, . Show that SOL:
We check all axioms of topology (T1) Ø, X ∈ T (T2) Ø∪ X ∈ T (T3) Ø∩X ∈ T ⇒ T is topology Definition: T =∅, X is minimal topology which is defined on X. If we remove either set, we have no topology. T =∅, X is called Trivial Topology on X
Example 1.3 Let X be a nonempty set and
be collection of all subset of X. Show that
SOL:
is a topology
3 By definition it is clear that all condition of topology will be satisfied so T is a topology This kind of topology is called discrete topology. This is the largest topology define on X Definition: On real line ℝ . Define a topology open sets are the empty and every set in ℝ with finite complement.
OR
Tfc = {Ø, All sub set of ℝ which has finite number of element in complement} complement} Tfc = {∅, ℝ-{finite set}} 03 Example: A= ℝ-{1, 2, 3} B= ℝ-{a, b, c} C= ℝ-{0} c c A = {1, 2, 3} B = {a, b, c} C C = {0}
Example 1.4 Prove that an open set in finite complement topology on SOL:
ℝ
For prove we check all axioms of topology T1 Ø, X ∈ Tfc ℝ-ℝ = Ø∈ Tfc & ℝ-Ø= ℝ ∈ Tfc T2 Arbitrary Union of element Let {U}i ∈I be collection arbitrary union of element of Tfc OR {U}i ∈I ⊂ Tfc There are two cases Either U{U}i ∈I = Ø (ii) or U{U}i∈I ≠ Ø (i) U{U}i ∈I = Ø∈ Tfc (ii) Let Uα ∈ U{U}i ∈I ; Uα ≠ Ø Uα has finite complement ⇒ Uα ⊂ ∪ Uα c c ⇒ Uα ⊃ ∪ Uα Rmarks ⇒∪α∈ I Uα Is finite A⊂B ⇒ U{U}i∈I ∈ Tfc T3 Intersection of element of Tfc belong to Tfc Then ⊃ Let {u1 ,u2 ,…,un } be finite collection of element of Tfc There are two cases n ∩α =1 Uα = Ø (ii) ∩nα =1 Uα ≠ Ø ∩nα =1 Uα = Ø∈ Tfc if any Uα is empty (i) Suppose Uα are nonempty ⇒ u1 = ℝ-F1 ,.. ,un= ℝ-Fn where F1 ,………….,Fn are finite sets n ∩α =1 u = (ℝ − F1 ) ∩ ℝ-F2 ∩ … ∩ ( ℝ-Fn ) =F1 c ∩ F2 c ∩…….∩ Fn c =(∪nα =1 Fα )c = ℝ Taking complement on both side (∪ Uα )c = ∪ Uαn =1 Fα viz finite where F1 …Fn are finite sets and union of finite set is a finite set ⇒∩nα =1 Uα ∈ Tfc (ii)
Defination:1.2 Finer and Coarser Let T1 and T2 be two topologies define on non empty empty set X
1
2
Coarse OR Weaker Finer OR Stronger If member of 1 is also member of 2 OR 1 ⊂ 2 If 1 ≠ 2 then 2 strictly finer and 1 strictly coarser Example
4
Let X={1,2] T1 ={∅,{1},{1,2}} T2 ={∅,{1},{2},{1,2}=X} T1 is coarser and T2 is finer
Definition: 1.3
Let X be top-sp and x ∈ and U is an open set containing x, U said t o be neighborhood of x X U
.
.
Theorem:1.4
Let X be a Top-sp and let A be a subset of X. then A is open in X ⇔ for each x ∈A. There is a nieghbourhood Uof x s.t xєU⊂A X
Proof: Consider A is open Let x∈A
A
. U
⇒ x∈A⊂A ⇒x∈U⊂A
A=U
Hence proved Conversely we proved that A is open. We are given that Where U is a neighborhood By union lemma A=U A is the union of open sets
⇒ A is open
∈⊂
proved.
1.1 Given X= {a,b} P(x) = {Φ {Φ, {a}, {b}, X} All topology on X are T1 = {Φ, X} T1 is trivial Top T2 ={ Φ,X,{a},{b},{a,b}} T2 is discrete Top T3 = Φ, X, a, a, b T4 ={ Φ,X,{b},{a,b}}
1.2
One of the three – three – point point set X={a,b,c},the trivial topology has two open sets and discrete topology has eight open sets .For each n=3,…,7 either find a topology on X consisting of n open sets or prove that no such topology exist SOL. Let X= {a,b,c} A topology on X always contain Ø and X (By T1 and T2), and since X ≠ Ø these are two distinct subsets that must be in T. If A is any third subset of X (A≠ Ø and A≠ X) then T= {Ø, A, X} can easily be seen to form a topology on X. There are six such subsets A in X= {a,b,c},this lead to six different topologies on X with precisely three element element . The possibilities for A are
5
{a},{b],{c},{a,b},{a,c} and {b,c} The power set P(x) of X={a,b,c} has eight elements .These are Ø,X and the six set listed above .A collection T of seven subsets is obtained by removing one of these from P(x). If removed set is Ø or X then T will not satisfy the T1 axiom. So removed set must be one of the six set listed above. If the removed set is a singleton then the T3 axiom will fail. If the removed set is a doubleton then T2 axiom will fail .this show that no matter which subset is removed to from T, T will not be topology. From this we conclude that there is no topology on this set X which has precisely seven elements
1.3
Prove that a topology T on X is the discrete topology SOL:
⇔ {x} ∈T for all x∈X
Let T on X is discrete top. Since T has all subset of X ⇒{x}∈T for all x∈X {x} is also subset of X Conversely: Suppose that {x}∈T for all xєX
⇒ each sub set of X belong to T Let x,y ∈T ⇒ {x},{y} ∈T {x}∪{y}={x,y} ∈T Now { }∈ be arbitrary collection of elements of X then ∪ { }∈
⇒T is discrete
∈T
1.4
SOL: Let X be a topological space X={a,b,c} P(x)={Φ,X, {a},{b},{c},{a,b},{a,c},{b,c} } Discrete Top 1 = {Φ,X, {a},{b},{c},{a,b},{a,c},{b,c} } Finite complement top 2 = { , Φ ,{b,c}, {a,c},{a,b},{c},{b},{a}}
If X is a finite set then discrete top is same as finite complement top. Let X be a finite set. Then the discrete Top is just power set of X. Since X is finite, any nonempty subset of X has a finite complement that is the finite complement Top on X is same as discrete Top on X 1.5
st
Find three Top on the five-point set X={a,b,c,d,e} s.t that 1 is finer then nd nd rd 2 and 2 finer then 3 ,without using either the trivial or the discrete top. st Find a top on X that is not complement to each of 1 three that you found SOL:
1 = , , , , , , 2 = {, , , , , , , 3 = , , T={Φ, b, X
T1 is finer then T2 , T2 is finer then T3 . All are not compareable to T
1.6
Define a Top on ℝ (by listed the open sets with it that contain the open set (0,2) and (1,3) and that contain as few set as possible
6
SOL:
T={Φ, ℝ,(0,2),(1,3),(0,3),(1,2)} We check all axioms of topology (T1) Φ,ℝ ∈T (T2) arbitrary union of elements belong to T (0,2)∪(1,3)=(0,3) ∈T (0,2)∪(0,3)=(0,3) ∈T (1,2)∪(1,3)=(1,3) ∈T (T3) Finite collection of elements of T belong to T (0,2)∩(1,3)=(1,2) ∈T (0,2)∩(0,3)=(1,2) ∈T (1,2)∩(1,3)=(1,2) ∈T
1.7
0
1
2
3
Let X be a set and assume p ∈X.show that collection T, consisting of Φ,X and all subsets of X containing p is a topology on X.This Top is called Particular point topology on X and denoted by SOL: Top define as
PPXP = Φ, X , A
⊂ Xp ∈ A
We verify all axioms of topology (T1) Φ,X ∈ (T2) arbitrary union let {}∈ ∈ and let U=∪∈ {}∈ There are two cases If each { }∈ is empty then U=Φ ⇒U ∈ If any ∈ is nonempty then p∈ ∈ ⇒ ∈ ⇒ U ∈ (T3) finite collection Let { } =1 ∈ and let V=∩ =1 { } There are two cases If any { } is empty then V=Φ ⇒V ∈ If non of { } is nonempty then p∈ { } ⇒ ∈ ⇒ V∈
1.8
Let X be a set and assume p ∉X.show that collection T, consisting of Φ,X and all subsets of X exclude p is a topology on X.This Top is called Exclude point topology on X and denoted by SOL: Top define as
= , , ⊂ ∉
We verify all axioms of topology (T1) Φ,X ∈ (T2) arbitrary union let {}∈ є and let U=∪∈ {}∈ There are two cases If each { }∈ is empty then U=Φ ⇒U ∈ If any ∈ is nonempty then p∉∈ ⇒ ∉
7
⇒ U ∈
(T3) finite collection Let { } =1 ∈ and let V=∩ =1 { } There are two cases If each { } is empty then V=Φ ;p ∉{ } ⇒V ∈ If non { } of is nonempty then p∉{ } ⇒ ∉ ⇒ V∈
1.9
Let T consist of Φ, ℝ and all interval ( −∞, p) for p∈ Top on ℝ SOL:
ℝ.Prove that T is a
We check all axioms of top (T1) Φ, ℝ ∈T (T2) arbitrary union Let {−∞, pi }i ∈I be collection of intervals in ℝ Then ∪i −∞, pi =−∞, pk where pk =max {pi } ⇒−∞, pk in ℝ and Open ⇒−∞, pk ∈ T (T3) finite intersection Let open interval−∞, p1 , −∞, p2 , … , −∞, pn where p1 , p2 ,…,pn єR Then ∩ni=1 −∞, pi ,= −∞, pl ,where pk =min{p1 , p2 ,…,pn } Then −∞, pl is in ℝ and open ⇒−∞, pl ∈T
Let X be a non empty set and be a collection of subsets of X. We say is basis (for a top) if following statements holds (B1) for each x in X. there is a B in s.t x∈B (B2) if 1 & 2 ∈ and x∈ 1 ∩ 2 ,∃ 3 in s.t x 3 ⊂1 ∩ 2 X
2
1
3
We call the sets in basis element paraphrasing the two conditions for a basis we have Every point in X is contain in a basis element Every point in the intersection of two basis element is contained in a basis element contained in the intersection
Example 1.5
ℝ
<
We check axioms of basis (B1) Every point of ℝ contained in an open interval ⇒r ∈ ( − 1, + 2) for all r∈ ℬ (B2) intersection Let 1 = 1 , 1 , 2 = 2 , 2 3 = 3 , 3 There are three cases
1 ∩ 2
8
Example1.8 show that
ℝ < ℝ
As we know ℝst define as ℝst ={Φ,all open interval And their union} We check all axioms of top (T1) ∅, ℝ ∈ ℝst (T2) Arbitrary Union Let {Ui } ∈ ℝ then ∪ {Ui } i∈I ∈ ℝst Union of open inerval ia open interval (T3) finite intersection ∩ of each case result is an open interval as in (B2) n ∩i=1 {Ui } ∈ ℝst
1 2 2 1 1 , 1 ∩ 2 , 2 = 2 , 2 ∈ 1 ∩ 2 2 1 1 2 1 , 1 ∩ 2 , 2 = 1 , 1 ∈
1 ∩ 2
2 1 1 , 1 ∩ 2 , 2 = (1 , 2 ) ∈ Example 1.7 Show that
∈
SOL: We check all axioms of top (T1) Φ, ℝ ∈ Td (T2) arbitrary union let {x},y ∈ ℝ {x}∪ y={x,y}∈ Td (T3) finite intersection {x}∩ y=Φ∈ Disrete top define as Td
= {Φ, X, all subset of X, and their
union}
2 1
Example 1.6
∈ We check all axioms of basis (B1) every x∈B where B⊂ (B2) let 1 ={x},2 = where1 , 2 are disjoint
1 ∩ 2 =Φ∈ 3 ⇒ is basis
,
∈
∩= . ′ ∈ ′ ⊂∩= We prove ths by induction method .Let’s just check for two subsets B1 , B2 By (B2) ∃B3 ∈ s.t x ∈ B3 ⊂B1 ∩ B2 We assume that claim is true for n-1 & prove for n −1 B Suppose B1 , B2 ,…, Bn є s.t x∈∩ni=1 Bi .⇒ x ∈∩ni=1 i −1 B By (B2) ∃ B∗ ∈ s.t x∈ B∗ ⊂∩ni=1 i ∗ Now xєB ∩ Bn By (B2) ) ∃ B ′ ∈ s.t x∈ B ′ ⊂B ∗ ∩ Bn −1 B it follow that x ∈ B ′ ⊂∩n B ⇒B∗ ⊂∩ni=1 i i=1 i Thus ,if result hold for n-1 ,then hold for n ∎
Theorem:1.4
The topology T generated by a basis Proof:
is a topology
We check all axioms of topology (T1) Φ,X∈T I,e every element in X contained in some basis element. so union of these basis is X (T2) Arbitrary union Let {Uα }α∈ I ∈T and let U=∪ {Uα }α∈ I There are two cases If all Uα are empty then U=Φ ∈T If Uα are nonempty then union of Uα equal to X
9
⇒V=∪ α =X∈T
(T3) finite intersection ∈T and let V=∩ni=1 Ui we show tht V∈ T Let There are two cases If one of {Ui } is empty then V=Φ ∈T Let {Ui } is the union of basis element. we show that V is the union of basis element Let‟s check for two U1 and U2 for each xєU1 ∩ U2 there are x ∈ B1 ⊂ U1 and x ∈ B2 ⊂ U2 , this is because U1 ,U2 ∈ T and x∈ U1 ,x ∈ U2 Now by (B2) ∃ B3 s.t x ∈ B3 ⊂B1 ∩ B2 Now we found B3 ∈ s.t xєB3 ⊂U⇒x∈ Bi ⊂ Ui for each i by lemma 1.7 x∈∩ni=1 Bi then ∃ Bx s.t x∈ Bx ⊂∩ni=1 Bi ⊂ now by union lemma V= ∪x ∈V Bx V is the union of basis element thus finite intersection of Tin T
{Ui }ni=1
See exercise 1.10 Let =[a,b⊂ ℝ |a
See exercise 1.14 Let X=Z , n∈Z
{n} ,if n is odd B(n)= {n-1,n,n+1};if n is even Is the basis for a topology? The resulting topology is called digital line topology Example at book
Theorem 1.9 let X be a set and
∈
∃
∈
∈ X
Proof:
Let U is open set in top generated by basis and x∈U As U=∪ and there is at least one basis element ′ in union that contain x ⇒x∈ ′ ⊂U
Conversely: Now suppose that U⊂X is s.t for each x∈U there ∃ a basis element ∈ s.t x∈ ⊂U By union lemma U=∪∈ ⇒U is a union of basis elements. thus U is open set in topology by ∎
ℝ
For x=(1 ,2 )and y=(1 ,2 ) are two point in ℝ we introduce Euclidean distance formula d(x , y)= (1 − 1 )2 + (2 − 2 )2 for each x in ℝ >0, define B(x, )={p∈ ℝ2 |d(x ,p)< } The set B(x, ) is called the open ball of radius centred at x
.
10
B(x, )|{x∈
ℝ , > 0} is basis for a topology on ℝ
We check axioms of basis (B1)since for each x ∈ ℝ2 contain in basis element B(x, 1)={1 ∈ ℝ2 |d(x ,1)< 1} (B2) intersection of two elements Let x∈ B(p, 1 )∩ B(p, 2 ) By lemma 1.11 there exist1 , 2 >0 s.t B(p, 1 )⊂ B(p, 1 ) and B(p, 2 )⊂ B(p, 2 ) Let =min{1 , 2 } Then B(x, )⊂ B(p, 1 ) ∩ B(p, 2 )⊂ B(p, 1 )∩ B(p, 2 ) ⇒ is basis for a topology on 2
ℝ
Suppose that xє By , r
y
Then d(x ,y)
r
⇒ dy , z
x z
B(x , )
B(y, r)
ℝ
ℝ
We check axioms of basis (B1) Since every point on real line contained in a i nterval and therefore contain in a interval B r∈B then r∈ [, + 1) (B2) If two interval in B intersect then a point in intersect of two interval in B that contain in the intersection [ [ ) ) [ [ ) ) [ [ ) )
1 2
1
2
2 1 1
2
ℝ ∈ℤ
(B1) let r∈ 1 and 1 ⊂1 ⇒r∈ 1 =(r-1,r+1) (B2) 1 , 2 and x∈ 1 ∩ 2 then ∃ 3 ⊂1 s.t x∈ 3 ⊂ 1 ∩ 2 ( ( ) ) 0 n 2 n+2 Which not hold
1
ℝ
2
2 1
11
e.g let
1 = 0 ,2 , 2 = (1 ,3) where 1 , 2 ⊂ 1 1 ∩ 2 = (1 , 2)∉ 1
ℝ
ℝ ≤
Let 1 = [1 ,2], 2 =[2 ,3] where 1 , 2 ⊂ 2 Z ∈ 1 ∩ 2 since carnality of basis element is finite {z} is finite .so a finite set cannot an infinite set ⇒2 is not basis (B1) Is satisfied because every element of C2 belong to any B (B2) Let B1 = [1 ,2], B2 =[2 ,3] where B1 , B2 ⊂ C2 B1 ∩ B2 = [ 2] ∈ C3 ⇒ [2] ∈ B3 ⊂ B1 ∩ B2 ⇒the intersection of two closed interval is either closed interval or s singleton (B2) will satisfy in every case
ℝ ∈ℝ
(B1) let r∈(-r-1 , r+1) ⇒ r∈B as B ⊂ 4 (B1) Is satisfied because every element of 4 belong to any B Let 1 = −1 ,1 , 2 = (−2 , 2) where 1 , 2 ⊂ 4 1 ∩ 2 = −1 , 1 ∈ 3 ⊂1 ∩ 2 where 3 ⊂ 5
⇒ ∈ 3 ⊂1 ∩ 2
(B2) is satisfy because the intersection of nested intervals is a nested interval of small length
∪
ℝ
(B1) Is satisfied because every element of C5 belong to any B (B2) let 1 = 1 ,2⋃{3} , 2 = (2 ,3)⋃{4} where 1 , 2 ⊂ 5 1 ∩ 2 = 3 = 1 ,2⋃3 ∈ 3 ⊂1 ∩ 2 where 3 ⊂ 5 (B2) is satisfy because a , b⋃b + 1is closed in ℝ . Intersection of two closed interval is a closed or singleton
⇒ ∈ 3 ⊂1 ∩ 2
ℝ
ℝl is lower limit topology and define as ℝl = {Φ, open interval & their union, set of form [a, band their union}
A is open in because [a,b) is open in ℝ ⇒A∈ B is not open in ℝ B∉ℝ C is not open in ℝ ⇒C∈ ℝ D is open in ℝ ⇒D∈ ℝ
Consider the following six topologies define on ℝ; Trivial top, Discrete top, Complement Top, Standard Top, Lower limit Top and Upper limit Top. Show how they compare to each other (finer, strongly finer, coarser, strictly coarser, non comparable) and justify your answer COMPARSION:
12
Trivial top contained in finite complement top, standard top, lower limit top, upper limit top and discrete top Finite complement top is also contain in standard top, standard top, lower limit top, upper limit top and discrete top ⇒ ⊆ ⊆ ⊆ ⊆ ⊆ & ⊆ ⊆ ⊆ ⊆ And Non ⊈ Comparable ⊈
Let B be collection of subsets of ℤ used in definition of digital line topology in example 1.10.Show that B is basis for a topology on ℤ SOL: let = ℤ , n∈
ℤ
{n} ,if n is odd B(n)= {n-1,n,n+1};if n is even Is the basis for a topology .The resulting topology is called digital line topology Now we show that the collection ={B(n)|n ∈ ℤ } is basis for topology on ℤ We check all axioms of basis (B1) (B1) Is satisfied because every element of ℤ belong to any B Let x∈ ℤ then x∈ Bx ={x} if x is odd And x∈ Bx ={x-1 , x , x+1} if x is even (B2) intersection There are three possibilities If m an n are odd ⇒Bm = m, Bn = n ⇒Bn ∩ Bm =Φ∈ BΦ ⊆ Bn ∩ Bm where BΦ ⊆ ℤ (ii)If m and n are even then two possibilities m, n consuctive even integer n
ℤ is a set , = … , − , − , , + , + , … , ℤ and b≠0. Prove that the collection of arithmetic progression = { , | a , b∈ ℤ and b≠ } topology on ℤ. the resulting Top is called arithmetic progression top on ℤ An arithmetic progression in
SOL:
Examples Let a=0 , b=1 ∈ ℤ 0,1 ={… , -2 , -1, 0 , 1 , 2 ,…} ∈ ℤ 2,3 ={… , -4 , -1, 0 , 2 , 5 ,…} ∈ ℤ
We check axioms of basis (B1) Clearly every element of belong to any A let z∈ ℤ ⇒z∈ 1 = 1 ,1 ∈ (B2) Let , ,1 =1 , ,,2 = 2 where 1 , 2 ⊆A Then z∈ , 1 ∩ , 2 = 1 ∩ 2 So z∈ 3
⊆ 1 ∩ 2 where 3= ,(1 2 )
Prove theorem 1.12
13
On the plane ℝ . Let ={(a,b)×(c,d) Show that basis for a top on ℝ
ℝ
We check axioms of basis (B1) [x ,y]∈B where B⊆ [x , y]∈(x-1,x+1)×(y-1 ,y+1) where [x , y] for an order pair and and( x1,x+1) and (y-1 ,y+1) are interval (B2) Let a
Show that T, generated by B is standard top on By (′ , ′ )×( ′ , ′ )⊆(a ,b)x(c ,d) ⇒′ ⊆ --------------------- (i) If ′ < < < < ′ ′ < < < < ′ Then ( ′ , )⨉(x, ′ ) and ( ′ , )⨉(y, ′ ) Now [x, y]∈(a, b)×(c, d)⊆ (′ , ′ ) × ( ′ , ′ ) ⇒ ⊆ ′ --------------------- (ii)
ℝ
By (i) and (ii) ⇒ = ′
∞, ) ⊆ ℝ We check axioms of basis (B1) Let r ∈ ℝ Then rє−∞, ⌊⌋=B⊆ (B1) satisfied (B2) Let 1 = −∞, 1 2 = −∞, 2 Then 1 ∩ 2 =−∞, 1 1 < 2 ⇒−∞, 1 ∈ 1 ⊆ 1 ∩ 2 (B2) satisfied so given collection are basis
×
ℝ
ℝ
∈ℝ
14
SOL: (B1) c ------------------ y+1 Let (x, y)∈{{x}×(y-1,y+1)}=B y ------------------ (x, y) 2 b ------------------- y-1 ⇒every pair in ℝ belong to any B (B2) x Let B1 = x ⨉y1 − 1, y1 + 1and (a) B1 = x ⨉y2 − 1, y2 + 1 if y1 < y2 ⇒ B1 ∩ B2 ⊆ B1 = x ⨉y1 − 1, y1 + 1
(B1 and B2 satisfied ⇒these are
basis
(b) ℝ From fig b ℝ2 ⊆ ℝ where circles represents ℝ2 and lines representsℝ
Closed Set A subset of a topological space X is closed if the set X-A is open OR If complement of a subset of topological space is open. The subset said to be closed
Example 1.13
ℝ
Show that the interval (0, 1) open in standard Top on [a, b] closed in standard topology on ℝ {c} is closed in standard Top on ℝ SOL: A =(-∞, 0] ∪ [1, ∞) Ac is closed ⇒ A is open
A=(0, 1)
c
B =(-∞,a) Bc is open c
[ 1
B=[a, b]
∪ (b, ∞) ⇒ B is closed
C =(-∞,c) C c is open c
] 0 [ a
∪ (c, ∞) ⇒ C is closed
] b
C={c}
. c
1.15 Definition (i) Closed ball: For each x∈ ℝ2 and > 0 define closed ball of radius centred at x to be set
(x, )= {y ∈ ℝ2 |d(x, y)≤ } (ii) closed Rectangular: If [a, b] and [c, d]are closed bounded interval in ℝ. Then the product [a, b]×[c, d]⊆ℝ2 is called closed rectangular
Theorem 1.16 Closed balls and closed rectangular are closed sets in standard top on Proof: As we know that closed balls in ℝ of centered x and radius to be set (x, )= {y ∈ ℝ2 |d(x, y)≤ } Where (x, y) is distance between x and y To show that it is closed in ℝ2 .we prove that ℝ2 − (x, ) is open Let a∈ (x, ) ,where d(x, a)>
ℝ
15 i,e d(x, a)> + where >0 ------- (i) so = d(x, a)- -----------------(ii) Clearly is +ve number We claim that B(a, ) ⊆ ℝ2 − (x, ) Let b∈ (x, ) ,then d(a, )< by triangular inequality d(x, a)≤ d(x, b) +d(a, b) d(x, b)≥ d(x, a) −d(a, b) d(x, b)> d(x, a) – by (i) d(x, b)> by (ii)
x a
⇒ b⊆ ℝ2 − (x, ) Thus B(a, ) ⊆ ℝ2 − (x, ). This show that a ∈ ℝ2 − (x, ) is enclosed by an open ball contain in ℝ2 − , ⇒ℝ2 − , is open ⇒(x, ) is closed Hence proved closed ball are closed set in standard top on ℝ2 Rectangular Part As we know closed rectangle define as A=[a, b]×[c, d]⊆ℝ2 To show A is closed set in standard Top We prove that ℝ2 − A is open We know that ℝ2 − A can be expressed as union of four open half planes {(x, y):x
b} , {(x, y):yd} Since each half plane is open set and union of open sets is open set ⇒ℝ2 − A is open set
⇒A is closed set
Example 1.16 Let set X= {a, b, c, d,} have topology T={Φ, X, {b}, {a, b}, {c, d}, {b, c, d}} Explained which subsets are open or closed SOL: A= {b} which is closed in T ⇒ A={b} is open B= {a} ={b, c, d} which is open in T ⇒ B is closed C= {a, b} ={c, d} which is both closed and open in T D= {b, c}
={a, c, d}
={a, d}
which is not closed nor open in T
16
We check Φ and X are closed As Φ =X where X is open in top space ⇒ X is closed & X c = Φ where Φ is open in top space ⇒ Φ is closed c
As we know in top space subsets are open as closed Let C∈X , where C is closed Now we prove that ∩ Ci is closed By De-Morgin Law (∩ Ci )c is open (∩ Ci )c =∪ Ci viz open in X ⇒ ∩ Ci is closed Let yi be finite closed set in X We prove that ∪ yi is closed By De-Morgin Law (∪ yi )c is open (∪ yi )c =∩ Ci viz open in X ⇒ ∪ yi is closed
Example Topology on ℝ generated by basis of sets {(n, n+1) sets are open SOL:
ℝ |n∈ ℤ }.show that
Hence each set consist of single point .thus single point sets are open
1.18 Definition Housdroff space
A topological space X is Housdroff if for every point of distint point x and y, There ∃ neighborhood U and V of x and y respectively s.t U∩V=Φ i.e Distinct point have distinct neighborhood
Example 1.17 The real line
ℝ with standard topology is Housdroff SOL:
Let two distinct points a, b with distinct neighborhood (open interval) + + , + 1) U= (a-1, ) and V= ( 2
2
Then U∩ =
⇒
Example 1.18 Show that Discrete Top is Housdroff SOL: As for each point there is a neighborhood Let x, y are distinct points and having neighborhood U={x} ,V={y} Then U∩ =
⇒
If X is Housdroff space, Then every single point subset of X is closed Proof: To show {x} is closed .we must show that X-{x} is open This set will be given by any point y in the set We can find neighborhood V of y s.t V⊂ X-{x} since y ∈ X-{x} ; x ≠
17
Let U and V be distinct open sets. Contain the two set But U⊂X-{x} and V is neighborhood of y Hence the set is open so {x} is closed ∎
C is closed ⇒X-C is open Now U-C=U ∩ (X-C) The intersection of t wo open sets is an open set⇒ U- C is open (ii) U is open ⇒X-U is closed Now C-U=C∩(X-U) (i)
The intersection of two closed set is closed set ⇒ C-X is closed
ℝ Proved in Theorem 1.6
≤ ≤ ] ∪
= , , , … ≤ ≤ ℝ
ℝ ( . ) SOL:
1
1
1
8
4
2
0…
fig 1.17
To show C is not closed in standard topology we prove that C c = ℝ2 − C is not open As {(0, y); 0≤y≤1} is limiting line but not belong to C ⇒ C is not closed OR c to prove C = ℝ − C is open then ∃ an open disk for every xєC c = ℝ − C and also in their union and intersection But in infinite comb open disks cannot exist ⇒ C c = ℝ2 − C is not open in R2 ⇒C is not cloed in the standard topology on R2 To show C is closed in vertical interval topology we prove that C c = ℝ2 − C is open As The collection {{a}×(b, c)⊂ ℝ2 }of vertical interval in plane is basis for topology on ℝ2 For arbitrary a, b, c similarly this topology are vertical line c ---------b --------So we can draw such lines in infinite c c 2 a ⇒these are open in C = ℝ − C ⇒ C is closed in vertical interval topology
1.28
Which sets are closed in finite complement topology on X SOL:
18 All infinite intervals are closed in ℝ and the finite set whose complement is infinite are open in ℝ
1.29
Which sets are closed set in exclude point topology on a set X SOL: Those subset of X which are contain point p (p∈X) are closed in EP
1.30
Which sets are closed set in particular point topology on a set X SOL: Those subset of X which are do not contain point p (p∉X) are closed in PP Or those subsets of X which are open in EP are closed in PP
1.31
Shows that a single point {n} is closed in digital line top
n is even
SOL: Suppose {n} is closed and prove that n is even Let A={n} As given {n} is closed ⇒ = {−∞, … , − 2, − 1} ∪n+1, n+2, …,∞} ⇒n-1 and n+ 1 both are odd ⇒ n is even Conversely Let n is even then prove that {n} is closed As A= {n} ⇒ = {−∞, … , − 2, − 1} ∪n+1, n+2, …,∞} viz open (∴infinite union of open set is open) For any number which is even when we take its complement, it will be open in T ⇒ Every single point {n} is closed in digital line top ⇔ n is even
1.32
Prove that intervals of form [a, b) are closed in lower limit topology on ℝ SOL:
1.33 1.34
As we know that [a, b) is open in ℝl because basis for ℝl Now we prove that it is also closed in ℝl Let A=[a, b) c Then A = {−∞, a) ∪[b, ∞) which is union of open interval ⇒ Ac = (−∞, a) ∪[b, ∞) is open
⇒ A is closed
Proved in theorem 1.17 On the five point set {a, b, c, d, e} consist two topologies (a)One is that Housdroff (other than discrete top) (b) One is Housdroff (other than trivial top) SOL: (a) We cannot find any topology that is Housdroff (other than discrete top) (b)
1.35
T={Φ, X, {a}} etc is Housdroff (other than trivial t op)
Show that ℝ in lower limit topology is housdroff SOL:
19
Lower limit topology define as l ={Φ, open interval and their union &the interval of form [a, b) and their union} As for each point there is a neighborhood Let x, y are distinct points and having neighborhood U=[x y),V=[y,z) Then U∩ V = Φ ⇒ Housdroff
ℝ
1.36
Show that
ℝ in the finite complement topology is not housdroff SOL:
Let a finite set A={x1 , x2 , x3 , … , xn } Now by the definition of finite complement topology Tfc Ac = ℝ -x1 , x2 , x3 , … , xn }∈ Tfc Let r1 ≠ r2 Then r1 ∈ O1 ∈ ℝ − {x1 , x2 , x3 , … , xn } & r2 ∈ O2 ∈ ℝ − {y1 , … , xn } where O1 and O2 are neighborhood of r1 and r2 respectively Clearly O1 ∩ O2 ≠Φ ⇒ space is not Housdroff
1.37
Prove that the arithmetic progression topology on (see Ex 1.15) SOL: Arithmetic progression define as , = … , − 2 , − , , + , + 2 , … Let x , y ∈ ℤ where x≠y Let x∈ 1 ,1 , y∈ 2 ,2 We prove that a 1 ,1 ∩ 2 ,2 = ∅ Now we let x is even and y is odd Then x∈ 1 ,1 = … , −2, − ,0, ,2, … y∈ 2 ,2 ={…,y-2, y, y+2,…} then 1 ,1 ∩ 2 ,2 = ∅ END END END
Muhammad Tahir Aziz Takleed (Bhakkar) BSc BZU Multan, MSc (Math) GCUF FB: www.fb.com/tahiraziztakleed Email:[email protected] [email protected] Ph: +92 301 7810067 +92 343 46974245 +92 313 6974245 +92 333 8839067
ℤ is Housdroff
20
Chapter # 2
Interior, Closure and Boundary 2.1 Definition: Interior and Closure of Sets: Let A be a subset of a topological space X, a point x∈A is said to be an interior point of A if x is in some open set contained in A OR The union of all open sets contained in A and denoted by “ A° ” or “Int(A)” It is the largest open subset of A And let X be a subset of topological space X. A point x∈ is said to be an closure point of A if there exist an open set U containing X such that U⊂ . In other words, an interior point of is said to be a closure of A OR The intersection of closed superset of A It is the smallest closed set containing A denoted by “ ” or “Cl(A)” From the above definitions we get a sandwich relation Int (A) ⊂ A ⊂ Cl(A)
Let X be a topological space and A and B be subsets of X and (i) (ii) (iii)
If U is open set in X and U ⊂A, then U ⊂IntA. If C is closed set in X and A C ,Then Cl(A) A. If A B then Int(A) Int(B)
(iv) (v) (vi)
If A B then Cl(A) Cl(B) A is open iff A=Int(A) A is closed iff A=Cl(A)
OR (iii) If A B then Int(A) Int(B) Let A ⊂ B then by definition of interior of A Int(A) ⊂A and Int(B) ⊂B ⇒ Int(A) ⊂ A ⊂B ⇒ Int(A)⊂B ⇒ Int(A) ⊂ IntB ∴ Int(B) is the largest open subset of B.
Proof:
(i) Suppose U is an open set in X and U ⊂A Since Int (A) =⋃ U (by def of Interior)
⇒ U⊂IntA (ii) Suppose C is an closed set in X and A⊂C Since Cl(A)= ∩ C (by def of closure) ⇒ Cl(A)⊂A (iii) Let xє A° then ∃ an open set U containing x s.t U⊂A, Since A⊂B ⇒U⊂B this show that x also contain in interior point of B i.e x∈ B° ⇒ A°⊂ B° Let A ⊂ B ---- (i) then by definition of closure of A (iv) A ⊂ClA-----(ii) and B ⊂ClB-----(iii) by previous theorem By (i) and (iii) A⊂B ⊂ClB ⇒ A ⊂ ClB ∴ A ⊂ClA-----(ii) Then Cl(A)⊂ ClB (v) Suppose A is open by definition of interior of A Int(A) ⊂A --------(i)
Also we can write A⊆A Then A⊂IntA ----------------(ii)
21
By i and ii ⇒ IntA=A Conversely suppose Int(A)=A As Int (A) is unoin of open subsets of A so Int(A) is open then A is open (vi) Suppose A is closed by definition of closure of A A ⊂ClA --------(i)
Also we can write A⊆A Then ClA⊂A ----------------(ii) By (i) and ii ⇒ ClA=A Conversly suppose Cl(A)=A
⇒A is closed because closure is always closed∎
ℝ Int(A)=(0,1)
and Cl(A)=[0, 1]
ℝ Int(A)=∅ becauese no open sets contain in A Cl(A)= ℝ A is infinite the only infinite closed set in this topology is ℝ
A=[0,1)
=(-∞, 0)∪[1,∞)
ℝ Int(A)=[0,1) because A is open in this topology Cl(A)=A because =(-∞, 0)∪[1,∞) is open in this topology therefore A closed in this top
Show that the Int( ℚ)=∅ and Cl(A)= ℝ in standard topology on SOL:
ℝ
We show that Int(ℚ)=∅ we suppose contrary that Int(ℚ)≠ ∅ And suppose that U in a non empty open set contain in ℚ. Let x∈U then there is an open interval s.t x∈ (a,b)⊂U⊂ ℚ but in every pair of real number there is an irrational number. Thus every interval contains element of ℝ - ℚ. Therefore U∉ ℚ this is contradiction ⇒ Int(ℚ)=∅ As Int(ℚ)=∅ we get ℝ in closure ⇒ Cl(ℚ)= ℝ
Definition 2.3: A subset A of a topological space X is called (see example 2.4 and 2.5)
if Cl(A)=X
22
ℝ In this topology closed set are either finite or ℝ itself Therefore ℝ is only closed set containing in infinite set. Thus ,if A is an infinite subset of R then Cl(A)= ℝ ⇒A is dense in ℝ
As given A⊂X , y∊X We suppose that there exist open set U s.t y∊U⊂A Thus U is open and contained in A It follow that U⊂ Int(A) if y∊U
⇒y∊Int(A)
Conversely
Let y ∊Int(A) and Int(A) =U ⇒U is open set s.t y ∊ U ⊂ A
Let X be a topological space, A⊂X and y ∊ X then we show that every open set that containing y intersect A
As we know that Cl(A)= ∪ ′ ⇒ y ∊ ′
Then ∩ ≠ ∅ Conversely suppose ∩ ≠ ∅ we show that y ∈ Cl(A) ⇒ ∊ ∴ y ∊A ′ ⇒ ∊ ∪ ( ) ∴ y ∊A ⇒ ∊ Cl(A) ∴ Cl(A)= ∪ ′ ∎
(iv)
∪ ∩
⊂ ⊂
∪ ∩
)
(i) Let {x∝ : ∝∈ I} be a collection of closed super set of A then Cl(A)=∩∝∈I x∝ R.H.S = X-Cl(A)=X- ∩∝∈I x∝ = ∩∝∈I x∝ c =∪∝∈I x∝ c =∪∝∈I (X − x∝ )
∴ ∝ is closed ⇒ X-∝ is open and ∪( X-∝) = (X − ∝)
23
=Int(X-A)=L.H.S (ii) Let {x∝ : ∝∈ I} be a collection of open subset of A then Int(A)=∪∝∈I x∝ R.H.S = X-Int(A)=X- ∪∝∈I x∝ = ∪∝∈I x∝ c =∩∝∈I x ∝ c =∩∝∈I (X − x∝ ) =Cl(X-A)=L.H.S
∴ ∝ is open ⇒ X-∝ is closed and ∩( X-∝)(X − ∝ )
(iii) As A ⊂ A ∪ B and B ⊂ A ∪ B so Int(A)⊂Int(A∪B) and Int(B)⊂Int(A∪B) ⇒ Int(A∪B)⊂Int(A) ∪ Int(B)-------------------------(1) Now we prove that Int(A∪B)≠Int(A) ∪ Int(B) By counter example Let A=[0, 1] ,B=[1, 2] and A∪ B = [0, 2] then Int(A)=(0, 1),Int(B)=(1, 2), Int(A)∪Int(B)=(0, 2)-{1}, Int(A∪B)= (0, 2) or A=Q, Int(A)=∅ and B=, , = ∅ and Int(A) ∪ Int(B)=∅then A∪B= ℝ and Int(A∪B) = ℝ ⇒ Int(A) ∪ Int(B) ≠ Int(A∪B) (iv) As A ∪ B ⊂ A and ⊂ B so Int(A∪B)⊂ Int(A) and Int(A∪B)⊂ Int(B) ⇒ Int(A∩B)⊂Int(A) ∩ Int(B)-------------------------(1) Let y∈ Int(A) ∩ Int(B) ⇒ y ∈ Int(A) and y ∈ Int(B) As we know Int(A)⊂A and Int(B)⊂B ⇒ y ∈ A and y ∈ B ⇒ y ∈ A ∪ B ⇒ y ∈ Int(A∩B) ⇒ Int(A) ∩ Int(B ⊂)Int(A ∩ B)-------------------------(2) By 1 and 2 ⇒Int(A) ∩ Int(B) = Int(A∩B)
ℝ As A= ℝ - ℚ be set of irrational numbers ⇒ ClA=Clℝ - ℚ) = ℝ -Int(A) = ℝ -∅= ℝ
⇒ A is dense
∴ Intℚ)=∅
Exercises for section 2.1 2.1
Determine Int(A) and Cl(A) in each case (a) A=(0, 1]in lower limit topology on ℝ SOL: Int(A)=(0, 1) Cl(A)=[0, 1]
(b) A={a} in X={a, b, c} with topology { ∅, X, {a}, {a, b}} SOL: Closed set in topology are {X, ∅, {b,c},{c} Int(A)={a} Cl(A)=X
The smallest closed set contain {a} is X so Cl A is X
(c) A={a, c} in X={a, b, c} with topology { ∅, X, {a}, {a, b}} SOL: Closed set in topology are {X, ∅, {b,c},{c} Int(A)=∅ Cl(A)=X
(d)
The smallest closed set contain {a, c} is X so Cl(A) is X
A={b} in X={a, b, c} with topology { ∅, X, {a}, {a, b}} SOL: Int(A)={ } Cl(A)=X
Closed set in topology are {X, ∅, {b,c},{c} The smallest closed set contain {b} is {b, c} so Cl(A) is {b, c}
24
(e)
A=(-1,1)∪ {}in standard topology on SOL:
ℝ
Int(A)=(-1, 1) Cl(A)=[-1,1]∪{2}
(f)
(g)
A=(-1,1)∪ {}in lower limit topolgy on SOL: A= {(0, x)∈
Int(A)=(-1, 1) Cl(A)=[-1,1]∪{2}
ℝ |x∈ ℝ} in ℝ the standard
Int(A)={ } Cl(A)=A
{(, ) ∈ ℝ | ∈ ℝ}
topology
Standard top on ℝ define as T={∅, ℝ2 , all open disks and their union. All set are closed and open so Cl(A) is A 2
SOL:
Int(A)=∅ Cl(A)=A
ℝ
ℝ
SOL:
{( , ) ∈ ℝ | ∈ ℝ}
ℝ
Int(A)=A Cl(A)=A
2.2
2.3
Let X be a topological space and A and B be subsets of X (a) If C is a closed set in X and A X then Cl(A) C (b) if A B then Cl(A) Cl(B) (c) A is closed if and only if A=Cl(A). SOL: solved in theorem 2.2 For m
m=2 , n=6
l2,6 = {2, 4, 5, 6} Int(l2,6 )={3, 4, 5} Cl(l2,6 )={2, 3,4,5,6} m=2, n=5 l2,5 ={2,3,4.5} Int(l2,5 )={3,4,5} Cl(l2,5 ) ={2,3,45,6} m=1 ,n=5
l1,5 = 1,2,3,4,5 Int(l1,5 )=1,2,3,4,5} CL(l1,5 )={0,1,2,3,4,5} m=1,n=6
l1,6 = 1,2,3,4,5,6
In(l1,6 ) ={1,2,3,4,5} Cl(l1,5 ) = {0,1,2,3,4,5.6]
25
2.4
Consider the particular point topology PP on a set X.Determine Int(A) and Cl(A) for sets A containing p and for sets A non-containing p SOL: If we first consider A contains p then Int(A)=A Cl(A)=X If consider A not contains p then Int(A)=∅ CL(A)=A
2.5
Consider the excluded point topology EP on a set X.Determine Int(A) and Cl(A) for sets A containing p and for sets A noncontaining p SOL: If we first consider A contains p then Int(A)=∅ Cl(A)=A If consider A not contains p then Int(A)= CL(A)=X
2.6
2.7
Prove that Cl( ℚ)= ℝ in standard topology on SOL:
Let B={
ℝ
Let A= ℚ Cl(A)=Cl(R-ℚ′ ) Cl(A)=R-Int(ℚ′ ) CL(A)= ℝ -∅ CL(A)= ℝ
∈ ℝ |a∈ ℤ, n∈ ℤ+}.Show that B is dense in ℝ SOL:
Given B={ ∈ ℝ |a∈ ℤ, n∈ ℤ+} 2 1 2 3 B={0, , 2 , 3 , … , } a and n are varing
22
2
The subgroup of ℝ are two types (i) Cyclic (ii) Dense
2
Cl(B) is the smallest closed set that consist B.It is clearly B is not cyclic
⇒ ClA is dense ⇒ ClB= ℝ
2.8
ℤ
ℤ
ℤ SOL:
We show that Cl(ℤ )= ℤ ⇒CL(ℤ )= ℤ ∪ ℤ′ --------------(i) Let x∈ ℤ.Then either x is even or odd if x is odd then by definition x is the limit point of ℤ if every open set containing x ∩ ℤ other then ℤ
26
2.9
Now {x} ∩ ℤ = {x} ⇒ x∉ ℤ′ If x is even then {x-1, x, x+1}∩ ℤ ≠ ∅ thus intersection will be other then x ⇒ x∈ ℤ′ So ℤ′ ={0, +2, +4, … } So by (i) we get CL(ℤ ) ={0, +1, +2, +3, +4, … } ⇒ Cl(ℤ )= ℤ Hence the set of odd integer is dense in digital line topology on ℤ For even numbers Let ={0, +2, +4, … } As we know Cl(E) = ∪ ′ ------------(ii) Let x ∈ ℤ then x is either even or odd If x is odd then {x} ∩ E=∅ ⇒x∉ ′ If x is even then {x-1, x, x-2}∩ E≠ ∅ ⇒x∉ ′ ∴ x is even and ′ is for odd ⇒ ′ =∅ So Cl(E)≠ ℤ Hence the set of even is not dense in digital line topology (b) In discrete both odd and even are both dense
ℝ
× ×
×
×
SOL:
In We prove that Cl( (a, b)×(c, d) )=[a, b]×[c, d] Cl( (a, b)×(c, d) ) is the smallest open set that contain (a, b)×(c, d) So Cl( (a, b)×(c, d) )=[a, b]×[c, d] Now we prove that Int( [a, b]×[c, d] )=(a, b)×(c, d) In Int( [a, b]×[c, d] ) is largest closed set that contain [a, b]×[c, d] So Int( [a, b]×[c, d] )=(a, b)×(c, d) 2
Closure
a
Interior
b
a
SOL: Already proved
SOL: Already proved
∩ ∪
∩ ∪
We know that A⊆ Cl(A) and B⊆ Cl(B) ⇒ A ∩B⊆A and A ∩B⊆B
b
27
Then Cl(A ∩B) ⊆ (A) and Cl(A ∩B⊆ClB ⇒ Cl(A) ∩ ClB⊆ClA ∩B). Let A={a} and B={b} in X=[a, b, c} with topology T={∅, X, {a}, {b},{a, b}} Then Cl(A)={a, c} Cl(B)={b, c} Cl(A ∩B)=∅ Cl(A) ∩ Cl(B)=∅ (b)
A⊆ClA and B⊆ClB A⊆A ∪B and B⊆A ∪B) ClA⊆ClA ∪B and ClB⊆ClA ∪B) C(A) ∪ (B⊆ ClA ∪B) A=(-1, 0) ans B=(1, 2) Cl(A)=[-1, 0] and Cl(B)=[1, 2] Cl(A)∩Cl(B)=[-1,0]∪[1, 2] A∪B=(-1,0) ∪(1,2) Cl(A∪B)=[-1, 2] ⇒ ClA ∪ (B⊆ ClA ∪B)
2.2
LIMITING POINTS
Let (X, T) be a topological space and A⊆X be a sub set of X. A point x⊆X is said to be limit point of A if every open set containing x intersect with A-{x} i.e U∩ A-{x}≠ ∅ Set of all limits points of A is denoted by
Note Closure of a set A must contain its limits points ={x∈X:Every open subset of X intersect with A-{x} i.e U∩ A-{x}≠ ∅
Example
Let X={a, b, c, d} T={∅, , , , {, }}
= ,
Find the derived set of A SOL:
take a∈X Step 1: A-{a}={d} Step 2 : choose their open subset of X , which contain „a‟ X∩A-{a}=X ∩{d}={d}≠ ∅ U∩A-{a}={a} ∩{d}=∅ so a∉ A Now take b∈X Choose a open set of X which contain „b‟ X∩A-{b}={a, d}≠ ∅ b ∈ Take c ∈ X Step 1: A-{c}={a, d} Step 2: Choose a open set of X which contain c X∩A-{c}=X ∩{a, d}={a, d} ≠ ∅
28 U∩A-{c}={c} ∩{a, d}=∅ so c ∉ take d ∈ X step 1: A-{d}={a} step 2: Choose a open set of X which contain d X∩A-{d}=X ∩{a}={a} ≠ ∅ so d ∈ ={a, d}
Let A={
∈ ℝ |n ∈ ℤ+} subset of ℝ with standard topology. Show that L(A)=0 Solve in book try to understand
Let A=(0, 1] subset of
ℝ with standard topology. What are the limits points of A
SOL: A=(0,1] Let y1 is not limiting point Let {0-1 , 0+2 }∩ A≠ ∅ other then 0 Let {1-1 , 1+2 }∩ A≠ ∅ other then 1 So limit points of A is [0, 1] ⇒ =[0, 1]
Let Q subset of SOL:
ℝ with standard topology. Show that L( ℚ)= ℝ
let x∈ ℝ there is a open set U containing x such that x ∈ (a, b) (a, b)-x ∩ ℚ ≠ ∅(infinitely many points other then x) ⇒ x∈ ℝ is limit point
⇒ every point of R is limit point of ℚ ⇒ Lℚ)= ℝ
∪ ′
′
Proof: Let x∈ Cl(A) = { x∈X : every open subset U containing X intersect with A : U ∩ A≠ ∅} If x ∈ A , x ∈ A∪ ′ ⇒ ClA⊆A ∪ ′
29
If x ∉ A⇒ A-{x}=A U∩A≠ ∅ ⇒ U∩A−{} ≠ ∅ ⇒x∈ ′ ⇒ x∈ ∪ ′ ⇒ ClA⊆A ∪ ′ ------------------(1) Conversely Let x∈ ∪ ′ If x ∈ ⊆ Cl(A)
⇒ x ∈ ⇒ ∪ ′ ⊆ ClA If x∉ A and x ∈ ′ By the definition of limiting point U∪A-{x}≠ ∅ U∪A≠ ∅ ⇒ x∈ Cl(A) ′ ⇒ ∪ ⊆ ClA ----------------(2) By (1) And (2) We have Cl(A)=A ∪ ′ ∎
′ ⊆ Proof:
Suppose A is closed by the theorem A is closed ⇔ A=ClA And by theorem if A= Cl(A) ⇔A ∪ ′ =A ⇔ ′ ⊆A Exercises for section 2.2
ℝ
We know that x said to be limit point of A if N(x) ∩ {A-{x} } ≠ ∅ then x ∈ ′ Now N(0) ∩ {A-{0} } ≠ ∅ ⇒ 0 ∈ ′ and N(1) ∩ {A-{1} } ≠ ∅ ⇒ 1 ∈ ′ ′ =[0, 1]
∅, , , , }
N(a)={ {a}, {a, b}, X} N(b)={ {a, b},X} N(c)={X} Now N(a) ∩ {A-{a} } = ∅ ⇒ 0 ∉ ′ N(b) ∩ {A-{b} } ≠ ∅ ⇒ b ∈ ′
30
N(c) ∩ {A-{c} } ≠ ∅ ⇒ 0 ∈ ′ ′ =[b, c]
∅, , , , }
N(a)={ {a}, {a, b}, X} N(b)={ {a, b},X} N(c)={X} Now N(a) ∩ {A-{a} } = ∅ ⇒ 0 ∉ ′ N(b) ∩ {A-{b} } ≠ ∅ ⇒ b ∈ ′ N(c) ∩ {A-{c} } ≠ ∅ ⇒ 0 ∈ ′ ′ =[b, c]
∅, , , , }
N(a)={ {a}, {a, b},{a, c}, X} N(b)={ {a, b},X} N(c)={X} Now N(a) ∩ {A-{a} } = ∅ ⇒ 0 ∉ ′ N(b) ∩ {A-{b} } = ∅ ⇒ b ∉ ′ N(c) ∩ {A-{c} } ≠ ∅ ⇒ 0 ∈ ′ ′ =[c]
∪ {} Now N(-1) ∩ {A-{-1} } ≠ ∅ ⇒ -1 ∈ ′ N(1) ∩ {A-{1} } ≠ ∅ ⇒ 1 ∈ ′ N(2) ∩ {A-{2} }= ∅ ⇒ 2 ∉ ′ ′ =[-1, 1]
∪ {} Now N(-1) ∩ {A-{-1} } ≠ ∅ ⇒ -1 ∈ ′ N(1) ∩ {A-{1} } ≠ ∅ ⇒ 1 ∈ ′ N(2) ∩ {A-{2} }= ∅ ⇒ 2 ∉ ′ ′ =[-1, 1]
{(, ) ∈ ℝ | ∈ ℝ}
ℝ
A={(, 0) ∈ ℝ2 | ∈ ℝ}. We know that open sets in ℝ2 are open disks. On putting different values x we get horizontal line i.e real line Hence limit point of A whole real line i.e ′ =ℝ-A or ′ =A
{(, ) ∈ ℝ | ∈ ℝ}
ℝ
A={(0, ) ∈ ℝ2 | ∈ ℝ}. We know that open sets in ℝ2 are open disks. On putting different values x we get vertical line i.e real line
31
Hence limit point of A whole real line i.e ′ =ℝ-A or ′ =A
{( , ) ∈ ℝ | ∈ ℝ}
ℝ
A={(, 0) ∈ ℝ2 | ∈ ℝ}. We know that open sets in ℝ2 are open disks. On putting different values x we get horizontal line i.e real line Hence limit point of A whole real line i.e ′ =ℝ-A or ′ =A
∈ ℤ+
∈ ℤ+
We check axioms of basis (B1) ∀ x ∈ ℤ+ ∃ Ɓ ∈ Ɓ s.t x ∈ Ɓ (B2) Let , ∈ Ɓ If mn then x∈ ∩ = ∈ Ɓ All axioms of basis satisfied so Ɓ is basis fot topology on
ℤ+
ℤ+
SOL: Let x, y are distinct points and having neighborhood U={x} ,V={y}.we cannot find open set whose intersection is empty Then U∩ ≠ ⇒ Not Housdroff
2.15
Determine the set of limit points [0, 1] in finite complement topology on SOL: -∞
A=[0, 1] [ ] 0 1 ′ =ℝ
+∞
By the definition of digital line topology B(n)={n} if n is odd and B(n)={n-1, n, n+1} if n i s even In exercise 2.8 dicussed
∈ℚ
ℝ We check axioms of basis
32
(B1) Since every point on real line contained in a i nterval and therefore contain in a interval B x∈B (B2) If two interval in B intersect then a point in intersect of two interval in B that contain in the intersection [
[
1 2
)
1
)
[
2
[
)
2 1 1
ℝ ℝ
)
[
2
[
1
2
)
)
2 1
Cl(A) =[0, 2) and Cl(B)= [ 2,3) in ℝ & Cl(A) =[0, 2] and Cl(B)= [ 2,3] in ℝ
+ ∈
∈ ℤ+
SOL: We fixed “m” and very “n” i.e if m=1. n=1,2,3,4,5,6,…… Then A={ 2, 3/2,1/3, ……} If m=2, n=1,2,3,4,…….. Then A={ 3/2, 1, 5/6,…..} Hence conquest limit point 0 and 2 and all element b/w the x ∈ Q ⇒ ={x∈Q :0≤ ≤ 2}
Let x∈ ℝ and U be an open neighborhood of x .U contains all but finally many points
of ℝ and thus all but finally many points of . Since x and U are general , converges to every points in ℝ.
ℝ
Apply definition of limit points
ℝ 1
∈ ℝ
≤
(x, sin( ) ) put the value of x ∈ 0, 1] we get the value which lie in interval (-1, 1)
33
⇒ limit point of S are -1 and 1and every point lie between them ⇒ =[-1, 1]
The Boundary of a Set Definition 2.13
Let A be a subset of a topological space X. The boundary of A, denoted by and define as set =Cl(A)-Int(A) or =X-(Int(A) ∪ Cl(A) )
Example 2.14 Let A=[-1, 1] in standard topology on Cl(A)=[-1,1]
[ -1 ( -1
Int(A)=(-1,1) ] =Cl(A)-Int(A) =[-1, 1]-(-1,1) ={-1, 1}
. Find boundary ] Cl(A) 1 ) Int(A) 1
|
|
Example 2.15 Concept mjj Cl(A)
A
Int(A)
∈ Let x∈
⇒ x∈ ClA and x∉ IntA ∴ =Cl(A)-Int(A) When x∈ ClA⇒ for every neighborhood of x intersect A When x∉ IntA ⇒ for every neighborhood of x is not subset of A then intersect X-A ⇒ every neighborhood of x intersect A and X-A Conversely Suppose every neighborhood of x intersect A and X-A ⇒ x∈ Cl(A) and x∈ Cl(X-A) As we know Cl(A)=X-IntA ⇒ x∉ IntA Thus x∈ ClA and x∉ IntA ⇒ x∈ Cl(A)-Int(A)
⇒ x∈ ∎
34
= Cl(A) ∩ − ∩ ∅ ∪ ∩ A ∅ ∩ A ∅
As we know that =Cl(A)-Int(A) =Closed set – open set The remaining set is closed set which is So is closed
=
(ii) Cl(X-A) ∩ Cl(A) = A =X-(Int(A) ∪ Cl(A) ) =X-(Int(A) U Int(X-A) ) = (Int A U Int X − A)c c = Int A ∩ ( (Int X − A)c by De Morgin Law = (X-Int(A) ) ∩ (X-Int(X-A) ) = Cl(A) ∩ Cl(X-A
∩ −
we know that Cl(A) is a close that contain A and Cl(X-A) is closed contain X-A (i) If A is open then Cl(X-A)=X-Int(A) Cl(A)=Int(A) ∪ ′ Then ′ belong to X Cl(X-A) ∩ Cl(A) =(X-Int(A)) ∩ (Int(A- ′ ) =X-(Int(A) U Int(X-A) ) = ′ = A (ii) If A is closed then Cl(A)=A=Int(A) ∪ ′ Cl(A)=(X-A) ∪ ′ Cl(X-A) ∩ Cl(A) = [(X-A) ∪ ′ ] ∩ [Int(A) ∪ ′ ] = ′ = A So Cl(X-A) ∩ Cl(A) = A
(iii) ∂A ∩ Int(A)=∅ Suppose on contrary ∂A ∩ Int(A)≠ ∅ , and let x ∈ Int(A) ∩ ∂A,then x ∈ Int(A) , so there exist an open set U containing x such that
U⊆A ⇒ U∩ Int(A)=∅ ⇒ x∉ ∂A This is contradiction. Hence ∂A ∩ Int(A)= ∅
∩
∅
We know that =Cl(A)-Int(A) Let x ∈ ⇒ x ∈ ClA but x∉ IntA ⇒ x∉ IntA ∩ A As x is an arbitrary point of A Now let x ∈ ( ) then X ∈ Cl(A) ∴ Cl(A)=A=Int(A) ∪ ′ But =Cl(A)-Int(A)
So x∉
As x is an arbitrary point of Int(A) ⇒ ∂A ∩ Int(A)=∅
∪
(iv) Cl(A) = ∂A ∪ Int(A) R.H.S =∂A ∪Int(A) = Cl(A) ∩ Cl(X-A) ∪ A =( A ∪ Cl(A) ) ∩( A ∪ ClX − A ) = Cl(A) ∩ X = Cl(A) = L.H.S
Let x ∈ ∂A ∪ Int(A) ⇒ x ∈ ∂A or x ∈ Int(A) As =Cl(A)-Int(A) and IntA⊆ ClA ⇒ x ∈ Cl(A)
35
So ∂A ∪ Int(A)⊆ ClA--------------(1) Now we prove that ClA ⊆ ∂A ∪ Int(A) Let y∈ Cl(A) We know that =Cl(A)-Int(A) If y ∈ IntA then y∉ A ⇒ y ∈ ∂A ∪ Int(A) If y ∉ IntA then y ∈ ∂A ⇒ y ∈ ∂A ∪ Int(A) ⇒ ClA ⊆ ∂A ∪ Int(A) -------------(2) From (1) AND (2) ∂A ∪Int(A)=Cl(A)
⊂ ⇔
Let ∂A⊆A Let x ∈ then x ∈ ClA but x∉ IntA ⇒as x ∈ ∂A ⇒ x ∈ A ⇒ x ∈ Cl(A) So A is closed Conversely Let A is closed Then by definition Cl(A)=A ⇒ A ∪ ∂A =A
⇒ ∂A ⊆ A
A⊆B ⇔ AUB=B
∩ ∅
If a is open then A=Int(A) As we know that ∂A ∩ Int(A)=∅ ⇒ ∂A ∩ A=∅ Conversely Let ∂A ∩ A=∅ Let x ∈ A then x ∉ ∂A ⇒ x∈ Int(A) As x is an arbitrary point of A So x ∈ A also x ∈ Int(A) Which show that A = Int(A)
⇒ A is open
∅
If A is open then X-A is closed so that X-A=Cl(X-A), if A closed then Cl(A)=A therefore by (ii) Cl(X-A) ∩ Cl(A) = A = A∩ X-A =∅ Conversely Let ∂A =∅ By (iv of this theorem) ∂A ∪Int(A)=Cl(A) As ∂A =∅ ⇒ ∅ ∪Int(A)=Cl(A)
⇒IntA=ClA As we know IntA⊂ A ⊆ClA ⇒IntA = A =ClA
This show that A is both open and close∎
Example 2.16
ℚ ℝ
36
by the Example 2.5 Intℚ=∅ and Clℚ=ℝ so ℚ= Clℚ- Intℚ =ℝ-∅ =ℝ Example 2.17
×
ℝ
Int(A)=∅ and Cl(A)=A so = Cl(A)- Int(A) =A
Example 2.18
Int(A)=[-1, 1] and Cl(A)=[-1, 1] so = Cl(A)- Int(A)
∅
Example 2.19
Int(A)=[-1, 1] and Cl(A)=[-1, 0) so = Cl(A)- Int(A) {-1}
NOTED: Interior and closure depend upon the topology on the set X containing A
2.24
Exercises for section 2.3
Determine A in each case (a) A=(0,1] in the lower limit on Int(A)=(0,1) Cl(A)=A = Cl(A)- Int(A) =(0,1]-(0,1)
(b)A={a} in X={a, b, c} with topology { ∅, X, {a}, {a, b}} SOL:
Closed set in topology are {X, ∅, {b,c},{c} The smallest closed set contain {a} is X so Cl(A) is X
Int(A)={a} Cl(A)=X = Cl(A)- Int(A) =X-{a} ={b,c}
(c) A={a, c} in X={a, b, c} with topology { ∅, X, {a}, {a, b} SOL: Closed set in topology are {X, ∅, {b,c},{c} The smallest closed set contain {a, c} is X so Cl A is X
Int(A)={a} Cl(A)=X
37
= Cl(A)- Int(A) =X-{a} ={b,c}
(d)A={b} in X={a, b, c} with topology { ∅, X, {a}, {a, b}} SOL: Closed set in topology are {X, ∅, {b,c},{c} The smallest closed set contain {b} is {b, c} so Cl(A) is {b, c}
Int(A)={ } Cl(A)=X = Cl(A)- Int(A) =X-{ } =X
(e) A=(-1,1) ∪ {}in standard topology on SOL: Int(A)=(-1, 1) Cl(A)=[-1,1]∪{2} = Cl(A)- Int(A) =[-1,1]∪{2} - (-1, 1) ={-1, 1,2}
(f) A=(-1,1) ∪ {}in lower limit topolgy on SOL:
(g) A= {(0, x)∈
Int(A)=(-1, 1) Cl(A)=[-1,1]∪{2} = Cl(A)- Int(A) =[-1,1]∪{2} - (-1, 1) ={-1, 1,2}
ℝ |x∈ } in ℝ the standard SOL:
Int(A)={ } Cl(A)=A = Cl(A)- Int(A) =A-{ } =A
{(, ) ∈ ℝ | ∈ ℝ} Int(A)=∅ Cl(A)=A = Cl(A)- Int(A) =A
(i)
Standard top on ℝ2 define as T={∅, ℝ2 , all open disks and their union. All set are closed and open so Cl(A) is A
ℝ
SOL:
{( , ) ∈ ℝ | ∈
Standard top on ℝ2 define as T={∅, ℝ2 , all open disks and their union. All set are closed and open so Cl(A) is A
ℝ} ℝ
Int(A)=A Cl(A=A = Cl(A)- Int(A) =∅
2.25
topology
(a) For m
38 Discussion of cases with example m=2 , n=6
l2,6 = {2, 4, 5, 6} Int(l2,6 )={3, 4, 5} Cl(l2,6 )={2, 3,4,5,6} m=2, n=5 l2,5 ={2,3,4.5} Int(l2,5 )={3,4,5} Cl(l2,5 ) ={2,3,45,6} m=1 ,n=5
l1,5 = 1,2,3,4,5 Int(l1,5 )=1,2,3,4,5} CL(l1,5 )={0,1,2,3,4,5} m=1,n=6
l1,6 = 1,2,3,4,5,6
In(l1,6 ) ={1,2,3,4,5} Cl(l1,5 ) = {0,1,2,3,4,5.6]
Int(A)={m, m+1, ...,n} Cl(A)={m-1,m, m+1,…,n} = Cl(A)- Int(A) ={m-1,m, m+1,…,n}-{m, m+1, ...,n} ={m-1} (ii) m is odd and n is even then Int(A)={m, m+1, ...,n-1} Cl(A)= {m-1,m, m+1,…,n} = Cl(A)- Int(A) ={m-1,m, m+1,…,n} - {m, m+1, ...,n-1} = {m-1} (iii) if m is even and n is odd then Int(A)={m+1,m+2, ...,n} Cl(A)={m, m+1,…,n+1} = Cl(A)- Int(A) = {m, m+1,…,n+1} - {m+1,m+2, ...,n} = {m,n+1} (iv) m and n are even then Int(A)={m+1, ...,n-1} Cl(A)={m, m+1,…,n} = Cl(A)- Int(A) = {m, m+1,…,n} - {m,+1,m+2, ...,n-1} = {m}
SOL: Solved in exercise 2.8
ℝ
{( , ) ∈ ℝ | ∈ ℝ Int(A)=∅ Cl(A)=A = Cl(A)- Int(A) =A
{( , ) ∈ ℝ | > 0,
Int(B)={ ( , )
SOL:
≠
∈ ℝ | > 0, > 0} ∪ {( , ) ∈ ℝ2 | > 0, < 0} Cl(A)= {( , ) ∈ ℝ2 | ≥. ∈ ℝ} B=Cl(B)-Int(B) 2 B={( , ) ∈ ℝ | ≥. ∈ ℝ} −{ ( , ) ∈ ℝ2 | > 0, > 0} ∪ {( , ) ∈ ℝ2 | > 0, < 0} 2
y-axsis
x-axsis
39
{( /, ) ∈ ℝ |
ℤ+
SOL: Int(C)={ } 1
Cl(C) ={[ , 0] ∈ ℝ2 | ∈ ℤ+ } = Cl(C)− Int(C) 1
={[ , 0] ∈ ℝ2 | ∈ ℤ+}−{ } 1 ={[ , 0] ∈ ℝ2 | ∈ ℤ+} =Cl(C)
{( , ) ∈ ℝ | ≤ − < SOL:
i.e
Int(D) ={( , ) ∈ ℝ2 |0 < 2 − 2 < 1} Cl(D) ={( , ) ∈ ℝ | ≤ − ≤ } D={( , ) ∈ ℝ| ≤ − ≤ }−{( , ) ∈ ℝ2 |0 < 2 − 2 < 1} equation of hyperbola
