Blank Template Efr´ Ef r´en en Chave Ch avess 25 de junio de 2014
2
´ Indice general 1. Cap. 1
5
1.1. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2. Ejercicios Cap. 2
5 9
2.1. Ejercicios 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2. Ejercicios 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3. Referencias
17
4. Anexos
19
3
4
´ INDICE GENERAL
Cap´ıtulo 1 Cap. 1 1.1.
Exercises
1. Let X = a,b,c,d,e,f . Determine whether (or) not each of the following collection of subsets of X is a topology on X
{
}
a ) τ 1 = X, , a , a, f , b, f , a,b,f
{ ∅ { } { } { } { }} Note that {a, f } {b, f } = {f } ∈ / τ . Hence, τ is not a topology. b) τ = {X, ∅, {a,b,f }, {a,b,d}, {a,b,d,f }} Again, note that {a,b,f } {a,b,d} = {a, b} ∈ / τ . Hence, τ is not a topology. c ) τ = {X, ∅, {f }, {e, f }, {a, f }} Note that {e, f } {a, f } = {a,e,f } ∈ / τ }. Hence, τ is not a topology. 2. Let X = {a,b,c,d,e,f }. Which of the following collections of subsets of X is a topology 1
1
2
2
2
3
3
3
on X ? (Justify your answers)
a ) τ 1 = X, , c , b,d,e , b,c,d,e , b
{ ∅ { } {
}{ } { }} Not a topology since {c} {b} = {b, c} ∈ / τ b) τ = {X, ∅, {a}, {b,d,e}, {a,b,d}, {a,b,d,e }} Not a topology since {b,d,e} {a,b,d} = {b, d} ∈ / τ c ) τ = {X, ∅, {b}, {a,b,c}, {d,e,f }, {b,d,e,f }} 1
2
2
3
It is a topology since
It contains X and . Any union of sets in τ 3 is in τ 3 Any intersection of sets in τ 3 is in τ 3
∅
3. Let X = a,b,c,d,e,f and τ is the discrete topology on X . Which of the following statements are true?
{
}
a ) X τ True
{a} ∈ τ True h ) a ∈ τ False i ) ∅ ⊆ X True j ) {a} ∈ X False k ) {∅} ⊆ X False l ) a ∈ X True
∈ b) {X } ∈ τ False c ) {∅} ∈ τ False d ) ∅ ∈ τ True e ) ∅ ∈ X False f ) {∅} ∈ X False
g )
5
CAP ´ ITULO 1. CAP. 1
6 m ) X
⊆ τ False n ) {a} ⊆ τ False
{X } ⊆ τ True o) a ⊆ τ False
n ˜ )
4. Let (X, τ ) be any topological space. Verify that the intersection of any finite number of members of τ is a member of τ . Soluci´on n A τ ”. Let S = n N : Let P n be the statement that ¨If Ai ni=1 τ = i i=1 P n is a true statement . From the definition of a topology, we have 1 N. Assume that k A τ . Consider k k S i.e. P k is a true statement i.e. whenever A τ = i i i=1 i=1 k k+1 k+1 k Ai i=1 τ . i=1 Ai = Ak+1. By assumption, we have B = τ . i=1 Ai i=1 Ai Now from the definition of a topology, we get B Ak+1 τ since B, Ak+1 τ . Hence, we +1 A τ whenever A k+1 τ . Hence, P is a true statement. Hence, k + 1 S . get ki=1 i i i=1 k+1 So by principle of mathematical induction we have 1 S , and k + 1 S whenever k S . Hence, S = N . Hence, the intersection of any finite number of members of τ is a member of τ .
{ } ∈
}
∈ { } ∈
∈
{ } ∈
⇒ { } ∈ ∈ ∈
∈ ⇒
∈
{ ∈
∈ ∈
∈
∈ ∈ ∈
5. Let R be the set of all real numbers. Prove that each of the following collections of subsets of R is a topology a ) τ 1 consists of R, and every interval ( n, n), for any positive integer
∅ Clearly, R, ∅ ∈ τ
−
1
First note that τ 1 is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let Ak = ( k, k). Note that Ak ’s are monotone A = A increasing sequence of sets. Hence, any finite union of the form m kl p l=1 where p = m´ax(k1 , k2 , . . . , km ). And A p τ 1 and hence any finite union again A where k N can be rewritten either as belongs to τ 1 . Any infinite union ∞ kl l l=1 A where k = k . m a finite union of the form l=1 Akl (or) an infinite union ∞ kl i j l=1 ∞ The claim now is that any infinite union where ki = k j is R i.e. l=1 Akl = R where ki = k j . To prove this, we prove the two way inclusion. Note that Akl R . A R. Further given any x R, by archimedian property n N Hence, ∞ kl l=1 such that x A n. Further since it is an infinite union, kl such that x A n A = A τ . Akl . Hence, ∞ τ 1 and An R. Further, An R = R 1 kl n l=1 Hence, any union is also in τ 1 . Now we need to prove the last claim that intersection of any two elements of τ 1 gives an element in τ 1. Consider Am, An τ 1 . By well-ordering principle, we get mn (or) m = n (or) mn. If m = n, then Am An = An τ 1 . If mn, then Am An = A m τ 1. If mn, then Am An = A n τ 1 . Further, An R = A n τ 1 = and An τ 1 . Hence, intersection of any two elements of τ 1 gives an element in τ 1 .
− ∈
∈
⊆
∈
∈
⊇
∈
∈
∈ ∅ ∅ ∈
∈
∃
∅
∈
∈ ∃ ∈ ∈ ⊆ ∈
∈
b) τ 2 consists of R, and every interval [ n, n], for any positive integer
∅ Clearly, R, ∅ ∈ τ
−
2
First note that τ 2 is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let Ak = [ k, k]. Note that Ak ’s are monotone A = A increasing sequence of sets. Hence, any finite union of the form m kl p l=1 where p = m´ax(k1 , k2 , . . . , km ). And A p τ 2 and hence any finite union again A where k N can be rewritten either as belongs to τ 2 . Any infinite union ∞ kl l l=1 A where k = k . m a finite union of the form l=1 Akl (or) an infinite union ∞ kl i j l=1 ∞ The claim now is that any infinite union where ki = k j is R i.e. l=1 Akl = R where ki = k j . To prove this, we prove the two way inclusion. Note that Akl R .
− ∈
∈
∈
7
1.1. EXERCISES
Hence, ∞ R . Further given any x R , by archimedian property n N l=1 Akl such that x A n. Further since it is an infinite union, kl such that x A n A R = R τ and A = A τ . . Further, Akl . Hence, ∞ A R k n n n 2 2 l=1 l Hence, any union is also in τ 2 . Now we need to prove the last claim that intersection of any two elements of τ 2 gives an element in τ 2. Consider Am, An τ 2 . By well-ordering principle, we get mn (or) m = n (or) mn. If m = n, then Am An = An τ 2 . If mn, then Am An = A m τ 2. If mn, then Am An = A n τ 2 . Further, An R = A n τ 2 = and An τ 2 . Hence, intersection of any two elements of τ 2 gives an element in τ 2 .
∈
⊆
∈
⊇
∃
∈
∈
∈ ∅ ∅ ∈
c ) τ 3 consists of R, and every interval [n,
∅ Clearly, R, ∅ ∈ τ
∅
∈
∈
∃ ∈ ∈ ⊆ ∈
∈
∞), for any positive integer
3
First note that τ 3 is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let Ak = [k, ). Note that Ak ’s are monotone A = A decreasing sequence of sets. Hence, any finite union of the form m kl p l=1 where p = m´ın(k1 , k2, . . . , km ). And A p τ 3 and hence any finite union again A where k N equals A where p = belongs to τ 3 . Any infinite union ∞ kl l p l=1 m´ın(k1 , k2 , . . .). Further, An R = R τ 3 and An = An τ 3 . Hence, any union is also in τ 3 . Now we need to prove the last claim that intersection of any two elements of τ 3 gives an element in τ 3. Consider Am, An τ 3 . By well-ordering principle, we get mn (or) m = n (or) mn. If m = n, then Am An = An τ 3 . If mn, then Am An = A n τ 3 . If mn, then Am An = A m τ 3 . Further, An R = A n τ 3 = and An τ 3 . Hence, intersection of any two elements of τ 3 gives an element in τ 3 .
∞
∈ ∈
∈ ∅
∈
∈ ∅ ∅ ∈
∈
∈
∈
∈
6. Let N be the set of all positive integers. Prove that each of the following collections of subsets of N is a topology. a ) τ 1 consists of N, and every set of the form 1, 2, . . . , n for any positive integer n.
∅
This is called the initial segment topology.
{
}
b) τ 2 consists of N, and every set of the form n, n + 1, . . . for any positive integer
∅
n. This is called the final segment topology.
{
}
8
CAP ´ ITULO 1. CAP. 1
Cap´ıtulo 2 Ejercicios Cap. 2 2.1.
Ejercicios 2.1
1. Prove that if a, b R with ab, then neither [a, b) nor (a, b] is an open subset of R. Also show that neither is a closed subset of R.
∈
a ) We shall first consider the interval [a, b).
We will prove that the interval [a, b) is not open. If [a, b) is open, then x [a, b), we have an interval (c, d) containing x such that (c, d) [a, b). We will show that the point a doesn’t lie in any open interval (c, d) containing a such that (c, d) [a, b). The proof goes by contradiction. If there exists (c, d) containing a, such that (c, d) [a, b), consider x = a+2 c . Note that x (c, d) but x / [a, b). Hence, (c, d) [a, b) for any open interval (c, d). Hence, [a, b) is not open. We will now prove that the interval [a, b) is not closed. Look at the complement in R and the argument is similar as above to prove that the complement in R is not open as well. Hence, [a, b) is not closed.
⊆
⊆
⊆
⊆
∈
∀ ∈ ∈
b) Now consider the interval (a, b] and the argument is almost the same as above.
∞) and (−∞, a] are closed subsets of R. a ) Consider R\[a, ∞) = (−∞, a). Consider any x ∈ ( −∞, a). Let = a − x and consider the subset (x − , x + ). Clearly, x ∈ (x − , x + ) ⊂ ( −∞, a). This is true for any x ∈ ( −∞, a). Hence, (−∞, a) is an open subset. Hence, [a, ∞) is a closed set of R. b) Consider R\(−∞, a] = (a, ∞). Consider any x ∈ (a, ∞). Let = x − a and consider the subset (x − , x + ). Clearly, x ∈ (x − , x + ) ⊂ (a, ∞). This is true for any x ∈ (a, ∞). Hence, (a, ∞) is an open subset. Hence, (−∞, a] is a closed set of R.
2. Prove that the sets [a,
3. Show, by example, that the union of an infinite number of closed subsets of R is not necessarily a closed subset of R. a ) Let An = [ n1 , 1
−
an open set of R.
1
]. Then, A = n
∞
n=2 An =
∞
[1,1
n=2 n
−
1 n
] = (0, 1) which is clearly
4. Prove each of the following statements. a ) The set Z of all integers is not an open subset of R.
Proof by contradiction. Assume that it is an open subset, then for each x Z, there exists an open interval, say (a, b) such that x (a, b) and (a, b) Z. Let = m´ın(x a, b x). By Archimedean property, choose a positive integer n 2
∈
− −
9
∈ ⊆ ≥
CAP ´ ITULO 2. EJERCICIOS CAP. 2
10
such that n1 . Then we have that x n1 (a, b) but x Hence, Z of all integers is not an open subset of R.
± ∈
±
1 n
is not an integer.
P of all prime numbers is a closed subset of R but not an open subset of R. We shall first prove that P is a closed subset of R. Consider C = R\P . We will prove that this set is open. Note that since primes are a subset of natural numbers, it is well-ordered. In particular, we can write P = { p ∈ N : p {isthe}k prime} i.e. we have p p p ···. This means C = (−∞, p ) ∞ ( p , p Note that (−∞, p ) and ( p , p ) are open sets. Since union of open sets is still an open set, we get that C is an open set. Hence, we have that P is a closed subset of R. We shall now prove that P is not an open subset of R. The proof goes by contradiction. Assume that it is an open subset, then given any p ∈ P , there exists and open subset (a, b) ⊂ P such that p ∈ (a, b). Let = m´ın({ p − a, b − p }). By Archimedean property, we have that there exists a positive integer n ≥ 2 such that . Now we have p ± ∈ (a, b) but p ± ∈ / P . CONTRADICTION. Hence, P is not an open subset of R.
b) The set
k
th
k
1 2 3
1
1
k=1
k
k+1
k+1
k
k
k
k
k
1
k
n
1
k
n
1
n
c ) The set I of all irrational numbers is neither a closed subset nor an open subset of R.
Note that if A is not open, then X A is not closed and similarly if A is not closed, then X A is not open. We proved as a proposition earlier that the set Q = R I is neither open nor closed. Hence, I is neither open nor closed.
\
\
\
5. If F is a non-empty finite subset of R, then show that F is closed in R but that F is not open in R. a ) Since F is a non-empty finite subset of R, we can write F = f 1 , f 2 , . . . , fn where n Z + and f 1 f 2 f n .
∈
{
···
}
We shall first that F is closed. Let F c = R F . Then we have F c = nprove −1 ( , f 1 ) , f 1 ) and (f k , f k+1 ) are k=1 (f k , f k+1 ) . Each of the set i.e. ( open sets. Hence, union of open sets is again open. Hence, F c is open. Hence, F is a closed set of R. We shall not prove that F is not open in R. We shall prove by contradiction. Assume F is open. This means that for any f k F , we have an open set say (a, b) such that f k (a, b) and (a, b) F . Let = m´ın( f k a, b f k , f k+1 f k , f k f k−1 ). By Archimedean property, we can choose a positive integer n 1 1 1 2, such that n1 . Consider f k . We have f k (a, b) but f k / F . n n n CONTRADICTION. Hence, F is not an open subset of R.
\ −∞
−∞ ∪
−
}
∈
⊆
±
∈
{ −
± ∈
− − ≥ ± ∈
6. If F is a non-empty countable subset of R, prove that F is not an open set. a ) F is a countable subset of R. We shall prove by contradiction that F is not open. Assume the F is a open subset of R. Then this means that for every x F , there
∈
exists an open interval (a, b) F . Hence, we now have (a, b) F . However note that (a, b) is an uncountable set whereas F is a countable set. But any subset of a countable set is again a countable set. CONTRADICTION. Hence, F is not an open set.
⊆
⊆
7. (7) a ) Let S = 0, 1, 21 , 31 , . . . , n1 , . . . . Prove that the set S is closed in the euclidean topology on R.
{
}
).
11
2.1. EJERCICIOS 2.1 c
c
∞ =
1 ,1 k+1 k
Let S = R S . Hence, we have S ( , 0) (1, ). Note k=1 that each of the interval in the union is an open interval. Hence, the set S c is also an open set. Hence, S is a closed set on R.
\
∪ −∞ ∪ ∞
b) Is the set T = 1, 21 , 31 , . . . , n1 , . . . closed in R?
{
}
∞ 1 1 Let T c = R T . Hence, we have T c = ( , 0] (1, ). We k=1 k+1 , k shall prove that T c is not open. Again, the proof is by contradiction. Assume that the set T c is open. Consider the point 0. Since T c is assumed to be open, there exists an open set (a, b) containing 0 and (a, b) T c . Let = m´ın( a, b). By Archimedean property, there exists an integer n 2 such that n1 . Hence, 1 (a, b). However, n1 / T c . This CONTRADICTS the fact the (a, b) T c . n Hence, T c is not open. Hence, T is not closed.
\
∪ −∞ ∪ ∞
⊆ ≥
∈
− ⊆
∈ √ √ √ √ c ) Is the set S = { 2, 2 2, 3 2, . . . , n 2, . . .} closed in R? ∞ √ √ √ We have S = (k 2, (k + 1) 2) ∪ (−∞, 2). S c
c
is a union of open intervals which are open sets. Hence, S is a open set. Hence, S is a closed set in R. k=1
c
8. (8) a ) Let (X, τ ) be a topological space. A subset S of X is said to be an F σ -set if it is the
union of a countable number of closed sets. Prove that all open intervals (a, b) and all closed intervals [a, b], are F σ -sets in R.
1 ∞ 1 We have (a, b) = k=1 a + k , b − k . Hence, any open interval is a F σ set. We have [a, b] = ∞ [a, b]. Hence, any closed interval is a F set. σ
k=1
b) Let (X, τ ) be a topological space. A subset S of X is said to be an G δ -set if it is the
intersection of a countable number of open sets. Prove that all open intervals (a, b) and all closed intervals [a, b], are Gδ -sets in R.
(a, b). Hence, any open interval is a G set. We have (a, b) = ∞ δ k=1
a We have [a, b] = ∞ k=1
−
1
, b + k
1
k
. Hence, any closed interval is a Gδ set.
c ) Prove that the set of rationals is an F δ set in R.
The rationals form a countable set. Hence, we can list the rationals as Q = q k ∞ propositions earlier that the set a is a closed k=1 . We proved in one of the ∞ set in R. Hence, we have Q = k=1 q k which means Q is a countable union of closed sets in R. Hence, the set of rationals is an F δ set in R.
{ }
{ }
{ }
d ) Verify that the complement of an F σ -set is a G δ -set and the complement of a G δ -set
is a F σ -set. First we shall prove that the complement of an F σ -set is a Gδ -set. Let A be an F where F are closed sets. This means that F σ -set. This means that A = ∞ k k k=1 ∞ c c c F k are open sets. We have A = k=1 F k , where F kc are open sets and hence Ac is a countable intersection of open sets. Hence, Ac is a Gδ -set. Now we shall prove that the complement of a Gδ -set is a F σ -set. Let A be a G where G are open sets. This means that Gδ -set. This means that A = ∞ k k=1 k ∞ c c c Gk are closed sets. We have A = k=1 Gk , where Gck are closed sets and hence Ac is a countable union of closed sets. Hence, Ac is a F σ -set.
CAP ´ ITULO 2. EJERCICIOS CAP. 2
12
2.2.
Ejercicios 2.2
1. In this exercise, you will prove that the disc (x, y) : x2 + y 2 < 1 is an open subset of R2 , and then that every open disc in the plane is an open set.
{
}
a ) Let (a, b) be any point in the disc D = (x, y) : x2 + y 2 < 1 . Put r =
{
} be the open rectangle with vertices at the points a ± ⊂ D.
Let R (a,b) that R(a,b)
1 r ,b 8
−
±
√ a
2 + b2 . 1−r
. Verify
8
All we need to show is that all the four vertices lies inside the disc, for which we need to show that the distance from the origin to each of the four vertices is less than unity. To show that all we need to show is that the distance of the farthest vertex from the origin is less than unity. The √ of2 the farthest vertex from √ 2 distance 1−r the origin is given by d = r + 2 8 = 8 + r 1 . Since 0 r1, we get 8
√
√ 2 8
−
≤ d1. Hence, we get that R ⊂ D.
≤
(a,b)
b) Using the previous part show that D =
(a,b) D R(a,b) .
∈
Consider any point (a, b) D. Clearly, (a, b) R(a,b) . Hence, any point in the disc is contained in an open rectangle centered about that point. Hence, D previous part, we have that for any (a, b) we (a,b)∈D R(a,b) . From the have R(a,b) D. Hence, D (a,b)∈D R(a,b) . Combining the two, we get D = (a,b)∈D R(a,b) .
∈
⊆
⊆
∈
⊇
c ) Deduce from above that D is an open set in R2 .
Every open rectangle is an open set. Hence, any arbitrary union of open sets is again an open set. From the previous part, we have that D is an arbitrary union of open rectangles. Hence, D is an open set. d ) Show that every disc (x, y) : (x
2
2
− a)
2 2
2
− b) c ,a,b,c ∈ R} is open in R . Let (m, n) be any point in the disc D = {(x, y) : (x − a) + (y − b) c }. Put r = (m − a) + (n − b) . Let R be the open rectangle with vertices at the points (a + m ± − , b + n ± − ). Now the same arguments as the above three gives us that the disc D = {(x, y) : (x − a) + (y − b) c } is open in R . {
2
+ (y
2
2 2
(a,b)
c r
c r
8
8
2
2 2
2
13
2.2. EJERCICIOS 2.2
2. In this exercise you will show that the collection of all open discs in R2 is a basis for a topology on R2 . [Later we shall see that this is the euclidean topology.] a ) Let D1 and D2 be any open discs in R2 with D1
in D1 D(a,b)
∩ D = ∅. If (a, b) is any point 2
∩ D , show that there exists an open disc D ⊆ D ∩ D .
(a,b) with
2
1
centre (a, b) such that
2
We are given that D1 D2 = . Let the center and radius of D 1 be (a1 , b1) and r1 respectively. Let the center and radius of D 2 be (a2, b22) and r2 respectively. Since, we have D1 D2 = , we have r1 + r2 (a1 a2 ) + (b1 b2 )2 . Let r =
∩ ∅ ∩ ∅ − − m´ın(r − (a − a ) + (b − b ) , r − (a − a ) + (b − b ) ). Consider the disc centered at (a, b) with radius r. This disc lies completely inside D ∩ D . 1
1
2
1
2
2
2
2
2
2
1
2
b) Show that
D1
∩D
=
2
D(a,b) .
(a,b) D1 D2
∈ ∩
Consider any (x, y) D 1 D2 . Note that (x, y) D (x,y) . (x, y) is one such (a, b) in the union. Hence (x, y) D (x,y) (a,b)∈D ∩D D (a,b) . Hence, we get D 1 D2 D1 D 2 , we have D(x,y) D1 D2 . a,b) . Now, for any (x, y) (a,b)∈D ∩D D ( Hence, we get (a,b)∈D ∩D D(a,b) D 1 D2 .
∈ ∩ ∈
1
⊆ ∈ ∩ ⊆ ∩ 1
2
1
2
∈
∩ ⊆ ⊆ ∩
2
c ) Using the above and the proposition proved earlier, prove that the collection of all open discs in R2 is a basis for a topology on R2 .
First note that we have R2 = D(a,b) (n) where D(a,b) (n) denotes an open unit disc of radius n centered at (a, b). The proof is trivial since any D(a,b) (n) R2 and also for any point (x, y) R 2 , we have n N such that n x2 + y 2 and hence (x, y) D (0,0) (n). Next note that from the part above, we have that intersection of any two discs is again a union of open discs. Hence, by the proposition proved earlier we have that the collection of all open discs in R 2 is a basis for a topology on R2 .
∈
∈
∈
√
⊆
3. Let be the collection of all open intervals (a, b) in R with ab and a and b rational numbers. Prove that is a basis for the euclidean topology on R.
B
B
CAP ´ ITULO 2. EJERCICIOS CAP. 2
14
⊆ R is an open set then given any x ∈ S we can find an open interval (a , b ) ⊆ S where a , b ∈ Q. Since S is an open set, we know that there exists a, b ∈ R such that x ∈ (a, b) ⊆ S . Further, since the rationals are dense in R, given any r ∈ R , there exists a sequence of rationals monotonously converging to r. For instance, there exists a sequence of monotonously increasing b ∈ Q such that l´ım →∞ b = b. Similarly, there exists a sequence of monotonously decreasing a ∈ Q such that l´ım →∞ a = a. This means that (a, b) = →∞ (a , b ). Hence, if x ∈ (a, b), then x ∈ (a , b ) for some n where a , b ∈ Q . Hence, given any open set S , for every x ∈ S , there exists an open interval with rational end points containing
a ) We shall prove that if S q
q
q
q
n
n
n
n
n
n
n
n
n
n
n
n
n
x lying within S . Hence, the set of open intervals with rational end points generate the same topology as the euclidian topology.
4. A topological space (X, τ ) is said to satisfy the second axiom of countability or to be second countable if there exists a basis for τ , where consists of only a countable number of sets.
B
B
a ) Using the previous exercise show that R satisfies the second axiom of countability.
We have = (aq , bq ) : aq , bq Q is a basis for the euclidean topology. We have the set of rational numbers to be countable. The set can be written as = aq bq ;aq ∈Q bq ∈Q (aq , bq ) which is again a countable set.
B {
∈ }
B
B
b) Prove that the discrete topology on an uncountable set does not satisfy the second
axiom of countability. Let be a basis for the discrete topology. For every x X , we have x to be an open set in the discrete topology. This means that we have x to be a union of elements in . This means that the singleton sets should be in the basis . Hence, we have x : x X . Since X is uncountable, we have x : x X to be an uncountable set. Hence, we have that to be an uncountable set. Hence, (X, τ ) satisfies the second axiom of countability.
B
{{ }
B
∈
∈ }
{ } { }
B {{ } ∈ } ⊆ B
B
c ) Prove that Rn satisfies the second axiom of countability, for each positive integer n.
We proceed by induction. Let P (n) be the statement that Rn satisfies the second axiom of countability. Let S = n N : P (n) is true . From the first part of this problem, we have 1 S i.e there exists a countable basis for the euclidean topology on R. Assume that k S . Note that Rk+1 = Rk R . By induction hypothesis, we have that there exists a countable basis k for the euclidean topology on Rk . From the question 6, which is proved later we have that k is a basis for R k R = R k+1 . Product of two countable sets is again a countable set. Hence, k is a countable basis for Rk R = Rk+1 . Hence, Rn satisfies the second axiom of countability, for each positive integer n.
∈
{ ∈ ∈
}
B
× B × B
B ×
B ×B
×
d ) Let (X, τ ) be the set of all integers with the finite-closed topology. Does the space
(X, τ ) satisfy the second axiom of countability? Let τ c = A X : X A τ i.e. τ c contains all the closed sets induced by the topology τ . In this case, we have τ c = A X : A is finite . Note that c τ is equivalent to τ since there is a clear bijection from τ to τ c as A τ iff X A τ c . We shall prove that τ c is a countable set. This would mean that τ is a countable set and since any basis τ we would have proved that the (X, τ ), with the finite-closed topology, satisfies the second axiom of countability. Since X is a countable set, list the element of X as x0, x1, . . . , xn , . . . . Any finite subset B of X is of the form B = xk , xk , . . . , xkn where n N and ki N
{ ⊆
\ ∈
\ ∈ }
{ ⊆
}
{
}
∈
B⊆
{
0
1
}
∈
∈
15
2.2. EJERCICIOS 2.2
for i
∈ {0, 1, . . . , n}. Let f (B) =
n
2kl . It is not hard to see that f : τ c
l=0
→N
is a bijection. Hence, τ c is a countable set. This means that the topology, τ , is also countable and hence any basis is also a countable set. Hence, (X, τ ), with the finite-closed topology, satisfies the second axiom of countability. 5. Prove the following statements. a ) Let m and c be real numbers, with m = 0. Then the line L = (x, y) : y = mx + c is a closed subset of R2 .
{
}
Consider L c = R 2 L. We shall prove that Lc is an open set. Consider (a, b) L c . We shall prove that there is a rectangle R L c such that (a, b) R. Let d denote −am−c 0. Consider the distance of the point (a, b) from the line L. We have d = b√ 1+m d d the open rectangle R with vertices (a 2 , b 2 ). We get (a, b) R and R L c . Hence, Lc is an open set. Hence, L is a closed set.
\
⊆
∈
∈
2
±
±
b) Let S1 be the unit circle given by S1 = (x, y) closed subset of R2 .
{
∈
2
∈ R
⊆
: x2 + y 2 = 1 . Then S1 is a
}
Consider T = R2 S1 . We shall prove that T is open. Consider any (a, b) T. Let r = abs(1 a2 + b2 ). Consider the open rectangle R with vertices at (a 2r , b 2r ). Clearly, we have (a, b) R T. This is true for any (a, b) T. Hence, T is an open set. Hence, S1 is a closed set of R2 .
±
\√ −
±
∈ ∈
∈ ⊆
c ) Let Sn be the unit n-sphere given by Sn = (x1 , x2 , . . . , xn , xn+1 )
{
∈ R
n+1
: x 21 + x22 +
·· · + x
2
n+1
= 1 .
}
Then Sn is a closed subset of Rn+1 . Consider T = Rn+1 Sn . We shall prove that T is open. Consider any (a1 , a2 , . . . , an+1 ) a21 + a22 + + a2n + a2n+1 . Consider the open rectangle R T. Let r = abs 1
\ ··· − with vertices at a ± ,a ± ,...,a ± ,a ± . Clearly, we have (a , a , . . . , a , a ) ∈ R ⊆ T. Hence, T is an open set. Hence, S is a closed r n+1
1
1
2
n
2
r n+1
n
r n+1
n+1
r n+1
n+1
subset of Rn+1 .
d ) Let B n be the closed unit n-ball given by
B n = (x1 , x2 , . . . , xn ) : x 21 + x22 +
{
Then B n is a closed subset of Rn.
2
· ·· + x ≤ 1}. n
n
∈
CAP ´ ITULO 2. EJERCICIOS CAP. 2
16
Consider T = R2 n B2n. We shall2 prove that T is open. Consider any (a1, a2, . . . , an) + an 1. Consider the open rectangle R with vertices T. Let r = a1 + a2 +
\
r
···
r
− ±
r
r
∈
at a1 n+1 , a2 n+1 , . . . , an n+1 , an+1 n+1 . Clearly, we have (a1 , a2 , . . . , an , an+1 ) R T . Hence, T is an open set. Hence, B n is a closed subset of Rn.
⊆
±
±
e ) The curve C = (x, y)
{
∈ R
2
±
: xy = 1 is a closed subset of R2 .
}
Consider T = R 2 C . We shall prove that T is open. Consider any (a, b) T . Let r be the minimum distance from the point (a, b) to the curve C . Consider the open rectangle R with vertices at (a 2r , b r2 ). Clearly, we have (a, b) R T . This is true for any (a, b) T . Hence, T is an open set. Hence, C is a closed set of R2 .
\
± ±
∈
B
∈ ∈ ⊆
B
6. Let 1 be a basis for the topology τ 1 on a set X and 2 be a basis for the topology τ 2 on a set Y . The set X Y consists of all ordered pairs (x, y), x X and y Y . Let be the collection of subsets of X Y consisting of all the sets B1 B2 where B1 1 and B2 is a basis for a topology on X Y . The topology so defined is 2 . Prove that called the product topology on X Y .
× B
∈ B
×
∈ ×
×
×
∈
B ∈ B
B and B are a basis for τ and τ respectively. Hence, we have ∪ ∈B B = X and ∪ ∈B B = Y . Also, if we have C, D ∈ B , then C ∩ D ∈ B . Similarly, if we have C, D ∈ B , then C ∩ D ∈ B . Now consider any element in B × B . It is of the form B × B where B ∈ B and B ∈ B . Now the union over all elements in B × B can be written as ∪ ∈B ∈B B ×B . We then have ∪ ∈B ∈B B ×B = ∪ ∈B (B × ∪ ∈B B ) = ∪ ∈B B × Y = ∪ ∈B (B ) × Y = X × Y . Also, if we have A, B ∈ B, then A = A × A and B = B × B where A , B ∈ B and A , B ∈ B . Then A ∩ B = ().
a ) All we need to do is to check the definitions of a basis. We are given that 1
2
B
1
1
2
1
B1
1
,B2
B1
1
1
1
2.3.
2
2
1
2
1
2
1
2
B1
B1
2
Ejercicios 2.3
1
1
1
,B2
1
1
2
1
2
2
B1
1
2
2
2
2
1
1
1
2
2
2
1
1
B
1
1
2
B2
2
2
∈
Cap´ıtulo 3 Referencias Tomado de http://adhvaithist.blogspot.com/2011/08/11-exercises.html
17
18
CAP ´ ITULO 3. REFERENCIAS
Cap´ıtulo 4 Anexos
19
