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Problem 1, Page 83
Let X be a topological space. Let A be a subset of X. Suppose that for each x ∈ A
there is an open set U containing x such that U ⊂ A. Show that A is open in X.
Proof
: Let Ux ⊂ A be an open set containing
S
S x. Because Ux ⊂ A,Swe have
U
⊂
A.
Because
x
∈
U
,
we
have
A
⊂
x
x
x∈A
x∈A .Thus we have A =
x∈A Ux .
By axiom of topology, A is open.
Problem 3, Page 83
Show that the collection Tc given in Example 4 of §12 is a topology on the set X.
Is the collection
T∞ = {U | X − U is infinite or empty or all of X}
a topology on X.
Proof : Tc is a topology on X.
1) Since X − X = ∅ and X − ∅ = X, we have X ∈ Tc and ∅ ∈ Tc .
S
2)
T For any {Uα }α∈I ⊂ Tc where I is any index set, we have X − Sα∈I Uα =
Tα∈I X − Uα by D’Morgan theorem.
S If for all α ∈ I Uα = ∅, then X − α∈I Uα =
X
−U
=
X.
It
follows
that
If there is atTleast one Uα 6= ∅, say
α
α∈I
α∈I Uα ∈ Tc . S
Uα0 , then X −Ui is countable.
It
follows
that
X
−
α∈I U
S
Sα = α∈I X −Uα ⊂ X −Ui
is countable. Thus α∈I Uα ∈ Tc . Above all, we have α∈I Uα ∈ Tc . T
3)
S For any {Uα }α∈I ⊂ Tc where I is a finiteTindex set, we
S have X − α∈I Uα =
X
−
U
.
If
there
is
a
U
=
∅,
then
X
−
U
=
α
α
α
α∈I
α∈I X − Uα = X. Thus
Tα∈I
U
∈
T
.
If
there
is
no
U
=
∅,
then
X
−
U
is
countable
α
c
α
α
α∈IT
S
T for each α. We have
X−
U
=
X
−
U
is
countable.
It
follows
that
α
α
α∈I
α∈I Uα ∈ Tc . Above
T α∈I
all, α∈I Uα ∈ Tc for all I finite.
So Tc satisfies all axiom of topology, it is a topology on X.
T∞ = {U | X − U is infinite or empty or all of X} is not a topology on X when
card(X) is infinite. It is a topology on X if card(X) is finite.
1) If card(X) is finite, then T∞ = {U | X − U is infinite or empty or all of
X} = {∅, X}. This is a topology on X.
2) If Card(X) is infinite. Choose a sequence of distinct points {xi }∞
i=1 ⊂ X.
Let Ui = X − ({xj }∞
∪ {x1 }). Then Ui ∈ T∞ for each i = 1, 2, 3, · · · . But
j=i+1
S∞
S∞
S∞
X − i=1 Ui = i=1 X − Ui = {x1 } is finite. So i=1 Ui is not in T∞ .It means that
T∞ does not satisfy axiom 3 of topology. Therefore, it is not a topology on X.
Problem 7, Page 83
Consider the following topologies on R
T1 = the standard topology.
T2 = the topology of Rk .
T3 = the finite complement topology.
T4 = the upper limit topology, having all sets (a, b] as a basis.
T5 = the topology having all set (−∞, a) = {x | x < a} as a basis.
1
2
Determine, for each of these topologies which of the others it contains.
Proof :
Let Bi be the basis for Ti
1) T2 is finer than T1 , since B1 ⊂ B2 and by Lemma 13.3.
2) T3 and T5 are not comparable.
For any x ∈ R, choose a basis element A = R − {y} where y < x is a real number. Then you cannot find an interval (−∞, a) such that x is in this interval and
(−∞, a) ⊂ A. Thus by lemma 13.3,T3 is not finer than T5 . Similarly, for any x ∈ R,
let A = (−∞, x + 1). Suppose there is an element B in B3 such that x ∈ B and
B ⊂ A. Then R − A ⊂ R − B, then B is not in B3 . Contradiction. Thus by lemma
13.3,T5 is not finer than T3 .
Above all, they are not comparable.
Problem 4, Page 92
A map f : X → Y is said to be an open map if for every open set U of X, the
set f (U ) is open in Y . Show that π1 : X × Y → X and π2 : X × Y → Y are
open maps.
Proof : By Theorem 15.2 in page 88 and S
Lemma 13.1 in page 80, we can see that
any open set in X ×Y S
has following form:
α is open in X and
α Uα ×Vα where US
S
S Vα
is open in Y . But π1 ( α Uα × Vα ) = α Uα is open in X. π2 ( α Uα × Vα ) = α Vα
is open in Y . Thus π1 and π2 are open maps.
Problem 6, Page 92
Show that the countable collection
B 0 = {(a, b) × (c, d) | a < b, c < d, and a, b, c, d are rational}
is a basis for R2 .
Proof : The basis for usual topology on R2 is B = {(a, b) × (c, d) | a < b, c < d, and
a, b, c, d are real numbers}. Let T 0 be the topology generated by B 0 on R2 , and let
T be the topology generated by B on R2 . The second one is the standard topology
on R2 .
1) T 0 is finer than T .
For each (x, y) ∈ R2 and each element (a, b) × (c, d) ∈ B containing (x, y), we can
know that a < x < b and c < y < d. Since Q is dense in R, we can find rational
a0 , b0 , c0 , d0 such that a < a0 < x < b0 < b and c < c0 < y < d0 < d. Then there is a
basis element (a0 , b0 )×(c0 , d0 ) ∈ B 0 , such that (x, y) ∈ (a0 , b0 )×(c0 , d0 ) ⊂ (a, b)×(c, d).
By Lemma 13.3 on page 81, we have T 0 is finer than T .
2) Since B 0 ⊂ B, it is easy to see T is finer than T 0 by Lemma 13.3.
Above all, T 0 =T .
3
Problem 8, Page 92
If L is a straight line in the plane, describe the topology L inherits as a subspace
of Rl × R and as a subspace of Rl × Rl . In each case it is a familiar topology.
Proof : By lemma 16.1, we just need to consider the intersection with basis of
the total space of the line L.
1) The total space if Rl ×R. B is the basis for the topology on it. Then by Theorem
15.2, the element in B has form [a, b) × (c, d) where a, b, c, d are real numbers.
a) If the line is vertical, i.e X = x0 .
For any element [a, b) × (c, d) ∈ B, we have L ∩ [a, b) × (c, d) = {x0 } × (c, d) if x0 ∈
[a, b) and ∅ if x0 ∈[a, b). Then the basis for subspace topology is {(c, d) | c, d ∈ R}.
This is the usual topology on line.
b) If the line is horizontal, i.e. Y = y0 .
For any element [a, b) × (c, d) ∈ B, we have L ∩ [a, b) × (c, d) = [a, b) × {y0 } if y0 ∈
(c, d) and ∅ if y0 ∈(c, d). Then the basis for subspace topology is {[a, b) | a, b ∈ R}.
This is the lower limit topology on line.
c) If the line is generic.
Let A0 be the basis for Rl and A be the basis for R. The basis for this topology is
A ∪ A0 .
2) The total space is Rl × Rl .
a) If the line is vertical, i.e X = x0 .
For any element [a, b)×[c, d) ∈ B, we have L∩[a, b)×[c, d) = {x0 }×[c, d) if x0 ∈ [a, b)
and ∅ if x0 ∈[a, b). Then the basis for subspace topology is {[c, d) | c, d ∈ R}. This
is the lower limit topology on line.
b) If the line is horizontal, i.e. Y = y0 .
For any element [a, b)×[c, d) ∈ B, we have L∩[a, b)×[c, d) = [a, b)×{y0 } if y0 ∈ [c, d)
and ∅ if y0 ∈[c, d). Then the basis for subspace topology is {[a, b) | a, b ∈ R}. This
is the lower limit topology on line.
c) If the line is Y = kX + a, where k > 0.
Let A0 be the basis for Rl and A be the basis for R. The basis for the topology on
L is A0 .
d)If the line is Y = kX + a, where k < 0.
The basis for the topology on L is {(a, b) or [a, b] | a, b are real numbers}. This is
the discrete topology on R.
Problem 10, Page 92
Let I = [0, 1]. Compare the product topology on I × I, the dictionary order topology on I ×I, and the topology I ×I inherits as a subspace of R×R in the dictionary
order topology.
Proof :Let T1 be product topology on I ×I, and T2 be the dictionary order topology
on I × I, and T3 the topology I × I inherits as a subspace of R × R in the dictionary
order topology. Let Bi be the topological basis for Ti .
1) T1 is incomparable with T2 .
4
Let A = {(a, y) | y > c} ∪ {(x, y) | a < x < b} ∪ {(b, y) | y < d} where a, b, c, d ∈ I.
Then A ∈ B2 . Then for (a, y0 ) ∈ A, we cannot find a set B in B1 such that
(a, y0 ) ∈ B ⊂ A. Thus, by Lemma 13.3 T2 is not finer than T1 .
Let B = [0, a) × (b, 1] ∈ B1 . Then for (0, 1), we cannot find a set A in B2 such that
(0, 1) ∈ A ⊂ B. Thus, by Lemma 13.3 T2 is not finer than T1 .
Above all, T1 is incomparable with T2 .
2) T3 is finer than T1 .
For any (x0 , y0 ), y0 6= 1 and 0, and A ∈ B1 containing (x0 , y0 ). We can find
B = {(x0 , y) | y ∈ (y0 − ², y0 + ²) ⊂ [0, 1]}, such that x ∈ B ⊂ A.
For any (x0 , y0 ), y0 6= 1, and A ∈ B1 containing (x0 , y0 ). We can find B = {(x0 , y) |
y ∈ (1 − ², 1] ⊂ [0, 1]}, such that x ∈ B ⊂ A.
For any (x0 , y0 ), y0 6= 0, and A ∈ B1 containing (x0 , y0 ). We can find B = {(x0 , y) |
y ∈ [0, y0 + ²) ⊂ [0, 1]}, such that x ∈ B ⊂ A.
Thus, by Lemma 13.3 T3 is finer than T1 .
3) T3 is finer than T2 .
For any (x0 , y0 ), and A ∈ B1 containing (x0 , y0 ). Let B = {(x0 , y) | y ∈ R} ∩ A,
Then B ∈ B3 x ∈ B ⊂ A.
Thus, by Lemma 13.3 T3 is finer than T1 .