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Solutions to Additional Exercises
Exercise (3.4) Prove the following: Let X be a topological space.
(1) X and ∅ are closed sets.
(2) The intersection of any collection of closed sets in X is closed.
(3) The union of any finite collection of closed sets in X is closed.
Proof: We shall apply Theorem (3.5) on page 39.
Theorem (3.5) Let X be a topological space.
(1) X and ∅ are open sets.
(2) The union of any collection of open sets in X is open.
(3) The intersection of any finite collection of open sets in X is open.
We also need Definition (3.7): A subset C in X is closed if X − C is open.
Now we are ready to prove Exercise (3.4). Since by Theorem 3.5(1), ∅ is open,
and since ∅ = X − X, by Definition (3.7), X is closed. Similarly, since by Theorem
3.5(1), X is open, and since X = X − ∅, by Definition (3.7), ∅ is closed. This proves
Exercise (3.4)(1).
To prove Exercise (3.4)(2), let C1 , C2 , · · · be a collection of closed sets in X. By
Definition (3.7), X − C1 , X − C2 , · · · is a collection of open sets in X. By DeMorgan
Law,
[
\
X − Ci = (X − Ci ).
i
i
Since each X − Ci is open and since, by Theorem (3.5)(2), the union of any collection
T
S
of open sets in X is open, X − i Ci = i (X − Ci ) is open in X. Now it follows by
T
Definition (3.7) that i Ci is closed in X. This proves Exercise (3.4)(2).
To prove Exercise (3.4)(3), let C1 , C2 , · · · , Cm be a finite collection of closed sets
in X. By Definition (3.7), X − C1 , X − C2 , · · · , Cm is a finite collection of open sets
in X. By DeMorgan Law,
X−
m
[
i=1
m
\
Ci =
(X − Ci ).
i=1
Since each X −Ci is open and since, by Theorem (3.5)(3), the intersection of any finite
S
Tm
collection of open sets in X is open, X − m
i=1 Ci = i=1 (X − Ci ) is open in X. Now
S
it follows by Definition (3.7) that m
i=1 Ci is closed in X. This proves Exercise (3.4)(3).
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Exercise (3.10) Let X be a topological space with closed subsets A and B such
that X = A ∪ B. Let f : A 7→ Y and g : B 7→ Y e continuous functions such that for
each x ∈ A ∩ B, f (x) = g(x). Define a new function F = f ∪ g : X 7→ Y by

 f (x)
F (x) = (f ∪ g)(x) =
 g(x)
for x ∈ A
.
for x ∈ B
(1) Prove that F is continuous.
(2) Give an example to show that the condition that A and B must be closed is
necessary.
Proof: (1) We shall first show that, for any set U ⊆ Y ,
F −1 (U ) = f −1 (U ) ∪ g −1 (U ).
Firstly, let x ∈ F −1 (U ) ⊆ X. Since X = A ∪ B, either x ∈ A or x ∈ B. If x ∈ A,
then f (x) = F (x) ∈ U , and so by the definition of f −1 , x ∈ f −1 (U ). It follows that
x ∈ f −1 (U ) ∪ g −1 (U ). If x ∈ B, then g(x) = F (x) ∈ U , and so by the definition
of g −1 , x ∈ g −1 (U ). It also follows that x ∈ f −1 (U ) ∪ g −1 (U ). Hence we have
F −1 (U ) ⊆ f −1 (U ) ∪ g −1 (U ).
Nest, we pick an arbitrary element x0 ∈ f −1 (U )∪g −1 (U ). Then either x0 ∈ f −1 (U )
or x0 ∈ g −1 (U ). If x0 ∈ f −1 (U ), then by the definition of F , F (x0 ) = f (x0 ) ∈ U , and
so by the definition of F −1 , x0 ∈ F −1 (U ). Similarly, if x0 ∈ g −1 (U ), then by the
definition of F , F (x0 ) = g(x0 ) ∈ U , and so by the definition of F −1 , x0 ∈ F −1 (U ) also.
This proves that F −1 (U ) ⊇ f −1 (U ) ∪ g −1 (U ), and so we have established the equality
F −1 (U ) = f −1 (U ) ∪ g −1 (U ).
Now we are ready to complete the proof of Exercise (3.10). Since we are given the
condition that both A and B are closed in X, we will make use of these two known
results
Exercise (3.9) Let X and Y be topological spaces. A function f : X 7→ Y is
continuous if and only if for each closed set C ⊆ Y , f −1 (C) is closed in X.
Theorem (3.12) Let A be a subset of a topological space X. A subset B ⊆ A is
closed in A if and only if B = A ∩ C for some set C ⊆ X closed in X.
Let C be a closed set in Y . Since f and g are continuous functions, by Exercise
(3.9), f −1 (C) is closed in A and g −1 (C) is closed in B. Since A is closed and since
f −1 (C) is closed in A, it follows by Theorem (3.12) that for some closed set D in
X, f −1 (C) = D ∩ A. Then by Exercise (3.4)(2) above that f −1 (C) = D ∩ A is
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closed in X. Similarly, since B is closed and since g −1 (C) is closed in B, it follows by
Theorem (3.12) and Exercise (3.4)(2) that g −1 (C) is closed in X.
By Exercise (3.4)(2), and by the equality we just established,
F −1 (C) = f −1 (C) ∪ g −1 (C)
is also closed in X. Therefore, by Exercise (3.9), F is continuous.
(2) Example: Let X = [0, 2] with A = [0, 1) and B = [1, 2], and let Y = [0, 3].
Define f : A 7→ Y by f (x) = x and g : B 7→ Y by g(x) = 3. Then both f and g
are continuous. As A ∩ B = ∅, it satisfies the condition that for each x ∈ A ∩ B,
f (x) = g(x). But it is well known that from Calculus that f ∪ g has a discontinuity
at x = 1.
Exercise (3.31) Show that the topology defined on Y in Definition (3.33) satisfies
the conclusions of Theorem 3.5.
Proof: We first establish the two equalities. Let f : X 7→ Y be a function. For any
collections of subsets Ui in Y ,
!
f −1
\
!
Ui =
i
\
f −1 (Ui ), f −1
i
−1
[
i
Ui =
[
f −1 (Ui ).
i
Pick an arbitrary x ∈ f ( i Ui ). By the definition of f −1 , f (x) ∈ i Ui . By the
definition of intersection, f (x) ∈ Ui , for any i. By the definition of f −1 , x ∈ f −1 (Ui )
T
for any i. By the definition of intersection, x ∈ i f −1 (Ui ). The other equality can
be proved similarly.
We are now ready to do Exercise (3.31). Let T 0 denote the topology defined on
Y by Definition (3.33).
Firstly, since f : X 7→ Y is a function, Y = f −1 (X). By Theorem (3.5)(1),
and since X is a topological space with topology T , X ∈ T . By Definition (3.33),
Y ∈ T 0 as X = f −1 (Y ) ∈ T . Similarly, ∅ ∈ T 0 as ∅ = f −1 (∅) ∈ T . This shows that
T 0 satisfies Theorem (3.5)(1).
Next, let Ui be a collection of open sets in T 0 . By Definition (3.33), each f −1 (Ui )
S
is in T . By Theorem (3.5)(2), i f −1 (Ui ) ∈ T . It now follows by the second equality
S
S
S
that f −1 ( i Ui ) = i f −1 (Ui ) ∈ T . By Definition (3.33), and since f −1 ( i Ui ) ∈ T ,
S
0
0
i Ui ∈ T . This proves that T satisfies Theorem (3.5)(2).
Finally, let U1 , U2 , · · · , Um be a finite collection of sets in T 0 . By Definition
T
−1
(3.33), each f −1 (Ui ) is in T . By Theorem (3.5)(3), m
(Ui ) ∈ T . By the
i=1 f
T
T
3
m
−1
(Ui ). By Definition (3.33), and since
first equality that f −1 ( m
i=1 f
i=1 Ui ) =
Tm
0
−1 Tm
f ( i=1 Ui ) ∈ T , i=1 Ui ∈ T . This proves that T 0 satisfies Theorem (3.5)(3).
T
T
Exercise (3.36) Show that if X is compact and ∼ is an equivalence relation on X,
then X/ ∼ is compact.
Proof: Define a function from X 7→ X/ ∼ by f (x) = [x]. Then by the definition
of an onto function, f is onto. According to Definition (3.33), a subset U ⊂ X/ ∼
if open iff f −1 (U ) is open in X. By the definition of continuous functions, f is a
continuous function from X onto Y . Now we apply Theorem (3.19) to finish the
proof.
Theorem (3.19) Let X be a compact topological space and f : X 7→ Y a continuous
function from X onto a topological space Y . Then Y is compact.
Since f : X 7→ X/ ∼ is a continuous function from X onto X/ ∼. By Theorem
(3.19), X/ ∼ is compact.
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