Bài tập topo

Topology
Hints and Solutions to Munkres’s Book
∗
DongPhD (Editor)
DongPhD Problem Books
Series
υo.1
Available at
∗
James R. Munkres “Topology”-Prentice Hall, 2 ed. | 2000 | ISBN: 0139254951
1
DongPhD 2
Exercises
“Say what you know, do what you must, come
what may.”- Sofia Kovalevskaya
Topological spaces, basis for a topology
Ex 13.1. 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.
Ex 13.2.
Ex 13.3.
Ex 13.4. (a) If {τ
α
} is a family of topologies on X, show that ∩{τ
α
}
is a topology on X. Is ∪{τ
α
} a topology on X?
(b) Let {τ
α

} be a family of topologies on X. Show that there is a
unique smallest topology on X containing all the collections {τ
α
},
and a unique largest topology contained in all {τ
α
}.
(c) If X = {a, b, c}, let τ
1
= {∅, X, {a}, {a, b}} and τ
2
= {∅, X, {a}, {c, b}}.
Find the smallest topology containing τ
1
and τ
2
, and the largest
topology contained in τ
1
and τ
2
.
Ex 13.5. Show that if A is a basis for a topology on X, then the
topology generated by A equals the intersection of all topologies on X
that contain A. Prove the same if A is a subbasis.
Ex 13.6. Show that the topologies of R
l
and R
K
- are not comparable.

Ex 13.7. Consider the following topologies on R:
τ
1
= the standard topology, τ
2
= the topology of R
K
- , τ
3
= the finite
complement topology, τ
4
= the upper limit topology, having all sets
(a, b] as basis, τ
5
= the topology having all sets (−∞, a) = {x|x < a}
as basis.
Determine, for each of these topologies, which of the others it contains.

DongPhD 3
Ex 13.8. (a) Show that the countable collection
B = {(a, b) | a < b, a and b rational}
is a basis that generates t he standard topology on R.
(b) Show that the collection
C = {[a,b)|a<b, a and b rational}
is a basis that generates a topology different from the lower limit
topology on R
Subspace topology
Ex 16.9. Show that if Y is a subspace of X, and A is a subset of
Y , then the topology A inherits as a subspace of Y is the same as the

topology it inherits as a subspace of X.
Ex 16.10.
Ex 16.11. Consider the set Y = [−1, 1] as a subspace of R. Which of
the following sets are open in Y ? Which are open in R?
A = {x|
1
2
< |x| < 1}
B = {x|
1
2
< |x| ≤< 1}
C = {x|
1
2
≤ |x| < 1}
D = {x|
1
2
≤ |x| ≤ 1}
E = {x|0 < |x| < 1 and
1
x
/∈ Z
+
}
Ex 16.12. 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 π
1
: X ×Y → X

and π
2
: X × Y → Y are open maps.
Ex 16.13.

DongPhD 4
Ex 16.14. Show that the countable collection
{(a, b) × (c, d)|a < b and c < d and a, b, c, d arerational}
Ex 16.15. Let X be an ordered set. If Y is a proper subset of Y that
is convex in X, does it follow that Y is an interval or a ray in X?
Ex 16.16.
Ex 16.17. Show that the dictionary order topology on the set R × R
is the same as the product topology R
d
× R, where R
d
denotes R in the
discrete topology. Compare this topology with the standard topology on
R
2
.
Closed Sets and Limit Points
Ex 17.18.
Ex 17.19.
Ex 17.20. Show that if A is closed in X and B is closed in Y , then
A × B is closed in X × Y .
Ex 17.21.
Ex 17.22.
Ex 17.23. Let A, B and A
α

denote subsets of a space X. Prove the
following:
(a) If A ⊂ B then A ⊂ B
(b) A ∪ B = A ∪ B
(c) ∪A
α
⊃ ∪A
α
, give an example where equality fails.
Ex 17.24.
Ex 17.25. Let A, B and A
α
denote subsets of a space X. Determine
whether the following equations hold; if an equality fails, determine
whether one of the inclusions ⊃ or ⊂ holds.

DongPhD 5
(a) A ∩ B = A ∩ B
(b) ∩A
α
⊃ ∩A
α
,
(c) A \ B = A \ B
Ex 17.26. Let A ⊂ X andB ⊂ Y . Show that in the space X × Y ,
A × B = A × B
Ex 17.27. Show that every order topology is Hausdorff.
Ex 17.28. Show that the product of two Hausdorff spaces is Hausdorff.
Ex 17.29. Show that a subspace of a Hausdorff space is Hausdorff.
Ex 17.30. Show that X is Hausdorff if and only if the diagonal A =

{(x, x)|x ∈ X} is closed in X × X.
Ex 17.31. In the finite complement topology on R, to what point or
points does the sequence x
n
=
1
n
converge?
Ex 17.32. (Kuratowski) Consider the collection of all subsets A of the
topological space X. The operations of closure A → A and complemen-
tation A → X \ A are func- tions from this collection to itself.
(a) Show that starting with a given set A, one can form no more than
14 distinct sets by applying these two operations successively.
(b) Find a subset A of R (in its usual topology) for which the maxi-
mum of 14 is obtained.
Continuous Functions
Ex 17.33. Prove that for functions f : R → R, the  − δ definition of
continuity implies the open set definition.
Ex 17.34. Suppose that f : X → Y is continuous. If x is a limit point
of the subset A of X, is it necessarily true that f(x) is a limit point of
f(A)?
Ex 17.35.

DongPhD 6
Ex 17.36.
Ex 17.37.
Ex 17.38. Find a function /f : R → R that is continuous at precisely
one point.
Ex 17.39. (a) Suppose that f : R → R is "continuous from the
right," that is,

lim
x→a
+
f(x) = f(a)
Show that f is continuous when considered as a function from R
l
to R.
(b) Can you conjecture what functions f : R → R are continuous
when considered as maps from R to R
l
? As maps from R
l
to R
l
?
Ex 17.40. Let Y be an ordered set in the order topology. Let f, g :
X → Y be continuous.
(a) Show that the set {x|f(x) < g(x)} is closed in X.
(b) Let h : X → Y be the function
h(x) = min{f(x), g(x)}.
Show that h is continuous.
Ex 17.41.
Ex 17.42. Let f : A → B and g : C → D be continuous functions.
Let us define a map f × g : A × C → B × D by the equation
(f × g)(a × c) = f(a) × g(c).
Show that f × g is continuous.
Ex 17.43.
Ex 17.44.
Ex 17.45. Let A ⊂ X; let f : A → Y be continuous; let Y be
Hausdorff. Show that if f may be extended to a continuous function

g : A → Y , then g is uniquely determined by f.

DongPhD 7
The Quotient Topology
Ex 22.46.
Ex 22.47. (a) Let p : X → Y be a continuous map. Show that if
there is a continuous map f : Y → Xsuch that p ◦ f equals the
identity map of Y , then p is a quotient map.
(b) If A ⊂ X, a retraction of X onto A is a continuous map r : X →
A such that r(a) = a for each a ∈ A. Show that a retraction is a
quotient map.
Ex 22.48. Let π
1
: R × R → R be projection on the first coordinate.
Let A be the subspace of R × R consisting of all points x × y for which
either x > 0 or y = 0 (or both); let q : A → Rbe obtained by restricting
π
1
. Show that q is a quotient map that is neither open nor closed.
Ex 22.49.
Ex 22.50. Let p : X × Y be an open map. Show that if A is open in
X, then the map q : A → p(A) obtained by restricting p is an open
map.
Compact Spaces
Ex 26.51. (a) Let τ and τ

be two topologies on the set X; suppose
that τ

⊃ τ. What does compactness of X under one of these

topologies imply about compactness under the other?
(b) Show that if X is compact Hausdorff under both τ and τ

, then
either τ and τ

are equal or they are not comparable.
Ex 26.52. (a) Show that in the finite complement topology on R,
every subspace is compact.
(b) If R has the topology consisting of all sets A such that R \ A is
either countable or all of R, is [0, 1] a compact subspace?
Ex 26.53. Show that a finite union of compact subspaces of X is com-
pact.
Ex 26.54.

DongPhD 8
Ex 26.55. Let A and B be disjoint compact subspaces of the Hausdorff
space X. Show that there exist disjoint open sets U and V containing
A and B, respectively.
Ex 26.56. Show that if f : X → Y is continuous, where X is compact
and Y is Hausdorff, then f is a closed map (that is, f carries closed
sets to closed sets).
Ex 26.57. Show that if Y is compact, then the projection π
1
: X ×Y →
X is a closed map.
Ex 26.58. Let f : X → Y ; let Y be compact Hausdorff. Then f is
continuous if and only if the graph of f is closed in X × Y .
Ex 26.59.
Ex 26.60.

Ex 26.61.
Ex 26.62. Let p : X → Y be a closed continuous surjective map such
that p
−1
({y}) is compact, for each y ∈ Y . (Such a map is called aperfect
map.) Show that if Y is compact, then X is compact.
Compact Subspaces of the Real Line
Ex 26.63. Prove that if X is an ordered set in which every closed
interval is compact, then X has the least upper bound property.
Ex 26.64.
Ex 26.65. Recall that R
K
denotes R in the K -topology.
(a) Show that [0, 1] is not compact as a subspace of R
K
(b) Show that R
K
is connected.
(c) Show that R
K
is not path connected.
Ex 26.66.
Ex 26.67. Let X be a compact Hausdorff space; let A
n
be a count-
able collection of closed sets of X. Show that if each set A, has empty
interior in X, then the union ∪A
n
has empty interior in X.

DongPhD 9
Local Compactness
Ex 29.68. Show that the rationale Q are not locally compact.
Ex 29.69. Let {X
α
} be an indexed family of nonempty spaces.
(a) Show that if

X
α
is locally compact, then each X
α
is locally
compact and X
α
is compact for all but finitely many values of
or.
(b) Prove the converse, assuming the Tychonoff theorem.
Ex 29.70. Let X be a locally compact space. If f : X → Y is con-
tinuous, does it follow that f(X) is locally compact? What if f is both
continuous and open? Justify your answer.
The Countability Axioms
Ex 30.71.
Ex 30.72.
Ex 30.73. Let X have a countable basis; let A be an uncountable subset
of X. Show that uncountably many points of A are limit points of A.
Ex 30.74. Show that every compact metrizable space X has a countable
basis.
Ex 30.75. (a) Show that every metrizable space with a countable
dense subset has a countable basis,

(b) Show that every metrizable Lindelof space has a countable basis.
Ex 30.76. Show that R
l
and I
2
0
are not metrizable.
Ex 30.77. Which of our four countability axioms does S
ω
satisfy?
What about S
ω
?
Ex 30.78.
Ex 30.79. Let A be a closed subspace of X. Show that if X is Lindelof,
then A is Lindelof. Show by example that if X has a countable dense
subset, A need not have a countable dense subset.

DongPhD 10
Ex 30.80.
Ex 30.81.
Ex 30.82. Let f : X → Y be a continuous open map. Show that if X
satisfies the first or the second countability axiom, then f(X) satisfies
the same axiom.
Ex 30.83. Show that if X has a countable dense subset, every collec-
tion of disjoint open sets in X is countable.
The Separation Axioms
Ex 31.84. Show that if X is regular, every pair of points of X have
neighborhoods whose closures are disjoint.
Ex 31.85. Show that if X is normal, every pair of disjoint closed sets

have neighborhoods whose closures are disjoint.
Ex 31.86. Show that every order topology is regular.
Ex 31.87.
Ex 31.88. Let f, g : X → Y be continuous; assume that Y is Haus-
dorff. Show that {x|f(x) = g(x)} is closed in X.
Ex 31.89. Let p : X → Y be a closed continuous surjective map. Show
that if X is normal, then so is Y .
Ex 31.90. Let p : X → Y be a closed continuous surjective map such
that p
−1
({y}) is compact, for each y ∈ Y . (Such a map is called aperfect
map.)
(a) Show that if X is Hausdorff, then so is Y .
(b) Show that if X is regular, then so is Y .
(c) Show that if X is locally compact, then so is Y .
(d) Show that if X is second-countable, then so is Y .

DongPhD 11
Normal Spaces
Ex 32.91. Show that a closed subspace of a normal space is normal.
Ex 32.92.
Ex 32.93. Show that every locally compact Hausdorff space is regular.
Ex 32.94. Show that every regular Lindelof space is normal.
Ex 32.95. Is R
ω
normal in the product topology? In the uniform topol-
ogy?
Ex 32.96. A space X is said to be completely normal if every subspace
of X is normal. Show that X is completely normal if and only if for
every pair A, B of separated sets in X (that is, sets such that A∩B = ∅

and B ∩ A = ∅), there exist disjoint open sets containing them.
The Urysohn Lemma
Ex 33.97. Examine the proof of the Urysohn lemma, and show that
for given r,
f
−1
(r) = ∩
p>r
U
p
\ ∪
q<r
U
q
,
p, q rational.
Ex 33.98.
Ex 33.99.
Ex 33.100. Recall that A is a "D
δ
set" in X if A is the intersection
of a countable collection of open sets of X.
Theorem. Let X be normal. There exists a continuous function f :
X → [0, 1] such that f(x) = 0 for x ∈ A, and f(x) > 0 for x /∈ A if
and only if A is a closed G
δ
set in X.
A function satisfying the requirements of this theorem is said to
vanish precisely on A.
Ex 33.101. Theorem (Strong form of the Urysohn lemma). Let X be

a normal space. There is a continuous function f : X → [0, 1] such
that f(x) = 0 for x ∈ A, and f(x) = 1 for x ∈ B, and0 < f(x) <
1otherwise, if and only if A and B are disjoint closed G
δ
sets in X.

DongPhD 12
Ex 33.102.
Ex 33.103. Show that every locally compact Hausdorff space is com-
pletely regular.
Ex 33.104. Let X be completely regular; let A and B be disjoint closed
subsets of X. Show that if A is compact, there is a continuous function
f : X → [0, 1] such that f(A) = {0} and f(B) = {1}
The Urysohn Metrization Theorem
Ex 33.105. Give an example showing that a Hausdorff space with a
countable basis need not be metrizable.
Ex 33.106. Give an example showing that a space can be completely
normal, and satisfy the first countability axiom, the Lindelof condition,
and have a countable dense subset, and still not be metrizable.
Ex 33.107. Let X be a compact Hausdorff space. Show that X is
metrizable if and only if X has a countable basis.
Ex 33.108. Let X be a locally compact Hausdorff basis, then X is
metrizable? Is it true that if X is metrizable, then X has a countable
basis?
Ex 33.109. Let X be a locally compact Hausdorff space. Let Y be the
one-point compactification of X. Is it true that if X has a countable
basis, then Y is metrizable? Is it true that if Y is metrizable, then X
has a countable basis?
Hints and Solutions
1

“ If I have been to see further, it was only be-
cause i stood on the shoulders of giants.” - Isaac
Newton
1
Source: moller/ and I just combine them


Munkres §17
Ex. 17.3. A × B is closed because its complement
(X × Y ) − (A × B) = (X − A) × Y ∪ X × (Y − B)
is open in the product topology.
Ex. 17.6.
(a). If A ⊂ B, then all limit points of A are also limit points of B, so [Thm 17.6] A ⊂ B.
(b). Since A ∪ B ⊂ A ∪ B and A ∪ B is closed [Thm 17.1], we have A ∪ B ⊂ A ∪ B by (a).
Conversely, since A ⊂ A ∪ B ⊂ A ∪ B, we have A ⊂ A ∪ B by (a) again. Similarly, B ⊂ A ∪ B.
Therefore A ∪ B ⊂ A ∪ B. This shows that closure commutes with finite unions.
(c). Since

A
α
⊃ A
α
we have

A
α
⊃
A
α
by (a) for all α and therefore


A
α
⊃

A
α
. In general
we do not have equality as the example A
q
= {q}, q ∈ Q, in R shows.
Ex. 17.8.
(a). By [Ex 17.6.(a)], A ∩ B ⊂ A and A ∩ B ⊂ B, so A ∩ B ⊂ A ∩ B. It is not true in general
that A ∩ B = A ∩ B as the example A = [0, 1), B = [1, 2] in R shows. (However, if A is open and
D is dense then A ∩ D = A).
(b). Since

A
α
⊂ A
α
we have

A
α
⊂
A
α
for all α and therefore


A
α
⊂

A
α
. (In fact, (a) is
a special case of (b)).
(c). Let x ∈ A − B. For any neighborhood of x, U − B is also a neighborhood of x so
U ∩ (A − B) = (U − B) ∩ A ⊃ (U − B) ∩ A = ∅
since x is in the closure of A [Thm 17.5]. So x ∈ A − B. This shows that A−B ⊂ A − B. Equality
does not hold in general as R − {0} = R − {0}  R − {0} = R.
Just to recap we have
(1) A ⊂ B ⇒ A ⊂ B (A ⊂ B, B closed ⇒ A ⊂ B)
(2) A ∪ B = A ∪ B
(3) A ∩ B ⊂ A ∩ B (A ∩ D = A if D is dense.)
(4)

A
α
⊃

A
α
(5)

A
α
⊂


A
α
(6)
A − B ⊂ A − B
Dually,
(1) A ⊂ B ⇒ Int A ⊂ Int B (A ⊂ B, A open ⇒ A ⊂ Int B)
(2) Int (A ∩ B) = Int A ∩ Int B
(3) Int (A ∪ B) ⊃ Int A ∪ Int B
These formulas are really the same because
X − A = X − Int A, Int (X − A) = X − A
Ex. 17.9. [Thm 19.5] Since A × B is closed [Ex 17.3] and contains A × B, it also contains the
closure of A × B [Ex 17.6.(a)], i.e. A × B ⊂ A × B.
Conversely, let (x, y) ∈ A × B. Any neighborho od of (x, y) contains a product neighborho od
of the form U × V where U ⊂ X is a neighborhood of x and V ⊂ Y a neighborhood of y. The
intersection of this product neighborhood with A × B
(U × V ) ∩ (A × B) = (U ∩ A) × (V ∩ Y )
is nonempty because U ∩ A = ∅ as x ∈ A and V ∩ B = ∅ as y ∈ B . Since thus any neighborhood
of (x, y) intersect A × B nontrivially, the point (x, y) lies in the closure of A × B [Thm 17.5]. This
shows that A × B ⊂ A × B.
1
2
Ex. 17.10 (Morten Poulsen).
Theorem 1. Every order topology is Hausdorff.
Proof. Let (X, ≤) be a simply ordered set. Let X be equipped with the order topology induced
by the simple order. Furthermore let a and b be two distinct points in X, may assume that a < b.
Let
A = { x ∈ X | a < x < b },
i.e. the set of elements between a and b.
If A is empty then a ∈ (−∞, b), b ∈ (a, ∞) and (−∞, b) ∩ (a, ∞) = ∅, hence X is Hausdorff.
If A is nonempty then a ∈ (−∞, x), b ∈ (x, ∞) and (−∞, x) ∩ (x, ∞) = ∅ for any element x in

A, hence X is Hausdorff. 
Ex. 17.11 (Morten Poulsen).
Theorem 2. The product of two Hausdorff spaces is Hausdorff.
Proof. Let X and Y be Hausdorff spaces and let a
1
× b
1
and a
2
× b
2
be two distinct points in
X × Y . Note that either a
1
= a
2
or b
1
= b
2
.
If a
1
= a
2
then, since X is Hausdorff, there exists open sets U
1
and U
2
in X such that a

1
∈ U
1
,
a
2
∈ U
2
and U
1
∩ U
2
= ∅. It follows that U
1
× Y and U
2
× Y are open in X × Y . Furthermore
a
1
× b
1
∈ U
1
× Y , a
2
× b
2
∈ U
2
× Y and (U

1
× Y ) ∩ (U
2
× Y ) = (U
1
∩ U
2
) × Y = ∅ × Y = ∅, hence
X × Y is Hausdorff.
The case b
1
= b
2
is similar. 
Ex. 17.12 (Morten Poulsen).
Theorem 3. Every subspace of a Hausdorff space is Hausdorff.
Proof. Let A be a subspace of a Haussdorff space X and let a and b be two distinct points in A.
Since X is Hausdorff there exists two open sets U and V in X such that a ∈ U, b ∈ V and
U ∩ V = ∅. Hence a ∈ A ∩ U, b ∈ A ∩ V and (A ∩ U) ∩ (A ∩ V ) = (U ∩ V ) ∩ A = ∅ ∩ A = ∅. Since
A ∩ U and A ∩ V are open in A, it follows that A is Hausdorff. 
Ex. 17.13 (Morten Poulsen).
Theorem 4. A topological space X is Hausdorff if only if the diagonal
∆ = { x × x ∈ X × X | x ∈ X }
is closed in X × X.
Proof. Suppose X is Hausdorff. The diagonal ∆ is closed if and only if the complement ∆
c
=
X × X − ∆ is open. Let a × b ∈ ∆
c
, i.e. a and b are distinct points in X. Since X is Hausdorff

there exists open sets U and V in X such that a ∈ U, b ∈ V and U ∩ V = ∅. Hence a × b ∈ U × V
and U × V open in X × X. Furthermore (U × V ) ∩ ∆ = ∅, since U and V are disjoint. So for
every point a × b ∈ ∆
c
there exists an open set U
a×b
such that a × b ∈ U
a×b
⊂ ∆
c
. By Ex. 13.1
it follows that ∆
c
open, i.e. ∆ closed.
Now suppose ∆ is closed. If a and b are two distinct points in X then a × b ∈ ∆
c
. Since ∆
c
is
open there exists a basis element U × V , U and V open in X, for the product topology, such that
a × b ∈ U × V ⊂ ∆
c
. Since U × V ⊂ ∆
c
it follows that U ∩ V = ∅. Hence U and V are open sets
such that a ∈ U , b ∈ V and U ∩ V = ∅, i.e. X is Hausdorff. 
Ex. 17.14 (Morten Poulsen). The sequence converges to every real number, by the following
result.
Theorem 5. Let X be a set equipped with the finite complement topology. If (x
n

)
n∈Z
+
is an
infinite sequence of distinct points in X then (x
n
) converges to every x in X.
Proof. Let U be a neighborhood of x ∈ X, i.e X − U is finite. It follows that x
n
∈ U, for all, but
finitely many, n ∈ Z
+
, i.e. (x
n
) converges to x. 
3
Ex. 17.21 (Morten Poulsen). Let X be a topological space. Consider the three operations on
P(X), namely closure A → A, complement A → X − A and interior A → A
◦
. Write A
−
instead
of A and A
c
instead of X − A, e.g. X − X − A = A
c−c
.
Lemma 6. If A ⊂ X then A
◦
= A

c−c
.
Proof. A
◦
⊃ A
c−c
: Since A
c
⊂ A
c−
, A
c−c
⊂ A and A
c−c
is open.
A
◦
⊂ A
c−c
: Since A
◦
⊂ A, A
c
⊂ A
◦c
and A
◦c
is closed, it follows that A
c−
⊂ A

◦c
, hence
A
◦
⊂ A
c−c
. 
This lemma shows that the interior operation can be expressed in terms of the closure and
complement operations.
(a). The following theorem, also known as Kuratowski’s Closure-Complement Problem, was first
proved by Kuratowski in 1922.
Theorem 7. Let X be a topological space and A ⊂ X. Then at most 14 distinct sets can be
derived from A by repeated application of closure and complementation.
Proof. Let A
1
= A and set B
1
= A
c
1
. Define A
2n
= A
−
2n−1
and A
2n+1
= A
c
2n

for n ∈ Z
+
. Define
B
2n
= B
−
2n−1
and B
2n+1
= B
c
2n
for n ∈ Z
+
.
Note that every set obtainable from A by repeatedly applying the closure and complement
operations is clearly one of the sets A
n
or B
n
.
Now A
7
= A
c−c
4
= A
◦
4

= A
−◦
3
. Since A
3
= A
−c
1
it follows that A
3
is open, hence A
3
⊂ A
7
⊂ A
−
3
,
so A
−
7
= A
−
3
, i.e. A
8
= A
4
, hence A
n+4

= A
n
for n ≥ 4. Similarly B
n+4
= B
n
for n ≥ 4.
Thus every A
n
or B
n
is equal to one of the 14 sets A
1
, . . . , A
7
, B
1
, . . . , B
7
, this proves the
result. 
(b). An example:
A = ((−∞, −1) − {−2}) ∪ ([−1, 1] ∩ Q) ∪ {2}.
The 14 different sets:
A
1
= ((−∞, −1) − {−2}) ∪ ([−1, 1] ∩ Q) ∪ {2}
A
2
= (−∞, 1] ∪ {2}

A
3
= (1, ∞) − {2}
A
4
= [1, ∞)
A
5
= (−∞, 1)
A
6
= (−∞, 1]
A
7
= (1, ∞)
B
1
= {−2} ∪ ([−1, 1] − Q) ∪ ((1, ∞) − {2})
B
2
= {−2} ∪ [−1, ∞)
B
3
= (−∞, −1) − {−2}
B
4
= (−∞, −1]
B
5
= (−1, ∞)

B
6
= [−1, ∞)
B
7
= (−∞, −1).
Another example:
A = { 1/n | n ∈ Z
+
} ∪ (2, 3) ∪ (3, 4) ∪ {9/2} ∪ [5, 6] ∪ { x | x ∈ Q, 7 ≤ x < 8 }.
Munkres §18
Ex. 18.1 (Morten Poulsen). Recall the ε-δ-definition of continuity: A function f : R → R is
said to be continuous if
∀ a ∈ R ∀ ε ∈ R
+
∃ δ ∈ R
+
∀x ∈ R : |x − a| < δ ⇒ |f(x) − f(a)| < ε.
Let T be the standard topology on R generated by the open intervals.
Theorem 1. For functions f : R → R the following are equivalent:
(i) ∀ a ∈ R ∀ ε ∈ R
+
∃ δ ∈ R
+
∀ x ∈ R : |x − a| < δ ⇒ |f(x) − f (a)| < ε.
(ii) ∀ a ∈ R ∀ ε ∈ R
+
∃ δ ∈ R
+
: f((a − δ, a + δ)) ⊂ (f (a) − ε, f(a) + ε).

(iii) ∀ U ∈ T : f
−1
(U) ∈ T .
Proof. “(i) ⇔ (ii)”: Clear.
“(ii) ⇒ (iii)”: Let U ∈ T . If a ∈ f
−1
(U) then f(a) ∈ U . Since U is open there exists
ε ∈ R
+
such that (f (a) − ε, f(a) + ε) ⊂ U. By assumption there exists δ
a
∈ R
+
such that
f((a − δ
a
, a + δ
a
)) ⊂ (f(a) − ε, f(a) + ε), hence (a − δ
a
, a + δ
a
) ⊂ f
−1
(U). It follows that
f
−1
(U) =

a∈f

−1
(U)
(a − δ
a
, a + δ
a
), i.e. open.
“(iii) ⇒ (ii)”: Let a ∈ R. Given ε ∈ R
+
then f
−1
((f(a) − ε, f (a) + ε)) is open and contains
a. Hence there exists δ ∈ R
+
such that (a − δ, a + δ) ⊂ f
−1
((f(a) − ε, f(a) + ε)). It follows that
f((a − δ, a + δ)) ⊂ (f (a) − ε, f(a) + ε). 
Ex. 18.2. Let f : R → {0} be the constant map. Then 2004 is a limit point of R but f(2004) = 0
is not a limit point of f (R) = {0}. (The question is if f(A

) ⊂ f (A)

in general. The answer is
no: In the above example R

= R so that f(R

) = f (R) = {0} but f(R)


= ∅.)
Ex. 18.6 (Morten Poulsen).
Claim 2. The map f : R → R, defined by
f(x) =

x, x ∈ R − Q
0, x ∈ Q,
is continuous only at 0.
Proof. Since |f (x)| ≤ |x| for all x it follows that f is continuous at 0.
Let x
0
∈ R − {0}. Since lim
x→x
0
,x∈Q
f(x) = 0 and lim
x→x
0
,x∈R−Q
f(x) = x it follows that f
is not continuous at x
0
. 
Ex. 18.7 (Morten Poulsen).
(a). The following lemma describes the continuous maps R
l
→ R.
Lemma 3. Let f : R → R. The following are equivalent:
(i) f : R → R is right continuous.
(ii) ∀ x ∈ R ∀ ε ∈ R

+
∃ δ ∈ R
+
: f([x, x + δ)) ⊂ (f(x) − ε, f(x) + ε).
(iii) For each x ∈ R and each interval (a, b) containing f (x) there exists an interval [c, d)
containing x such that f([c, d)) ⊂ (a, b).
(iv) f : R
l
→ R is continuous.
Proof. ”(i) ⇔ (ii)”: By definition.
”(ii) ⇒ (iii)”: Let x ∈ R. Assume f (x) ∈ (a, b). Set ε = min{f(x) − a, b − f(x)} ∈ R
+
. Then
(f(x)−ε, f(x)+ε) ⊂ (a, b). By (ii) there exists δ ∈ R
+
such that f ([x, x+δ)) ⊂ (f(x)−ε, f (x)+ε),
hence [c, d) = [x, x + δ) does the trick.
”(iii) ⇒ (ii)”: Let x ∈ R and ε ∈ R
+
. By (iii) there exists [c, d) such that x ∈ [c, d) and
f([c, d)) ⊂ (f(x) − ε, f (x) + ε). Set δ = d − x ∈ R
+
. Then [x, x + δ) = [x, d) ⊂ [c, d), hence
f([x, x + δ)) ⊂ (x − ε, x + ε).
”(iii) ⇔ (iv)”: Clear. 
1
2
(b). The continuous maps from R to R
l
are the constant maps, c.f. Ex. 25.1.

A map f : R
l
→ R
l
is continuous if and only if for any x and ε > 0 there exists δ > 0 such
that f([x, x + δ)) ⊂ [f (x), f (x) + ε). Hence f : R
l
→ R
l
is continuous if and only if f : R → R is
right continuous and increasing.
Ex. 18.8. Let Y be an ordered set. Give Y × Y the product topology. Consider the set
∆
−
= {y
1
× y
2
| y
1
> y
2
}
of points below the diagonal. Let (y
1
, y
2
) ∈ ∆
−
so that y

1
> y
2
. If y
2
is the immediate predecessor
of y
1
then
y
1
× y
2
∈ [y
1
, ∞) × (−∞, y
2
) = (y
2
, ∞) × (−∞, y
1
) ⊂ ∆
−
and if y
1
> y > y
2
for some y ∈ Y then
y
1

× y
2
∈ (y, ∞) × (−∞, y) ⊂ ∆
−
This shows that ∆
−
is open.
(a). Since the map (f, g): X
∆
−→ X × X
f×g
−−−→ Y × Y is continuous, the preimage
(f, g)
−1
(∆
−
) = {x ∈ X | f(x) > g(x)}
is open and the complement {x ∈ X | f(x) ≤ g(x)} is closed.
(b). The map
min{f, g}(x) =

f(x) f(x) ≤ g(x)
g(x) f (x) ≥ g(x)
is continuous according to [1, Thm 18.3].
Ex. 18.10. Let (f
j
: X
j
→ Y
j

)
j∈J
be an indexed family of continuous maps. Define

f
j
:

X
j
→

Y
j
to be the map that takes (x
j
) ∈

X
j
to (f
j
(x
j
)) ∈

Y
j
. The commutative
diagram


X
j
π
k

Q
f
j
//

Y
j
π
k

X
k
f
k
//
Y
k
shows that π
k
◦

f
j
= f

k
◦ π
k
is continuous for all k ∈ J . Thus

f
j
:

X
j
→

Y
j
is continuous
[1, Thm 18.4, Thm 19.6].
Ex. 18.13. Let f, g : X → Y be two continuous maps between topological s paces where the
codomain, Y , is Hausdorff. The equalizer
Eq(f, g) = {x ∈ X | f(x) = g(x)} = (f, g)
−1
(∆)
is then a closed subset of X for it is the preimage under the continuous map (f, g): X → Y × Y
of the diagonal ∆ = {(y, y) ∈ Y × Y | y ∈ Y } which is closed since Y is Hausdorff [1, Ex 17.13].
(This is [1, Ex 31.5].)
It follows that if f and g agree on the subset A ⊂ X then they also agree on A for
A ⊂ Eq(f, g) =⇒ A ⊂ Eq(f, g)
In particular, if f and g agree on a dense subset of X, they are equal: Any continuous map into
a Hausdorff space is determined by its values on a dense subset.
Munkres §22

Ex. 22.2.
(a). The map p: X → Y is continuous. Let U be a subspace of Y such that p
−1
(U) ⊂ X is open.
Then
f
−1
(p
−1
(U)) = (pf)
−1
(U) = id
−1
Y
(U) = U
is open because f is continuous. Thus p: X → Y is a quotient map.
(b). The map r : X → A is a quotient map by (a) because it has the inclusion map A → X as
right inverse.
Ex. 22.3. Let g : R → A be the continuous map f (x) = x × 0. Then q ◦ g is a quotient map, even
a homeomorphism. If the compostion of two maps is quotient, then the last map is quotient; see
[1]. Thus q is quotient.
The map q is not closed for {x ×
1
x
| x > 0} is closed but q(A ∩ {x ×
1
x
| x > 0} = (0, ∞) is not
closed.
The map q is not open for R × (−1, ∞) is open but q(A ∩ (R × (−1, ∞))) = [0, ∞) is not open.

Ex. 22.5. Let U ⊂ A be open in A. Since A is open, U is open in X. Since p is open,
p(U) = q(U ) ⊂ p(A) is open in Y and also in p(A) because p(A) is open [Lma 16.2].
SupplEx. 22.3. Let G be a topological group and H a subgroup. Let ϕ
G
: G × G → G be the
map ϕ
G
(x, y) = xy
−1
and ϕ
H
: H × H → H the corresponding map for the subgroup H. Since
these maps are related by the commutative diagram
G × G
ϕ
G
//
G
H × H
ϕ
H
//
?

OO
H
?

OO
and ϕ

G
is continuous, also ϕ
H
is continuous [Thm 18.2]. Moreover, any subspace of a T
1
-space is
a T
1
-space, so H is a T
1
-space. Thus H is a topological group by Ex 22.1.
Now, by using [Thm 18.1.(2), Thm 19.5], we get
ϕ
G
(H × H) = ϕ
G
(H × H) ⊂ ϕ
G
(H × H) = H
which shows that H is a subgroup of G. But then H is a topological group as we have just shown.
SupplEx. 22.5. Let G be a topological group and H a subgroup of G.
(a). Left multiplication with α induces a commutative diagram
G
p

f
α
//
G
p


G/H
f
α
/H
//
G/H
which shows [Thm 22.2] that f
α
/H is .¸
(b). The saturation gH = f
g
(H) of the point g ∈ G is closed because H is closed and left
multiplication f
g
is a homeomorphism [SupplEx 22.4].
(c). The saturation UH =

h∈H
Uh =

g
h
(U) is open because U is open and right multiplication
g
h
is a homeomorphism [SupplEx 22.4].
1
2
(d). The space G/H is T

1
by point (b). The map G × G
ϕ
→ G: (x, y) → xy
−1
is continuous
[SupplEx 22.1] and it induces a commutative diagram
G × G
p×p

ϕ
//
G
p

G/H × G/H
ϕ/H
//
G/H
where p × p is a quotient map since it is open as we have just shown. (It is not true that the
product of two quotient maps is a quotient map [Example 7, p. 143] but is is true that a product
of two open maps is an open map.) This shows [Thm 22.2] that ϕ/H is continuous and hence
[SupplEx 22.1] G/H is a topological group.
SupplEx. 22.7. Let G be a topological group.
(a). Any neighb orhood of U of e contains a symmetric neighborhood V ⊂ U such that V V ⊂ U.
By continuity of (x, y) → xy, there is a neighborhood W
1
of e such that W
1
W

1
⊂ U. By continuity
of (x, y) → xy
−1
, there is a neighborhood W
2
of e such that W
2
W
−1
2
⊂ W
1
. (Any neighborhood
of e contains a symmetric neighborhood of e.) Now V = W
2
W
−1
2
is a symmetric neighborhood of
e and V V ⊂ W
1
W
1
⊂ U.
(b). G is Hausdorff.
Let x = y. Since the set {xy
−1
} is closed in G, there is a neighborhood U of e not containing
xy

−1
. Find a symmetric neighborhood V of e such that V V ⊂ U. Then V x ∩ V y = ∅. (If not,
then gx = hy for some g, h ∈ V and xy
−1
= g
−1
h ∈ V
−1
V = V V ⊂ U contradicts xy
−1
∈ U.)
(c). G is regular.
Let A ⊂ G be a closed subspace and x a point outside A. Since e is outside the closed set
xA
−1
, a whole neighborhood U of e is disjoint from xA
−1
. Find a symmetric neighborhood V
of e such that V V ⊂ U. Then V x ∩ V A = ∅. (If not, then gx = ha for some g, h ∈ V and
xA
−1
 xa
−1
= g
−1
h ∈ V
−1
V = V V ⊂ U contradicts xA
−1
∩ U = ∅)

(d). G/H is regular for any closed subgroup H ⊂ G.
The point xH is closed in G/H because the subspace xH is closed in G. Let x ∈ G be a point
ouside a saturated closed subset A ⊂ G. Find a neighborhood V of e such that V x and V A are
disjoint. Then also V xH ∩ V A = ∅ since A = AH is saturated. Since p: G → G/H is open
[SupplEx 22.5.(c)], p(V x) is an open neighborhood of p(x) disjoint from the open neighborhood
p(V A) of p(A). This shows that G/H is regular.
Munkres §26
Ex. 26.1 (Morten Poulsen).
(a). Let T and T

be two topologies on the set X. Suppose T

⊃ T .
If (X, T

) is compact then (X, T ) is compact: Clear, since every open covering if (X, T ) is an
open covering in (X, T

).
If (X, T ) is compact then (X, T ) is in general not compact: Consider [0, 1] in the standard
topology and the discrete topology.
(b).
Lemma 1. If (X, T ) and (X, T

) are compact Hausdorff spaces then either T and T

are equal
or not comparable.
Proof. If (X, T ) compact and T


⊃ T then the identity map (X, T

) → (X, T ) is a bijective
continuous map, hence a homeomorphism, by theorem 26.6. This proves the result. 
Finally note that the set of topologies on the set X is partially ordered, c.f. ex. 11.2, under
inclusion. From the lemma we conclude that the compact Hausdorff topologies on X are minimal
elements in the set of all Hausdorff topologies on X.
Ex. 26.2 (Morten Poulsen).
(a). The result follows from the following lemma.
Lemma 2. If the set X is equipped with the finite complement topology then every subspace of X
is compact.
Proof. Suppose A ⊂ X and let A be an open covering of A. Then any set A
0
∈ A will covering all
but a finite number of points. Now choose a finite number of s ets from A covering A − A
0
. These
sets and A
0
is a finite subcovering, hence A compact. 
(b). Lets prove a more general result: Let X be an uncountable set. Let
T
c
= { A ⊂ X | X − A countable or equal X }.
It is straightforward to check that T
c
is a topology on X. This topology is called the countable
complement topology.
Lemma 3. The compact subspaces of X are exactly the finite subspaces.
Proof. Suppose A is infinite. Let B = {b

1
, b
2
, . . .} be a countable subset of A. Set
A
n
= (X − B) ∪ {b
1
, . . . , b
n
}.
Note that {A
n
} is an open covering of A with no finite subcovering. 
The lemma shows that [0, 1] ⊂ R in the countable complement topology is not compact.
Finally note that (X, T
c
) is not Hausdorff, since no two nonempty open subsets A and B of X
are disjoint: If A∩B = ∅ then X −(A∩B) = (X −A)∪(X −B), hence X countable, contradicting
that X uncountable.
Ex. 26.3 (Morten Poulsen).
Theorem 4. A finite union of compact subspaces of X is compact.
Proof. Let A
1
, . . . , A
n
be compact subspaces of X. Let A be an open covering of

n
i=1

A
i
. Since
A
j
⊂

n
i
1
A
i
is compact, 1 ≤ j ≤ n, there is a finite subcovering A
j
of A covering A
j
. Thus

n
j=1
A
j
is a finite subcovering of A, hence

n
i=1
A
i
is compact. 
1

2
Ex. 26.5. For each a ∈ A, choos e [Lemma 26.4] disjoint open sets U
a
∈ a and V
a
⊃ B. Since A
is compact, A is contained in a finite union U = U
1
∪ · · · ∪ U
n
of the U
a
s. Let V = V
1
∩ · · · V
n
be
the intersection of the corresponding V
a
s. Then U is an open set containing A, V is an open set
containing B, and U and V are disjoint as U ∩ V =

U
i
∩ V ⊂

U
i
∩ V
i

= ∅.
Ex. 26.6. Since any closed subset A of the compact space X is compact [Thm 26.2], the image
f(A) is a compact [Thm 26.5], hence closed [Thm 26.3], subspace of the Hausdorff space Y .
Ex. 26.7. This is just reformulation of The tube lemma [Lemma 26.8]: Let C be a closed subset
of X × Y and x ∈ X a p oint such that the slice {x} × Y is disjoint from C. Then, since Y is
compact, there is a neighborhood W of x such that the whole tube W × Y is disjoint from C.
In other words, if x ∈ π
1
(C) then there is a neighborhood W of x which is disjoint from π
1
(C).
Thus The tube lemma says that π
1
: X × Y → X is closed when Y is compact (so that π
1
is an
example of a perfect map [Ex 26.12]). On the other hand, projection maps are always open [Ex
16.4].
Ex. 26.8. Let G ⊂ X × Y be the graph of a function f : X → Y where Y is compact Hausdorff.
Then
G is closed in X × Y ⇔ f is continuous
⇐: (For this it suffices that Y be Hausdorff.) Let (x, y) ∈ X × Y be a point that is not in the
graph of f. Then y = f(x) so by the Hausdorff axiom there will be disjoint neighborhoods V  y
and W  f (x). By continuity of f, f(U) ⊂ W ⊂ Y − V . This means that (U × V ) ∩ G = ∅.
⇒: Let V b e a neighborhood of f(x) for some x ∈ X. Then G ∩ (X × (Y − V )) is closed in X × Y
so [Ex 26.7] the projection π
1
(G ∩ (X × (Y − V ))) is closed in X and does not contain x. Let U
be a neighborhood of X such that U × Y does not intersect G ∩ (X × (Y − V )). Then f(U) does
not intersect Y − V , or f(U ) ⊂ V . This shows that f is continuous at the arbitrary point x ∈ X.

Ex. 26.12. (Any perfect map is proper; see the January 2003 exam for more on proper maps.)
Let p: X → Y be closed continuous surjective map such that p
−1
(y) is compact for each y ∈ Y .
Then p
−1
(C) is compact for any compact subspace C ⊂ Y .
For this exercise we shall use the following lemma.
Lemma 5. Let p : X → Y be a closed map.
(1) If p
−1
(y) ⊂ U where U is an open subspace of X, then p
−1
(W ) ⊂ U for some neighborhood
W ⊂ Y of y.
(2) If p
−1
(B) ⊂ U for some subspace B of Y and some open subspace U of X, then p
−1
(W ) ⊂
U for some neighborhood W ⊂ Y of B.
Proof. Note that
p
−1
(W ) ⊂ U ⇔

p(x) ∈ W ⇒ x ∈ U

⇔


x ∈ U ⇒ p(x) ∈ W

⇔ p(X − U) ⊂ Y − W
⇔ p(X − U) ∩ W = ∅
(1) The point y does not belong to the closed set p(X − U). Therefore a whole neighborhood
W ⊂ Y of y is disjoint from p(X − U), i.e. p
−1
(W ) ⊂ U .
(2) Each point y ∈ B has a neighborhood W
y
such that p
−1
(W
y
) ⊂ U. The union W =

W
y
is
then a neighborhood of B with p
−1
(W ) ⊂ U . 
We shall not need point (2) here.
Let C ⊂ Y be compact. Consider a collection {U
α
}
α∈J
of open sets covering of p
−1
(C).

For each y ∈ C, the compact space p
−1
(y) is contained in a the union of a finite subcollection
{U
α
}
α∈J(y)
. There is neighborhood W
y
of y such that p
−1
(W
y
) is contained in this finite union. By
compactness of C, finitely many W
y
1
, . . . , W
y
k
cover Y . Then the finite collection

k
i=1
{U
α
}
α∈J(y
i
)

cover p
−1
(C). This shows that p
−1
(C) is compact.
Ex. 26.13. Let G be a topological group and A and B subspaces of G.
3
(a). A closed and B compact ⇒ AB closed
Assume c ∈ AB =

b∈B
Ab. The regularity axiom for G [Suppl Ex 22.7] implies that there are
disjoint open sets W
b
 c and U
b
⊃ Ab separating c and Ab for each point b ∈ B. Then A
−1
U
b
is
an open neighborhood of b. Since B is compact, it can be covered by finitely many of these open
sets A
−1
U
b
, say
B ⊂ A
−1
U

1
∪ · · · ∪ A
−1
U
k
= A
−1
U
where U = U
1
∪ · · · ∪ U
k
. The corresponding open set W = W
1
∩ · · · ∩ W
k
is an open neighborhood
of c that is disjoint from AB since W ∩ AB ⊂

W ∩ U
i
⊂

W
i
∩ U
i
= ∅.
(b). H compact subgroup of G ⇒ p: G → G/H is a closed map
The saturation AH of any closed subset A ⊂ G is closed by (a).

(c). H compact subgroup of G and G/H compact ⇒ G compact
The quotient map p : G → G/H is a perfect map because it is a closed map by (b) and has compact
fibres p
−1
(gH) = gH. Now apply [Ex 26.12].
Munkres §27
Ex. 27.1 (Morten Poulsen). Let A ⊂ X b e bounded from above by b ∈ X. For any a ∈ A is
[a, b] compact.
The set C = A ∩ [a, b] is closed in [a, b], hence compact, c.f. theorem 26.2. The inclusion map
j : C → X is continuous, c.f. theorem 18.2(b). By the extreme value theorem C has a largest
element c ∈ C. Clearly c is an upper bound for A.
If c ∈ A then clearly c is the least upper bound. Suppose c /∈ A. If d < c then (d, ∞) is an open
set containing c, i.e. A ∩ (d, ∞) = ∅, since c is a limit point for A, since c ∈ C ⊂ A. Thus d is not
an upp e r bound for A, hence c is the least upper bound.
Ex. 27.3.
(a). K is an infinite, discrete, closed subspace of R
K
, so K can not be contained in any compact
subspace of R
K
[Thm 28.1].
(b). The subspaces (−∞, 0) and (0, +∞) inherit their standard topologies, so they are connected.
Then also their closures, (−∞, 0] and [0, +∞) and their union, R
K
, are also connected [Thm 23.4,
Thm 23.3].
(c). Since the topology R
K
is finer than the standard topology [Lem ma 13.4] on R we have
U is connected in R

K
Ex 23.1
⇒ U is connected in R
Thm 24.1
⇔ U is convex
for any subspace U of R
K
.
Let now f : [0, 1] → R
K
be a path from f(0) = 0 to f(1) = 1. The image f([0, 1]) is convex
since it is connected as a subspace of R
K
[Thm 23.5], and connected subspaces of R
K
are convex
as we just noted. Therefore the interval [0, 1] and the its subset K is contained in f([0, 1]). The
image f([0, 1]) is also compact in the subspace topology from R
K
[Thm 26.5]. Thus the image is
a compact subspace of R
K
containing K; this is a contradiction (see (a)). We conclude that there
can not exist any path in R
K
from 0 to 1.
Ex. 27.5. I first repeat Thm 27.7 in order to emphasize the similarity between the two statements.
Theorem 1 (Thm 27.7). Let X be a compact Hausdorff space with no isolated points. Then X
contains uncountably many points.
Proof. Let A = {a

1
, a
2
, . . .} be a countable subset of X. We must find a point in X outside A.
We have X = {a
1
} for {a
1
} is not open. So the open set X − {a
1
} is nonempty. By regularity
[Lemma 26.4, Lemma 31.1], we can find an open nonempty set U
1
such that
U
1
⊂ U
1
⊂ X − {a
1
} ⊂ X
We have U
1
= {a
2
} for {a
2
} is not open. So the open set U
1
− {a

2
} is nonempty. By regularity
[Lemma 26.4, Lemma 31.1], we can find an open nonempty set U
2
such that
U
2
⊂ U
2
⊂ U
1
− {a
2
} ⊂ U
1
Continuing this way we find a descending sequence of nonempty open sets U
n
such that
U
n
⊂ U
n
⊂ U
n−1
− {a
n
} ⊂ U
n−1
for all n.
Because X is compact, the intersection


U
n
=

U
n
is nonempty [p. 170] and contained in

(X − {a
n
}) = X −

{a
n
} = X − A. 
Theorem 2 (Baire category theorem). Let X be a compact Hausdorff space and {A
n
} a sequence
of closed subspaces. If Int A
n
= ∅ for all n, then Int

A
n
= ∅.
1
2
Proof. (See Thm 48.2.) Let U
0

be any nonempty subspace of X. We must find a point in U
0
outside

A
n
.
We have U
0
⊂ A
1
for A
1
has no interior. So the open set U
0
− A
1
is nonempty. By regularity
[Lemma 26.4, Lemma 31.1], we can find a nonempty open set U
1
such that
U
1
⊂ U
1
⊂ U
0
− A
1
⊂ U

0
We have U
1
⊂ A
2
for A
2
has no interior. So the open set U
1
− A
2
is nonempty. By regularity
[Lemma 26.4, Lemma 31.1], we can find a nonempty open set U
2
such that
U
2
⊂ U
2
⊂ U
1
− A
2
⊂ U
1
Continuing this way, we find a descending sequence of nonempty open sets U
n
such that
U
n

⊂ U
n
⊂ U
n−1
− A
n
⊂ U
n−1
for all n.
Because X is compact, the intersection

U
n
=

U
n
is nonempty [p. 170] and contained in
U
0
∩

(X − A
n
) = U
0
−

A
n

.

Ex. 27.6 (The Cantor set).
(a). The set A
n
is a union of 2
n
disjoint closed intervals of length 1/3
n
. Let p and q be two points
in C. Choose n so that |p − q| > 1/3
n
. Then there is point r between them that is not in A
n
, so
not in C. As in [Example 4, p. 149], this shows that any subspace of C containing p and q has a
separation.
(b). C is compact because [Thm 26.2] it is closed subspace of the compact space [0, 1].
(c). C is constructed from any of the A
n
by removing interior points only. Thus the boundary of
A
n
is contained in C for all n. Any interval of length > 1/3
n+1
around any point of A
n
contains
a boundary point of A
n+1

, hence a point of C. Thus C has no isolated points.
(d). C is a nonempty compact Hausdorff space with no isolated points, so it contains uncountably
many points [Thm 27.7].