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General Topology
Lecture Notes
William S. Neal
February 24, 2014
Contents
1 Lecture 01 31.1 History of Topology . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Set
Theoretic Preliminaries . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . 31.3 Relations and Functions . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4
Cardinality and Operations on Sets . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . 6
2 Lecture 02 72.1 Logic and Techniques of Proof . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Topology on
a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 92.3 Open and Closed Sets . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Lecture 03 123.1 Set Theoretic Preliminaries . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Interior
and Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . 133.3 Exterior and Boundary . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . 16
4 Lecture 04 184.1 Interior, Closure, Exterior, and Boundary
Points . . . . . . . . . . . . . . . . . . . . . . . 184.2 Cluster
Points and Isolated Points . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . 21
5 Lecture 05 235.1 The Isolated Points of a Space . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . 235.2
Introduction to Connectedness . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . 255.3 Introduction to Separation
Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. 26
6 Lecture 06 286.1 Dense Subsets . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 286.2 Nowhere
Dense Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 306.3 Separable Spaces . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.4
Equivalent Statements About Open and Closed Sets . . . . . . . . .
. . . . . . . . . . . . 326.5 More on Cluster Points and Isolated
Points . . . . . . . . . . . . . . . . . . . . . . . . . . 32
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7 Lecture 07 337.1 Basis for a Topology . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 337.2 The
Digital Line Topology . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . 37
8 Lecture 08 388.1 More Properties of a Basis . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 388.2
Calculating the Interior and Closure of Unions and Intersections .
. . . . . . . . . . . . . 398.3 Properties of the Digital Line . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
9 Lecture 09 439.1 More Properties of the Digital Line . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . 439.2 Relations
on a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . 459.3 Introduction to the Order Topology . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
10 Lecture 10 4810.1 More Properties of Posets and Simply
Ordered Sets . . . . . . . . . . . . . . . . . . . . . 4810.2
Properties of the Order Topology . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . 50
11 Lecture 11 5311.1 Compositions, Inverses, and Restrictions of
Functions . . . . . . . . . . . . . . . . . . . . 5311.2 Finite
Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . 55
12 Lecture 12 5712.1 The Cofinite Topology . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.2
Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . 6012.3 Introduction to the
Cocountable Topology . . . . . . . . . . . . . . . . . . . . . . .
. . . . 62
13 Lecture 13 6213.1 Properties of the Cofinite and Cocountable
Topologies . . . . . . . . . . . . . . . . . . . . 6213.2 More
Countability Conditions . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . 6413.3 The Order Topology on Q . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
14 Lecture 14 6814.1 More on Local Bases . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 6814.2 The
Order Topology on Q and R . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . 6814.3 Introduction to the Subspace Topology
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
15 Lecture 15 7315.1 Properties of the Subspace Topology . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . 7315.2 Convex
Subsets and Subbases . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . 77
16 Lecture 16 7816.1 Inherited Properties of Subspaces . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.2
Functions Acting on Sets . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . 8116.3 Introduction to Continuity . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1 Lecture 01
1.1 History of Topology
Remark 1.1. The original source for the following historical
excerpts is the 6th edition ofAn Introduction to the History of
Mathematics by Howard Eves.
Topology started as a branch of geometry, but is now more
properly considered, along with geometry,algebra, and analysis, as
a fundamental division of mathematics (Eves, 617).
From the [geometric] point of view, topology may be regarded as
the study of those properties ofgeometric figures that remain
invariant under single-valued continuous mappings possessing
single-valuedcontinuous inverses (Eves, 617).
The notion of a geometric figure as made up of a finite set of
joined fundamental pieces graduallygave way to the concept [that]
an arbitrary set can constitute a topological space. This latter,
and verygeneral, viewpoint of topology has become known as set
topology (Eves, 619).
The earlier [geometric] viewpoint has become known as algebraic
topology (Eves, 619).
Remark 1.2. The functions mentioned in the second historical
excerpt are bijections where both thefunction and its inverse are
continuous. These functions have a special name. They are called
homeo-morphisms.
Remark 1.3. Set topology is more commonly called general
topology.
Remark 1.4. General topology has evolved into the study of the
properties of abstract spaces. Ofparticular interest are those
properties which are preserved under homeomorphisms. Such
properties arecalled topological properties.
Fundamentally, general topology is concerned with abstract
spaces formed from arbitrary sets. Ac-cordingly, the foundation
underlyining topology, and indeed all of modern mathematics, is set
theory.
1.2 Set Theoretic Preliminaries
Remark 1.5. We accept as a fundamental intuitive fact the
primative notion of a set : A set is acollection of objects called
the elements or members of the set.Let A be a set. If the element x
is in A, we write x A. If x is not in A, we write x / A.Definition
1.6. Let A and B be sets. Then A is a subset of B, denoted A B, if
every element of Ais also an element of B. That is, if x A x B,
then A B.Remark 1.7. We accept as an axiom from set theory the
existence of a set = {} which has no ele-ments. is called the empty
set or null set.
If A is any arbitrary set, then A.
The empty set satisfies the defintion of subset vacuously. There
are no elements in that must becontained in the other set.
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Definition 1.8. Let A and B be sets. Then A = B if and only if A
B and B A. In notation,A = B A B and B A.Remark 1.9. (Explanation
of Notation)A B means x A, x BA B means A B or A = BA ( B means A B
and A 6= B
Definition 1.10. If A ( B, then A is called a proper subset of
B.
Lemma 1.11. Every set is a subset of itself.
Proof. Let A be a set and let x A. Since x A x A, A A.Lemma
1.12. (Transitivity of Subsets)If A B and B C, then A C.Proof.
Suppose that A B and B C.Let x A. Then as A B, x B.Now as B C, x
C.Hence, A C.
So we can represent this fact in one statement: A B C.Definition
1.13. Let A and B be sets.The Cartesian product of A and B is the
set AB = {(a, b) : a A and b B}.Example 1.14. Let A = {a, b} and B
= {1, 2, 3}.Then AB = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b,
3)}.
In particular, (2, b) / AB but (2, b) B A.
1.3 Relations and Functions
Definition 1.15. Let A and B be sets. A relation between A and B
is any subset R A B. Inparticular, R contains ordered pairs.
Remark 1.16. If (a, b) R, this is also denoted aRb.Definition
1.17. Let A and B be sets. A function between A and B is a nonempty
relation ABwith the following property:
If (a, b) and (a, c) , then b = c.Remark 1.18. A function is
also called a map or mapping.
Definition 1.19. Let be a function between sets A and B. The
domain of the function isthe set dom = {a A : b B such that (a, b)
}. The range of the function is the setrng = {b B : a A such that
(a, b) }.Definition 1.20. Let be a function between sets A and B.
If dom = A, then is a function fromA into B, denoted : A B. In this
case, the set A is called the domain of the function and the setB
is called the codomain of the function.
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Remark 1.21. Let : A B. If (a, b) , this is denoted as (a) =
b.Definition 1.22. The function : A B is called an injection, or an
injective function, if everyelement in the domain A is mapped into
a unique element in the codomain B. If is injective, then[(a) =
(b)] [a = b].Example 1.23. The function : Z Z defined by (a) = 2a
is injective.Proof. Suppose (a) = (b).Then 2a = 2b. So a = b. is an
injection.
Remark 1.24. An injection is sometimes called a 1 1 function
(read one to one).Definition 1.25. The function : A B is called a
surjection, or a surjective function, if eachelement in the
codomain B is mapped onto by at least one element from the domain
A. If is asurjection, then [b B] [a A such that (a) = b].Example
1.26. The function : Z Z defined by (a) = 2a is not
surjective.Proof. Notice 1 Z and 6 a Z such that 1 = (a).Example
1.27. The function : Z 2Z defined by (a) = 2a is surjective.Proof.
Let b 2Z. Then b = 2a for some a Z.That is, b 2Z a Z such that b =
(a). is a surjection.
Remark 1.28. A surjection is sometimes called an onto
function.
Remark 1.29. A necessary condition for : A B to be a surjection
is rng = B.Definition 1.30. A bijection is a function : A B that is
both injective and surjective.Remark 1.31. A bijection is sometimes
called a 1 1 correspondence.Example 1.32. The function : Z 2Z
defined by (a) = 2a is a bijection.Proof. We have already
demonstrated that is a surjection.Suppose (a) = (b).Then 2a = 2b.
So a = b.Hence, is an injection. is a bijection as it is both
injective and surjective.
Remark 1.33. A necessary condition for a function : A B to be a
bijection is that the number ofelements in the domain and codomain
must be equal. That is, |A| = |B|.
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1.4 Cardinality and Operations on Sets
Definition 1.34. The cardinality of a set is the number of
distinct elements in the set. The cardinalityof the set A is
denoted by |A|.Example 1.35. The set A = {1, 1, 1, 2} has two
distinct elements. So |A| = 2.Definition 1.36. A set A is finite if
a bijection : A {1, 2, . . . , n} for some positive integer n.
Inthis case, |A| = n.Definition 1.37. A set which is not finite is
infinite.
Definition 1.38. A set A is countably infinite if a bijection :
A Z+. In this case, |A| = 0.Definition 1.39. A set is countable if
it is either finite or countably infinite.
Definition 1.40. An infinite set which is not countably infinite
is uncountable.
Definition 1.41. Let A and B be sets. The union of the sets A
and B, denoted AB, is the set thatcontains those elements that are
either in A or in B, or in both.
Remark 1.42. Notice that A A B and B A B. Further, if A B, then
A B = B. So A = A.Definition 1.43. Let A and B be sets. The
intersection of the sets A and B, denoted A B, is theset containing
those elements that are simultaneously in both A and B.
Remark 1.44. Notice A B A and A B B. Further, if A B, then A B =
A. So A = .Definition 1.45. Sets are called disjoint if their
intersection is the empty set.
Definition 1.46. Given a countable set A, the power set of A,
denoted P(A), is the set of all subsetsof the set A.
Remark 1.47. If A is a finite set with cardinality |A| = n, then
the cardinality of the power set of Ais |P(A)| = 2n. If |A| = 0,
then |P(A)| = 20 = c, where c is a transfinite number which stands
forcontinuum.
Example 1.48. Let A = {a, b, c}.P(A) = {, {a}, {b}, {c}, {a, b},
{a, c}, {b, c}, {a, b, c}}.Notice |A| = 3 and |P(A)| = 23 =
8.Definition 1.49. A collection of sets is called a family of
sets.
Remark 1.50. Notice that P(A) is a family of sets containing the
subsets of A.
Definition 1.51. Let I be a non-empty set, called an indexing
set, such that for each index i I therecorresponds a set Ai. Then
the family A = {Ai : i I} is an indexed family of sets.Definition
1.52. Let {Ai : i I} be an indexed family of sets. The arbitrary
union of sets in theindexed family of sets over the index i is the
set
iI Ai = {x : x Ai for at least one i I}.
Definition 1.53. Let {Ai : i I} be an indexed family of sets.
The arbitrary intersection of sets inthe indexed family of sets
over the index i is the set
iI Ai = {x : x Ai for every i I}.
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Remark 1.54. If |I| = 0, theniI Ai =
i=1Ai and is called a countable union. Similarly,
i=1Ai
is called a countable intersection.If |I| = n, then iI Ai =
ni=1Ai and is called a finite union. Similarly, ni=1Ai is called a
finiteintersection.
Remark 1.55. (Arbitrary Unions and Intersections)If x iI Ai,
then x Ai for at least one index i I. In notation, we have
x iI Ai i I such that x Ai.If x / iI Ai, then x / Ai for any
index i I. In notation, we have
x / iI Ai 6 i I such that x Ai.If x iI Ai, then x Ai for every
index i I. In notation, we have
x iI Ai x Ai i I.If x / iI Ai, then at lest one i I such that x
/ Ai. In notation, we have
x / iI Ai i I such that x / Ai.Definition 1.56. Let A X. The
complement of A in X is the set X \A = {x X : x / A}.Remark 1.57.
Notice that if A X and x / A, then x X \A.Theorem 1.58. (De Morgans
Laws)Let {Ai : i I} be an indexed family of sets where each Ai X.
Then
(1) X \iI Ai = iI(X \Ai)(2) X \iI Ai = iI(X \Ai)
Proof. (1) Let x X \iI Ai. Then x / iI Ai. So i I such that x /
Ai. Thus, x X \Ai for atleast one index i I. Hence, x iI(X
\Ai).Conversely, let x iI(X \ Ai). Then x X \ Ai for at least one
index i I. So i I such thatx / Ai. Thus x /
iI Ai. Hence, x X \
iI Ai.
(2) Let x X \ iI Ai. Then x / iI Ai. So 6 i I such that x Ai.
Thus, x X \ Ai i I.Hence, x iI(X \Ai).Conversely, let x iI(X \ Ai).
Then x X \ Ai i I. So 6 i I such that x Ai. Thus,x / iI Ai. Hence,
x X \iI AI .2 Lecture 02
2.1 Logic and Techniques of Proof
Definition 2.1. An implication is a compound statement of the
form p q. When written out, it hasthe form If p, then q. In the
implication, the statement p is called the hypothesis and the
statementq is called the conclusion.
Remark 2.2. To prove an implication directly, we assume that the
hypothesis is true and show thatthis logically leads to the
conclusion.
Remark 2.3. Given an implication, we can form other statements,
namely, the converse, the contra-positive, and the inverse.
Definition 2.4. Given an implication p q, the converse of the
implication is the compound statementq p.
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Remark 2.5. Notice that in the converse we have a reversal of
roles. What was the hypothesis in theinplication is now the
conclusion in the converse and what was the conclusion in the
implication is nowthe hypothesis in the converse.
Definition 2.6. Given an implication p q, the contrapositive of
the implication is the compoundstatement q p.Definition 2.7. Given
an implication p q, the inverse of the implication is the compound
statementp q.Remark 2.8. We can group the four statements in pairs
that are logically equivalent:The implication is logially
equivalent to its contrapositive: p q q pThe converse is logically
equivalent to the inverse: q p p q
Notice that the inverse is the contrapositive of the
converse.
Remark 2.9. (Indirect Proof of an Implication by its
Contrapositive)We can prove an implication indirectly by proving
its contrapositive. To do this, we assume the negationof the
conclusion and show that this leads logically to the negation of
the hypothesis.
Remark 2.10. (Proof by Contradiction)Proof by contradiction is
based on the logical equivalence p q (p q) Contradiction. Soto
prove an implication by contradiction, we assume that both the
hypothesis and the negation of theconclusion are true and show that
this leads to a contradiction.
Remark 2.11. (Proof by Cases)Proof by cases is based on the
logical equivalence (pq) r p rq r. We use this techniqueof proof
when we must consider two or more cases in the hypothesis.
Example 2.12. Suppose we wish to prove that A1 A2 B.Then for any
arbitrary element x A1 A2, the element can be in the set A1 or the
set A2.So we must prove that both cases lead to the conclusion that
the element is also in the set B.That is, we prove both x A1 x B
and x A2 x B.When both implications are true, the implication (x A1
x A2) B is also true.Hence, A1 A2 B.Definition 2.13. A
biconditional is a statement of the form p q. When written out, it
has theform p if and only if q. The phrase if and only if is
abbreviated iff.
Remark 2.14. The biconditional p q is logically equivalent to
the implication p q and itsconverse q p: [p q] [p q q p]So to prove
a biconditional directly, we must prove that both the implication
and its converse are true.
Remark 2.15. (Indirect Proof of the Biconditional by its
Contrapositive)Just as we can prove an implication indirectly by
proving its contrapositive, we can prove a biconditionalindirectly
by proving its contrapositive. The contrapositive of a
biconditional p q is the statementp q. It is formed by taking the
contrapositive of both the implication and its converse.
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2.2 Topology on a Set
Definition 2.16. A topology on a set X is a collection of
subsets of X such that(1) , X .(2) If {Ui : i I} is an indexed
family of subsets of X with Ui , then
iI Ui .
(3) If U1, U2, . . . , Un , thenni=1 Ui .
Definition 2.17. A set X together with a topology specified on X
is called a topological space andis often denoted (X, ).
Remark 2.18. In mathematics, a space is a set which admits some
type of structure. In a topologicalspace, the structure of interest
is the topology itself.
Remark 2.19. Given a topological space X, the empty set and the
entire set are in the space. Arbitraryunions of elements in the
space remain in the space and finite intersections of elements in
the space remainin the space.
Definition 2.20. (Indiscrete Topology)Let X be any arbitrary set
and let = {, X}. Then is the indiscrete topology on X.Proof. (1)
Clearly, , X .(2) X = X .(3) X = . is a topology on X.
Remark 2.21. The indiscrete topology is also called the trivial
topology.
Definition 2.22. (Discrete Topology)Let X be a countable set and
let =P(X). Then is the discrete topology on X.
Proof. (1) Since X and X X, , X .(2) An arbitrary union of
subsets of X is itself a subset of X.(3) A finite intersection of
subsets of X is itself a subset of X. is a topology on X.
Definition 2.23. (Particular Point Topology)Let X be any set and
let p X be a specified point. Then = {} {U X : p U} is called
theparticular point topology.
Proof. (1) since {} . Similarly, X since X {U X : p U} .(2) Let
{Ui : i I} be an indexed family of sets such that Ui i I.Notice Ui
= or Ui {U X : p U}.Suppose p Ui for at least one index i.Then p iI
Ui. Hence, iI Ui {U X : p U} .Otherwise,
iI Ui = .
In either case,iI Ui .
(3) Let U1, . . . , Un .Again, either Ui = or Ui {U X : p
U}.Suppose p Ui for every index i.Then p ni=1 Ui. Hence, ni=1 Ui {U
X : p U} .
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Otherwise,ni=1 Ui = .
In either case,ni=1 Ui .
is a topology on X.
Example 2.24. (Excluded Point Topology)Let X be any set and let
p X be a specified point. Then = {X} {U X : p / U} is called
theexcluded point topology.
Proof. (1) X since X {X} . Similarly, since {U X : p / U} .(2)
Let {Ui : i I} be an indexed family of sets such that Ui i I.Notice
Ui = X or Ui {U X : p / U}.Suppose p / Ui for any index i.Then p /
iI Ui. Hence, iI Ui {U X : p / U} .Otherwise,
iI Ui = X .
In either case,iI Ui .
(3) Let U1, . . . , Un .Again, either Ui = X or Ui {U X : p /
U}.Suppose p / Ui for at least one index i.Then p / ni=1 Ui. Hence,
ni=1 Ui {U X : p / U} .Otherwise,
ni=1 Ui = X .
In either case,ni=1 Ui .
is a topology on X.
Definition 2.25. (Comparable Topologies)Let and be two
topologies defined on the same underlying set X. If , then is said
to befiner than . Equivalently, is said to be coarser than . If ( ,
then is said to be strictlyfiner than . If either or , then the two
topologies are said to be comparable. Otherwise,the two topologies
are incomparable.
Remark 2.26. Evidently, the indiscrete topology is the coarsest
topology possible while the discretetopology is the finest topology
possible.
Remark 2.27. Recall from the previous lecture that if A B, then
A B = A.So if , then = .Equivalently, if 6= , then 6 .Similarly, if
6= , then 6 .So if the intersection is not one of the topologies,
then the two topologies are incomparable.
Example 2.28. Let p X be the specified point, let P be the
particular point topology on X, and letE be the excluded point
topology on X. Then
P E = [{} {U X : p U}] [{X} {U X : p / U}]= {, X}
So P and E are incomparable.
Example 2.29. Let X = {a, b} and let a be the specified
point.The indiscrete topology on X is I = {, {a, b}}.The discrete
topology on X is D = {, {a}, {b}, {a, b}}.
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The particular point topology on X is P = {, {a}, {a, b}}.The
excluded point topology on X is E = {, {b}, {a, b}}.Notice I ( P (
D and I ( E ( D.
Remark 2.30. Notice that if in example 2.29 X = {a}, then I = P
= E = D. In particular, sinceI = D, any topology defined on a
singleton set is indistinguishable from the indiscrete
topology.
Definition 2.31. (Sierpinski Space)A topological space
consisting of a two point set and either the particular point
topology or the excludedpoint topology is called a Sierpinski
space.
Remark 2.32. A Sierpinski space is the smallest topological
space that is neither indiscrete nor discrete.
2.3 Open and Closed Sets
Definition 2.33. Let (X, ) be a topological space. A subset U X
is open in X if U .Definition 2.34. Let (X, ) be a topological
space. A subset A X is closed in X if (X \ A) .That is, a subset A
X is closed if its complement X \A is open.Remark 2.35. Within a
topological space, the complement of an open set is closed and the
complementof a closed set is open.
Example 2.36. The subset (0, 1) R is called an open interval in
R because there exists a topology on R, called the standard
topology on R, such that (0, 1) = {x R : 0 < x < 1} .
Similarly, [0, 1] ={x R : 0 x 1} is called a closed interval in R
since its complement R\ [0, 1] = (, 0) (1,) .Remark 2.37. Let (X, )
be a topological space. Then
(1) and X are open.(2) Arbitrary unions of open sets are
open.(3) Finite intersections of open sets are open.
Theorem 2.38. Let (X, ) be a topological space. Then(1) and X
are closed.(2) Arbitrary intersections of closed sets are
closed.(3) Finite unions of closed sets are closed.
Proof. (1) Since X \ = X , is closed. Similarly, since X \X = ,
X is closed.(2) Let {Ui : i I} be an indexed family of sets such
that Ui i I. Then X \Ui is closed i I.Since
iI(X \ Ui) = X \
iI Ui and
iI Ui ,
iI(X \ Ui) is closed.
(3) Let U1, . . . , Un . Then X \ Ui is closed i {1, . . . , n}.
Sinceni=1(X \ Ui) = X \
ni=1 Ui andn
i=1 Ui ,ni=1(X \ Ui) is closed.
Remark 2.39. In the first statement of Theorem 2.38, we are
confronted right away with a fundamentalfact about open and closed
sets. Not only are the two not opposites, but they are also not
mutuallyexclusive. In particular, and X are both open and closed in
any topological space X.In general, a subset of the underlying set
in a given topological space can be
(1) both open and closed.(2) open but not closed.(3) closed but
not open.(4) neither open nor closed.
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Example 2.40. Let X = {a, b} and let a be the specified
point.The particular point topology on X is P = {, {a}, {a, b}}.{a}
is open in (X, p) but is not closed since {a, b} \ {a} = {b} / PThe
excluded point topology on X is E = {, {b}, {a, b}}.{b} is open in
(X, E) but is not closed since {a, b} \ {b} = {a} / E .Remark 2.41.
Notice that in a Sierpinski space the only sets that are both open
and closed are andX. As we will see, this is a characteristic of
connected spaces. A Sierpinski space is thus the smallestnontrivial
connected space.
Remark 2.42. In contrast, every subset in a discrete topological
space (X, ) is both open and closedsince the complement of any
subset is again a subset of X. As we will see, a discrete
topological spaceis completely disconnected.
3 Lecture 03
3.1 Set Theoretic Preliminaries
Theorem 3.1. Let A B and let C be a set. Then(1) A C B C.(2) A C
B C.
Proof. (1) Let x A C. Then x A or x C.Suppose x A. Then since A
B, x B. Thus x B C.Now suppose x C. Then x B C. In either case, if
A B, then A C B C.(2) Let x A C. Then x A and x C. Since x A and A
B, x B. Thus x B C. If A B, then A C B C.Theorem 3.2. (Distributive
Laws)Let A, B, and C be sets. Then
(1) A (B C) = (A B) (A C)(2) A (B C) = (A B) (A C)
Proof. (1) Let x A (B C). Then x A or x B C.Suppose, x A. Then x
A B. It is also true that x A C. Thus, x (A B) (A C).Now suppose x
B C. Then x B and x C. Since x B, x A B. Similarly, since x C,x A
C. Thus, x (A B) (A C).Hence, in either case, A (B C) (A B) (A
C).Conversely, let x (A B) (A C). Then x A B and x A C.Since x A B,
x A or x B and since x A C, x A or x C.Suppose x A. Then x A (B
C).Now suppose x / A. Then x B and x C. That is, x B C. Thus x A (B
C).Hence, (A B) (A C) A (B C).(2) Let x A (B C). Then x A and x B
C.Since x B C, x B or x C.Suppose x B. Then since x A, x A B. Thus,
x (A B) (A C).Now suppose x C. Then since x A, x A C. Thus, x (A B)
(A C).
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Hence, A (B C) (A B) (A C).Conversely, let x (A B) (A C). Then x
A B or x A C.Suppose x A B, then x A and x B. Since x B, x B C.
Thus, x A (B C).Now suppose x A C. Then x A and x C. Since x C, x B
C. Thus, x A (B C).Hence, (A B) (A C) A (B C).Theorem 3.3. Let A B
X. Then(1) B \A = B (X \A).(2) X \B X \A.Proof. (1) Let x B \A.Then
x B and x / A.As x / A, x X \A.So x B (X \A).Hence, B \A B (X
\A).Conversely, let x B (X \A).Then x B and x X \A.As x X \A, x /
A.So x B \A.Hence, B (X \A) B \A. B \A = B (X \A).(2) Let x X
\B.Then x / B. Since A B, x / A. Thus, x X \A.Hence, X \B X \A.
3.2 Interior and Closure
Definition 3.4. Let (X, ) be a topological space and let A X.
The interior of A is the setInt(A) =
{Ui : Ui A and Ui }.That is, Int(A) is the union of all open
sets contained in A.
Remark 3.5. Since Int(A) is an arbitrary union of open sets, it
is open. It is in fact the largest (i.e.most inclusive) open set
contained in A. So if U is an open set such that U A, then U Int(A)
A.Definition 3.6. Let (X, ) be a topological space and let A X. The
closure of A is the set
Cl(A) ={Ci : A Ci and X \ Ci }.
That is, Cl(A) is the intersection of all closed sets containing
A.
Remark 3.7. Since Cl(A) is an arbitrary intersection of closed
sets, it is closed. It is in fact thesmallest (i.e. most exclusive)
closed set containing A. So if C is a closed set such that A C,
thenA Cl(A) C.Remark 3.8. The closure of A is sometimes denoted
A.
Remark 3.9. Let (X, ) be a topological space. For any subset A
X, Int(A) A Cl(A).Theorem 3.10. Let (X, ) be a topological space
and let A X. Then(1) A is open iff A = Int(A)(2) A is closed iff A
= Cl(A).
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Proof. (1) Suppose A is open. Then A is an open set contained in
itself. So A Int(A). SinceInt(A) A, we have Int(A) A Int(A). Hence,
A = Int(A).Conversely, suppose A = Int(A). Then since the interior
of a set is always open, A is open.(2) Suppose A is closed. Then A
is a closed set containing itself. So Cl(A) A. Since A Cl(A),
wehave Cl(A) A Cl(A). Hence, A = Cl(A).Conversely, suppose A =
Cl(A). Then since the closure of a set is always closed, A is
closed.
Definition 3.11. (Neighborhood of a Point)Let (X, ) be a
topological space and let x X. Any subset A X with x Int(A) is
called aneighborhood of x. If A is open, then A is called an open
neighborhood of the point x X.
3.3 Exterior and Boundary
Definition 3.12. Let (X, ) be a topological space and let A X.
The exterior of A is the setExt(A) = X \ Cl(A).
Remark 3.13. Notice Ext(A) is open as it is the complement of a
closed set.Further, since Ext(A) = X \ Cl(A), we have
X \ (X \ Cl(A)) = X \ Ext(A)Cl(A) = X \ Ext(A)
Definition 3.14. Let (X, ) be a topological space and let A X.
The boundary of A is the setA = X \ [Int(A) Ext(A)].
Remark 3.15. Notice Int(A)Ext(A) is open as it is a union of
open sets. So A = X\[Int(A)Ext(A)]is closed as it is the complement
of an open set.
Remark 3.16. Let (X, ) be a topological space and let A X.
We are accustomed to considering the subset A as a collection of
points in the set X.
We can now begin to consider the subset A as a geometric object
in the topological space X thathas a boundary, an interior, and an
exterior.
For simplicity, lets suppose that the space resembles, at least
in the near vicinity of the subset A, a2-dimensional plane that is
very similar to the real plane.
Suppose that the boundary of the subset A is the closed curve
shown below.
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Remark 3.17. (The Subset A as a Geometric Object)
The interior of A is everything inside the boundary (not
including the boundary itself).
The exterior of A is everything outside the boundary (not
including the boundary itself).
The boundary can indeed be described as what remains when both
the interior and exterior of A areremoved.
The closure of A can be described as what remains when the
exterior is removed: it is the boundaryand everything inside the
boundary.
Definition 3.18. Let X be a set.An algebra over X is a
collection F of subsets of X closed under(1) Arbitrary Unions(2)
Arbitrary Intersections(3) Complements of all subsets in F
Remark 3.19. The collection F = {, X} is the smallest collection
of subsets that constitutes ananlgebra over X. This is the trivial
algebra over X.
Remark 3.20. In contrast, a topology over X is a collection of
subsets of X closed under(1) Arbitrary Unions(2) Finite
Intersections(3) Complements of closed sets (i.e. double
complements of subsets in ).
Remark 3.21. As we already know, = {, X} is the smallest
collection of subsets of X that constitutesa topology over X. It is
called the trivial topology.
Remark 3.22. Notice that not every topology is an algebra but
every algebra is a topology. However,as a topology every subset A F
is both open and closed. This generally undesireable condition
resultsin a disconnected space.As we will see, the less restrictive
condition of requiring that a topology be closed under finite
intersections
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and double complements gives an orientation to the space.In this
way topology is much more like geometry than algebra. Indeed, it
gives us a different way ofthinking about the geometry of the space
X.
Remark 3.23. Let (X, ) be a topological space and let A X.
Recall that A = X\(Int(A)Ext(A)).A is a closed set since its
complement is a union of open sets and arbitrary unions of open
sets areopen.Int(A) is the largest subset of A that is open (i.e.
in the topology).Ext(A) is the subset of X \A that is open.Cl(A) is
the smallest superset of A that is closed. Hence, its complement is
open.
Theorem 3.24. Let (X, ) be a topological space and let A X. Then
A = Cl(A) \ Int(A).Proof.
A = X \ [Int(A) Ext(A)]= (X \ Int(A)) (X \ Ext(A))= (X \ Int(A))
Cl(A)
Since Int(A) Cl(A) X, Cl(A) (X \ Int(A)) = Cl(A) \ Int(A).Hence,
A = Cl(A) \ Int(A).Remark 3.25. Theorem 3.24 not only gives us
another way to compute the boundary of a subset Abut also gives us
another interpretation of the boundary: the boundary of a set A is
that portion of theclosure of A that completely encloses the subset
of A that is open.
3.4 Examples
Example 3.26. Let I = {,R}.Consider (a, b) R where (a, b) = {x R
: a < x < b}.(a, b) / .R \ (a, b) = (, a] [b,] / .So (a, b)
is neither open nor closed in (R, I).The only open set contained in
(a, b) is . So Int((a, b)) = .The only closed set that contains (a,
b) is R. So Cl((a, b)) = R.Hence,
(a, b) = Cl((a, b)) \ Int((a, b)) = R \ = RExt((a, b)) = R \
Cl((a, b)) = R \ R = R \ (a, b) = R \ R =
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Example 3.27. Let be the indiscrete topology on X and let A be a
proper nonempty subset of X.Subset Interior Closure Boundary
Exterior XA X X X X X
Example 3.28. Let E = {R} {U R : 0 / U}.Consider (a, b) R. There
are two cases.Case 1: 0 (a, b)
(a, b) is not open since 0 (a, b) and (a, b) ( R.(a, b) is
closed since 0 / R \ (a, b) R.So (a, b), where a < 0 < b, is
closed but not open in (R, E).Cl((a, b)) = (a, b)Int((a, b)) = (a,
b) \ {0} = (a, 0) (0, b)Hence,
(a, b) = (a, b) \ [(a, 0) (0, b)] = {0}Ext((a, b)) = R \ (a, b)
= (, a] [b,)R \ (a, b) = R \ {0} = (, 0) (0,)
Case 2: 0 / (a, b). We will consider the subcase a > 0.(a, b)
is open since 0 / (a, b) and (a, b) ( R.(a, b) is not closed since
0 R \ (a, b) and R \ (a, b) ( R.So (a, b), where a > 0, is open
but not closed in (R, E).Int((a, b)) = (a, b)Cl((a, b)) = (a, b)
{0}Hence,
(a, b) = [(a, b) {0}] \ (a, b) = {0}Ext((a, b)) = R \ [(a, b)
{0}] = (, 0) (0, a] [b,]R \ (a, b) = R \ {0} = (, 0) (0,)
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Example 3.29. Let be the excluded point topology on X with p X
the specified point.Let U be open but not closed.Let C be closed
but not open.There are no subsets that are neither open nor
closed.The only subsets that are both open and closed are and
X.
Subset Interior Closure Boundary Exterior XU U U {p} {p} X \ [U
{p}]C C \ {p} C {p} X \ CX X X
4 Lecture 04
4.1 Interior, Closure, Exterior, and Boundary Points
Lemma 4.1. Let (X, ) be a topological space and let A X. Then(1)
Cl(A) = A Int(A)(2) A Cl(A)(3) Int(A) A = (4) Int(A) = Cl(A) \
AProof. (1) A = Cl(A) \ Int(A)So (A) Int(A) = [Cl(A) \ Int(A)]
Int(A) = Cl(A).(2) A A Int(A) = Cl(A)(3) Int(A) A = Int(A) Cl(A) \
Int(A) = (4) A = Cl(A) \ Int(A) Cl(A)So Cl(A) \ A = Cl(A) \ [Cl(A)
\ Int(A)] = Int(A).
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Theorem 4.2. Let (X, ) be a topological space and let A X.
Then(1) x Int(A) iff an open neighborhood U of x such that U A.(2)
x Cl(A) iff open neighborhood U of x, U A 6= .Proof. (1) Let x
Int(A). Then Int(A) is itself an open neighborhood of x where
Int(A) A.Conversely, suppose an open neighborhood U of x such that
U A.Then U is an open set contained in A. So U Int(A).As x U and U
Int(A), x Int(A).(2) We will prove the contrapositive: x / Cl(A)
iff an open neighborhood U of x such that U A = .Suppose x / Cl(A).
Then x X \ Cl(A).Notice that X \ Cl(A) is an open neighborhood of
x.Now since A Cl(A), X \ Cl(A) X \A.So [X \ Cl(A)] A (X \A) A =
.Since [X \ Cl(A)] A, we have [X \ Cl(A)] A = .Conversely, suppose
an open neighborhood U of x such that U A = .Then X \ (U A) = X \ .
Equivalently, X \ U X \A = X. So
A [(X \ U) (X \A)] = X A(A (X \ U)) = A
A (X \ U) = A
Hence, A (X \ U).Now since U is open, X \ U is closed. That is,
X \ U is a closed set containing A.So Cl(A) X \ U . Equivalently, U
X \ Cl(A).As x U and U X \ Cl(A), x X \ Cl(A).Thus x / Cl(A).Remark
4.3. Theorem 4.2 states the following:(1) A point is in Int(A) iff
it has an open neighborhood contained in A.(2) A point is in Cl(A)
iff every open neighborhood of it intersects A. That is, every open
neighborhoodof it contains a point of A.
Remark 4.4. (Interior Points)x Int(A) an open neighborhood U of
x such that U Ax / Int(A) open nighborhood U of x, U 6 A
Remark 4.5. (Closure Points)x Cl(A) open neighborhood U of x, U
A 6= x / Cl(A) an open neighborhood U of x such that U A =
Remark 4.6. A closure point is more commonly called an adherent
point.
Theorem 4.7. Let (X, ) be a topological space and let A X.
Then(1) Int(X \A) = X \ Cl(A).(2) Cl(X \A) = X \ Int(A).Proof. (1)
Since A Cl(A), we have X \ Cl(A) X \A.As Cl(A) is closed, X \ Cl(A)
is open. So X \ Cl(A) is an open set contained in X \A.Hence, X \
Cl(A) Int(X \A).
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Conversely, let x Int(X \A). Notice that Int(X \A) is open
neighborhood of x.Since Int(X \A) X \A, Int(X \A) A (X \A) A =
.Since Int(X \A) A, Int(X \A) A = .So x / Cl(A). Thus x X \
Cl(A).Hence, Int(X \A) X \ Cl(A). Int(X \A) = X \ Cl(A).(2) Since
Int(A) A, we have X \A X \ Int(A).As Int(A) is open, X \ Int(A) is
closed. So X \ Int(A) is a closed set containing X \A.Hence, Cl(X
\A) X \ Int(A).Conversely, let x X \ Int(A). We need to show x Cl(X
\A).Suppose to the contrary that x / Cl(X \A). Then x X \ Cl(X
\A).By part (1), X \ Cl(X \A) = Int(X \ (X \A)) = Int(A). That is x
Int(A).But this contradicts the fact that x X \ Int(A). So, by
contradiction, x Cl(X \A).Hence, X \ Int(A) Cl(X \A). Cl(X \A) = X
\ Int(A).Corollary 4.8. Let (X, ) be a topological space and let A
X. Then(1)Ext(A) = Int(X \A).(2) A = Cl(A) Cl(X \A).Proof. (1)
Ext(A) = X \ Cl(A)= Int(X \A)
(2)
A = Cl(A) \ Int(A)= Cl(A) X \ Int(A)= Cl(A) Cl(X \A)
Remark 4.9. Notice that by part (1) of corollary 4.8, Ext(A) is
the union of all open sets contained inX \ A and is therefore the
largest (i.e. most inclusive) open set contained in X \ A. So if U
is an openset such that U X \A, then U Ext(A).Theorem 4.10. Let (X,
) be a topological space, let A X, and let x X. Then(1) x Ext(A)
iff an open neighborhood U of x such that U X \A.(2) x A iff open
neighborhood U of x, U A 6= and U (X \A) 6= .Proof. (1) Let x
Ext(A). Then Ext(A) is an open neighborhood of x where Ext(A) X \
(A).Conversely, suppose an open neighborhood U of x such that U X
\A.Then as U is an open set contained in X \A, U Ext(A).Since x U
and U Ext(A), x Ext(A).(2) Let x A. Then x Cl(A) Cl(X \A).That is,
x Cl(A) and x Cl(X \A).As x Cl(A), open neighborhood U of x, U A 6=
.
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As x Cl(X \A), open neighborhood U of x, U (X \A) 6=
.Conversely, suppose open neighborhood U of x, U A 6= and U (X \A)
6= .Then as it is true that open neighborhood U of x, U A 6= , x
Cl(A).Similarly, since open neighborhood U of x, U (X \A) 6= , x
CL(X \A).That is, x Cl(A) Cl(X \A) = A.Remark 4.11. Theorem 4.10
states the following:(1) A point is in Ext(A) iff it has an open
neighborhood contained in X \A.(2) A point is on A iff every open
neighborhood ot it intersects A and X \ A. That is, every
openneighborhood of it contains a point of A and X \A.Remark 4.12.
(Exterior Points)x Ext(A) an open neighborhood U of x such that U X
\Ax / Ext(A) open neighborhood U of x, U 6 X \ARemark 4.13.
(Boundary Points)x A open neighborhood U of x, U A 6= and U (X \A)
6= x / A an open neighborhood U of x such that U A = or U (X \A) =
Remark 4.14. Let (X, ) be a topological space and let A X. Then
X = A (X \ A)= A (X \ [X \ (Int(A) Ext(A)])= Int(A) A Ext(A)
Exercise 4.15. We have already demonstrated that Int(A) A =
.Show Int(A) Ext(A) = and A Ext(A) = .
4.2 Cluster Points and Isolated Points
Definition 4.16. Let (X, ) be a topological space and let A X.
Then a point x X is a clusterpoint of A iff every open neighborhood
U of x intersects A in at least one point other than x itself.That
is, x is a cluster point of A iff open neighborhood U of x, U (A \
{x}) 6= .Remark 4.17. In light of Theorem 4.2, we have x is a
cluster point of A iff x Cl(A \ {x}).Remark 4.18. A cluster point
is also called a limit point or accumulation point.
Definition 4.19. Let (X, ) be a topological space, let A X, and
let U be an open neighborhood ofthe point x X. If U is not a
singleton, then U \ {x} is called a deleted neighborhood of the
pointx.
Exercise 4.20. Prove that a point x is a cluster point of a set
A iff every deleted neighborhood of xintersects A.
Definition 4.21. The set of all cluster points of a set A is
called the derived set of A and is denotedby A.
Example 4.22. Let X be a countable set and let D =P(X). Then A
X, A = .
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Proof. In the discrete topology on X, every singleton set is
open. So {x} is an open neighborhood of xwhere {x} (A \ {x}) = for
any subset A X. Hence, x is not a cluster point of the set A. Since
thisis true x X, A has no cluster points. That is, A = .Theorem
4.23. Let (X, ) be a topological space and let A X. Then Cl(A) = A
A.Proof. Let x Cl(A). Since A Cl(A), we must consider two cases: x
A or x / A.Suppose x A. Then x A A.Now suppose x / A. Then since x
Cl(A), open neighborhood U of x, U A 6= .Further, as x / A, U
intersects A in a point different from x. That is, U (A \ {x}) 6=
.So x is a cluster point of A. That is, x A. Thus, x A A.Hence, in
either case, Cl(A) A A.Conversely, let x A A. Then x A or x
A.Suppose x A. Then as A Cl(A), x Cl(A).Now suppose x A. Then x is
a cluster point of A and so open neighborhood U of x, U(A\{x}) 6=
.That is, y X such that y U (A \ {x}).Since A \ {x} A, we have U (A
\ {x}) U A.As y U (A \ {x}) U A, we have U A 6= . So x Cl(A).Hence,
in either case, A A Cl(A). Cl(A) = A A.Corollary 4.24. Let (X, ) be
a topological space. A subset A X is closed iff A A.Proof. Suppose
A is closed. Then A = Cl(A) = A A.So A A.Conversely, suppose A A.
Then A A A A = A.That is, Cl(A) A. Since Cl(A) = A A, we have A =
Cl(A).Hence A is closed.
Example 4.25. Let (X, ) be a topological space and let A X. Then
Cl(A) = A A = A A.However, in general, A 6= A.We will give a
counterexample in a Sierpinski space:Let X = {a, b} with = {, {a},
X}.Consider A = {b}. A is closed since X \A = {a, b} \ {b} = {a}
.So Cl(A) = A, Int(A) = , and A = A \ = A.Notice that the only open
neighborhood of b is X.Since A X = {b}, 6 an open neighborhood of b
that intersects A in a point different from b.So b is not a cluster
point of A and thus b / A.Now {a} is an open neighborhood of a such
that {a} A = .So a / Cl(A) = A A and thus a / A.Hence, A =
.Clearly, A 6= A.Definition 4.26. Let (X, ) be a topological space
and let A X. A point x A is an isolated pointof A iff an open
neighborhood U of x such that U A = {x}.Example 4.27. Let X be a
countable set with D = P(X) and let A X. Then every point in A isan
isolated point of A.
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Proof. In the discrete topology on X, every singleton set is
open. So x A, {x} is an open neighbor-hood of x such that {x} A =
{x}. That is, every point in A is an isolated point of A.Theorem
4.28. Let (X, ) be a topological space and let A X. A point x A is
either a clusterpoint of A or an isolated point of A but not
both.
Proof. Let x A and suppose x is a cluster point of A.Then open
neighborhood U of x, U (A \ {x}) 6= .So y X such that y U (A \
{x}).Since A \ {x} A, U (A \ {x}) U A.So y U A open neighborhood U
of x.Thus 6 an open neighborhood U of x such that U A = {x}.Hence x
is not an isolated point of A.Now suppose x is not a cluster point
of A.Then an open neighborhood U of x such that U (A \ {x}) = .
So
[U (A \ {x}] {x} = {x}U A = {x}
Hence x is an isolated point of A.
Example 4.29. Let I = {,R}.Consider [a, b] R.[a, b] is neither
open nor closed in this space.Notice r R, the only open
neighborhood of r is R.Since r R, R ([a, b] \ {r}) 6= , every point
r R is a cluster point of [a, b].That is, [a, b] = R. So [a, b] has
no isolated points.Cl([a, b]) = R and [a, b] = R.Notice that in
this case [a, b] = [a, b].
Exercise 4.30. Let E = {R} {U R : 0 / U}.Consider [a, b] where a
> 0.Determine the cluster points and isolated points of [a,
b].
5 Lecture 05
5.1 The Isolated Points of a Space
Definition 5.1. Let A and B be sets such that A 6 B. The
relative complement of A in B is theset B \A = {x B : x / A}. That
is, B \A = B \ (B A).Remark 5.2. If A 6 B, the relative complement
of A in B is the absolute complement of A B in B.Theorem 5.3. Let
(X, ) be a topological space and let A X. A point x X is a cluster
point of Aiff deleted neighborhood U \ {x} of x, (U \ {x}) A 6=
.Proof. Suppose x is a cluster point of A.Then open neighborhood U
of x, U (A \ {x}) 6= .So y 6= x X such that y U (A \ {x}).
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That is, y U and y A \ {x}.As A \ {x} A, y A.As y 6= x, y U \
{x}.So y (U \ {x}) A.Hence, (U \ {x}) A 6= .Conversely, suppose
deleted neighborhood U \ {x} of x, (U \ {x}) A 6= .Then y 6= x X
such that y (U \ {x}) A.That is, y U \ {x} and y A.As U \ {x} U , y
U .As y 6= x, y A \ {x}.So y U (A \ {x}).That is, U (A \ {x}) 6=
.Hence, x is a cluster point of A.
Example 5.4. Let E = {R} {U R : 0 / U}.Consider [a, b] where a
> 0.Notice that any subset that contains 0 is closed.So the only
open neighborhood of 0 is R.Clearly, R ([a, b] \ {0}) 6= or,
equivalently, (R \ {0}) [a, b] 6= .Hence 0 is a cluster point of
[a, b].Notice r [a, b], {r} is an open neighborhood of r such that
{r} [a, b] = {r}.That is, every point in [a, b] is an isolated
point of [a, b].Cl([a, b]) = [a, b] {0}.Notice that in this case
[a, b] = [a, b] = {0}.Example 5.5. Let |X| 2, let I = {, X}, and
let x X.The only open neighborhood of the point x is the entire
space X.As |X| 2, X (X \ {x}) 6= since X must contain at least one
point distinct from x.So x is a cluster point of X.As the point x
was chosen abitrarily, every point x X is a cluster point of
X.Hence there are no isolated points of X.
Example 5.6. Let |X| 2 and let P = {} {U X : p U}.Notice {p} is
an open neighborhood of p such that {p} X = {p}.Hence p is an
isolated point of X.Now let x 6= p X. Then every open neighborhood
U of x contains the point p.That is, p U (X \ {x}).Thus U (X \ {x})
6= .Hence every point x 6= p X is a cluster point of X.Example 5.7.
Let |X| 2 and let E = {X} {U X : p / U}.Notice that the only open
neighborhood of p is the entire space X.As |X| 2, X \ {p} 6= .So X
(X \ {p}) 6= .Hence, p is a cluster point of X.Now let x 6= p
X.Then {x} is an open neighborhood of x such that {x} X = {x}.Hence
every point x 6= p X is an isolated point in X.
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Example 5.8. Let X be a countable set and let D =P(X).Then x X,
{x} is an open neighborhood of the point x such that {x} X =
{x}.Hence every point x X is an isolated point of X.Remark 5.9. As
we will see, the number of isolated points in a given space is a
topological invariant.That is, any space that is topologically
equivalent (i.e. homeomorphic) to a given space must have thesame
number of isolated points.
5.2 Introduction to Connectedness
Definition 5.10. Let X be any set. A partition of the set X is
any collection {Xi : i I} of nonemptysubsets of X such that(1) X
=
iI Xi and
(2) Xi Xj = i 6= j.Lemma 5.11. Let A,B,C and D be sets such that
A B and C D. Then A C B D.Proof. Let x A C. Then x A and x C.As x A
and A B, x B.As x C and C D, x D.That is, x B D.Hence A C B
D.Example 5.12. Let (X, ) be a topological space and let A X.We
have already demonstrated that A Int(A) = .Now Ext(A) A = Ext(A) [X
\ (Int(A) Ext(A))] = andExt(A) Int(A) = Int(X \A) Int(A).As Int(X
\A) X \A and Int(A) A, Int(X \A) Int(A) (X \A) A = .That is, Ext(A)
Int(A) .Since Ext(A) Int(A), Ext(A) Int(A) = .
Remark 5.13. Let (X, ) be a topological space and let A X.Then X
= A Int(A) Ext(A).So if A 6= , Int(A) 6= , and Ext(A) 6= , then {A,
Int(A), Ext(A)} is a partition of X.Definition 5.14. Let (X, ) be a
topological space. A separation of X is a pair U, V of
nonemptyproper open subsets of X that form a partition of X.
Definition 5.15. A topological space X is connected iff 6 a
separation of X. Otherwise, the space isdisconnected.
Theorem 5.16. A topological space X is connected iff the only
subsets of X that are both open andclosed are and X.Proof. We will
prove the contrapositive: A topological space X is disconnected iff
a nonempty propersubset of X that is both open and closed.Suppose X
is disconnected.Then nonempty, proper, open subsets U and V that
partition X.That is, X = U V and U V = .
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Notice X \ U = V is open. So U is both open and closed.Hence, a
nonempty proper subset U ( X that is both open and
closed.Conversely, suppose U is a nonempty proper subset of X that
is both open and closed.Then as U is closed, X \ U is open.
Further, X \ U 6= .Since X = U X \ U , U and X \ U is a separation
of X.Hence X is disconnected.
Remark 5.17. As we will see, connectedness is also a topological
invariant of a given topological space.
Example 5.18. We have already demonstrated that there are no
nonempty proper subsets that areboth open and closed in either the
particualr point topology or the excluded point topology. So
spaceswith these topologies are connected spaces.
Remark 5.19. Notice that a Sierpinski space is indeed the
smallest nontrivial connected space as itis a two point set with
either the particular point topology or the excluded point
topology. In eithertopology, the space has one cluster point and
one isolated point.
Example 5.20. Let X be a countable set and let D = P(X). Then
every nonempty proper subsetU ( X is both open and closed. Hence a
discrete topological space is completely disconnected.
5.3 Introduction to Separation Conditions
Definition 5.21. (T0 Space)A topological space X for which an
open neigborhood for at leat one point in each pair x1, x2
ofdistinct points not containing the other is called a T0, or
Kolmogorov, space. That is, x1 6= x2 X, open neighborhoods U1 of x1
and U2 of x2 such that either x2 / U1 or x1 / U2.Definition 5.22.
(T1 Space)A topological space X for which open neigborhoods for
both points in each pair x1, x2 of distinctpoints not containing
the other is called a T1, or Frechet, space. That is, x1 6= x2 X,
openneighborhoods U1 of x1 and U2 of x2 such that x2 / U1 and x1 /
U2.Remark 5.23. Notice that not every T0 space is T1. However,
every T1 space is T0. The contrapositiveis also true. That is, if a
space is not T0, then it is not T1.
Example 5.24. Let P = {} {U X : p U}.Notice that every open set
contains the point p.So 6 an open neighborhood U of x 6= p X such
that p / U .Hence the space X is not T1.However, pair of distinct
points x1 6= x2 6= p X, {x1, p} is an open neighborhood of x1 not
containingx2 and {x2, p} is an open neighborhood of x2 not
containing x1.Further, x 6= p X, {p} is an open neighborhood of p
not containing x.Hence the space is T0.
Theorem 5.25. A topological space X is T1 iff every singleton
set is closed.
Proof. Let X be a T1 space.Then x 6= y X an open neigborhood Uy
of y such that x / Uy.Let {Ui : i I} be the collection of all open
neighborhoods of all points y 6= x (i.e. all points y X \{x})such
that x / Ui.
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TheniI Ui is open as it is a union of open sets.
Further,iI Ui = X \ {x}.
That is, {x} is closed.This is true x X.Hence, every singleton
set in a T1 space is closed.Conversely, suppose that all singleton
sets are closed.Let x, y X such that x 6= y.Then X \ {x} is an open
neighborhood of y not containing x.Similarly, X \ {y} is an open
neighborhood of x not containing y.This is true x 6= y X.Hence the
space X is T1.
Example 5.26. Let E = {X} {U X : p / U}.Then x 6= p X, {x} is
open but not closed.Hence the space X is not T1.
Exercise 5.27. Show that a topological space X with the excluded
point topology is T0.
Example 5.28. Let I = {, X}.Then x X, the only open neighborhood
of x is the entire space X.So 6 an open neigborhood that does not
contain any given point.Hence the space X is not T0 and therefore
is also not T1.
Corollary 5.29. Every finite point set in a T1 space is
closed.
Proof. Let X be a T1 space and let F = {x1, . . . , xn} be a
finite point set in the space X.Then F =
ni=1{xi} where {xi} is closed i {1, . . . , n}.
Hence F is closed.
Definition 5.30. (T2 Space)A topological space X for which
disjoint open neigborhoods for both points in each pair x1, x2
ofdistinct points is called a T2, or Hausdorff, space. That is, x1
6= x2 X, open neighborhoods U1of x1 and U2 of x2 such that U1 U2 =
.Remark 5.31. Some definitions of a Hausdorff space allow for any
neighborhoods of the points. Thatis, the neighborhoods need not
necessarily be open.
Remark 5.32. A T1 space is not necessarily Hausdorff since it is
not required that the neighborhoodsbe disjoint. However, every
Hausdorff space is T1. The contrapositive is also true. That is, if
a space isnot T1, then it is not Hausdorff.
Example 5.33. Let X be a countable space and let D =P(X).Then
every singleton set is both open and closed.So x 6= y X, {x} is an
open neighborhood of x and {y} is an open neighborhood of y such
that{x} {y} = .Hence the space X is T2 (Hausdorff).
Remark 5.34. In a Hausdorff space, every distinct pair of points
can be separated by disjoint neigh-borhoods (not necessarily
open).Today, we see this as a separation condition that may or may
not be satisfied by a given space. Oneof the earliest definitions
of a topological space included what we now consider the Hausdorff
condition.This definition was stated by none other than Felix
Hausdorff.
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Remark 5.35. In a T1 or Frechet space, every distinct pair of
points can be separated by open neigh-borhoods that are not
necessarily disjoint.
Remark 5.36. In a T0 or Kolmogorov space, at least one of the
points in any distinct pair of pointscan be separated from the
other by an open neighborhood.
Remark 5.37. In the early days of topology, these separation
conditions were taken as axioms. TheT is for the German word
Trennungsaxiom, which means separation axiom.The trend in modern
mathematical literature is to move away from the use of the T terms
in favor ofmore descriptive terms that do not refer to an axiom
since we no longer take these separation conditionsas axioms. This
is especially true for the Hausdorff condition forward.
Remark 5.38. As we will see, the separation condition statisfied
by a given topological space is also atopological invariant.
Remark 5.39. The topological invariants which we have explored
in this lecture are summarized below:
Indiscrete Topology PPT EPT Discrete Topology
# of Isolated Points 0 1 |X| 1 |X|Connectedness Connected
Connected Connected Disconnected
Separation Condition Satisfied None T0 T0 T2
Exercise 5.40. Determine the subsets A X for which Cl(A) = X in
the(1) indiscrete topology(2) particular point topology (PPT)(3)
excluded point topology (EPT)(4) discrete topologyFor each topology
assume |X| 2. For the discrete topology assume |X| 0.
6 Lecture 06
6.1 Dense Subsets
Example 6.1. Let E = {X} {U X : p / U}. Then the space X is
T0.Proof. Notice that the only open neighborhood of the excluded
point p is the entire space X. So 6 anopen neighborhood U of p not
containing a point x distinct from p. Hence the space X is not
T1.However, x 6= p X, {x} is an open neighborhood of x such that p
/ {x}.Further, x 6= y 6= p X, {x} is an open neighborhood of x and
{y} is an open neighborhood of y suchthat x / {y} and y / {x}.Hence
the space X is T0, or Kolmogorov.
Definition 6.2. Let (X, ) be a topological space. A proper
subset D ( X is said to be dense, oreverywhere dense, in X iff
Cl(D) = X.
Example 6.3. Let I = {, X} and let A be a nonempty proper subset
A ( X.Subset Interior Closure A XX X X
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So every nonempty proper subset A ( X is dense in X.
Example 6.4. Let |X| 0, let D =P(X), and let A be a nonempty
proper subset A ( X.Subset Interior Closure A A AX X X
So 6 a proper subset U ( X such that Cl(U) = X. Hence there are
no dense subsets in a discretetopological space.
Example 6.5. Let P = {} {U X : p U} and let U and C be nonempty
proper subsets of Xsuch that U,X \ C P .
Subset Interior Closure U U XC CX X X
So every nonempty proper open subset U ( X is dense in X.
Example 6.6. Let E = {X} {U X : p / U} and let U and C be
nonempty proper subsets of Xsuch that U,X \ C E .
Subset Interior Closure U U U {p}C C \ {p} CX X X
So the only proper subset that is dense in X is the open set X \
{p}.Theorem 6.7. Let (X, ) be a topological space. A proper subset
D ( X is dense in X iff everynonempty open subset U X intersects
D.Proof. Suppose D is dense in X.Then Cl(D) = X.So x X, x
Cl(D).Hence open neighborhood U of x, U D 6= .That is, every
nonempty open seubset U X intersects D.Conversely, suppose that
every nonempty open subset U X intersects D.Then x X, open
neighborhood U of x, U D 6= .That is, every point x X is in
Cl(D).Hence Cl(D) = X.Thus D is dense in X.
Remark 6.8. Notice that theorem 6.7 states that if D ( X is
dense in the space X, then every nonemptyopen subset U X contains
at least one point x D.
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6.2 Nowhere Dense Subsets
Definition 6.9. (Nowhere Dense Set)Let (X, ) be a topological
space. A nonempty subset A X is nowhere dense in X iff Int(Cl(A)) =
.That is, a subset A X is nowhere dense in X if there are no
nonempty open sets contained in Cl(A).Exercise 6.10. Determine the
validity of the following conjecture: if A X is nowhere dense in
X,then nonempty proper open subset U ( X, U A = .Remark 6.11. The
complement of a dense subset is not always a nowhere dense subset.
For example,a nonempty proper subset A and X \A are both dense in X
with the indiscrete topology. However, thecomplement of a nowhere
dense subset is a dense subset.
Theorem 6.12. Let (X, ) be a topological space. If A X is
nowhere dense in X, then X \A is densein X. The converse is true if
X \A is open.Proof. Suppose A is nowhere dense in X. Then
Int(Cl(A)) = . So 6 a nonempty open set containedin Cl(A). As
Int(A) A Cl(A), Int(A) = . Since Int(A) = X \ Cl(X \ A) = , Cl(X \
A) = X.Hence, X \A is dense in X.Now suppose that X \ A is both
open and dense in X. Then since X \ A is open, A is closed. SinceX
\ A is dense in X, Cl(X \ A) = X. So A = Cl(A) Cl(X \ A) = A X = A.
Now sinceA = Cl(A) \ Int(A) = A \ Int(A) = A, Int(A) = . So
Int(Cl(A)) = Int(A) = . Hence, A is nowheredense in X.
Theorem 6.13. Subsets of nowhere dense sets are nowhere
dense.
Proof. Let A be nowhere dense and let B A. As B A Cl(A), Cl(B)
Cl(A).So Int(Cl(B)) Cl(B) Cl(A). Hence, Int(Cl(B)) Int(Cl(A)) = as
A is nowhere dense. ThusInt(Cl(B)) = . That is, B is nowhere
dense.Example 6.14. Let I = {, X} and let A ( X such that A 6=
.
Subset Closure Int(Closure) A X XX X X
So 6 a nonempty proper subset A ( X such that Int(Cl(A)) = .
Hence there are no nowhere densesubsets in an indiscrete space.
Example 6.15. Let |X| 0, let D =P(X), and let A ( X such that A
6= .Subset Closure Int(Closure) A A AX X X
So 6 a nonempty proper subset A ( X such that Int(Cl(A)) = .
Hence there are no nowhere densesubsets in a discrete space.
Example 6.16. Let P = {} {U X : p U} and let U and C be nonempty
proper subsets of Xsuch that U,X \ C P .
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Subset Closure Int(Closure) U X XC C X X X
So every nonempty proper closed subset C ( X is nowhere dense in
X.
Example 6.17. Let E = {X} {U X : p / U} and let U and C be
nonempty proper subsets of Xsuch that U,X \ C E .
Subset Closure Int(Closure) U U {p} UC C C \ {p}X X X
So the only proper subset that is nowhere dense in X is the
closed set {p}.
6.3 Separable Spaces
Definition 6.18. A topological space X is separable iff it
contains a countable dense subset.
Remark 6.19. The term separable is known to have originated from
Maurice Frechet but it is notclear why he chose this term to denote
a space with a countable dense subset. In the collected works
ofFelix Hausdorff, Hausdorff comments that the term was well
established but that he was unsure of thereason for the choice of
the term.
Remark 6.20. The condition of separability is related to the
countability of a space. In fact, separabilityis considered a
countability condition that may or may not be satisfied by a given
space. As we will see,the countability condition satisfied by a
given space is another topological invariant.
Example 6.21. In the indiscrete topological space X, every
singleton set {x} is dense in X. So afinite dense subset in X.
Hence an indiscret space is separable.
Example 6.22. In the discrete topological space X, there are no
dense subsets. Hence a discretetopological space is not
separable.
Example 6.23. Let P = {} {U X : p U}. Then the singleton set {p}
is dense in X. Hence atopological space with the particular point
topology is separable.
Example 6.24. Let E = {X} {U X : p / U}. The only dense subset
in this space is X \ {p}. So atopological space with the excluded
point topology is separable only if the underlying set is
countable.
Remark 6.25. The topological invariants which we have explored
so far are summarized below:
Indiscrete Topology PPT EPT Discrete Topology
# of Isolated Points 0 1 |X| 1 |X|Connectedness Connected
Connected Connected Disconnected
Separation Condition None T0 T0 T2Countability Condition
Separable Separable Separable only if |X| 0 Not Separable
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6.4 Equivalent Statements About Open and Closed Sets
Remark 6.26. (Open Sets)A subset A X is open in a topological
space (X, ) iff
(1) A (2) A = Int(A) = Cl(A) \ A(3) A A = (4) x A an open
neighborhood U of x such that U A(5) X \A is closed
Remark 6.27. (Closed Sets)A subset A X is closed in a
topological space (X, ) iff
(1) X \A (2) A = Cl(A) = A A = A A(3) A A(4) A A(5) x A open
neighborhood U of x, U A 6= (6) X \A is open
Theorem 6.28. Let (X, ) be a topologial space and let A X. Then
A = iff A is both open andclosed.
Proof. Suppose A = .Then Cl(A) \ Int(A) = .As Int(A) Cl(A),
Int(A) = Cl(A) \ = Cl(A).Since Int(A) A Cl(A) and Int(A) = Cl(A), A
= Int(A) and A = Cl(A).That is, A is both open and
closed.Conversely, suppose that A is both open and closed.The A =
Int(A) and A = Cl(A).So A = Cl(A) \ Int(A) = A \A = .Remark 6.29.
(Properties of A)Let (X, ) be a topological space and let A X. The
following are properties of the boundary of A.
(1) A = X \ (Int(A) Ext(A))(2) A = Cl(A) \ Int(A)(3) A = Cl(A)
Cl(X \A)(4) A Cl(A)(5) A Int(A) = (6) A Int(A) = Cl(A)(7) A is
closed.(8) A = iff A is both open and closed.
6.5 More on Cluster Points and Isolated Points
Theorem 6.30. Let X be a T1 (Frechet) space and let A X. A point
x X is a cluster point of Aiff every open neighborhood of x
contains infinitely many points of A.
Proof. Let x X be a cluster point of A. We need to show that
every open neighborhood of x containsinfinitely many points of
A.
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Suppose to the contrary that an open neighborhood U of x that
contains only a finite number of pointsof A.Then U A \ {x} = {x1, .
. . , xm} is closed as every finite point set is closed in a T1
space.So X \ {x1, . . . , xm} is open and U X \ {x1, . . . , xm} is
open as it is a finite intersection of open sets.Further U X \{x1,
. . . , xm} is an open neighborhood of x such that (U X \{x1, . . .
, xm})A\{x} = .But this contradicts the fact that x is a cluster
point of A.Hence, by contradiction, every open neighborhood of x
contains infinitely many points of A.Conversely, suppose that every
open neighborhood U of x contains infinitely many points of A.Then
U A \ {x} also contains infinitely many points of A.That is, open
neighborhood U of x, U A \ {x} 6= .Hence x is a cluster point of
A.
Theorem 6.31. Let (X, ) be a topological space where |X| 2. Then
x X is an isolated point ofX iff {x} is open.Proof. Suppose x X is
an isolated point of X.Then an open neighborhood U of x such that U
X = {x}.So U X \ {x} = .As |X| 2, X \ {x} 6= .Hence U =
{x}.Conversely, suppose {x} is open.Then {x} is an open
neighborhood of x such that {x} X = {x}.Hence, x is an isolated
point of X.
Exercise 6.32. Let B = {U X : p U} and let U1, U2 B.Show that if
x U1 and x U2, then U3 B such that x U3 U1 U2.Consider two cases: x
= p and x 6= p.
7 Lecture 07
7.1 Basis for a Topology
Remark 7.1. Recall that a proper subset A of a topological space
X is dense in X iff every nonemptyopen subset U X contains at least
one point x A.Example 7.2. We give a counterexample to the
conjecture: If A X is nowhere dense in X, then nonempty proper open
subset U X, U A = .Counterexample: Let |X| > 2, let P = {} {U X
: p U}, and let x 6= p X.Then {x} is nowhere dense in X.However,
{x, p} is a nonempty proper open subset of X such that {x, p} {x} =
{x} 6= .Remark 7.3. The previous conjecture asks If A is nowhere
dense in a space X, is A a set such that everyopen set distinct
from the entire space X has no points in common with it? From the
counterexample,we see that the answer is No.
Remark 7.4. The contrapositive of theorem 6.7 is also a true
statement: A proper subset D ( X isnot dense in X iff at least one
open subset U X such that U D = .Let A be nowhere dense in X.Then
Int(Cl(A)) = .
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So Cl(A) 6= X.Hence A is not dense in X.However, a nowhere dense
set is more than a set which is simply not dense in X. It is a set
for whichthe only open set contained in its closure is the empty
set.In addition, we know that the complement of a nowhere dense set
is always a dense set, while thecomplement of a dense set is
nowhere dense only if it is open.
Remark 7.5. Let {Ai : i I} be an indexed family of subsets of
X.Recall that the arbitrary union is
iI Ai = {x : x Ai for at least one i I}.
If the indexing set I = Z+, we have a countable unioniZ+ Ai
=
i=1Ai = {x : x Ai for at least one i Z+}
If the indexing set I = {1, 2, . . . , n}, we have a finite
unioniI Ai =
ni=1Ai = A1 A2 An
There are two degenerate cases:(1) Suppose I = {1}. Then iI Ai =
1i=1Ai = A1.(2) Suppose I = . Then iAi = {x : x Ai for at least one
i }.Since 6 i , x / Ai for any index i. So
iAi = .
The degenerate case (2) is called the empty union.
Definition 7.6. (Basis)Let (X, ) be a topological space. A basis
for is a subcollection B such that every member of is a (possibly
empty) union of some members of B. That is, a basis B is a
collection {Bi : i I} ofopen subsets of X such that open subset U X
J I such that U = jJ Bj .Remark 7.7. Clearly, can be its own
basis.
Remark 7.8. A basis is sometimes called a base. The elements of
a basis are called basis elements,base elements, or basic open
sets.
Lemma 7.9. Let (X, ) be a topological space and let A X.(1) If A
Int(A), then A is open.(2) If Cl(A) A, then A is closed.Proof. (1)
Suppose A X such that A Int(A).Then since Int(A) A, A =
Int(A).Hence A is open.(2) Suppose A X such that Cl(A) A.Then since
A Cl(A), A = Cl(A).Hence A is closed.
Theorem 7.10. Let (X, ) be a topological space and let B = {Bi :
i I} be a basis for . ThenU X is open iff x U Bi B such that x Bi U
.Proof. Suppose U X is open. Then U = jJ Bj for some J I. We
consider two cases.Case 1: J 6= .Let x U = jJ Bj .Then x Bj for at
least one index j J .Further, Bj
jJ Bj = U j J .
So x Bj U .
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That is, x U Bi B such that x Bi U .Case 2: J = .Then U =
jBj = and the result follows vacuously.
Conversely, suppose U X satisfying the condition that x U Bi B
such that x Bi U .Let x U .Then Bi B such that x Bi U .As Bi U X, X
\ U X \Bi where X \Bi is closed.So Cl(X \ U) X \Bi.As x Bi, x / X
\Bi.So x / Cl(X \ U) = X \ Int(U).Hence x Int(U).Thus U Int(U).That
is, U is open.
Remark 7.11. In any topology, the entire set X is open. Hence,
by Theorem 7.10, x X Bi B,where B is a basis for the topology, such
that x Bi X. That is, x X, there is at least one basiselement
containing x.
Corollary 7.12. Let (X, ) be a topological space and let B = {Bi
: i I} be a basis for . ThenX =
iI Bi.
Proof. Since x X at least one Bi B such that x Bi X, x iI Bi x
X. Hence,
X iI Bi. The reverse inclusion is obvious since Bi X i I. Thus X
= iI Bi.Definition 7.13. A collection A = {Ai : i I} of subsets of
a space X is said to cover X, or to be acovering of X, iff X =
iI Ai. It is called an open covering of X if its elements are
open subsets of
X.
Remark 7.14. Notice that a basis for a given topological space X
is an open covering of X. This isone importnat property of a
basis.
Corollary 7.15. Let (X, ) be a topological space, let B = {Bi :
i I} be a basis for , and let x X.If B1, B2 B such that x B1 B2,
then B3 B such that x B3 B1 B2.Proof. Each set Bi B is open. So B1
B2 is open as it is a finite intersection of open sets. Hence,
byTheorem 7.10, x B1 B2 at least one basis element, call it B3,
such that x B3 B1 B2.Example 7.16. Let B = {U X : p U} and let U1,
U2 B.Suppose x U1 and x U2.Then x U1 U2.We consider two cases.Case
1: x = pThen {p} B such that p {p} U1 U2 since p U1 and p U2.Case
2: x 6= pThen {x, p} B such that x {x, p} U1 U2 since x, p U1 and
x, p U2.So if x U1 U2, where U1, U2 B, then there exists a third
element in B, call it U3, such thatx U3 U1 U2.
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Theorem 7.17. Let (X, ) be a topological space. Suppose that B
is a collection of open subsets of Xsuch that open U X and x U ,
there is an Bx B such that x Bx U . Then B is a basis for .
Proof. We need to show that every open subset U X is a (possibly
empty) union of some elements ofB.If U = , then U is the empty
union of elements in B.Now suppose U 6= .Then x U Bx B such that x
Bx U .Since Bx
xU Bx x U , x
xU Bx.
That is, U xU Bx.Now Bx
xU Bx, Bx U .
SoxU Bx U .
Hence U =xU Bx.
B is a basis for by definition.
Remark 7.18. As an alternate proof, notice that
xU
Bx =xU
Bx
= Int(U)
As U is open, U = Int(U) =xU Bx.
Definition 7.19. Let X be a set and let B = {Bi : i I} be a
collection of subsets of X such that(1) x X, x Bi for at least one
index i I(2) If x B1 B2, then B3 B such that x B3 B1 B2.
Then the topology generated byB is defined as follows: A subset
U X is open iff x U Bi Bsuch that x Bi U .Proof. (1) Notice
satisfies the condition for openness vacuously. Now x X Bi B such
thatx Bi X. Hence, X .(2) Let {U : J} be an indexed family of sets
in and let V =
J U.
Suppose x V .Then x U for at least one J where U is open.So Bi B
such that x Bi U.Hence x V Bi B such that x Bi V .Thus V =
J U .
(3) Suppose U1, U2 and let x U1 U2.Then x U1 and x U2.As x U1,
B1 B such that x B1 U1.As x U2, B2 B such that x B2 U2.That is, x
B1 B2 U1 U2.So B3 B such that x B3 B1 B2 U1 U2.Hence U1 U2 .Let U1,
. . . , Un . We will show by induction that
n=1 U .
The case n = 1 is the statement U1 , which is true.Now suppose
the case n 1 is true.
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That is, suppose U1 Un1 .Then
n=1 U = (U1 Un1) Un is the intersection of two open sets.
Hencen=1 U .
Exercise 7.20. Let B = {U X : p U}.(1) Show B satisfies the two
conditions for a topology generated by a basis.(2) Show that the
topology generated by B is the particular point topology on X.
Exercise 7.21. Let (X, ) be a topological space and let B = {Bi
: i I} be a collection of opensubsets of X such that(1) X =
iI Bi.
(2) If x B1 B2, then B3 B such that x B3 B1 B2.Show that is
finer than the topology generated by B.
7.2 The Digital Line Topology
Definition 7.22. (Digital Line Topology)Let X = Z and n Z define
Bn by
Bn =
{{n} if n is odd{n 1, n, n+ 1} if n is even
Then B = {Bn : n Z} generates a topology on Z called the digital
line topology.Proof. (1) Clearly, n Z, Bn B containing n.(2) If Bm
Bn 6= for distinct m,n Z, then at least one of the m,n Z must be
even. Lets choosem to be even. Then Bm = {m 1,m,m+ 1}.Suppose n is
odd. Then Bn = {n}. So Bm Bn 6= implies n = m 1 or n = m+ 1. In
either case, ifn Bm Bn, then Bn B such that n Bn Bm Bn.Now suppose
n is also even. Then Bn = {n 1, n, n + 1}. Let k Bm Bn. Then k = n
+ 1 = m 1or k = n 1 = m + 1. So Bk = {k}. In either case, if k Bm
Bn, then Bk sucht that k Bk Bm Bn.Remark 7.23. Explicitly, the
digital line topology is
= {U Z : n U Bn B such that n Bn U}.The topological space (Z, )
is called the digital line.
Remark 7.24. For the remainder of this lecture, Z is assumed to
have the digital line topology.
Example 7.25. Let Z \ 2Z = O. That is, O is the set of all odd
integers. Then O is dense in Z.Proof. Notice that every Bn B
contains at least one odd integer. So every nonempty open subsetU
Z, which is a union of some Bn B, contains at least one odd
integer. That is, nonempty opensubset U Z, U O 6= . Hence, O is
dense in Z.Example 7.26. The digital line Z is separable.
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Proof. We will show that the dense subset O Z is countable.Let :
Z+ O be defined by
(n) =
{n if n is oddn 1 if n is even
Then m O, where m > 0, n 2Z+ such that m = n 1.Similarly, m
O, where m < 0, n Z+ \ 2Z+ such that m = n.So m O n Z+ such that
m = (n).Hence is surjective.Now suppose (m) = (n).Then m,n Z+ are
either both even or both odd.Suppose m,n Z+ are both even.Then (m)
= (n) implies m 1 = n 1. So m = n.Now suppose m,n Z+ are both
odd.Then (m) = (n) implies m = n. So m = n.In either case, if (m) =
(n), then m = n.That is, : Z+ O is a bijection.Thus O is countably
infinite. Hence it is countable.As Z contains a countable dense
subset, it is separable.
Example 7.27. Let 2Z = E. Then E is nowhere dense in Z.
Proof. Notice O =i=1{mi} where mi is an odd integer.
That is, O is an arbitrary union of open sets and is therefore
open.As O is both dense and open, Z \O = E is nowhere dense in
Z.
8 Lecture 08
8.1 More Properties of a Basis
Example 8.1. Let B = {U X : p U}. Then B is a basis for the
particular point topology on X.Proof. We have already demonstrated
that if x U1 U2, where Ui B, then U3 B such thatx U3 U1 U2.Let x X.
Then {x, p} B.So x X Ux = {x, p} B such that x Ux X.Hence B
satisfies the two conditions for a topology geberated by a basis.So
B is a basis for B = {V X : x V Ui B such that x Ui V }.Let P = {}
{U X : p U}.Then nonempty U P , U B B.Further, B vacuously.So P
B.Now let V B. Then V =
jJ Uj where Uj P j J .
Hence B P . B = P .
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Lemma 8.2. Let (X, ) be a topological space and let B = {Bi : i
I} be a collection of open subsetsof X such that(1) X =
iI Bi
(2) If x B1 B2, then B3 B such that x B3 B1 B2.Then the topology
is finer than the topology generated by B.
Proof. Notice B and that B satisfies the two conditions for a
topology generated by a basis.So B is a basis for B = {U X : x U Bi
B such that x Bi U}.Let U B.Then U =
jJ Bj where Bj j J .
So U .Hence B .Remark 8.3. Notice that in order for the reverse
inclusion, B, to be true, U and x U ,it must be true that Bi B such
that x Bi U .Theorem 8.4. Let and be two topologies on the same
underlying set X with bases B and B,respectively. If , then x X and
B B such that x B B B such that x B B.Proof. Let x X.Then as B is a
basis for , B B such that x B X.Notice B B .That is, B is open in
(X, ).As B is a basis for , x B B B such that x B B.Hence x X and B
B such that x B B B such that x B B.
8.2 Calculating the Interior and Closure of Unions and
Intersections
Lemma 8.5. Let (X, ) be a topological space and let A B X.
Then(1) Int(A) Int(B)(2) Cl(A) Cl(B)Proof. (1) Since Int(A) A and A
B, Int(A) B.That is, Int(A) is an open set contained in B.Since
Int(B) is the largest open set contained in B, Int(A) Int(B).(2)
Since A B and B Cl(B), A Cl(B).That is, Cl(B) is a closed set
containing A.Since Cl(A) is the smallest closed set containing A,
Cl(A) Cl(B).Theorem 8.6. Let (X, ) be a topological space and let A
X and B X. Then(1) Int(A) Int(B) Int(A B)(2) Cl(A B) Cl(A)
Cl(B)Proof. (1) Since Int(A) A A B, Int(A) is an open set contained
in A B. As Int(A B) is thelargest open set contained in A B, we
must have Int(A) Int(A B).Now since Int(B) B AB, Int(B) is an open
set contained in AB. As Int(AB) is the largestopen set containted
in A B, we must have Int(B) Int(A B). Int(A) Int(B) Int(A B).(2)
Since AB A Cl(A) and AB B Cl(B), both Cl(A) and Cl(B) are closed
sets containing
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AB. As Cl(AB) is the smallest closed set containing AB, we must
have Cl(AB) Cl(A) andCl(A B) Cl(B). Thus, Cl(A B) Cl(A)
Cl(B).Remark 8.7. The inclusions can be strict. That is, its
possible for Int(A) Int(B) ( Int(A B) andCl(A B) ( Cl(A) Cl(B)
.Remark 8.8. Recall from the previous lecture that the set O = Z \
2Z of all odd integers is both openand dense. As a result, the set
E = 2Z of all even integers is closed and nowhere dense.
Example 8.9. As O is open, Int(O) = O. As O is dense in Z, Cl(O)
= Z.As E is closed, Cl(E) = E. As E is nowhere dense in Z,
Int(Cl(E)) = Int(E) = .So Int(O) Int(E) = O = O while Int(O E) =
Int(Z) = Z.Indeed, Int(O) Int(E) ( Int(O E).Now Cl(O) Cl(E) = Z E =
E while Cl(O E) = Cl() = .Indeed, Cl(O E) ( Cl(O) Cl(E).Remark
8.10. Notice O is open but not closed while E is closed but not
open.
Lemma 8.11. Let (X, ) be a topological space and let A X and B
X. Then(1) Int(A B) = Int(A) Int(B)(2) Cl(A B) = Cl(A) Cl(B)Proof.
(1) As A B A, Int(A B) Int(A). As A B B, Int(A B) Int(B).So Int(A
B) Int(A) Int(B).Conversely, let x Int(A) Int(B). We need to show
that x Int(A B). Suppose to the contrarythat x / Int(A B). Then
open neighborhood U of x, U 6 A B. But if U 6 A B, then eitherU 6 A
or U 6 B. Since Int(A) Int(B) Int(A) A and Int(A) Int(B) Int(B) B,
ineither case, x / Int(A) Int(B). This is a contradiction. So x
Int(A B) by contradiction. Thus,Int(A) Int(B) Int(A B). Int(A B) =
Int(A) Int(B).(2) As A A B, Cl(A) Cl(A B). As B A B, Cl(B) Cl(A
B).So Cl(A) Cl(B) Cl(A B).Conversely, let x Cl(A B). Then open
neighborhood U of x, U (A B) 6= . That is,(U A) (U B) 6= . So
either U A 6= or U B 6= . In the former case x Cl(A) and in
thelatter case x Cl(B). In either case, x Cl(A) Cl(B). Thus, Cl(A
B) Cl(A) Cl(B). Cl(A B) = Cl(A) Cl(B).Theorem 8.12. Let (X, ) be a
topological space and let A1, . . . , An be a finite collection of
subsets ofX. Then(1) Int(
ni=1Ai) =
ni=1 Int(Ai)
(2) Cl(ni=1Ai) =
ni=1Cl(Ai)
Proof. We will use induction:(1) Let P (n) : Int(
ni=1Ai) =
ni=1 Int(Ai).
Then P (1) : Int(A1) = Int(A1), which is true.Suppose P (n 1) :
Int(n1i=1 Ai) = n1i=1 Int(Ai) is true.Then Int(
ni=1Ai) = Int([
n1i=1 Ai] An).
By lemma 8.11, Int([n1i=1 Ai] An) = Int(
n1i=1 Ai) Int(An).
By the induction hypothesis, Int(n1i=1 Ai) =
n1i=1 Int(Ai).
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So Int(ni=1Ai) = [
n1i=1 Int(Ai)] Int(An) =
ni=1 Int(Ai).
Hence, P (n 1) implies P (n). By induction, Int(
ni=1Ai) =
ni=1 Int(Ai) is true n Z+.
(2) Let P (n) : Cl(ni=1Ai) =
ni=1Cl(Ai).
Then P (1) : Cl(A1) = Cl(A1), which is true.Suppose P (n 1) :
Cl(n1i=1 Ai) = n1i=1 Cl(Ai) is true.Then Cl(
ni=1Ai) = Cl([
n1i=1 Ai] An).
By lemma 8.11, Cl([n1i=1 Ai] An) = Cl(
n1i=1 Ai) Cl(An).
By the induction hypothesis, Cl(n1i=1 Ai) =
n1i=1 Cl(Ai).
So Cl(ni=1Ai) = [
n1i=1 Cl(Ai)] Cl(An) =
ni=1Cl(An).
Hence, P (n 1) implies P (n). By induction, Cl(
ni=1Ai) =
ni=1Cl(Ai) is true n Z+.
8.3 Properties of the Digital Line
Remark 8.13. Recall from theorem 6.31, that in a topological
space X with |X| 2, a point x X isan isolated point iff the
singleton containing that point, {x}, is open.Example 8.14. Every
odd integer is an isolated point of the digital line Z.
Proof. Notice n O, {n} is open as it is a basic open set.As |Z|
> 2, every n O is an isolated point of Z.Example 8.15. Every
even integer is a cluster point of the digital line Z.
Proof. Let n E.Then open neighborhood U of n, Bn = {n 1, n, n+
1} U .So {n 1, n+ 1} U Z \ {n}.That is, open neighborhood U of n, U
Z \ {n} 6= .Hence every n E is a cluster point of Z.Remark 8.16.
Recall that a point in a topological space is either an isolated
point or a cluster point,but not both.So the set of all isolated
points of the digital line Z is the set O of all odd integers while
the set of allcluster points of the digial line, the derived set Z,
is the set E of all even integers.Hence the number of isolated
points in Z is |O| = 0. That is, there is a countably infinite
number ofisolated points in the digital line Z.
Remark 8.17. Recall that subsets of a nowhere dense set are also
nowhere dense. So any subset of Eis nowhere dense in Z. In
particular, every singleton containing an even integer is nowhere
dense in Z.
Example 8.18. Every singleton set containing an even integer is
closed but not open. Further, it is itsown boundary.
Proof. Let m E = 2Z and let Be = {Bn : n E}.Using interval
notation, {n Z : n m 1} = (,m 1].
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Notice that (,m 1] is open since(,m 1] = {m 3,m 2,m 1} {m 5,m
4,m 3}
= Bm2 Bm4 =
nm
Bn where Bn Be
So Z \ {m} = (,m 1] [m+ 1,) is open as it is a union of open
sets.Hence {m} is closed. So Cl({m}) = {m}.As {m} is nowhere dense,
Int(Cl({m})) = Int({m}) = .Since {m} 6= Int({m}), {m} is not
open.
{m} = Cl({m}) \ Int({m})= {m} \ = {m}
Exercise 8.19. Show Be = {Bn : n 2Z} is an open cover but is not
a basis for the digital line.Example 8.20. Every singleton set
containing an odd integer is open but not closed.
Proof. Let m O = Z \ 2Z and let Be = {Bn : n 2Z}.Clearly, {m} is
open as it is a basis element.Notice (,m 2] = nm+2Bn where Bn Be is
open.So Z \ {m 1,m,m+ 1} = (,m 2] [m+ 2,) is open as it is a union
of open sets.Further, it is the largest open set not containing m
since any open set containing the even integer m 1must contain Bm1
= {m2,m1,m} and any open set containing the even integer m+1 must
containBm+1 = {m,m+ 1,m+ 2}.So {m 1,m,m+ 1} is the smallest closed
set containing m.That is, Cl({m}) = {m 1,m,m+ 1}.Since {m} 6=
Cl({m}), {m} is not closed.Example 8.21. Let m O. Then the boundary
of {m} is the set containing the pair of consecutiveeven integers
about m.
Proof.
{m} = Cl({m}) \ Int({m})= {m 1,m,m+ 1} \ {m}= {m 1,m+ 1}
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Remark 8.22. Since it is not true that every singleton set is
closed, the digital line Z is not T1 (Frechet).Hence it is also not
T2 (Hausdorff).
Exercise 8.23. Show that the digital line Z is T0
(Kolmogorov).
Example 8.24. {O,E} is a partition but is not a separation of
the digital line Z. Further, E is theboundary of both sets.
Proof. Clearly, O and E are nonempty subsets of Z such that O E
= and O E = Z.Hence {O,E} is a partition of Z. However, it is not a
separation since E is not open.
O = Cl(O) \ Int(O)= Z \O= E
E = Cl(E) \ Int(E)= E \ = E
Example 8.25. The digital line Z is connected.
Proof. Suppose {U, V } is a separation of Z.Then U and V are
nonempty open subsets of Z such that U V = and U V = Z.Since {O,E}
is not a separation of Z, neither U nor V consists enitrely of odd
integers.That is, m,n E such that m U and n V .As U is open, Bm =
{m 1,m,m+ 1} U .Notice m+2 / V since otherwise Bm+2 = {m+1,m+2,m+3}
V , which would require that m+1 V .This is not possible since U V
= and m+ 1 U .So m+ 2 U and hence Bm+2 = {m+ 1,m+ 2,m+ 3} U .But
then m+ 4 / V .We can continue ad infinitum so that V contains no
even integers.But this contradicts the fact that n E such that n V
.Hence, by contradition, there does not exist a separation of Z.
The digital line Z is connected.
Exercise 8.26. Let (a, b) = {n Z : a < n < b} and let B =
{(a, b) : a, b Z}. Show that the topologygenerated by B is the
discrete topology on Z.
9 Lecture 09
9.1 More Properties of the Digital Line
Example 9.1. Be = {Bn : n 2Z} is an open cover but is not a
basis for the digital line.
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Proof. Clearly, Be is a collection of open sets in the digital
line Z.Let m,m+ 2 2Z.Then Bm = {m 1,m,m+ 1} and Bm+2 = {m+ 1,m+
2,m+ 3}.Notice Bm Bm+2 = {m+ 1}.So m,m+ 2 2Z, m,m+ 1,m+ 2 Bm Bm+2
where Bm, Bm+2 Be.Hence
n2ZBn = Z.
That is, Be is an open covering of Z.However, Be is not a basis
for the digital line since for the odd integer m+ 1, m+ 1 Bm Bm+2
but6 B Be such that m+ 1 B Bm Bm+2.Example 9.2. The digital line Z
is a T0 (Kolmogorov) space.
Proof. Consider two consecutive integers m and m+ 1. Notice that
one is even while the other is odd.Without loss of generality, we
may assume m is odd. Then Bm = {m} is an open neighborhood of msuch
that m+ 1 / Bm. However, the smallest open set containing m+ 1 is
Bm+1 = {m,m+ 1,m+ 2}where m Bm+1.Now consider m,m+ 2 Z. Notice that
these integers are either both even or both odd.Suppose both are
even. Then Bm = {m1,m,m+1} is an open neighborhood of m such that
m+2 / Bmand Bm+2 = {m+ 1,m+ 2,m+ 3} is an open neighborhood of m+ 2
such that m / Bm+2.Now suppose both are odd. Then Bm = {m} is an
open neighborhood of m such that m+ 2 / Bm andBm+2 = {m+ 2} is an
open neighborhood of m+ 2 such that m / Bm+2.More generally, if m,n
Z such that |m n| 2, then Bm is an open neighborhood of m such
thatn / Bm and Bn is an open neighborhood of n such that m / Bn.So
m 6= n Z, an open neighborhood of at least one of the integers not
containing the other.Hence Z is T0 (Kolmogorov).
Remark 9.3. The topological invariants which we have explored so
far are summarized below for thedigital line topology (DLT):
# of Isolated Points 0Connectedness Connected
Separation Condition T0Countability Condition Separable
Remark 9.4. As the digital line Z is connected, the only sets
which are both open and closed are and Z. We now consider whether
there exist subsets that are neither open nor closed. We will begin
bylooking at a necessary condition for an open set which contains
an even integer.
Example 9.5. Let A Z. If A is open, then for every even integer
m A, the consecutive odd integersm 1,m+ 1 A.Proof. Suppose A is
open and let m A where m is an even integer.Then as A is is open,
Bm = {m 1,m,m+ 1} B such that m Bm A.But then m 1,m+ 1 A.This is a
contradiction if either m 1 / A or m+ 1 / A.So A must contain both
m 1 and m+ 1.Example 9.6. Let m,m+ 1 Z. Then {m,m+ 1} is neither
open nor closed.
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Proof. Notice one of the integers m and m+ 1 is even.So {m,m+ 1}
does not contain both consecutive odd intege