Complex Analysis
Complex Analysis
Complex Analysis
Example 1.2 1. Listing method: The set A whose elements are the letter a, e, i, o and
u is written as A = {a, e, i, o, u}. Note that the elements are separated by commas
and enclosed in braces {}.
2. Rule Method - stating those properties which characterize the elements in the set.
Definition 1.3 The set B whose elements are the positive integers is written as B = {x :
x is an integer, x > 0} which reads B is the set of x such that x is an integer and x is
greater than zero. A letter x, is used to denote an arbitrary member of the set; the colon is
read as ’such that’ and the comma as ’and’.
Definition 1.4 Two sets A and B are equal written A = B, if they consist of the same
elements, i.e. if each member of A belongs to B and each member of B belongs to A. The
negation of A = B is written as A 6= B.
Definition 1.6 A set can be finite or infinite. A set is finite if it consists of n different
elements, where n is some positive integer; otherwise a set is infinite. In particular, a set
which consists of exactly one element is called a singleton set.
Example 1.8 COnsider the sets A = {1, 3, 5, 7, ...}, B = {5, 10, 15, 20, ...}, and C = {x :
x is prime, x > 2}={3, 5, 7, 11, ...}. Then C ⊂ A since every prime number greater than 2
is odd. On the other hand, B 6⊂ A since 10 ∈ B but 10 ∈ / A.
2 J. Baculta-Nalzaro
Example 1.9 We will let N enote the set of positive integers, Z denote the set of integers,
Q denote the set of rational numbers and R denoe the set of real numbers. Accordingly,
N⊂Z⊂Q⊂R
Definition 1.10 Two sets A and B are equal if and only if A ⊂ B and B ⊂ A. In case
that A ⊂ B but A =
6 B, we say that A is a proper subset of B or B contains A properly.
Theorem 1.11 Let A, B and C be any sets. Then (i)A ⊂ A; (ii) if A ⊂ B and B ⊂ A
then A = B; and (iii) if A ⊂ B and B ⊂ B then A ⊂ C.
Definition 1.12 All sets unde investigation are subsets of a fixed set called as the uni-
versal set or universe of discourse and denote it by U . The empty or null set is a
set which contains no elements. This set, denoted by ∅, is considered finite and a subset of
every other set.. Thus, for any set A, ∅ ⊂ A ⊂ U .
Example 1.13 1. In plane geometry, the universal set consists of all points in the plane.
2. Let A = {x : x2 = 4 , is odd }. Then A is empty, i.e. A = ∅.
3. Let B = {∅}. Then B 6= ∅ for B contains one element.
To help clarify these situations, we use the words ”class”, ”collection” and ”family”
synonymously with set. Usually we use class for a set of sets, and the collections
or family for a set of classes. The subclass, subcollection and subfamily have meanings
analogous to a subset.
Example 1.15 The members of the class {{2, 3}, {2}, {5, 6}} are the sets {2, 3}, {2} and
{5, 6}.
Example 1.16 Consider any set A. The power set of A, denoted by P(A) is the class of all
subsets of A. In particular, if A = {a, b, c}, then
P(A) = {A, {a, b}, {a, c}, {b, c}, {a}, {b}, {c}, ∅}.
Definition 1.17 The word space shall mean a non-empty set which possesses some type
of mathematical structure, e.g. vector space, metric space or topological space. In such a
situation, we call the elements in a space point.
Definition 1.18 The union of two sets A and B, denoted by A∪B, is the set of all elements
which belong to A or B, i.e.
A ∪ B = {x : x ∈ A or x ∈ B}
Definition 1.20 If A ∩ B = ∅, that is, if A and B do not have any elements in common,
then A and B are said to be disjoint or non-intersecting. A class A is called a disjoint
class of sets if each pair of distinct sets in A is disjoint.
Complex Analysis 3
Definition 1.21 The relative complement of a set B with respect to a set A, or simply
the difference of A and B, denoted by A\B, is the set of elements which belong to A but
which do not belong to B. In other words,
A\B = {x : x ∈ A, x ∈
/ B}
Ac = {x : x ∈ U, x ∈
/ A}.
4. Identity Laws: A ∪ ∅ = A; A ∪ U = A; A ∪ U = U ; A ∩ ∅ = ∅
Remark 1.24 Each of the above laws follows an analogous logical law. For example,
A ∩ B = {x : x ∈ A and x ∈ B} = {x : x ∈ B and x ∈ A} = B ∩ A
1. A ∩ B = A
2. A ∪ B = B
3. B c ⊂ Ac
4. A ∩ B c = ∅
5. B ∪ Ac = U
Definition 1.26 Let A and B be two sets. The product set of A and B, written A × B,
consists of all ordered pairs ha, biwhere a ∈ A and b ∈ B, i.e.,
A × B = {ha, bi : a ∈ A, b ∈ B}
2 TOPOLOGICAL SPACES:DEFINITIONS
Definition 2.1 Let X be a nonempty set τ ⊆ P (X). Then τ is a topology in X if and
only if
1. X and ∅ belongs to τ .
2. The union of any number of sets in τ belongs to τ .
3. The intersection of any two sets in τ belongs to τ .
The members of τ are are called open τ -open sets, or simply open sets, and X together
with tau, i.e. the pair (X, τ ) is called a topological space.
Example 2.4 Let D denote the class of all subsets of X. Observe that D satisfies the
axioms for a topology on X. This topology is called the discrete topology; and X together
with its discrete topology, i.e., the pair (X, D), is called a discrete topological space or
simply discrete space.
Example 2.5 The class I = {X, ∅}, consisting of X and ∅ alone, is itself a topology on
X. It is called the indiscrete topology, and X together with its indiscrete topology, i.e.,
(X, τ ) is called an indiscrete topological space or simply indiscrete space.
Example 2.6 Let X = {0, 1} and consider the two topologies: τ1 = 2 = P (X); τ2 =
{∅, {0}, X}. Let S = {X, τ2 }. Then S is called the Sierpinski space.
Example 2.8 Let X = {a, b, c} and consider that two topologies: τ1 = {∅, X, {a}, {a, b}};
τ1 = {∅, X, {a}, {c}, {a, b}, {a, c}}. Then τ0 ⊂ τ1 . Thus, τ1 is finer than τ0 .
Example 2.10 Let X = {l, m, n, o, p} and consider the topologies: τ1 = {X, ∅, {m}};
τ2 = {X, ∅, {l}}. Then τ1 ∪ τ2 = {X, ∅, {l}, {m}} is not a topology since {l} ∪ {m} =
{l, m} ∈
/ τ1 ∪ τ2 . Therefore, ∪α∈A τα need not be a topology.
Example 2.12 Consider the following classes of subsets of X = {a, b, c, d, e} and the topol-
ogy τ1 = {X, ∅, {a}, {c, d}, {a, c, d}, {b, c, d, e}}. Then the closed sets in X are: ∅, X, {b, c, d, e}, {a, b, e}, {b, e},
and {a}.
Theorem 2.13 1. The intersection of any family of closed sets is a closed set.
2. The union of finitely many closed sets is a closed set.
\ [
Proof: (1) Suppose that Ai is closed for all i. Note that ( Ai )c = Aci is open since Aci
\ i i
is open ∀. Hence, Ai is closed.
i
n
[ n
\
(2) Suppose that Ai is closed ∀i. Note that ( Ai ) c = Aci is open since Aci is open
i=1 i=1
n
[
∀i. Hence, is closed.
i=1
3 ELEMENTARY CONCEPTS
Example 3.1 Consider Example 2.12. Then {b, c, d, e} and {a, b, e} are closed sets in X
and {b, c, d, e} ∩ {a, b, e} = {b, e} is closed. Also, {b, e} and {a} are closed sets in X and
{b, e} ∪ {a} = {a, b, e} is also closed in X.
Example 3.3 Let X = {a, b, c, d, e} and the topology τ = {∅, X, {a, b}, {c, d}, {e}, {a, b, e}, {c, d, e}, {a, b, c, d}}.
Consider A = {c, e}. Note that for a ∈ X, the nbd {a, b} of {a} and {b} intersection to the
set A = {c, e} is empty. Thus, {a} and {b} are not adherent points of A. However, for every
nbd of the sets c, d and e, its intersection to A is not empty. Hence, c, d and e are adherent
points of A. It follows that A = {c, e, d}.
Example 3.5 Ccnsider Example 3.7. Note that the singleton {e} is a closed set since its
complement set {a, b, c, d} is open. Consider A = {e}. Then A = {e} since the nbds of
e intersection with the set A is not empty. (nbds {a, b}, {a, b, c, d}, {c, d} of a, b, c and d
respectively has empty intersection with the set A.)
Theorem 3.6 A is the smallest closed set set containing A; i.e. A = ∩{F |F is closed and F ⊃
A}
Example 3.7 Let X = {a, b, c, d, e} and the topology τ = {∅, X, {a, b}, {c, d}, {e}, {a, b, e}, {c, d, e}, {a, b, c, d}}.
Consider A = {c, e}. Then A = {c, e, d}. Take note that X, ∅, {c, d, e}, {a, b, e}, {a, b, c, d}, {c, d}, {a, b}
and {e} are the closed sets in X and the smallest closed set containing A is {c, d, e} which
is the closure of A.
Example 3.9 Consider the set X = {a, b, c} and the topology τ1 = {∅, X, {b}, {b, c}, {a}, {a, b}}.
Let A = {b, c}. For a ∈ X, the nbd {a} of a intersection ({b, c}\a) is empty. For b ∈ X, the
nbd {a, b} of b intersection ({b, c}\b) = {c} is empty. Thus, a and b are not cluster points
of A. However, for the nbds of c which are X and {b, c}, their intersection with ({b, c}\c) is
not empty. Thus c is a cluster point of A and A0 = {c}.
Example 3.11 Consider Example 3.17. Note that A = {b, c} and A = {b, c} union A0 = {c}
is {b, c}. Thus, it follows from the above theorem.
Definition 3.12 Let A ⊂ X. The interior of A is the largest open set contained in A;
that is
Int(A) = ∪{U |U open and U ⊂ A}
Example 3.13 Let X = {a, b, c} and the topology τ = {∅, X, {b}, {b, c}, {a}, {a, b}}.
Consider A = {a, b}. The open sets contained in A are: ∅, {a}, and {a, b}. Note that
∅ ∪ {a} ∪ {a, b} ={a, b}. Thus, Int(A) = {a, b}.
Theorem 3.14 Int(A) = (Ac )c for any set A. In particular, A is open if and only if
A = Int(A).
Proof: ⇒ Assume that A is open. Since A ⊆ A, A is the largest open set contained in itself.
Hence, A = Int(A).
c c c c
⇐ Assume that A = Int(A) = (Ac ) . Then A = (Ac ) , that is, Ac = A . Since Ac = A is
a closed set, it follows that Ac is closed. Hence, A is open.
Complex Analysis 7
Example 3.15 Let X = {a, b, c} and the topology τ = {∅, X, {b}, {b, c}, {a}, {a, b}}. Con-
sider A = {a, b}. Note that A is open and Int(A) = {a, b}. Hence, A = Int(A).
Example 3.17 Consider the set X = {a, b, c} and the topology τ1 = {∅, X, {b}, {b, c}, {a}, {a, b}}.
Let A = {b, c}. Note that A = {b, c} and Ac = {a}. Thus, F r({b, c}) = {b, c} ∩ {a} = X =
F r({a}).
Proof: Suppose there exist a closed dense subset of X such that D 6= X. Then X = D = D.