ME(EE) 550
Foundations of Engineering Systems Analysis
Chapter One: Metric Space & Topological Space
Topology is a vast subject and it has several different branches such as pointset topology, algebraic topology, differential topology, and combinatorial topology.
This chapter deals with point-set topology only. To bring in the notion of topology,
the metric spaces are first introduced in this chapter and later followed by basic
concepts of point-set topology. After introducing the preliminaries of set-theoretic,
topological and algebraic concepts, this chapter begins with generalization of open
sets and continuity of functions that are presented in the setting of metric spaces
in the first section. This chapter should be read along with Chapter 3 of Naylor &
Sell. Both solved examples and exercises in Naylor & Sell would be very useful.
1
Metric Spaces
This section introduces the concept of metric spaces that forms the notion of quantification in various engineering applications.
1.1
Basic Concepts of Metric Spaces
Definition 1.1. (Metric) A metric on a nonempty set S is a function d : S×S → R,
where R , (−∞, ∞) is the real line, such that the following properties hold for all
x, y, z ∈ S:
1. Strict positivity: d(x, y) > 0 if x 6= y and d(x, x) = 0.
2. Symmetry: d(x, y) = d(y, x).
3. Triangular inequality: d(x, z) ≤ d(x, y) + d(y, z).
The metric d on S is called a distance function, where d(x, y) is the distance between
the two points x and y in (S, d); and the pair (S, d) is called a metric space.
Remark 1.1. If the condition of strict positivity, i.e., d(x, y) > 0 if x 6= y and
d(x, x) = 0, is relaxed to that of positivity, i.e., d(x, y) ≥ 0 if x 6= y and d(x, x) = 0,
then d is called a pseudometric. A pseudometric can be converted to a metric by
forming equivalence classes E(x) = {y ∈ S : d(x, y) = 0}.
Remark 1.2. It follows from Definition 1.1 that
|d(x, z) − d(y, z)| ≤ d(x, y) ∀x, y, z ∈ S
1
Example 1.1. For the real line R and the complex plane C, the usual (or standard)
metric is defined as: d(x, y) = |x − y|. For the Euclidean space Rn and the unitary
space Cn , where n ∈ N, the usual (also called Euclidean) metric is defined as:
v
u n
uX
d(x, y) = t
|xk − yk |2
k=1
Definition 1.2. (Subspace) Let (X, d) be a metric space. Then, (Y, d) is called a
subspace of (X, d) if ∅ Y
X, then (Y, d) is called a proper subspace of (X, d).
Definition 1.3. (Product Space) Let (X, dX ) and (Y, dY ) be two metric spaces.
Let Z = X × Y be the product set, i.e., Z , (x, y) : x ∈ X, y ∈ Y and let dZ
be a metric defined on Z. Then, the metric space (Z, dZ ) is called a product space
of the metric spaces (X, dX ) and (Y, dY ).
Remark 1.3. A metric dZ on the product set Z = X × Y can be defined in many
ways in terms of the metrics dX and dY . Two examples are given below.
p1
1. dpZ (u1 , u2 ) = (dX (x1 , x2 ))p + (dY (y1 , y2 ))p
for 1 ≤ p < ∞
2. d∞
Z (u1 , u2 ) = max dX (x1 , x2 ), dY (y1 , y2 ))
Definition 1.4. (Isometry) Let (X, dX ) and (Y, dY ) be two metric spaces. A bijective mapping T of (X, dX ) onto (Y, dY ) is said to be an isometry (i.e., a distance
preserving mapping) if dX x, x̃) = dY (T (x), T (x̃) for every pair of points (x, x̃) in
(X, dX ). In that case, the metric spaces (X, dX ) and (Y, dY ) are said to isometric
to each other.
Remark 1.4. The property of isometry yields an equivalence relation as it satisfies
the conditions of reflexivity, symmetry, and transitivity. (See Chapter 00).
1.2
Analysis in Metric Spaces
Definition 1.5. (Ordering and Ordered Sets) Let S be a non-empty set. A (strict)
order on S is a relation, denoted by ≺, with the following two properties (also see
Chapter 00):
1. If x ∈ S and y ∈ S, then one and only one of the following statements:
(x ≺ y), (x = y), (y ≺ x) is true.
2. If x, y, z ∈ S, then x ≺ y and y ≺ z ⇒ x ≺ z
Example 1.2. The relation ≺ in Definition 1.5 could be < on the real line R and
on a set.
Definition 1.6. (Bounded above and below) Let S be an ordered set and E ⊆ S.
If there exists an α ∈ S such that x ≤ α ∀x ∈ E, then E is said to be bounded
above. Similarly, if there exists a β ∈ S such that x ≥ β ∀x ∈ E, then E is said to
be bounded below.
Definition 1.7. (Supremum and Infimum) Let S be an ordered set and let E ⊆ S
be bounded above. Then, α is called the least upper bound or supremum of E
(denoted as α = sup E) if the following two conditions hold:
2
1. α ∈ S is an upper bound of E.
2. If x < α, then x is not an upper bound of E.
Similarly, let S be an ordered set and let F ⊆ S be bounded below. Then, β
is called the greatest lower bound or infimum of F (denoted as β = inf F ) if the
following two conditions hold:
1. β ∈ S is a lower bound of F .
2. If x > β, then x is not a lower bound of F .
Remark 1.5. Let S ⊆ R. Then, the following results hold for the usual metric d,
where d(x, y) = |x − y| ∀x, y ∈ R.
• If S has a maximum, then max S = sup S; and if S has a minimum, then
min S = inf S.
• Let S be a nonempty set. If S is not bounded from above, then sup S = ∞;
and if S is not bounded from below, then inf S = −∞.
• If S is the empty set, i.e., S = ∅, then sup S = −∞ and inf S = ∞.
Example 1.3. Let S ⊆ Q, the set of rational numbers, such that S = {x ∈ Q :
√
0 < x < 2}. Then, in the usual metric, S has neither a supremum in Q nor a
maximum; however, 0 ∈ Q is the infimum of S although S has no minimum, and
√
2 is the supremum of S although S has no maximum.
Definition 1.8. (Usual Notions in Analysis) Let (S, d) be a metric space and let
E ⊆ S be noneempty. We introduce the following concepts on (S, d).
• An (open) ε-neighborhood of a point x ∈ E, for a given ε ∈ (0, ∞), is the set
Bεd (x) , {y ∈ S : d(x, y) < ε}
and Bεd (x) is also called an open ball in (S, d) with radius ε with center at
x. Note that the set Bεd (x) is always nonempty because it is guaranteed that
x ∈ Bεd (x).
• A point x ∈ S is a limit point of E if every neighborhood of x contains a
point y 6= x such that y ∈ E.
• If x ∈ E and x is not a limit point of E, then x is an isolated point of E .
• A set E is closed in S if every limit point of E is contained in E.
• A point x is an interior point of E if there exists a neighborhood Bεd (x) ⊆ E.
• The complement of E in S, denoted as E c and also as S \ E, is the set of all
points y ∈ S such that y ∈
/ E.
• A set E is open in S if every point of E is an interior point of E.
• A set E is perfect in S if E is closed in S and every point of E is a limit
point of E.
• A set E is bounded if there exists a real number M ∈ (0, ∞) and a point
y ∈ S such that d(x, y) < M ∀x ∈ E.
3
• A set E is dense in S if every point of S is a limit point of E or a point of E
(or both).
Example 1.4. The set of rational numbers Q is dense in R but Q is neither closed
nor open in R.
Example 1.5. The set { k1 : k ∈ N} is not open, not closed, and not perfect in R
but it is a bounded set. As a matter of fact, the set { k1 : k ∈ N} is bounded both
above and below because its maximum is 1 and its infimum is 0.
Example 1.6. Every nonempty set S is open in itself and also closed in itself; S
is also dense in itself.
Example 1.7. The empty set ∅ is both open and closed in every nonempty set.
Definition 1.9. (Diameter and Distance) Let (S, d) be a metric space and let
∅ ( E ⊆ S, i.e., E is a nonempty subset of S. Then, the diameter of E is defined
as:
diameter(E) = sup{d(x, y) : x, y ∈ E}
and the distance of a point x ∈ S from E is defined as:
distanceE (x) = inf{d(x, y) : y ∈ E}
Example 1.8. Let S be the set of all bounded sequences of complex numbers.
Let x , {x1 , x2 , x3 , · · · } be a sequence of complex numbers such that, for all j =
1, 2, 3, · · · , we have |xj | ≤ Cx ∈ [0, ∞), where Cx may depend on x but not on j.
If, on this space, there is a metric defined as:
d∞ (x, y) = sup |xj − yj |,
j∈N
then this space is called the ℓ∞ -space.
Example 1.9. Let S , C[a, b] be the set of all real-valued continuous functions
defined on the closed interval [a, b]
R. A metric on the space C[a, b] could be
chosen as:
d(x, y) = max |x(t) − y(t)|.
t∈[a,b]
Example 1.10. Let Lp [a, b] be the space of pth -power integrable functions on the
interval [a,b], where p ∈ [1, ∞). The distance function is defined as:
d(x, y) =
Z
b
a
dt |x(t) − y(t)|p
p1
.
Example 1.11. A metric on Rm×n , the space of (m × n) real matrices could be
chosen as:
r
d(A, B) = T race (A − B)T (A − B)
Example 1.12. The discrete metric on a nonempty space is defined as:
(
0 if x = y
do (x, y) =
1 if x 6= y
4
Theorem 1.1. Let (X, d) be a metric space. Every neighborhood of x ∈ X is a
d-open set.
Proof. Let x ∈ X and ε > 0 be arbitrary. For each y ∈ Bεd (x), there exists a
positive real θ such that d(x, y) = ε − θ, implying that θ ≤ ε. Then, it follows by
triangular inequality that
t ∈ Bθd (y) ⇒ t ∈ Bεd (x) because d(x, t) ≤ d(x, y) + d(y, t) < ε − θ + θ = ε
Therefore, y is an interior point of Bεd (x). So, Bεd (x) is a d-open set.
Theorem 1.2. Let (X, d) be a metric space. Then, S ⊆ X is d-open if and only if
its complement X \ S is d-closed.
Proof. (if part) Let X \ S be d-closed. If x ∈ S, then x ∈
/ X \ S, which implies that
x is not a limit point of the d-closed set X \ S. Hence, there exists a neighborhood
Bεd (x) such that (X \ S) ∩ Bεd (x) = ∅, i.e., Bεd (x) ⊆ S. So, x is an interior point of
S and hence S is d-open.
(only if part) Let S be d-open and let x be a limit point of X \ S. Then, every
neighborhood of x contains a point of X \ S other than x itself, which implies that
x is not an interior point of S because this neighborhood is not a subset of S. Since
S is d-open (i.e., every point of S is an interior point of S), it follows that x ∈ X \ S.
Hence, X \ S contains each of its limit points. Therefore, X \ S is d-closed.
Corollary 1.1. Let (X, dX ) be a metric space. Then, S ⊆ X is dX -closed if and
only if its complement X \ S is dX -open.
Theorem 1.3. An arbitrary union of open sets is open. A finite intersection of
open sets is open.
Proof. Let F = {Fα } be an arbitrary (i.e., finite or countable or uncountable)
collection of open sets on a metric space. Then, we will show that
S
(a) α Fα is an open set, and
Tn
(b) k=1 Fk is an open set for any given n ∈ N.
S
(Part a) Let U = α Fα . If x ∈ U , then x ∈ Fα for at least one member Fα
in F . Since Fα is open, x is an interior point of Fα , i.e., there exists an ε > 0 such
that Bε (x) ⊆ Fα ⊆ U . Hence, U is open.
Tn
(Part b) Let V = k=1 Fk for some n ∈ N. Let us assume that V is nonempty;
otherwise, V is open because the empty set is open in all metric spaces. If x ∈ V ,
then x ∈ Fk for each k ∈ {1, 2, · · · , n}. Since each Fk is open, x is an interior point
of each Fk , i.e., there exists open balls Bεk (x) ⊆ Fk . Let ε = min{ε1 , · · · , εn }.
Since each εk > 0, it follows that ε > 0. Since Bε (x) ⊆ V , V is an open set.
Corollary 1.2. An arbitrary intersection of closed sets is closed. A finite union of
closed sets is closed.
Proof. The proof follows from Theorem 1.2 and Theorem 1.3 after application of
de Morgan’s laws that are stated below:
\
α∈I
Aα
c
=
[
α∈I
Aα
c
and
[
α∈I
Aα
c
where I is a finite, countable or uncountable index set.
5
=
\
α∈I
Aα
c
Remark 1.6. A set which is a countable union of closed sets is called an Fσ , where
F stands for closed and σ for sum. Similarly, a set which is a countable intersection
of open sets is called a Gδ , where G stands for open and δ for durchschnitt. Thus,
by deMorgan Theorem, complement of an Fσ is a Gδ and vice versa. In general,
an infinite intersection of open sets may or may not be open; similarly, an infinite
union of closed sets may or may not be closed. Examples follow.
T
S
Examples: k∈N 1 − k1 , 4 + k1 = 1, 4 and k∈N 1 + k1 , 4 − k1 = 1, 4 .
Definition 1.10. Let S be a (not necessarily non-empty) set in a metric space
(X, dX ). Then,
(a) The interior S o of S in (X, dX ) is the largest dX -open subset of S, i.e., the
S
union of all open sets that are subsets of S, or S o , {U : U ⊆ S, and U is open}.
(b) The closure S of S in (X, dX ) is the smallest closed superset of S, i.e.,
T
the intersection of all closed sets that are supersets of S, or S , {V : V ⊇
S, and V is closed}
(c) The boundary ∂S of S in (X, dX ) is the set difference S \ S o .
o
Remark 1.7. Note that S o ⊆ S and (S)o = (S)o . The boundary ∂S of S in
(X, dX ) is a closed set because it is intersection of two closed sets S and X \ S o .
Remark 1.8. Let (S, d) be a metric space. Then, S is always dense in (S, d), and
A ( S is dense in (S, d) if A = S.
Example 1.13. Let X = R and the metric d(x, y) = |x − y|.
Let S = (0, 1) ∩ Q ∪ (2, 3]. Then, we have:
S o = (2, 3) because (Q)o = ∅.
S = [0, 1] ∪ [2, 3] because Q = R.
∂S = [0, 1] ∪ {2, 3} because ∂Q = R.
Definition 1.11. (Convergence) A sequence {xk } converges in a metric space (S, d)
if there exists a point x ∈ S, called the limit point of the sequence, and the following
condition holds: ∀ε > 0 ∃ n ∈ N such that d(xk , x) < ε ∀k ≥ n. Equivalently, we
say d(xk , x) → 0 as k → ∞.
Remark 1.9. In Definition 1.11, n is a function of ε; smaller is ε, larger is the
positive integer n.
If x ∈
/ S, we cannot say that the sequence {xk } converges in S even though
d(xk , x) → 0 as k → ∞. In that case, we relax Definition 1.11 as follows.
Definition 1.12. (Cauchy Convergence) A sequence {xk } converges in Cauchy
sense in a metric space (S, d) if the following condition holds:
∀ε > 0 ∃ n ∈ N such that d(xk , xℓ ) < ε ∀k, ℓ ≥ n. Equivalently, d(xk , xℓ ) → 0 as
k, ℓ → ∞.
Remark 1.10. (Non-convergence – Negation of Definition 1.11) A sequence {xk }
does not converge to x ∈ S in a metric space (S, d) if the following condition holds:
There exists ε > 0 such that ∀ n ∈ N there exists a positive integer k ≥ n with
d(xk , x) ≥ ε.
6
Remark 1.11. (Non-Cauchy Convergence – Negation of Definition 1.12) A sequence {xk } does not converge in Cauchy sense in a metric space (S, d) if there exists
ε > 0 such that ∀ n ∈ N there exists positive integers k, ℓ ≥ n with d(xk , xℓ ) ≥ ε.
Remark 1.12. A convergent sequence is always Cauchy convergent because of the
non-negativity and triangular inequality properties of a metric, but the converse is
not true in general as seen in the following example.
Example 1.14. Let S = (0, 1]. Then, the sequence { k1 : k ∈ N} does not converge
in S but it is a Cauchy sequence in S.
Definition 1.13. (Uniform Convergence) A sequence {fk } of functions mapping
(X, dX ) → (Y, dY ), where k ∈ N, converges at a point x ∈ X if the following
condition holds: ∀ε > 0 ∃ n(x, ε) ∈ N such that dY (fk (x), f (x)) < ε ∀k ≥ n.
If this convergence holds for all x ∈ X, then {fk } converges to f pointwise on X.
The sequence {fk } converges to f uniformly on X if the following condition holds:
∀ε > 0 ∃ n(ε) ∈ N such that dY (fk (x), f (x)) < ε ∀x ∈ X ∀k ≥ n.
Remark 1.13. (Weierstrass test of uniform convergence)
Let limk→∞ fk (x) → f (x) for all x ∈ X and let Mk , supx∈X dY (fk (x), f (x))
Then, fk → f uniformly on X if and only if Mk → 0 as k → ∞.
Definition 1.14. (Continuity) Let (X, dX ) and (Y, dY ) be two metric spaces. A
function f : X → Y is said to be
at a point x ∈X if
( dX − dY ) continuous
∀ε > 0 ∃ δ(ε, x) > 0 such that dX (x, x̃) < δ ⇒ dY (f (x), f (x̃)) < ε ∀x̃ ∈ X.
If f is continuous at every x ∈ X, then f is continuous in X.
Remark 1.14. Definition 1.14 of continuous functions can be interpreted in the
following way. A function f is continuous at a point z0 if both of the following two
conditions are satisfied:
1. limz→z0 f (z) exists and f (z0 ) is defined.
2. limz→z0 f (z) = f (z0 ).
It is noted that the statement (2) above actually contains the statement (1) because
the existence of both sides of the equation limz→z0 f (z) = f (z0 ) in the statement
(2) is needed for their equality. In fact, the statement (2) is equivalent to Definition 1.14. For example, if f : R → R, continuity of f at z0 can be interpreted as
existence and equality of the left hand and right hand limits of f (z) as z → z0 ; in
addition, this limit must be equal to f (z0 ). However, this notion is not valid if,
for example, f : C → C because the limit of f (z) as z → z0 in the complex plane
cannot be interpreted simply as the existence and equality of left hand and right
hand limits. Note that Definition 1.14 is valid for f : C → C and other metric
spaces.
Remark 1.15. (Discontinuity – Negation of Definition 1.14) A function f : X → Y
is said to be ( dX − dY ) discontinuous
at a point x ∈ X if there exists ε > 0such
that ∀δ > 0 there exists x̃ ∈ X with dX (x, x̃) < δ and dY (f (x), f (x̃)) ≥ ε .
If f is discontinuous at a point x ∈ X, then f is said to be discontinuous in X.
7
Theorem 1.4 (Alternative notion of continuity). Following Definition 1.14, a function f : X → Y is (dX − dY ) continuous if and only if, for every dY -open set U ,
the inverse image f −1 (U ) , {x ∈ X : f (x) ∈ U } is a dX -open set.
Proof. (only if part) Let f be continuous and let U be dY -open. Having x ∈ f −1 (U ),
there exists ε > 0 such that the neighborhood Bε (f (x)) ⊆ U . By definition of
continuity, there exists δ > 0 such that x̃ ∈ Bδ (x) ⇒ f (x̃) ∈ Bε (f (x)) . This
implies that Bδ (x) ⊆ f −1 (U ). Thus, the set f −1 (U ) is dX -open.
(if part) Let x ∈ X and ε > 0. As f satisfies the condition of the theorem,
we have an open set U , Bε (f (x)) such that x ∈ f −1 (U ). The hypothesis is
that f −1 (U ) is dX -open, which implies that ∃ δ > 0 such that Bδ (x) ⊆ f −1 (U ) .
Therefore, x̃ ∈ Bδ (x) ⇒ f (x̃) ∈ Bε (f (x)) .
Remark 1.16. The converse of Theorem 1.4 may not be true, i.e., (dX − dY )
continuity does not imply that, for every dX -open set U , the image f {U } is dY open. To see this fact, let f : R → R be defined as f (x) = c ∀x ∈ R, where c ∈ R
is a constant. Then, f is a continuous function in the usual topology. Now, let
U = (a, b) be an open interval in R. Obviously, the image f {(a, b)} = {c} is not an
open set in the usual topology.
Theorem 1.5 (Convergence and continuity). Let (X, dX ) and (Y, dY ) be metric
spaces. Let a sequence {xk } converge to a point x ∈ X. Then, a function f : X → Y
is (dX − dY ) continuous at a point x ∈ X if and only if the sequence {f (xk )}
converges to the point f (x) ∈ Y .
Proof. (only if part) Let f be continuous at x ∈ X and ε > 0 be selected arbitrarily.
Then, it follows that ∃ δ > 0 such that
∀x̃ ∈ X
dX (x, x̃) < δ ⇒ (dY (f (x), f (x̃) < ε .
Following Definition 1.11, ∃ n ∈ N such that ∀k ≥ n, dX (xk , x) < δ. Thus,
dY (f (x), f (x̃) < ε and the sequence {f (xk )} converges to the point f (x) ∈ Y .
(if part) We use the proof by contradiction in this part. Let us assume that
f is not continuous at a point x ∈ X. Then, ∃ ε > 0 such that ∀δ > 0 ∃ x̃ ∈ X
with dX (x, x̃) < δ and dY (f (x), f (x̃) ≥ ε. Let xk be a value of x̃ corresponding to,
say, δ = k1 . Then, xk → x, but d(f (x), f (xk )) ≥ ε for all k; hence, f (xk ) does not
converge to f (x).
Remark 1.17. Let (X, dX ) and (Y, dY ) be metric spaces and f : X → Y . Then,
the following two conditions are equivalent.
1. f is continuous at x.
2. lim f (xk ) = f (lim xk ) for every sequence {xk } such that lim xk = x.
In other words, the function f : X → Y is (dX − dY ) continuous if and only if
it preserves convergent sequences. (See Naylor & Sell p. 74.)
Definition 1.15. (Uniform Continuity) Let (X, dX ) and (Y, dY ) be two metric
spaces. A function f : X → Y is said to be uniformly
(dX − dY ) continuous in X
if ∀ε > 0 ∃ δ(ε) > 0 such that dX (x, y) < δ ⇒ dY (f (x), f (y)) < ε ∀x, y ∈ X .
Note that, for a given ε > 0, a single parameter δ can be chosen for all x.
8
Remark 1.18. It follows from Definition 1.14 that δ is a function of both ε and
x for continuity at the point x. It follows from from Definition 1.15 that δ is a
function of ε and is invariant for any x in the domain of the function.
Example 1.15. Uniform continuity implies continuity, but the converse is not true
in general as seen in the following example.
Let S = (0, 1]. Then, a function f : S → [1, ∞) with f (x) = x1 is continuous in
S but it is not uniformly continuous in S.
Definition 1.16. (Lipschitz Continuity) Let (X, dX ) and (Y, dY ) be two metric
spaces. Then, a function f : X → Y is said to be Lipschitz continuous (more
precisely, Lipschitz dX − dY continuous) in X if there exists a constant K ∈ (0, ∞)
such that dY (f (x), f (y)) ≤ K dX (x, y) ∀x, y ∈ X. The parameter K is called
Lipschitz constant.
Example 1.16. Lipschitz continuity implies uniform continuity. This is seen by
ε
choosing δ = K
.
Definition 1.17. (Hölder Continuity) Let (X, dX ) and (Y, dY ) be two metric
spaces. Then, a function f : X → Y is said to be Hölder continuous (more precisely, Hölder dX − dY continuous) in X if there exists constants K ∈ (0, ∞) and
α ∈ (0, ∞) such that dY (f (x), f (y)) ≤ K dX (x, y)α ∀x, y ∈ X.
Definition 1.18. (Absolute Continuity)
be absolutely continuous on [a, b] if ∀ ε
Pn
i=1 |f (x̃i ) − f (xi )| < ε for every finite
{(xi , x̃i ) : i = 1, · · · , n} in [a, b] under the
A function f : [a, b] → R is said to
> 0 there exists δ(ε) > 0 such that
collection of pairwise disjoint intervals
Pn
constraint i=1 |x̃i − xi | < δ.
Definition 1.19. (Total Variation) Given a function f : [a, b] → R, let P be the
set of all finite collections {(xi , x̃i ) : i = 1, · · · , n} of disjoint intervals in [a, b].
Then, the total variation Vab (f ) of the function f on [a, b] is defined as:
Vab (f ) , sup
n
nX
i=1
|f (x̃i ) − f (xi )|
o
over all partitions P ∈ P. If Vab (f ) is finite, then f is of bounded variation on [a, b].
The set of all functions of bounded variation on [a, b] is denoted as BV [a, b].
Definition 1.20. (Monotonicity) A sequence {xk } of real numbers is said to be:
• monotonically increasing if xk ≤ xk+1 ; and strictly monotonically increasing
if xk < xk+1 .
• monotonically decreasing if xk ≥ xk+1 ; and strictly monotonically decreasing
if xk > xk+1 .
Definition 1.21. (lim sup and lim inf) The limit superior and limit inferior for a
sequence {xk } are respectively defined as:
lim sup xn = inf sup xk and lim inf xn = sup inf xk
n
n
k≥n
k≥n
A sequence is said to have a limit if its limit superior is equal to the limit
inferior; otherwise the sequence does not have a limit. Often lim sup is denoted as
lim and lim inf as lim.
9
Let (S, d) be a metric space and a point x ∈ S is called an adherent point to
A ⊆ S if every neighborhood of x in (S, d) (i.e., open set containing x) contains
a point in A (which may be x itself). In contrast, a point x ∈ S is called a limit
point (also called a point of accumulation) of A if every neighborhood of x in (S, d)
contains a point of A\ {x}. It is obvious that every limit point is an adherent point,
but the converse may not be true.
Let us further point out the subtle difference between the notions of limit point
and adherant point in the context of lim sup and lim inf in the real space R and the
usual metric d(x, y) = |x − y|. We paraphrase the definitions of a limit point and
an adherent point as explained below.
The limit point of a sequence {xk } is x if, any given ǫ > 0, all but finitely many
points of {xk } are located within the ǫ-neighborhood of x. Another condition,
where x is called an adherent point, is to have infinitely many points of {xk } within
the ǫ-neighborhood of x for any given ǫ > 0. The condition for an adherent point
is weaker than that of a limit point because there is a possibility of existence of
infinitely many terms of {xk } outside the ǫ-neighborhood of x.
Example 1.17. The sequence {(−1)k } does not have a limit because its lim sup
= 1 and lim inf = -1. However, it has two adherent points, 1 and −1.
One should be able to prove the following results..
1. lim(xn ) = −lim(−xn )
2. lim(xn ) = −lim(−xn )
3. lim(xn ) ≤ lim(xn )
4. lim(xn ) + lim(yn ) ≤ lim(xn + yn ) ≤ lim(xn ) + lim(yn ) ≤ lim(xn + yn )
provided that no sum is in the form of (∞ − ∞).
5. If xn > 0 and yn ≥ 0 for all n, then lim(xn yn ) ≤ lim(xn ) lim(yn ) provided
that the product on the right is not in the form of (0 × ∞).
1.3
Complete Metric Spaces
Definition 1.22. (Complete Metric Space) A metric space is called complete if every Cauchy sequence converges in the metric space. If at least one Cauchy sequence
does not converge in the metric space, then the space is called incomplete.
Example 1.18. We cite a few examples of incomplete metric spaces.
• The set (0, 1] is not complete in the metric space (R, d), where the d is the
usual metric, i.e., d(x, y) , |x − y|, but [0, 1] is complete in (R, d).
• The set Q of rational numbers is not complete in (R, d). Also note that Q
is not complete in (Q, d) although Q is closed in (Q, d) because every metric
space is closed in itself.
• Let P [a, b] be the set of all polynomials that are functions of t belonging to
the closed interval [a, b] and let a metric d be defined on the space P [a, b]
as: d(x, y) = maxt∈[a,b] |x(t) − y(t)|. The metric space P [a, b], d is not
10
complete, because a Cauchy sequence of polynomials converges uniformly on
[a, b] to a continuous function that could be a power series and not necessarily
a polynomial.
• Let C1 [0, 1] be the set of all continuous real-valued valued functions on [0, 1]
R1
and the metric d be defined as: d(x, y) , 0 dt|x(t) − y(t)| ∀x, y ∈ C1 [0, 1].
The metric space (C1 [0, 1], d) is not complete.
Theorem 1.6. Let (X, d) be a complete metric space and let Y ⊆ X be arbitrarily
chosen such that (Y, d) is an arbitrary subspace of (X, d). Then, (Y, d) is complete
if and only if Y is closed in (X, d).
Proof. The proof is given in Naylor & Sell (see p.116).
Let (X, d) be a complete metric space and let (Y, d) be an arbitrary subspace
of (X, d). Let Y be the closure of Y in (X, d). Obviously, Y is is closed in (X, d);
therefore, it follows from Theorem 1.6 that (Y , d) is a complete metric space. Since
Y is dense in (Y , d), the closure of Y fills up any holes that may exist in (Y, d).
For example, the completion of the incomplete subspace of rational numbers (Q, d)
is (R, d), where d is the usual metric, i.e., d(x, y) = |x − y|. In the process of
completion, all holes (i.e., the irrational numbers) are filled up. A formal definition
of completion of an incomplete metric space is presented below.
Definition 1.23. (Completion) Let (X, d1 ) be a metric space. A metric space
(Y, d2 ) is defined to be a completion of (X, d1 ) if the following two conditions are
satisfied.
1. The metric space (Y, d2 ) is complete.
2. The metric space (X, d1 ) is isometric to a dense subspace of (Y, d2 ).
Theorem 1.7. Every metric space has an essentially unique completion. In other
words, if (X, d) is a metric space, then it has a completion and if (Y1 , d1 ) and
(Y2 , d2 ) are two completions of (X, d), then (Y1 , d1 ) and (Y2 , d2 ) are isometric to
each other.
Proof. The proof is given in Naylor & Sell (see pp.122-124).
Definition 1.24. (Contraction) Let (X, d) be a metric space and let T be a mapping from (X, d) into (X, d). Then T is said to be a contraction if there exists a
real constant λ ∈ [0, 1) such that, for all x, y ∈ X, d(f (x), f (y)) ≤ λ d(x, y).
Theorem 1.8. Let (X, d) be a complete metric space and let T : X → X be a
contraction. Then, there is a unique fixed point of T , i.e., there exists one and only
one point xf ∈ X such that T (xf ) = xf . Furthermore, if x0 ∈ X is arbitrary and
xn is recursively defined by x1 = T (x0 ), x2 = T (x1 ), · · · , xn = T (xn−1 ), then
xn → xf as n → ∞.
Proof. The proof is given in Naylor & Sell (see pp.126-127).
11
2
Introduction to Point-set Topology
Only rudimentary concepts of point set topology are presented in this section to
allow generalization of open sets and continuity of functions beyond metric spaces
described in Section 1.
2.1
Basic Concepts of Topological spaces
Definition 2.1. (Topological Space of Open Sets) Let X be a set and ℑ be a
nonempty collection of subsets of X having the following properties:
1. ∅ ∈ ℑ and X ∈ ℑ
Tn
2. If Sk ∈ ℑ, k = 1, 2, · · · , n, then k=1 Sk ∈ ℑ
(finite intersection)
S
3. If Sα ∈ ℑ, then α Sα ∈ ℑ (arbitrary uniion)
Then, ℑ is a topology of X and (X, ℑ) is a topological space. Each member of
ℑ is called a ℑ-open subset of X.
Definition 2.2. (Closed Set) Let (X, ℑ) be a topological space. Then, the complement of every ℑ-open set in X is said to be ℑ-closed in X. That is, if S ∈ ℑ,
then S c , X \ S is ℑ-closed in X. In other words, S is ℑ-open in X if and only if
S c is ℑ-closed in X.
In view of Definition 2.2, an alternative form of Definition 2.1 is given below.
Definition 2.3. (Topological Space of Closed Sets) Let X be a set and ℑ be is a
nonempty collection of subsets of X having the following properties:
1. ∅ ∈ ℑ and X ∈ ℑ
S
2. If Sk ∈ ℑ, k = 1, 2, · · · , n, then nk=1 Sk ∈ ℑ
(finite union)
T
3. If Sα ∈ ℑ, then α Sα ∈ ℑ (arbitrary intersection)
Then, ℑ is a topology of X and (X, ℑ) is a topological space. Each member of
ℑ is called a ℑ-closed subset of X.
e be two topologies of a
Definition 2.4. (Coarse and Fine Topology) Let ℑ and ℑ
e
e
e ⊃ ℑ,
nonempty set X. If ℑ ⊇ ℑ, then ℑ is said to be finer (or stronger) than ℑ; if ℑ
e is said to be strictly finer than ℑ. Similarly, If ℑ
e ⊆ ℑ, then ℑ
e is said to be
then ℑ
e ⊂ ℑ, then ℑ
e is said to be strictly coarser than ℑ.
coarser (or weaker) than ℑ; if ℑ
e
e ⊆ ℑ or
A topology ℑ is said to be comparable with another topology ℑ if either ℑ
e ⊇ ℑ; otherwise the topologies ℑ
e and ℑ are incomparable.
ℑ
Example 2.1. If X is a set, the power set 2X (i.e., the collection of all subsets) of
X is a topology of X; it is the finest topology of X, called the discrete topology,
where every set is both open and closed. The collection consisting of X and ∅
only is the coarsest topology of X, which is called the indiscrete topology or trivial
topology.
12
Definition 2.5. (Basis for a Topology) A basis for a topological space (X, ℑ) is a
collection B of ℑ-open subsets of X (that are called basis elements) such that
1. For each x ∈ X, there exists at least one basis element B ∈ B containing x,
i.e., x ∈ B.
2. If B1 , B2 ∈ B and x ∈ B1 ∩ B2 , then there exists B3 ∈ B such that x ∈ B3
and B3 ⊆ B1 ∩ B2 .
If B satisfies the above two conditions, then we define the topology ℑ generated
by B as follows: A subset U of X is said to beℑ-open in X, i.e., U ∈ ℑ, if
∀x ∈ U ∃ B ∈ B such that x ∈ B and B ⊆ U . As stated above, each basis
element is a ℑ-open set.
Example 2.2. If X is a non-empty set, the collection of all singleton subsets of X
is a basis for the discrete topology on X.
Remark 2.1. Let B be a basis for a topological space (X, ℑ). Then, ℑ is the
collection of all unions of elements of B. This claim is justified as follows.
Given a collection of elements of B, these elements belong to ℑ and their unions
also belong to ℑ. Conversely, given U ∈ ℑ, let us choose an element Bx of B for
S
each x ∈ U such that x ∈ Bx ⊆ U . Then, U = x∈U Bx , which implies that
U equals the union of elements of B. In this setting, every open set U in X can
be expressed as a union of basis elements. However, this expression for U is not
unique. Thus, the usage of the term basis in topology differs from its usage in linear
algebra, where the representation of a given vector as a linear combination of basis
vectors is unique.
Remark 2.2. Let (X, ℑ) be a topological space and let C be a collection of ℑ-open
sets such that, for each ℑ-open set U and each x ∈ U , there is a member C ∈ C
with x ∈ C ⊆ U . Then, C is a basis for the topological space (X, ℑ).
e be two topological spaces. Then, the following
Remark 2.3. Let (X, ℑ) and (X, ℑ)
two statements are equivalent.
e is finer than ℑ.
1. ℑ
2. For each x ∈ X and each basis element B ∈ B containing x, there exists a
e ∈ Be such that x ∈ Be ⊆ B.
basis element B
Definition 2.6. (Usual or standard Topology) The topology U generated by the
collection of all open intervals on the real line R, i.e., (a, b) , {x ∈ R : a < x < b}
is called the usual or standard topology on R.
In view of the fact that the topology generated by a basis B can be described
as a collection of arbitrary unions of elements of B, what happens if one starts
with a certain (but not necessarily all) collection of elements of of B and take finite
intersections and arbitrary unions? This leads to the notion of subbasis.
Definition 2.7. (Subbasis) A subbasis Be of a basis B for a topological space (X, ℑ)
e generated
is a collection of some elements of B, whose union is X. The topology ℑ
by Be is defined to be the collection of all unions of finite intersections of the elements
e is coarser than ℑ.)
e (Note: The topology ℑ
of B.
13
Definition 2.8. (Subspace Topology) Let (X, ℑ) be a topological space and let
Y ⊆ X (not necessarily a ℑ-open set). Then, the collection ℑY , {Y ∩ U : U ∈ ℑ}
is a topology on Y, called subspace topology (also called relative topology or induced
topology). The ℑY -open sets are intersections of ℑ-open sets with Y . With this
topology, Y is called a subspace of X.
Remark 2.4. Let B be a basis of the topological space (X, ℑ) and let Y ⊆ X.
Then, the collection BY , {B ∩ Y : B ∈ B} is a basis for the subspace topology
(Y, ℑY ). An explanation follows.
Given a ℑ-open set U and y ∈ U ∩ Y , choose a basis element B ∈ B such that
y ∈ B ⊆ U . Then, y ∈ B ∩ Y ⊆ U ∩ Y and it follows from Remark 2.2 that BY is
a basis for the subspace topology (Y, ℑY ).
Remark 2.5. Let (Y, ℑY ) be a subspace of (X, ℑ). If U is ℑY -open and Y is
ℑ-open, then U is ℑ-open. This is so because since U is ℑ-open, it follows that
U = Y ∩ V for some ℑ-open set V . Further, since Y and V are ℑ-open, so is
U =Y ∩V.
Definition 2.9. (Product Topology) Let (X, ℑX ) and (Y, ℑY ) be topological spaces.
Then, the product topology ℑX×Y on X ×Y is defined to be the topology generated
by {UX × UY : UX ∈ ℑX and UY ∈ ℑY }.
Let us check that BX×Y , {UX × UY : UX ∈ ℑX and UY ∈ ℑY } is a basis of
the product topology ℑX×Y on X × Y . Following Definition 2.5, the first condition
is trivial since X × Y is a basis element. The second condition is also satisfied
because the intersection of any two basis elements U1 × V1 and U2 × V2 is another
basis element due to the following fact of the set theory:
(U1 × V1 ) ∩ (U2 × V2 ) = (U1 ∩ U2 ) × (V1 ∩ V2 ).
The set on the right is a basis element because U1 ∩ U2 and V1 ∩ V2 are ℑX -open
and ℑY -open, respectively. Note that the collection BX×Y may not be a topology
on X × Y because, for example, the union of two rectangles may not be a rectangle;
however, each element of BX×Y is ℑX × ℑY -open.
Definition 2.10. (Projection) Let (X, ℑX ) and (Y, ℑY ) be topological spaces. Let
the functions πX : X × Y → X and πY : X × Y → Y be respectively defined as:
πX (x, y) = x and πY (x, y) = y. Then, πX and πY are called the projections onto
its first and second factors, respectively.
−1
Remark 2.6. If U is ℑX -open, then the inverse image πX
(U ) is the set U × Y
that is ℑX×Y -open. Similarly, if V is ℑY -open, then the inverse image πY−1 (V ) is
the set X × V that is also ℑX×Y -open. Then, it follows from Definitions 2.7 and 2.9
−1
that the collection S , {πX
(U ) : U ∈ ℑX } ∪ {πY−1 (V ) : V ∈ ℑY } is a subbasis for
the product topology ℑX×Y on X × Y .
Remark 2.7. Let (Y, ℑY ) be a subspace of a topological space (X, ℑX ) and (Y, ℑY )
and let Z ⊆ Y . Let Z̄ be the ℑX -closure of Z (i.e., the smallest ℑX -closed set
containing Z, or the intersection of all ℑX -closed sets that contain Z). Then, the
ℑY -closure of Z equals Z̄ ∩ Y .
14
2.2
Continuity (Revisited)
Definition 2.11. (Continuity of a function) Let (X, ℑX ) and (Y, ℑY ) be two topological spaces. A function f : X → Y is said to be (ℑX − ℑY ) continuous if, for
each ℑY -open subset V of Y , the inverse image f −1 (V ) is a ℑX -open subset of X.
Note that if the topology of the range space Y is given by a basis B, then
continuity of f implies that the inverse image of every basis element is ℑX -open.
S
An arbitrary ℑY -open set V is a union of basis elements, i.e., V = α Bα .
Theorem 2.1. Let (X, ℑX ) and (Y, ℑY ) be two topological spaces, and let f : X →
Y .Then, the following conditions are equivalent.
1. f is ℑX − ℑY continuous.
2. f (Ā) ⊆ f (A) ∀ A ⊆ X.
3. f −1 (B) is ℑX -closed if B is ℑY -closed.
4. For each x ∈ X and each (ℑY -open) neighborhood V of f (x), there is a
(ℑX -open) neighborhood U of x such that f (U ) ⊆ V .
If the condition (4) holds for a point x ∈ X, then f is ℑX − ℑY continuous at the
point x.
Proof. We will show that (1) ⇒ (2) ⇒ (3) ⇒ (1) and (1) ⇒ (4) ⇒ (1).
To show that (1) ⇒ (2), we assume ℑX −ℑY continuity of f and let A ⊆ X. We
will show that if x ∈ Ā, then f (x) ∈ f (A). Let V be a (ℑY -open) neighborhood of
f (x). Then, f −1 (V ) is a ℑX -open set containing x and therefore f −1 (V ) intersects
A in some point x̃. Then, V intersects f (A) in f (x̃), which implies that f (x) ∈ f (A).
To show that (2) ⇒ (3), let B ∈ Y be ℑY -closed and A = f −1 (B). We will
show that A is ℑX -closed, i.e., Ā = A. Let V be a ℑY -neighborhood of f (x). Then,
f −1 (V ) is a ℑX -open set containing x and therefore f −1 (V ) intersects A in some
point x̃. From set theory, it follows that f (A) = f (f −1 (B)) ⊆ B. Hence, if x ∈ Ā,
then f (x) ∈ f (Ā) ⊆ f (A) ⊆ B̄ = B.
To show that (3) ⇒ (1), let V be ℑY -open. By setting B = Y \ V , i.e., B is
ℑY -closed, it follows that
f −1 (B) = f −1 (Y ) \ f −1 (V ) = X \ f −1 (V )
Since V is ℑY -open and f is ℑX − ℑY continuous, f −1 (V ) is ℑY -open. Hence,
−1
f (B) is ℑX -closed.
To show that (1) ⇒ (4), let x ∈ X and let V be a (ℑY -open) neighborhood
of f (x). Then, the set U = f −1 (V ) is a (ℑX -open) neighborhood of x such that
f (U ) ⊆ V .
To show that (4) ⇒ (1), let V be ℑY -open and let x ∈ f −1 (V ). Then, f (x) ∈
V so that, by hypothesis, there is a (ℑX -open) neighborhood Ux of x such that
f (Ux ) ⊆ V ⇒ Ux ⊆ f −1 (V ). It follows that f −1 (V ) can constructed as a union
of ℑX -open sets Ux so that it is open.
Definition 2.12. (Homeomorphism) Let (X, ℑX ) and (Y, ℑY ) be two topological
spaces. A bijective function f : X → Y is said to be homeomorphic if it is bicontinuous (i.e., f is (ℑX − ℑY ) continuous and f −1 is (ℑY − ℑX ) continuous).
In other words, a bijective function f : X → Y is homeomorphic provided that
15
the statement f (U ) is ℑY -open if and only if U is ℑX -open is true. Thus, two
topological spaces (X, ℑX ) and (Y, ℑY ) are said to be homeomorphic to each other
if there exists a bijective bi-continuous mapping from X to Y .
x
Example 2.3. Let the function f : (−1, 1) → R be defined as f (x) 7→ 1−x
2.
2y
√
.
It
is
evident
that
f
is
a
The bijective function f yields f −1 (y) 7→
2
1+
1+4y
homeomorphism.
Example 2.4. A bijective function can be continuous without being a homeomorphism. For example, consider the function f : [0, 1) → S 1 , where S 1 , {(x, y) ∈
R2 : x2 + y 2 = 1}, defined as f (t) 7→ (cos 2πt, sin 2πt). Although f is bijective and
continuous, the inverse function f −1 is not continuous. To see this, verify that the
image under f of the open set U = [0, 41 ) is not open in S 1 because the point f (0)
is not contained in any open set V of R2 such that V ∩ S 1 ⊂ f (U ).
Remark 2.8. The topological space (R, U) is homeomorphic to the subspace relativized by any open interval in R. For example, test a function f : (0, 1) → R
with f (x) = tan−1 π(x − 12 ) , which is both bijective and bi-continuous. You
2x−1
, which is also both bijective and
may also try with another function f (x) = x(x−1)
bi-continuous.
Definition 2.13. (Imbedding) Let (X, ℑX ) and (Y, ℑY ) be two topological spaces,
and f : X → Y be an injective continuous function. Let Z be the image set f (X),
considered as a subspace of Y . By restricting the range of f , if the bijective function
f˜ : X → Z happens to be a homeomorphism of X with Z, then f : X → Y is called
an imbedding of X in Y .
2.3
Separation and Connectedness
This section introduces the concepts of separation and connectedness and then we
deal with compactness in the next section. We see that a space is “separated” if it
can be broken into disjoint open sets. Otherwise, the space is said to connected. A
formal definition follows.
Definition 2.14. Let (X, ℑ) be a topological space. A separation of X is a pair
U, V of disjoint nonempty ℑ-open subsets of X such that U ∪ V = X. The space
X is said to be connected if there does not exist a separation of X. Equivalently,
connectedness in a topological space can be defined as follows.
A topological space X is connected if and only if the subsets that are both open
and closed in X are solely the empty set ∅ and the set X itself.
Let a nonempty proper subset A
X be both open and closed in X. Then the
sets, U = A and V = X \ A, constitute a separation of X because they are open,
disjoint, and nonempty and also their union is X. Conversely, if U and V form a
separation of X, then U 6= ∅, U 6= X, and U is both open and closed in X.
Example 2.5. The set Q of rational numbers is not connected because the only
connected subspaces of Q are the singleton sets as there are irrational numbers
between any two distinct rational numbers. If S is a subspace of Q containing two
distinct rational numbers p and q, we can choose an irrational θ between p and q.
Then, S is the union of two disjoint non-empty sets, S ∩ (−∞, θ) and S ∩ (θ, ∞).
16
Now we introduce the concept of Hausdorff space that is a restricted version of
the general notion of a topological space. In a metric space (e.g., the real line R
with the metric d(x, y) , |x − y|), a sequence can have at most one limit, i.e., if
a sequence converges in a metric space, then it cannot converge to more than one
point. However, in a topological space, this assertion may not hold as seen in the
definition below and the following example.
Definition 2.15 (Topological convergence of a sequence). Let (X, ℑ) be a topological space. A sequence {xk } of points in X converges to a point x in X provided
that, corresponding to each ℑ-neighborhood U of x, there exists n ∈ N such that
xk ∈ U ∀k ≥ n. In other words, for the sequence {xk } to converge to x, each
ℑ-neighborhood of x must contain all but finitely many points of {xk }.
Example 2.6. Let a sequence of points {xk } in an arbitrary topological space
(X, ℑX ) converge to a point x ∈ X. Then, corresponding to each neighborhood Ux
of x, there is a positive integer Nx such that xn ∈ Ux for all n ≥ Nx . For example,
let X = {a, b, c} and its topology be ℑX = {∅, {a, b}, {b, c}, {b}, X}. Note that the
set {b} is ℑX -open but {b} is not ℑX -closed, because the complement of {b} in X
is {a, c} that is not ℑX -open. Now, setting xn = b for all n, a sequence of points
{b, b, b, · · · } in X obviously converges to the point b. The sequence {b, b, b, · · · }
also converges to the point a and to the point c, which would not be true if {a, c}
was a ℑX -open set. The rationale is that if {a, c} was a ℑX -open set, then both
{a} and {c} become ℑX -open sets by the finite intersection property of topological
spaces; in that case, the sequence {b, b, b, · · · } would only converge to the point
b and not to any one of the points a and c. We also observe that if {a, c} is a
ℑX -open set, then {b} becomes ℑX -closed.
Topologies in which a sequence may converge to more than one point, or in
which a singleton set is not closed, are not of much interest to physicists and
engineers. To address this issue, Felix Hausdorff put a restrictive condition on
topological spaces by a separation of the topological space by disjoint neighborhoods
of distinct points as formally defined below.
Definition 2.16. (Hausdorff Space) A topological space (X, ℑX ) is called Hausdorff (or T2 ) if, for each pair of distinct points x, x̃, i.e., x 6= x̃ in X, there exist
e of x and x̃, respectively, that are mutually disjoint, i.e.,
neighborhoods U and U
e = ∅. Note: A neighborhood of a point x ∈ X is a ℑX -open set containing x.
U ∩U
Example 2.7. An example of a Hausdorff space is the usual topology (R, U).
Remark 2.9. Let (X, ℑX ) be a Hausdorff space. Every finite subset of X is ℑX closed.
Remark 2.10. Let (X, ℑX ) be a Hausdorff space. Every sequence of points in X
converges to at most one point of X.
The following results are presented in this context.
Proposition 2.1. Every finite set of points in a Hausdorff space is closed.
17
Proof. It suffices to show that every singleton set {x} in the space X is closed. If
x̃ ∈ X and x̃ 6= x, then there exist respective neighborhoods Ux and Ux̃ that are
mutually disjoint because of the Hausdorff property. Since Ux̃ does not intersect
{x}, the point x̃ cannot belong to the closure of the set {x}. Consequently, the
closure of the set {x} is {x} itself, which implies that {x} is a closed set.
Proposition 2.2. Let (X, ℑX ) be a Hausdorff space. Then, a sequence of points
in X converges to at most a single point.
Proof. Let a sequence {xk } converge to two distinct points x and x̃ in X, i.e., x̃ 6= x.
Then, there exist Ux and Ux̃ that are mutually disjoint neighborhoods of x and x̃,
respectively. Now since Ux contains xk for all but finitely many values of k, the set
Ux̃ will contain at most finitely many points of the sequence {xk }. Therefore, {xk }
cannot converge to x̃.
Remark 2.11. If a sequence {xk } of points in a Hausdorff space X converges to
a point x ∈ X, then we say that x is the limit of the sequence {xk } and denote it
as xn → x.
For the sake of completeness, we present the definitions of T0 , T1 , T3 , and
T4 spaces that are also encountered in engineering analysis. As stated earlier in
Definition 2.16, a Hausdorff pace is also called a T2 space.
Definition 2.17. (T0 space) A topological space (X, ℑX ) is called T0 if, for each
pair of distinct points x and x̃ (i.e., x 6= x̃) in X, there exists a ℑX -neighborhood
of x or x̃ such that it does not contain the other.
Definition 2.18. (T1 space) A topological space (X, ℑX ) is called T1 if, for each
pair of distinct points x and x̃ (i.e., x 6= x̃) in X, there exist a ℑX -neighborhood of
x that does not contain x̃ and a ℑX -neighborhood of x̃ that does not contain x.
Definition 2.19. (T3 space) A topological space (X, ℑX ) is called T3 or regular if
every one-point set is ℑX -closed and if, for each x ∈ X and any ℑX -closed F such
that x ∈
/ F , there exist ℑX -open subsets U and V such that x ∈ U , F ⊂ V , and
T
U V = ∅.
Definition 2.20. (T4 space) A topological space (X, ℑX ) is called T4 or normal
if every one-point set is ℑX -closed and if, for any two disjoint ℑX -closed subsets
e such that F ⊂ U , Fe ⊂ Ũ , and
F and Fe, there exist ℑX -open subsets U and U
Te
U U = ∅.
Remark 2.12. T4 ⊆ T3 ⊆ T2 ⊆ T1 ⊆ T0 .
2.4
Compactness
While connectedness is a natural property of a topological space, the notion of
compactness is not so natural as seen below.
Definition 2.21. Let X be a topological space and let S ⊆ X. Let C , {Cα :
α ∈ I} be a collection of subsets of X, where I is a nonempty index set that can
be finite or countable or uncountable. Then, C is said to be a covering of S if
S
S ⊆ α∈I Cα . If C1 is a covering of S and C2 is a covering of S such that C2 ⊆ C1 ,
then C2 is said to be a subcovering of C1 .
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Definition 2.22. Let (X, ℑX ) be a topological space. A covering C of S ⊆ X
is said to be a ℑX -open covering of S if every member of C is a ℑX -open set. A
covering C of a set S is called finite if card(C) is finite, i.e., if C has finitely many
members.
Example 2.8. Consider the real space with the usual topology
R, U that
n
o is
k
generated by the usual metric d(x, y) = |x − y|. Then, C1 , 0, k+1 : k ∈ N is
o
n
4k+3
: k ∈ N is an open
an open covering of (0, 1). It is also true that C2 , 0, 4(k+1)
covering of (0, 1). Since C2 ⊂ C1 , it follows that C2 is an open subcovering of C1 .
Definition 2.23. A set S ⊆ X in a topological space (X, ℑX ) is said to be ℑX compact if every ℑX -open covering of S has a finite subcovering.
Proposition 2.3. The topological space R, U is not compact. Hence, any open
interval (a, b) ⊂ R, where a < b, is not compact in the relativized topology.
Proof. Let us construct C , {(−k, k) : k ∈ N} as a U-open covering of R. For
every x ∈ R, there exists nx ∈ N such that nx > |x|, which implies that x ∈
S
(−nx , nx ). So, x ∈ n∈N (−n, n). Now, let (−n1 , n1 ), (−n2 , n2 ), · · · , (−nk , nk ) be
any arbitrary finite collection of members of C and let n∗ = max(n1 , n2 , · · · , nk ).
S
Then, n∗ ∈
/ n∈N (−n, n). Therefore, there is no finite collection of members of C
which is a covering of R. So, R, U is not compact.
The second assertion follows from a homeomorphism between R, U and
(a, b), U(a,b) , where (a, b) is an open interval in R and U(a,b) is the usual topology
U relativized to (a, b).
2.5
Continuity and Compactness
This subsection establishes the connections between continuity and compactness in
metric spaces.
Lemma 2.1. An infinite subset of a compact set K must have a limit point in K.
Proof. We prove the lemma by contradiction. Given an infinite subset A of the
compact set K, let no point of K be a limit point of A. Then, each q ∈ K would
have a neighborhood Vq that contains at most one point of A, namely, q if q ∈ A.
Therefore, no finite subcollection of {Vq } can cover A, and the same is true for K
because A ⊆ K. This contradicts the compactness of K.
Theorem 2.2. Let (X, dX ) be a compact metric space and (Y, dY ) be an arbitrary
metric space. Let f : X → Y be a continuous mapping. Then.
(i) f (X) is compact.
(ii) f is uniformly continuous on X.
Proof. We start with Part (i). Let {Vα } be a dY -open cover of f (X), Since f is
−1
dX − dY )-continuous, each of the
n sets f (Vα ) is dx -open. Since Xo is compact,
−1
there exists a finite open cover f (Vαk ), k ∈ {1, 2, · · · , n}, n ∈ N} of X.
Sn
−1
n Since f f (E) ⊆ E for every
o E ⊆ Y , it follows that f (X) ⊆ k=1 Vαk , i.e.,
Vαk , k ∈ {1, 2, · · · , n}, n ∈ N} of Y . This completes the proof of Part (i).
19
We prove Part (ii) by contradiction. Let us assume that f is not uniformly continuous although f is pointwise continuous. Then, for an arbitrarily chosen ε > 0
and every k ∈ N, there are two sequences {xk } and {x̃k } such that dX (xk , x̃k ) < k1
but dY (f (xk ), f (x̃k ) ≥ ε. By Lemma 2.1, the sequence {xk } must have a subsequence {xkn } that, in turn, must converge to a point x ∈ X. Hence,
ε ≤ dY (f (xk ), f (x̃k ) ≤ dY (f (x), f (x̃kn ) + dY (f (x), f (xkn )
which implies that at least one of the following two inequalities dY (f (x), f (x̃kn ) ≥ 2ε
and dY (f (x), f (xkn ) ≥ 2ε must hold for arbitrary n ∈ N. This contradicts the
continuity assumption of f .
2.6
Frequently Used Results in Compact Subspaces of Rn
This subsection presents a few important results on compact subspaces of Rn .
1. Let (X, ℑ) be a compact topological space and K ⊆ Y ⊆ X. Then K is
compact relative to ℑX if and only if K is compact relative to ℑY .
Sketch of the proof : See Rudin, Principles of Mathematical Analysis, p. 37.
2. Compact subsets of a metric space are closed,
Sketch of the proof : See Rudin, Principles of Mathematical Analysis, p. 37.
3. Closed subsets of a compact space are compact in the respective relativized
topology.
Sketch of the proof : See Rudin, Principles of Mathematical Analysis, pp.
37-38.
4. (Heine-Borel Theorem): For X ⊆ R, (X, U) is compact if and only if X is
bounded and U-closed.
Sketch of the proof : See Rudin, Principles of Mathematical Analysis, p. 40.
2.7
Concepts of ε-net and total boundedness
Definition 2.24. (ε-net) Let S be a set in a metric space (X, d). Given ε > 0, a
subset Sε ⊆ S is called an ε-net in S if the following two conditions are satisfied.
1. ∀y ∈ S there exists x ∈ Sε such that d(x, y) < ε.
2. Sε is finite.
Definition 2.25. (Totally bounded) A set S in a metric space (X, d) is called
totally bounded if ∀ε > 0 there exists an ε-net Sε in S.
Remark 2.13. For all spaces, totally bounded ⇒ bounded.
dimensional spaces, bounded ⇒ totally bounded.
Only for finite-
Theorem 2.3. Let (S, d) be a metric space. Then the following statements are
equivalent.
1. The metric space (S, d) is compact.
2. The metric space (S, d) is sequentially compact.
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3. The metric space (S, d) is closed and totally bounded.
4. Every infinite subset of S has an accumulation point.
Proof. See Naylor & Sell (pp. 142-145).
2.8
Metrizable Topology
Definition 2.26. If d is a metric on a set S, then the collection of ε-balls Bεd (x)
for all x ∈ S and all ε > 0 is a basis for a topology on S (see Definition 2.5), called
the metric topology induced by d. That is, a set U is open in this metric topology
if and only if, for each x ∈ U , there exists an ε > 0 such that Bεd (x) ⊆ U .
Definition 2.27. Let (S, ℑ) be a topological space. Then, (S, ℑ) is said to be
metrizable if there exists a metric d on S, which induces the topology (S, ℑ).
Example 2.9. Let S be a set with at least two elements. Then, the indiscrete
topology of S is not metrizable; however, the discrete topology of S is metrizable,
regardless of the cardinality of S.
Theorem 2.4. Every regular (i.e., T3 ) topological space with a countable basis is
metrizable.
Proof. The proof is given in Munkres, pp. 215-217.
Example 2.10. Let S = {0, 1} and let ℑ1 = {∅, S}, ℑ2 = {∅, {0}, S}, ℑ3 =
{∅, {1}, S}, and ℑ4 = {∅, {0}, {1}, S} be topologies on S. Then, the topology
(S, ℑ4 ) is metrizable but (S, ℑ1 ), (S, ℑ2 ) and (S, ℑ3 ) are not metrizable.
Remark 2.14. Different metric spaces may generate the same topological space.
For example, in a finite-dimensional vector space, all norms are topologically equivalent, i.e., a norm that serves as a metric generates the same topological space as
another norm. This topic is discussed in more details in Chapter 3.
Definition 2.28. A topological space X, ℑ is said to have a countable basis at
x ∈ X if there is a countable collection {Un } of (ℑ-open) neighborhoods of the
point x such that any (ℑ-open) neighborhood U of x contains at least one of the
sets Un . The topological space (X, ℑ) is said to be first countable if it has a
countable basis at each of its points, and is said to be second countable if the entire
space has a common countable basis for all of its points.
Remark 2.15. Second countability ⇒ first countability; if B is a countable basis
for (X, ℑ), then a subset of B consisting of those basis elements containing the point
x is a countable basis at x. For example, the usual topology of Rn , n ∈ N, satisfies
second countability; the uniform topology of the (countably infinite-dimensional)
space Rω satisfies first countabilitybut it does not satisfy second countability.
Remark 2.16. The topology of an uncountable product of R, i.e., RJ , where the
index set J is uncountable, is not metrizable. Also note that a metrizable space
always satisfies the first countability axiom but the converse is not true.
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Definition 2.29. (Density and Separability) Let (S, ℑ) be a topological space.
Then, A ⊆ S is said to be dense in (S, ℑ) if A = S. The space (S, ℑ) is said to be
separable if there exists a countable dense subset A ⊆ S.
Remark 2.17. In general, separability is a weaker condition than second countability. However, for a metrizable topology, separability is equivalent to second
countability.
Proposition 2.4. Separability is a topological property. That is, if two topological
spaces (X, ℑX ) and (Y, ℑY ) are homeomorphic, then it follows that
(X, ℑX ) is separable ⇔ (Y, ℑY ) is separable .
Proof. It is given that (X, ℑX ) and (Y, ℑY ) are homeomorphic, i.e., there exists
a bijective function ϕ : X → Y such that ϕ is ℑX − ℑY continuous and ϕ−1 is
ℑY − ℑX continuous. First, let us assume that the space (X, ℑX ) is separable, i.e.,
there exists a countable Ω ⊆ X such that Ω = X. To show that the space (Y, ℑY )
is separable, we proceed as follows.
Since Ω is countable and ϕ is bijective, the image ϕ(Ω) is countable. To show
that ϕ(Ω) = Y , we make use of continuity of ϕ. For any ℑY -open set W , its inverse
T
image ϕ−1 (W ) is ℑX -open. Since Ω = X, there exists z ∈ Ω ϕ−1 (W ) such that
T
T
T
ϕ(z) ∈ ϕ(Ω) W , which implies that ϕ(Ω) W is nonempty. Therefore, ϕ(Ω) W
is dense in (Y, ℑY ). So, (X, ℑX ) is separable ⇒ (Y, ℑY ) is separable .
Similarly, starting with the bijective function ϕ−1 : Y → X and assuming that
the space (Y, ℑY ) is separable, we establish that (Y, ℑY ) is separable ⇒ (X, ℑX )
is separable .
Remark 2.18. Completeness is not a topological property although it is a prop
erty of metric spaces. As a counterexample, let R, U be the usual topology
and (0, 1), U(0,1) be its subspace topology. While R, U is homeomorphic to
(0, 1), U(0,1) and R, U is a complete space, (0, 1), U(0,1) is not a complete space.
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