Complete Metric Space: 1 Sequence

Complete Metric Space
The aim of this chapter is to study one of the properties of metric space. The notion of distance
between points of an abstract set leads naturally to the discussion of convergence of sequences
and Cauchy sequences in the set. Unlike the situation of real numbers, where each Cauchy
sequence is convergent, there are metric spaces in which Cauchy sequences fail to converge. A
metric space in which every Cauchy sequence converges is called a ‘complete metric space’. This
property plays a vital role in analysis when one wishes to make an existence statement. We shall
see that a metric space need not be complete and hence we shall find conditions under which
such a property can be ensured.
1 Sequence
Let (X, d) be a metric space. A sequence of points in X is a function x : N → X. In other
words, a sequence assigns to each n ∈ N a uniquely determined element of X. If f (n) = xn , it is
customary to denote the sequence by the symbol hxn i, or by x1 , x2 , · · · , xn , · · · .
(i) Let hxn i be a sequence in the set X. If there exists n0 ∈ N such that xn = a for all n > n0 ,
then the sequence hxn i is called eventually constant.
(ii) If xn = n for all n ∈ N, then the sequence hxn i is called constant sequence. Obviously, a
constant sequence is a special case (n0 = 1) of an eventually constant sequence.
(iii) A sequence hxn i is said be frequently in a set S if for each positive integer m ∈ N, there
exists a positive integer n ≥ m such that xn ∈ S.
(iv) Suppose X is a non-empty set and x = hxn i is a sequence in X. For each m ∈ N, the set
{xn : n ∈ N, n ≥ .m} is called the mth tail of the sequence hxn i.
1.1 Convergent Sequence

The concept of a convergent sequence plays an important role to investigate the closedness
of a set, the continuity of a function and several other properties. In this section, we give
an introduction to the convergence of sequences and the properties under which a sequence is
convergent in arbitrary metric spaces.
Definition 1 [Convergence of a Sequence] Let (X, d) be a metric space and A be a non-

empty subset of X. A sequence hxn i in X is said to be convergent if there is a point a ∈ X
such that for each real ε > 0, ∃ a k ∈ N such that d(xn , a) < ε ∀ n ≥ k. Equivalently for each
open sphere S(a, ε) centred at a, ∃ a positive integer k such that
xn ∈ S(a, ε); ∀n ≥ k. (1)
Here a is called the limit of the sequence hxn i and we write lim xn = a.
n→∞
1
2 Small Overview On Complete Metric Space
(i) Therefore a sequence hxn i in a matric space (X, d) converges to a if and only if the
hd(xn , a)ii in Ru converges to zero.
(ii) A singleton set in a metric space is, of course, included in every ball centred at its only
point. A sequence that has a singleton set for a tail is said to be eventually constant; such
sequences must converge in any metric space to which they belong. Thus, an eventually
constant sequence, and hence a constant sequence, is convergent.
(iii) A constant sequence, in which all terms are the same, is a special case of an eventually
constant sequence and converges to its single value in any metric space to which it belongs.
(iv) In a discrete metric space X, it is very difficult for a sequence to converge. Let (X, d) be a
discrete metric space, and hxn i be a sequence in X, which converges to a in X. Let ε = 1/2.
Since hxn i converges to a, ∃m ∈ N such that d(xn , ) < ε = 1/2 for all n ≥ m. This means
that xn = a for n ≥ m, i.e., the sequence is of the form hx1 , x2 , · · · , xm−1 , a, a, · · · i. Each
singleton set {a} is open and the only way that {a} can include a tail of a sequence hxn i
is if the sequence is eventually constant with xn = a for all sufficiently large n ∈ N. Thus
in a discrete metric space, a sequence can converge to a point only if it is an eventually
constant sequence.
(v) The sequence h1/ni of inverses of the natural numbers converges to 0 because for each
r ∈ R+ there exists k ∈ N such that 1/k < r and then the ball B[0; r) includes the k th tail
of the sequence h1/ni.
(vi) Let X denote the space of all sequences of numbers with metric d defined by
∞
X 1 |xj − yj |
d(x, y) = j
; x = hxj i, y = hyj i ∈ X.
2 1 + |xj − yj |
j=1
(n)
Let hx(n) i = hhxj ii be a sequence in X which converges to, say, x ∈ X. In other words,
∞ (n)
X 1 |xj − xj |
d(x(n) , x) = → 0 as n → ∞.
2j 1 + |x(n) − xj |
j=1 j
This means that, for ε > 0, there exists an integer n0 (ε) ∈ N such that
∞ (n)
X 1 |xj − xj |
< ε, whenever n ≥ n0 (ε)
2j 1 + |x(n) − xj |
j=1 j
(n)
1 |xj − xj |
⇒ < ε, whenever n ≥ n0 (ε); j = 1, 2, · · · .
2j 1 + |x(n) − xj |
j
(n)
Since ε > 0 is arbitrary, it follows that lim xj = xj for each j. Consider any ε > 0.
n→∞
∞
1/2j < ε/2. Consequently,
P
There exists an integer m such that
j=m+1
m (n) ∞ (n)

(n)
X 1 |xj − xj | X 1 |xj − xj |
d x ,x = +
2j 1 + |x(n) − xj | 2j 1 + |x(n) − xj |
j=1 j j=m+1 j
m (n)
X 1 |xj − xj | ε
< j (n)
+ .
2 1 + |x − xj | 2
j=1 j
Dr.Prasun Kumar Nayak & Dr.Himadri Shekhar Mondol Home Study Materiel
(n)
As the first sum contains only finitely many terms and lim xj = xj for each j, there
n→∞
exists n0 (ε) ∈ N such that
m (n)
X 1 |xj − xj |
< ε/2, whenever n ≥ n0 (ε).
2j 1 + |x(n) − xj |
j=1 j
Hence, d(x(n) , x) < ε whenever n ≥ n0 (ε). Thus, convergence in the space of all sequences
is co-ordinatewise convergence. The metric space is known as Frechet space.
Result 1 (Criteria for Convergence): Suppose X is a metric space, a ∈ X and hxn i is a

sequence in X. Then following are some equivalent statements, any one of which might be used
in a definition of convergence in metric spaces:
\ DD E E
(i) (closure criterion I): xn : n ∈ S | S ⊆ N, S = infinite = hai.
\ DD E E
(ii) (closure criterion II): x∈ xn : n ∈ S | S ⊆ N, S = infinite .

(iii) (distance criterion) : dist a, {xn : n ∈ S} = 0 for every infinite subset S of N.
(iv) (ball criterion): Every open ball centred at a includes a tail of hxn i.
(v) (open set criterion): Every open subset of X that contains a includes a tail of hxn i.
Theorem 1 In a metric space, every convergent sequence has an unique limit.
Proof: Let (X, d) be a metric space and hxn i be a convergent sequence in X. If possible, let
a and b be two limits of hxn i, i.e., a 6= b, so d(a, b) > 0. Since lim xn = a and lim xn = b, so
n→∞ n→∞
corresponding to an ε > 0, ∃ n1 , n2 ∈ N such that
d(xn , a) < ε/2, for n ≥ n1 and d(xn , b) < ε/2, for n ≥ n2 .
Using the triangle inequality, we have
d(xn , xn ) ≤ d(xn , a) + d(a, b) + d(xn , b)

or , d(xn , xn ) − d(a, b) ≤ d(xn , b) + d(xn , a). (2)
Also, interchanging xn , a and xn , b in Eq.(2), we obtain
d(a, b) − d(xn , xn ) ≤ d(xn , b) + d(xn , a). (3)
Combining (2) and (3), we get

h i h i
− d(xn , b) + d(a, xn ) ≤ − d(xn , xn ) − d(a, b) ≤ d(xn , b) + d(xn , a)

⇒ d(x , x ) − d(a, b) ≤ d(xn , b) + d(xn , a)

n n

⇒ 0 ≤ d(xn , xn ) − d(a, b) ≤ ε, for n ≥ m = max{n1 , n2 }

⇒ 0 = d(xn , xn ) → d(a, b).
∴ d(a, b) = 0 and so a = b. Thus the limit of the sequence hxn i is unique.

Note: A sequence hxn i which is not convergent is called divergent.
1.2 Convergence of Subsequences

Every subsequence of a convergent sequence converges to the same limit as the parent sequence.
A sequence that does not converge, however, may have many convergent subsequences with
various limits. We discover in this section what those limits are.
Definition 2 Suppose X is a metric space, x ∈ X and hxn i is a sequence in X that converges

to x. Suppose hxmn i is a subsequence of hxn i. Then the sequence hxmn i also converges to x.
Theorem 2 Let hxn i be a convergent sequence in a metric space (X, d) such that
xn → x as n → ∞. If hxnk i is any subsequence of hxn i then xnk → a as k → ∞.
Proof: Since hxn i converges to a for any ε > 0, ∃ a natural number m such that
d(xn , a) < ε; ∀ n ≥ m.
Since hnk i is a strictly monotonically sequence of natural number so ∃ a k1 ∈ N such that

nk ≥ m, ∀ k ≥ k1 . Form these, we get
d(xnk , a) ≤ d(xnk , xn ) + d(xn , a) < ε, ∀ k ≥ k1 .
∴ The sub-sequence hxnk i converges to a. Thus every subsequence of a convergent sequence is

convergent.
Result 2 If a subsequence in a metric space (X, d) is convergent, then the sequence itself need
not be convergent. for example
Consider hxn = (−1)n ; n ∈ Ni in Ru . Let the subsequence hx2n i of the sequence hxn i as
x2n = 1; ∀n ∈ N, then x2n → 1 as n ⇒ ∞ in Ru . However hxn i is not convergent sequence.
1.3 Bounded Sequence

Definition 3 In a matric space a sequence is said to be bounded if the range of the sequence
forms a bounded set.
Theorem 3 In a matric space every convergent sequence is bounded.
Proof: Let hxn i be a convergent sequence to a. Then forε = 1, ∃ a natural number k such
that
d(xn , a) < 1, ∀ n ≥ k.
Let r = max{1, d(xn , x) : 1 ≤ n < k}, then d(xn , x) ≤ r, ∀ n ∈ N. Now
d(xm , xn ) ≤ d(xm , x) + d(x, xn )

≤ r + r = 2r.
i.e., d(xm , xn ) ≤ 2r, ∀ m, n ∈ N. The diameter of the range set of the sequence is bounded by
2r. Thus hxn i is bounded.
Result 3 The property of convergence of a sequence in a metric space (X, d) depends on the
set X as well as the metric d.
(i) When different metrics are placed on a given set, the sequences that converge may differ
and limits may differ too. For example, when R is endowed with the discrete metric, the
sequence h1/ni does not converge. However, the formal definition of convergence (unlike
the distance criterion and the ball criterion) does not rely on a fixed metric but only on the
open sets produced by the metric. It follows that, on any given set, metrics that produce
the same topology produce also the same convergent sequences with the same limits.
(ii) Convergence depends on set X: Let hxn i be a sequence in the usual metric space Ru ,
1
where xn = , ∀ n ∈ N. Then hxn i converges to 0 in Ru . However if we take X = (0, 1]
n
with usual matric, then the same sequence does not converge in X, as ‘0’ does not belongs
to X.
(iii) Convergence depends on the metric d: We know the set C[0, 1] of all continuous
function over [0, 1] forms a matric space with respect to d1 as well as d∞ where
Z 1
d1 (x, y) = |x(t) − y(t)|dt
0
and d∞ (x, y) = sup |x(t) − y(t)| x, y ∈ [0, 1].
t∈[0,1]
Let us consider the sequence hxn i in the space C[0, 1], where xn (t) = e−nt , t ∈ [0, 1], n ∈ N.
Then the sequence hxn i converges to zero in C[0, 1] with respect to the metric d1 , since
Z 1 Z 1
d1 (xn , 0) = |xn (t) − 0|dt = |e−nt − 0|dt
0 0
1 1 1
− e−nt 0 = (1 − e−n ) → 0.

=
n n
On the other hand the same sequence does not converge to zero, with respect to the metric
d∞ , since
d∞ (x, y) = sup |xn (t) − 0| = sup |e−nt − 0|

t∈[0,1] t∈[0,1]
−nt
= sup e =190 as n → ∞
t∈[0,1]
Theorem 4 (Characterization of limit points of a set in terms of convergent sequence)

: Let (X, d) be a metric space, A ⊆ X and a ∈ X.
(i) Then a is a limit point of A if and only if ∃ a sequence hxn i of points of A,
none of which equals to a such that lim xn = a.
n→∞
(ii) The set A is closed if and only if every convergent sequence of points of A has
its limit in A.
Proof: Let (X, d) be a metric space, A ⊆ X.

(i) Suppose a ∈ X is a limit point of A. Then S(x, r) − {a} ∩ A 6= φ for each r > 0. Let us
construct a sequence hxn i where,
n
1 o
xn ∈ S a, − {a} ∩ A ; for each n ∈ N.
n
Then xn 6= a; ∀n ∈ N. We now prove that hxn i converges to a. Let ε > 0 be given by

Archimidean property ∃ a natural number k ∈ N such that k1 < ε. Then for n > k,
1 1 1
d(xn , a) < < < ε; as xn ∈ S(a, )
n k n
i.e., d(xn , a) < ε ∀ n ∈ N.
Thus {xn } converges to a.

Conversely, suppose ∃ a sequence hxn i in A none of which equals to a such that xn → a
as n → ∞. We shall prove that a is a limit point of A. Let r > 0 be arbitrary. Since hxn i
converges to a, ∃ a natural number k such that d(xn , a) < r, ∀ n ≥ k i.e.,
d(xk , a) < r ⇒ xk ∈ S(a, r).
Since xk 6= a and xk ∈ A,, so

xk ∈ S(a, r) − {a} ∩ A

⇒ (S(a, r) − {a} ∩ A 6= φ.
Since r > 0 is arbitrary, a is a limit point of A.
(ii) Suppose that A is closed and hxn i is a sequence of points of A which converges to a point
a ∈ X. We have to show that a ∈ A.
(a) If the range set of the sequence hxn i is infinite, then it follows from part (i) that a is
a limit point of this set. Since A is closed, we have a ∈ A.
(b) If the range set of the sequence hxn i is finite, then xn = a for all n ≥ n0 , since hxn i
is a convergent sequence. Since each term of the sequence hxn i belongs to A, so is a.
Conversely, assume that every convergent sequence of points of A converges to a point of

A. We are to show that A is closed by showing that it contains all its limit points.
Let a be a limit point of A. Then by part (i), there is a sequence hxn i of points of A, none
of which equals to a, such that xn → a. By hypothesis a ∈ A, so A is closed.
Theorem 5 Let hxn i be a sequence in a matric space (X, d) which converges to x in

X and A be the range of hxn i
(i) If A is a finite set then xn = x for infinitely many n.
(ii) If A is an infinite set then x is a limit point of A.
Proof:
(i) Suppose range of A is finite and A = {y1 , y2 , · · · , ym }. We first show that a ∈ A. If possible
let a 6∈ A, then x 6= yi for i = 1, 2, 3, · · · , m. Then d(yi , a) > 0 for i = 1, 2, 3, · · · , m. Let
r = min{d(yi , a) : 1 ≤ i ≤ m}. Then yi 6∈ S(a, r) for all 1 ≤ i ≤ m ⇒ d(xn , a) ≥ r for all
n, which is a contradiction, as sequence hxn i converges to a.
∴ a ∈ A. With out loss of generality suppose a = y1 . Then d(y1 , yi ) > 0 for 2 ≤ i ≤ m.
Suppose r1 = min{d(y1 , yi ); 2 ≤ i ≤ m}. So, yi 6∈ S(y1 , r1 ) for i = 1, 2, 3, · · · , m; d(y1 , yi ) >

r1 . Then r1 > 0. Again since hxn i converges to x = y1 , for r1 > 0, ∃ a natural number k
such that
d(xn , a) < r1 ; ∀n ≥ k
⇒ d(xn , a) = 0; ∀n ≥ k [Since d(xn , a) = d(yi , y1 ) either = 0 or ≥ r1 ]
⇒ xn = a; ∀n ≥ k
i.e. xn = a for infinitely many values of n.
(ii) Suppose A is an infinite set. Let r > 0. Since hxn i converges to a for r > 0, ∃ a natural
number k such that
d(xn , a) > r ⇒ xn ∈ S(a, r), ∀n ≥ k.
Since A is an infinite set, ∃ a natural number m ≥ k such that xm 6= a. So

xm ∈ S(a, r) − {a}

⇒ xm ∈ S(a, r) − {a} ∩ A

⇒ S(a, r) − {a} ∩ A 6= φ
Since r > 0 is arbitrary, so x is a limit point of A.
Theorem 6 If hxn i and hyn i are two sequences in a metric space (X, d) such that hxn i
converges to a and hyn i converges to b then d(xn , yn ) → d(a, b) as n → ∞.
Proof: Since hxn i and hyn i converge to a and b respectively, for any ε > 0, ∃ natural numbers
k1 and k2 such that
d(xn , a) < ε/2; ∀ n ≥ k1 ; and d(yn , b) < ε/2; ∀ n ≥ k2 .
Let k = max{k1 , k2 } ∈ N, then

d(xn , yn ) − d(a, b) ≤ d(xn , a) + d(yn , b)

ε ε
< + = ε; ∀n ≥ k.
2 2
Thus d(xn , yn ) → d(a, b) as n → ∞.
1.4 Convergence in Product Spaces

Let us turn now to convergence in product spaces.
Theorem 7 Let (X, d1 ) and (Y, d2 ) be two metric spaces. A sequence h(xn , yn )i in
the product metric space X × Y converges to (a, b) if and only if the sequence hxn i
converges to a in X and hyn i converges to b in Y .
Proof: Consider the product metric d : X × Y → R given by

d (x1 , y1 ), (x2 , y2 ) = d1 (x1 , x2 ) + d2 (y1 , y2 ); (x1 , y1 ), (x2 , y2 ) ∈ X × Y.
Clearly, (X × Y, d) is a metric space. Assume that first xn → a in X and yn → b in Y . Then,

for each ε > 0, ∃ positive integers n1 , n2 ∈ N such that
d1 (xn , a) < ε/2, ∀n ≥ n1 and d1 (yn , b) < ε/2, ∀n ≥ n2 .
Let n0 = max{n1 , n2 } ∈ N. Then

d (xn , yn ), (a, b) = d1 (xn , a) + d2 (yn , b)
< ε/2 + ε/2 = ε, ∀n ≥ n0 .
Thus (xn , yn ) → (a, b). Conversely, let (xn , yn ) → (a, b) in X × Y . Then, for each ε > 0, ∃ a
positive integer n0 ∈ N such that

d (xn , yn ), (a, b) < ε, ∀n ≥ n0
⇒ d1 (xn , a) + d2 (yn , b) < ε, ∀n ≥ n0
⇒ d1 (xn , a) < ε, d2 (yn , b) < ε, ∀n ≥ n0
⇒ xn → a and yn → b.
1.5 Cauchy Sequence

In real analysis, we must have come across the concept of a Cauchy sequence where it is proved
that a sequence is a Cauchy sequence if and only if it is convergent. Such an equivalence is
no longer true when these notions are extended to metric spaces. Nevertheless, the notion of
Cauchy sequence is important in metric spaces. For an arbitrary topological space, the concept
of Cauchy sequence does not exist.
Definition 4 Let (X, d) be a metric space. A sequence hxn i of elements of X is said to be a
fundamental or Cauchy sequence if for any ε > 0, ∃ k ∈ N such that
d(xm , xn ) < ε ∀ m, n ≥ k. (4)
Equivalently, let hxn i be a sequence and let Tn = {xk : k ≥ n}, n ∈ N. Then hxn i is a Cauchy
sequence if and only if lim δ(Tn ) = 0, where δ(Tn ) is the diameter of Tn in the real line R.
n→∞
A sequence hxn i in K (R or C) is a Cauchy sequence in the sense familiar from elementary

analysis if and only if it is a Cauchy sequence according to definition (4) in the sense of the
usual metric on K (R or C).
Example 1 A sequence hxn i ⊂ R is a Cauchy sequence in R. Examine whether the
sequence h(xn , sin xn )i is a Cauchy sequence in R2 .
Solution: Let x, y ∈ R, with x 6= y. By Lagrange mean value theorem,
| sin x − sin y| = |x − y|| cos ξ|, min{x, y} < ξ < max{x, y}
⇒ | sin x − sin y| ≤ |x − y|, ∀ x, y ∈ R.
√ hxn i is a Cauchy sequence in R for any ε > 0 ∃ a natural number k such that |xm − xn | <
Since
ε/ 2 ∀ m, n ≥ k. Now
p
(x − xn )2 + (sin xm − sin xn )2
p m
≤ (xm − xn )2 + (xm − xn )2
√ √ ε
= 2|xm − xn | < 2 · √ = ε; ∀m, n ≥ k.
2
Thus h(xn , sin xn )i is a Cauchy sequence in R2 .
Example 2 Let X = C[0, 1] be a metric space with the metric d∞ defined by
d∞ (f, g) = sup |f (t) − g(t)|, for all f, g ∈ X.

t∈[0,1]
nt
Prove that hxn i in X given by xn (t) = , ∀t ∈ [0, 1] is a Cauchy sequence.
n+t
Solution: For m ≥ n, the function
mt nt (m − n)t2
g(t) = xm (t) − xn (t) = − =
m+t n+t (m + t)(n + t)
being continuous on [0, 1], assumes its maximum at some point t0 ∈ [0, 1]. So
(m − n)t20
d∞ (xm , xn ) = sup |xm (t) − xm (t)| =
t∈[0,1] (m + t0 )(n + t0 )
t20 1
≤ ≤ → 0, for large m, n.
n + t0 n
Moreover, the sequence hxn i converges to some limit. Indeed, let x(t) = t, then
nt t2 1
|xn (t) − x(t)| = − t = ≤ → 0, as n → ∞.

n+t n+t n
Therefore, hxn i converges to the limit x, where x(t) = t for all t ∈ [0, 1].
Theorem 8 In a metric space every convergent sequence is a Cauchy sequence.
Proof: Let (X, d) be a metric space. Suppose hxn i be a convergent sequence in X converging
to a. Then for any ε > 0, ∃ a natural number k such that d(xn , a) < ε/2, ∀ n ≥ k. Now, by
triangle inequality
d(xm , xn ) ≤ d(xm , a) + d(a, xm )

ε ε
< + = ε, ∀ m, n ≥ k.
2 2
Thus hxn i is a Cauchy Sequence.
Result 4 A Cauchy sequence may not be convergent as may be seen in the following examples:
(i) Let X = Q be the set of all rational numbers and d(x, y) = |x − y|, for all x, y ∈ X be the
usual metric on X. Consider the sequence hxn i represented in decimal
√ system, such that
2
xn = 1.a1 a2 · · · an is the largest rational number satisfying xn < 2. Then we have the
following sequence of rational numbers
x1 = 1.4, x2 = 1.41, x3 = 1.414, x4 = 1.4142, · · · .

1
∴ d(xm , xn ) = |xm − xn | = 0.00 · · · 0am+1 · · · an < ; ∀n ≥ m.
10m
Therefore, d(xm , xn ) → 0 as m → ∞.
√ Hence the sequence hxn i is a Cauchy sequence and
converges to the irrational number 2 6∈ Q.
(ii) Let X = (0, 1] be a metric space with the usual metric and xn be the sequence in X where
xn = n1 where n ∈ N. Then hxn i is a Cauchy sequence, since for each ε > 0
1 1 1
|xm − xn | = | − |<ε ∀ m, n ≥ [ ] + 1
m n ε
But hxn i does not converge to any point in X as 0 does not belongs to X. Since h1/ni
tends to 0 in R and 0 is the unique limit of the sequence h1/ni in R, no point of (0, 1] can
be the limit of h1/ni.
Theorem 9 Let hxn i be a Cauchy sequence in a metric space (X, d). Then hxn i is
convergent iff it has a convergent sub-sequence.
Proof: If hxn i is a convergent then, by Theorem 2 every sub-sequence of hxn i is convergent.

Conversely: Suppose hxnk i ba a convergent sub-sequence of hxn i such that hxnk i → a as
k → ∞. Let ε > 0 be given. Since hxn i is a Cauchy sequence ∃ a natural number N1 such that
ε
d(xm , xn ) < ∀ m, n ≥ N1
2
Let N = max{N1 , nk1 } and k2 be a natural number such that nk2 ≥ N . So
d(xn , a) ≤ d(xn , xnk2 ) + d(xnk2 , a)

ε ε
< + ∀n ≥ N
2 2
i.e. d(xn , a) < ε ∀ n ≥ N.
Hence the sequence hxn i is a convergent.
Example 3 If hxn i and hyn i are two Cauchy sequences in a metric space (X, d), Show
that hd(xn , yn )i is an convergence sequence of reals.
Solution: Since hxn i and hyn i are Cauchy sequence in (X, d) for ε > 0 ∃ natural number k1 , k2
such that
ε ε
d(xm , xn ) < ; ∀m, n ≥ k1 and d(ym , yn ) < ∀ m, n ≥ k2 .
2 2
Let k = max{k1 , k2 } ∈ N. Then Then

d(xm , ym ) − d(xn , yn ) ≤ d(xm , xn ) + d(ym , yn )
ε ε
≤ + = ε, ∀m, n ≥ k.
2 2
So, hd(xn , yn )i is a Cauchy Sequence of real numbers. Since every Cauchy sequence in R is
convergent, hd(xn , yn )i is a convergent sequence in R.
Theorem 10 Let (X, d) and (Y, ρ) be metric spaces and f : X → Y be a uniformly

continuous function. If hxn i is a Cauchy sequence in X, then hf (xn )i is a Cauchy
sequence in Y .
Proof: Since f is uniformly continuous, corresponding to an ε > 0, ∃ a δ > 0 such that

0 0 0
d(x, x ) < δ; x, x ∈ X ⇒ ρ f (x), f (x ) < ε.
In particular, we have

d(xm , xn ) < δ ⇒ ρ f (xm ), f (xn ) < ε. (5)
Since hxn i is a Cauchy sequence in X, given δ > 0, ∃n0 ∈ N, such that
d(xm , xn ) < δ, ∀m, n ≥ n0 . (6)
Therefore, from Eqs. (5) and (6), it follows that

ρ f (xm ), f (xn ) < ε, ∀m, n ≥ n0 . (7)
Hence hf (xn )i is a Cauchy sequence in Y .
Example 4 Let hxn i and hyn i be sequences in a metric space (X, d) such that hyn i is
Cauchy and d(xn , yn ) → 0 as n → ∞. Prove that
(i) hxn i is a Cauchy sequence in X
(ii) hxn i converges to a ∈ X if and only if hyn i converges to a.
Solution: (i) Let ε > 0 be chosen arbitrary. Since hyn i is a Cauchy sequence, ∃ n1 ∈ N such
that d(ym , yn ) < 3ε ; for all m, n > n1 .
Since d(xn , yn ) → 0 as n → ∞, ∃ n2 ∈ N such that n2 > 3ε , and
d(xn , yn ) < ε/3; for all n > n2 .
Let n0 = max{n1 , n2 } d(xn , yn ) < 3ε ; for all n > n2 by the triangle inequality, we have
d(xn , yn ) ≤ d(xm , ym ) + d(ym , yn ) + d(yn , xn )

ε ε
< + d(ym , yn ) + ; for all n, m ≥ n2
3 3
ε ε ε
< + + = ε ; for all n, m > n2
3 3 3
Thus hxn i is a Cauchy Sequence.
(ii) By the triangle inequality, we have
d(yn , a) ≤ d(yn , xn ) + d(xn , a)

lim d(yn , a) ≤ lim d(yn , xn ) + lim d(xn , a)
n→∞ n→∞ n→∞
≤ 0 + 0 = 0.
Thus yn → a as n → ∞.
2 Equivalence of Metrices
There are many metrices on a given set. For example, we have three different metrices on
R2 . Amoung these metrices, some of them have the same nature in relation to convergence of
sequences and continuity of function. This leads to the notion of equivalent metrices on a set.
Definition 5 [Equivalent metric space]: Let d and d∗ be two metrices on the set X such that
for every hxn i in X and x ∈ X
lim xn = x in (X, d) iff lim xn = x in (X, d∗ ).

n→∞ n→∞
Then the two metrices d and d∗ on the same underlying set X are said to be equivalent.
Using the above definition, we have the following theorem giving the condition for two metrices
on a set X to be equivalent.
Theorem 11 (Necessary condition) Let d and d∗ are two metrices on a nonempty

set X and if there exists a constant K such that
1 ∗
d (x, y) ≤ d(x, y) ≤ Kd∗ (x, y); ∀x, y ∈ X, (8)
K
then the metrices d and d∗ are equivalent.
Proof: Let hxn i be a sequence in X such that xn → x in (X, d). Using the left part of the
inequality Eq. (8), we have
1 ∗
d (xn , n) ≤ d(xn , x) → 0 as n → ∞
K
shows that xn → x in (X, d∗ ). Conversely, if xn → x in (X, d∗ ), then the right part of the
inequality (8), we have
d(xn , n) ≤ K d∗ (xn , x) → 0 as n → ∞
shows that xn → x in (X, d). Hence, from the definition d and d∗ on X are equivalent.
d(x, y)
Example 5 Prove that the metric space (X, d) and (X, d∗ ), where d∗ (x, y) =
1 + d(x, y)
are equivalent.
Solution: Let xn → x in the metric d so that d(xn , x) → 0 as n → ∞. Then obviously,

d∗ (xn , x) → 0 as n → ∞ so that xn → x in the metric d∗ .
Conversely, let xn → x in the metric d∗ so that d∗ (xn , x) → 0 as n → ∞. So corresponding
to an ε > 0, ∃n0 ∈ N such that
d∗ (xn , x) < ε, for n ≥ n0

d(xn , x)
⇒ < ε, for n ≥ n0
1 + d(xn , x)
ε
⇒ d(xn , x) < , for n ≥ n0 .
1−ε
Choosing ε < 1, d∗ (xn , x) → 0 as n → ∞ implies d(xn , x) → 0 as n → ∞. Therefore two
metrices d and d∗ are equivalent. On the other hand, we observe that there is no constant
K > 0 such that d(x, y)/K ≤ d∗ (x, y); ∀x, y ∈ R.
Example 6 Let c be the set of all convergent sequences with the following metrices
d and d∗ . If x = hxn i and y = hyn i are in c, let
∞
X 1 |xn − yn |
d(x, y) = sup |xn − yn |; d∗ (x, y) = .
n∈N 2n 1 + |xn − yn |
n=1
Prove that d and d∗ on c are not equivalent.
Solution: To prove that the two metrices d and d∗ on c are not equivalent, we produce a
sequence which converges to a limit in one metric but does not converge in the other metric.
Consider the sequence hen i, in c whose respective terms are sequences
e1 = (1, 0, 0, · · · ), e2 = (0, 1, 0, · · · ), · · · , en = (0, 0, 0, · · · , 0, 1, · · · ),
where 1 is in the nth place. Let e0 = (0, 0, 0, · · · , 0). In the metric space c of convergent sequences
1 1 1
d(en , e0 ) = 1; d∗ (en , e0 ) = = n+1 → 0 as n → ∞.
2n 1 + 1 2
Thus the sequence hen i converges with respect to the metric d but not with respect to the metric
d∗ . Hence the metrices d and d∗ are not equivalent.
Example 7 Let C[0, 1] be a space of all continuous linear functionals defined on [0, 1].
Consider the metrices d and d∗ defined on C[0, 1] by
Z 1
∗
d(f, g) = sup |f (t) − g(t)|; d (f, g) = |f (t) − g(t)|dt,
0≤t≤1 0
for all f, g ∈ C[0, 1]. Prove that they are not equivalent.
Solution: To prove that the two metrices d and d∗ on C[0, 1] are not equivalent, we produce
a sequence which converges to a limit in one metric but does not converge in the other metric.
Consider the sequence hfn (x)i in C[0, 1], defined by
fn (x) = xn ; for all x ∈ [0, 1].
Also, let f (x) = 0 for all x ∈ [0, 1]. Then
d(fn , f ) = sup |fn (t) − f (t)| = 1, ∀n ∈ N,

0≤t≤1
Z 1
1
d∗ (fn , f ) = tn dt = → 0 as n → ∞.
0 n+1
Therefore, lim d(fn , f ) 6= 0. Thus the sequence hfn (x)i converges to f with respect to the
n→∞
metric d∗ but not with respect to d. Therefore the metrices d and d∗ are not equivalent.
However, for any sequence hfn (x)i in C[0, 1] and any f ∈ C[0, 1], d(fn , f ) → 0 as n → ∞, it
follows that d∗ (fn , f ) → 0 as n → ∞. Observe that
Z 1
∗
d (fn , f ) = |fn (t) − f (t)|dt ≤ sup |f (t) − g(t)| = d(fn , f ).
0 0≤t≤1
3 Complete Metric Space

We now turn to some aspects of convergence in metric space. In our study of Cauchy sequences
in R, we have noted that every Cauchy sequence converges in the real line. In the general case
of metric spaces, this need not happen by following examples
(i) Let us consider X = (0, 1) as a subspace of R with usual metric. Then X is a metric space
in its own right. Then h1/ni is a Cauchy sequence in X but not convergent in X. Hence
(0, 1) is not complete with respect this metric.
1 1 1
(ii) In the metric space (Q, d), the sequence defined by 1 + + + · · · + , whose elements
1! 2! n!
become closer and closer to each other without converging to an element of Q, even though
it is a Cauchy sequence.
That is there are metric spaces in which not all Cauchy sequences of points converge to a
point of the metric space. Hence, it is worthwhile to single out those spaces in which Cauchy
sequence converges to a point of the metric space.
Definition 6 A metric space (X, d) is said to be complete if and only if, every Cauchy sequence
of elements of X converges to some element of X in the space.
(i) The definition of complete metric space at once suggests that, if d(xn , xm ) −→ 0 as m, n →
∞, then ∃ a x0 ∈ X such that d(xn , x0 ) −→ 0 as n → ∞.
(ii) A metric space (X, d) is complete if and only if every Cauchy sequence in X has a conver-
gent subsequence.
Many metric spaces, with or without ordering, have this property. We define completeness as
universal closure. For example,
(i) the usual metric Ru is a complete.
(ii) the usual metric Cu is a complete.
(iii) Every finite-dimensional normed linear space is complete.
Below, we shall give several examples of complete metric space.
Example 8 The set of integers Z with the usual metric is a complete metric space.
Solution: Let hxn i be a Cauchy sequence of integers, that is, each term of the sequence belongs
to Z = {· · · , −2, −1, 0, 1, 2, · · · }. Then the sequence must be of the form {x1 , x2 , x3 , · · · , xn , x, x, x, · · · }.
Indeed, if we choose ε = 1/2, then
xn , xm ∈ Z and |xn − xm | < 1/2 ⇒ xn = xm .
Hence the sequence {x1 , x2 , x3 , · · · , xn , x, x, x, · · · } converges to x.
Example 9 Any set X with discrete metric forms a complete metric space.
Solution: Let (X, d) be a discrete metric space and hxn i be a Cauchy sequence in (X, d). Thus
d(xm , xn ) = 0 if xm = xn and d(xn , xm ) = 1 if xm 6= xn . Then for ε = 1/2, ∃ a natural number
k such that
d(xm , xn ) < 1/2 ∀ m, n ≥ k
⇒ d(xm , xn ) = 0 ∀ m, n ≥ k, since d is discrete metric
⇒ xk = xk+1 = xk+2 = · · · = x (say)
∴ d(xn , x) = 0, ∀n ≥ k i.e., hxn i is of the form {x1 , x2 , · · · , xk , x, x, · · · }. Hence hxn i converges
to x, i.e., every Cauchy sequence is convergent in a discrete metric space. Thus the discrete
metric space (X, d) is a complete. From this we see that a sequence in a discrete metric space
is Cauchy if and only if it is eventually constant.
In particular,, the set (0, 1) with the discrete metric is a complete metric space.
Example 10 A Euclidian n-space Rn is a complete metric space with respect to usual
metric.
Solution: We know Rn is a metric space with respect to the metric d where, d(x, y) =
( n )1/2
X
(ai − bi )2 for all x = (a1 , a2 , · · · , an ); y = (b1 , b2 , · · · , bn ) and x, y ∈ Rn . Let us
i=1
(m) (m) (m)
suppose the coordinate sequences hxm i be Cauchy in Rn , where xm = a1 , a2 , · · · , an for
m = 1, 2, 3, · · · .
∴ For each ε > 0 ∃ a k ∈ N such that
ε
d(xm , xr ) < ∀m, r ≥ k
2
( n )1
2
X (m) (m) 2 ε
i.e., (ai − bi ) < ∀ m, r ≥ k
2
i=1
n ε 2
(m) 2
n o
(m)
X
⇒ ai − bi < ∀ m, r ≥ k.
2
i=1
Since each term in the above inequality are positive so

ε 2
(m) 2
n o
(m)
ai − bi < ∀ m, r ≥ k for i = 1, 2, 3, · · · , n
ε 2
(m) (m)
⇒ ai − bi < ∀ m, r ≥ k for i = 1, 2, 3, · · · , n.
2
(m) (1) (2)
∴ ai or ai , ai , · · · is a Cauchy sequence in R for i = 1, 2, · · · , n. Since R is complete,
(m)
hai i converges to ai (say) in R for i = 1, 2, · · · , n. So,
(r)
lim a = ai for i = 1, 2, · · · , n
r→∞ i
Let x = a1 , a2 , · · · , an , then x ∈ Rn . We now prove that hxm i converges to x.

n n
2 X (m) 2 X (m) (r) 2
∴ d(xm , x) = ai − ai = lim ai − ai
r→∞
i=1 i=1
ε 2
= ∀m≥k
2
ε
∴ d(xm , x) ≤ <ε ∀m≥k
2
Hence the Cauchy sequence hxm i converges to x ∈ Rn . Thus Rn is a complete metric space.
Result 5 Since the completeness of the metric is not destroyed by replacing an equivalent
metric, the above is true for any one of the three equivalent metrices for Rn .
Theorem 12 The space C[a, b] of all continuous real-valued functions defined on [a, b]
with the metric d∞ (f, g) = max |f (t) − g(t)| is a complete metric space.
t∈[a,b]
Proof: Let hfn i be a Cauchy sequence of elements in C[a, b]. Then for each ε > 0, there exists
a k ∈ N such that
d∞ (fm , fn ) < ε; ∀ m, n ≥ k

⇒ sup fm (t) − fn (t) < ε; ∀ m, n ≥ k
t∈[a,b]

⇒ fm (t) − fn (t) < ε; ∀ m, n ≥ k and for t ∈ [a, b].
∴ By Cauchy’s general principle of convergence, hfn i converges uniformly. Let f be the limit
function. Since each hfn i is a continuous function on [a, b] and hfn i converges uniformly to f on
[a, b], f is a continuous function on [a, b] i.e., f ∈ C[a, b]. Again since hfn i converges uniformly
to f on [a, b], for any ε > 0, ∃ a k ∈ N such that
fn (t) − f (t) < ε

∀ n ≥ k, ∀ t ∈ [a, b]
2
ε
⇒ sup fn (t) − f (t) ≤ ∀n≥k
t∈[a,b] 2
ε
⇒ d(fn , f ) ≤ < ε ∀ n ≥ k.
2
Thus hxn i converges to x on C[a, b]. Hence C[a, b] is a complete metric metric space and it is
defined as the space of continuous functions on I.
Example 11 The Hilbert space (lp , d), p ≥ 1 is complete.
Solution: The Hilbert space lp (1 ≤ p < ∞) of all sequences hxn i of real or complex numbers
∞
|xk |p < ∞ with the metric given by
P
such that
k=1
p 1/p
"∞ #
X
d(x, y) = ξi − ηi ; x = {ξi }, y = {ηi }.

i=1
Let htn i, where tn = hxni i∞

i=1 ,
be a Cauchy sequence in lp . Let ε > 0 be a real number. Then
there exists an n0 ∈ N such that
d(tn , tm ) < ε, for all n, m ≥ n0 . (9)
This shows that |xni − xm n ∞
i | < ε for all n, m ≥ n0 and consequently hxi ii=1 is Cauchy in K ( R or
n ∞
C). Since these spaces are complete, hxi ii=1 converges to a point xi ∈ K. Also for each k ∈ N,
the statement (9) gives
k
X
|xni − xm p
i | ≤ ε for all n, m ≥ n0
i=1
k
X
⇒ |xni − xi | ≤ εp for all n ≥ n0 , as m → ∞. (10)
i=1
We need to prove that t = hx1 , x2 , · · · i is in lp . The inequalities (10) above and Minkowski’s
inequality show that
#1/p
k k n p p o 1/p
" " #
X p X n0
xi ≤ xi − xi + xni 0

i=1 i=1
" k
#1/p k
" #1/p
p
n0 p
X X
n0
≤ x − x + x

i i i
i=1 i=1
" k #1/p
X p
≤ ε+ xni 0 .
i=1
k i
p 1/p
Dh X E
Since tn0 = hxni 0 i∞
i=1 is in lp , the above inequality shows that xi is bounded and it
i=1
k
X p
is monotonically increasing, and hence the series xi is convergent. Thus, t is in lp . Also, it
i=1
is obvious from (10) that htn i converges to t, and the completeness of lp -space is now established.
Example 12 Let (X, d) be a metric space and d∗ be the standard bounded metric
on X defined by d∗ (x, y) = min{1, d(x, y)}. Show that (X, d) is complete if and only if
(X, d∗ ) is complete.
Solution: First assume that (X, d) is complete and hxn i is a Cauchy sequence in (X, d∗ ). Let
ε > 0 be a real number. Therefore, for 0 < ε0 < min{ε, 1}, there exists a p ∈ N such that
d∗ (xn , xm ) < ε0 for all n, m ≥ p

0
⇒ d(xn , xm ) < ε < ε for all n, m ≥ p. (11)
Thus, hxn i is a Cauchy sequence in (X, d). Since (X, d) is complete, hxn i converges to a point
x ∈ X. Letting m tend to infinity in Eq. (11), it follows that
d(xn , x) < ε for all n ≥ p.
Now the completeness d(X, d∗ ) follows from the inequalities
d∗ (xn , xm ) ≤ d(xn , x) < ε for all n ≥ p.
Conversely, let hxn i is a Cauchy sequence in (X, d). Let ε > 0 be a real number, and
0 < ε0 < min{ε, 1}. Then there exists a p ∈ N such that
d(xn , xm ) < ε0 for all n, m ≥ p

∗ 0
⇒ d (xn , xm ) < ε < ε for all n, m ≥ p. (12)
Thus, hxn i is a Cauchy sequence in (X, d∗ ). Since (X, d∗ ) is complete, hxn i converges to a point
x ∈ X. Letting m tend to infinity in Eq. (12), it follows that
d∗ (xn , x) < ε0 < ε for all n ≥ p

d(xn , x) ≤ ε for all n ≥ p
and the complete is complete.
Example 13 Let X be the set of all convergent real (or complex ) sequences with
the function d : X × X → R is defined by
n o
d(x, y) = sup |xn − yn | ; x = {xn }, y = {yn } ∈ X.
n
Prove that (X, d) is a complete metric space.
Solution: In Example (??), we see that (X, d) is a metric space. Let htn i, where tn =
hxn1 , xn2 , · · · , xnk , · · · i for all n ∈ N is an element of X be a Cauchy sequence in X. Let
ε > 0 be a real number. Then ∃ a p ∈ N such that
d(tn , tm ) < ε; ∀n, m ≥ p

⇒ sup |xnk − xmk | < ε; ∀n, m ≥ p. (13)
k∈N
Therefore, for each k ∈ N, the sequence hxnk i∞ n=1 is a Cauchy sequence in R( or C), and using
the completeness of R( or C), the sequence hxnk i∞ n=1 converges to a point xk ∈ R ( or C). Thus
we get a sequence t = hx1 , x2 , · · · i. When we get m → ∞ in Eq. (13), it follows that
|xnk − xk | < ε, for all n ≥ p. (14)
Now, by inequalities
|xk | ≤ |xk − xpk | + |xpk |, for all k ∈ N

⇒ sup |xk | ≤ sup |xk − xpk | + sup |xpk |
k∈N k∈N k∈N
≤ ε + sup{|xpk |}.
k∈N
Since tp is bounded, it follows that t ∈ X. Taking supremum over k in Eq. (14), shows that
htn i converges to t, and (X, d) is complete.
Example 14 Let D[a, b] denote the set of all functions f on [a, b] which have continuous
derivatives at all points of I = [a, b]. For f, g ∈ D[a, b], define
d(f, g) = |f (a) − g(b)| + sup{|f 0 (x) − g 0 (x)| : x ∈ I}.
Show that d is a metric for D[a, b] and that the space (D[a, b], d) is complete.
Solution: Let hfn i be a Cauchy sequence in D[a, b]. Then for a given ε > 0, there exists m0 ∈ N
such that
d(fm , fn ) < ε; m, n ≥ m0
0
⇒ |fm (a) − fn (b)| + sup{|fm (x) − fn0 (x)| : x ∈ I} < ε,
0
⇒ |fm (x) − fn0 (x)| < ε; m, n ≥ m0 and all x ∈ I.
Hence hfn0 i is an uniformly convergent sequence of continuous functions and must therefore
converge to the continuous function φ, say. Then by a well known theorem on analysis, the
sequence hfn i converges uniformly to f such that
lim f 0 (x) = f 0 (x) = φ(x); a ≤ x ≤ b.

n→∞ n
Thus hfn i converges to a continuously differentiable function f so that f ∈ D[a, b]]. It follows
that the space, D[a, b], is complete.
Theorem 13 [Cantor intersection theorem] : Let (X, d) be a metric space. A nec-

essary and sufficient condition that the metric space X be complete is that every
nested sequence hFn i of non-empty closed subsets of X with δ(Fn ) → 0 as n → ∞ be
∞
T
such that F = Fi contains exactly one point.
i=1
Proof: Let the metric space (X, d) be complete. Consider a sequences of closed intervals hFn i
such that
F1 ⊃ F2 ⊃ · · · and δ(Fn ) → 0 as n → ∞.
Let us construct a sequence han i in a0 by selecting a point an ∈ Fn ∀ n ∈ N (since each Fn is

non-empty). Since δ(Fn ) → 0, for every ε > 0, ∃ a positive integer m0 ∈ N such that δ(Fm0 ) < ε.
Again since sequence hFn i is nested (monotone decreasing), for every positive integer, we have
n, n + p ≥ m0 ⇒ Fn+p , Fn ⊂ Fm0 ⇒ an , an+p ∈ Fm0

⇒ d(an , an+p ) ≤ δ(Fm0 ) < ε
⇒ d(an , an+p ) → 0 as n → ∞.
∴ han i is a Cauchy sequence in (X, d). Since the metric space (X, d) is complete, an → a0 for
some a0 ∈ X.
∞
T

We now prove that a0 ∈ Fn . Now the sub-sequence an , an+1 , an+2 , · · · of han i is contained
n=1
in Fn and still converges to a0 . But Fn being a closed sub-space of (X, d), a0 ∈ Fn . Since this
∞
T
is true for each n ∈ N, a0 ∈ Fn .
n=1
∞
a∗0 Fn . Then a0 , a∗0 ∈ Fn for each n ∈ N, so
T
Suppose ∈
n=1
0 ≤ d(a0 , a∗0 ) ≤ δ(Fn ) → 0 as n → ∞

∴ d(a0 , a∗0 ) = 0 ⇒ a0 = a∗0 .
Conversely, let F consists of a single point for every nested sequence hFn i of non empty closed
subsets Fn of X such that δ(Fn ) → 0. We are to show
that X is complete.
Let han i be any Cauchy sequence in X. Let Hn = xn , xn+1 , xn+2 , · · · . Since hxn i is Cauchy,
for a given ε > 0, ∃ a m0 ∈ N such that
d(xn , xm ) < ε for n, m ≥ m0 .
It follows that δ(Hn ) < ε for n ≥ m0 and consequently, δ(Hn ) → 0 as n → ∞. Also,
H1 ⊃ H2 ⊃ · · · ⇒ H̄1 ⊃ H̄2 ⊃ · · ·
and δ(Hn ) = δ(H̄n ). Hence
δ(H̄n ) < ε for n ≥ m0 ⇒ δ(H̄n ) → 0 as n → ∞.
Hence hH̄n i is a nested sequence of closed subsets of non empty sets in X whose diameters tend
∞
T
to zero. Then by hypothesis, ∃ a unique point x0 such that x0 ∈ H̄n .
n=1
We claim that the Cauchy sequence hxn i converges to x0 . Since δ(H̄n ) → 0 for a given ε > 0,
∃ a m0 ∈ N such that δ(H̄m0 ) < ε and so
xn , x0 ∈ H̄m0 ⇒ d(xn , x0 ) < ε for n ≥ m0 .
This implies that hxn i converges to x0 . Thus every Cauchy sequence in X converges to a point
in X. Hence X is complete.
Example 15 The real line R is a complete metric space, with respect to usual metric
for R.
Solution: Let hxn i be a Cauchy sequence of real numbers R. We define the sequence hnk i of
positive integers by induction as follows: nk+1 is the smallest integer greater than nk such that

xn − xm ≤ 1/2k+1 ; n, m ≥ nk .

h i
Let Ik be the closed interval xnk − 2−k , xnk + 2−k . Then we have, Ik+1 ⊂ Ik . For we have

xnk − xnk+1 ≤ 1/2k+1 .

∞
T
Also, the length of Ik → 0 as k → ∞. Hence by nested interval theorem, Ik contains exactly
k=1
one point, say a ∈ R. Thus a ∈ Ik for all k ∈ N so that

a − xnk ≤ 1/2k ; ∀k ∈ N.

Again for n ≥ nk , we have

k+1
x − x n ≤ 1/2 < 1/2k .

nk
Hence for all n ≥ nk , we have

|a − xn | = a − xnk + xnk − xn

≤ a − xnk + xnk − xn

1 1 1
< k
+ k = k−1 .
2 2 2
∴ lim xn = a. Thus every Cauchy sequence in R converges to a point in R and consequently
n→∞
R is complete. The Euclidean metric on R is derived from the absolute-value function, which
in turn depends on the ordering of R. The fact that the ordering of R is complete is crucial in
establishing that R is universally closed.
∞
T
Example 16 Show by an example that is (X, p) is not complete then Fn may be
n=1
empty where {Fn } is a sequence of non-empty closed sets in X with d(Fn ) → 0 as
n → ∞.
Solution:
Let X = (0, 2]. Then with respect to usual metric X is not a complete metric space,
Since n1 is a Cauchy sequence in X but it is not convergent in X. Let Fn = (0, n1 ]
Then hFn i is a monotonically decreasing sequence of non-empty closed set with δ(Fn ) = n1 → 0
∞
T
as n → ∞. But Fn = φ.
n=1
Theorem 14 Let (X, d) be a metric space. If for any monotonically decreasing se-
∞
T
quence hFn i of non-empty closed sets-with δ(Fn ) → 0 as n → ∞ imply Fn contains
n=1
exactly one point. Then (X, d) is complete.
n o
Proof: Let {xn } be a Cauchy sequence in (X, d). Suppose Gn = xn , xn+1 , xn+2 , · · · , then
G 1 ⊇ G 2 ⊇ G 2 ⊇ G 3 ⊇ · · · ⇒ G1 ⊇ G2 ⊇ G3 ⊇ · · · .
Suppose Fn = Gn , ∀ n. Since closure of a set is closed set, Fn is a closed set for all n. So, hFn i
is a monotonically decreasing sequence of non-empty closed set. We now prove that d(Fn ) → 0
as n → ∞. Let ε > 0 be given. Since hxn i is a Cauchy sequence, for the above chosen ε ∃ a
natural number k such that
ε
d(xm , xn ) < ; ∀ m, n ≥ k
2
ε
⇒ sup {d(xm , xn ) : m, n ≥ k} ≤
2
ε
⇒ δ(Gk ) ≤ < ε ⇒ δ(Gk ) ≤ ε, as δ(A) = δ(A)
2
i.e., δ(Fk ) < ε.
Since hFn i is monotonically decreasing, δ(Fn ) < ε∀ n ≥ k. This implies δ(Fn ) → 0 as n → ∞.
∞
T
So by given condition ∃ a point x ∈ X such that x ∈ Fn , i.e., x ∈ Fn ; ∀ n. We shall prove
n=1
hxn i converges to x.
Since δ(Fn ) → 0 as n → ∞ for any given ε > 0, ∃ a natural number m0 such that δ(Fn ) <
ε, ∀ n ≥ m0 . In particular, δ(Fm0 ) < ε, so
d(xn , x) < δ(Fm0 ) < ε, ∀ n ≥ m0
as xn ∈ Fm0 , ∀ n ≥ m0 and x ∈ Fm0 .
Thus {xn } converges to x. Hence (X, d) is complete.
Example 17 Let (X, d1 ) and (Y, d2 ) be two complete metric spaces. Prove that the
product space Z = X × Y with metric
q
d(x, y) = d21 (x1 , y1 ) + d22 (x2 , y2 )
is complete, where x = (x1 , x2 ) and y = (y1 , y2 ).
Solution: From Example ??, we see that d is a metric for Z. Let hzn i be a Cauchy sequence
in Z. Then for given ε > 0, ∃ a n0 (ε) ∈ N such that
d(zm , zn ) < ε, whenever m, n ≥ n0 (ε)
⇒ d2 (zm , zn ) < ε2 , whenever m, n ≥ n0 (ε)
⇒ d21 (xm , xn ) + d22 (ym , yn ) < ε2 , whenever m, n ≥ n0 (ε)
⇒ d21 (xm , xn ) < ε2 and d22 (ym , yn ) < ε2 , whenever m, n ≥ n0 (ε)
⇒ d1 (xm , xn ) < ε and d2 (ym , yn ) < ε, whenever m, n ≥ n0 (ε).
It follows that hxn i and hyn i are Cauchy sequence in the space X and Y respectively. Since
these spaces are complete, the sequences hxn i and hyn i converge respectively to points x ∈ X
and y ∈ Y . It follows that hzn i converges to z = (x, y) ∈ Z and consequently the product metric
space Z = X × Y is complete.
Result 6 The Cantor’s intersection theorem may not valid if any any of the following two
conditions is not satisfied:
(i) hFn i is a sequence of closed sets
(ii) δ(Fn ) → 0 as n → ∞.
The following examples cite that no condition in the Cantor’s intersection theorem can be
dropped without losing the conclusion of the theorem.
(i) Let X = R be a complete metric space. Consider the sequence of open intervals Fn =
1
0, , for all n ∈ N. Obviously
n
(a) F1 ⊃ F2 ⊃ F3 ⊃ · · · , i.e., hFn i is a nest of open intervals

(b) δ(Fn ) = n1 − 0 = n1 → 0 as n → ∞.

∞
\
But the intersection Fn = φ.
n=1
1 1
(ii) In the real line R, let us consider the sequence of open intervals Fn = − , , for all
n n
n ∈ N. Obviously
(a) F1 ⊃ F2 ⊃ F3 ⊃ · · · , i.e., hFn i is a nest of open intervals

(b) δ(Fn ) = n1 − − n1 = n2 → 0 as n → ∞.

∞
\
But the intersection Fn = {0}, which contains exactly one point.
n=1
h 1 1i
(iii) In the real line R, let us consider the sequence of closed intervals Fn = − 1 − , 1 + ,
n n
for all n ∈ N. Obviously
(a) F1 ⊃ F2 ⊃ F3 ⊃ · · · , i.e., hFn i is a nest of closed intervals

1 1 2
(b) δ(Fn ) = 1 + − −1− = 2 + → 2(6= 0) as n → ∞.

n n n
∞
\
But the intersection Fn = [−1, 1], which contains uncountably infinite number of
n=1
points.
h 1 i h 1i
(iv) In the real line R, let us consider the sequence of sets Fn = −1− , −1 ∪ 1, 1 + ,
n n
for all n ∈ N. Obviously
(a) Since union of a finite number of closed sets in closed, it follows that each Fn are
closed in the real line R
1 1 2
(b) δ(Fn ) = sup{|α − β|; α, β ∈ Fn } = 1 + − −1− = 2 + → 2(6= 0) as

n n n
n → ∞.
∞
\
But the intersection Fn = {−1, 1}, which is not a singleton set.
n=1
(v) In the discrete metric space (R, d) let us define Fn = {x ∈ R : x > n}; for all n ∈ N.
Obviously,
(a) hFn i is a nested sequence of non-empty close sets in (R, d)
(b) δ(Fn ) = 1(6= 0) for each n.
∞
\
n=1
(vi) Let Q be the set of rational numbers, and d the usual metric on Q and (Q, d) is not
n 1o
complete. Let Fn = x ∈ Q+ : 2 < x2 ≤ 2 + ; for all n ∈ N.
n
(a) hFn i is a sequence of non-empty closed sets in Q
(b) δ(Fn ) → 0.
∞
\
n=1
1
(vii) Consider the two dimensional Euclidean plane R2 . Let Fn = S (0, 0), − {(0, 0)}, for
n
all n ∈ N.
(a) hFn i is a nested sequence of non-empty
(b) δ(Fn ) → 0 as n → ∞.
∞
\
The point (0, 0) is a limit point of Fn , and it is not in Fn . Thus the intersection Fn = φ.
n=1
Theorem 15 (Completeness and continuity) Suppose that (X, dX ) and (Y, dY ) are met-
ric spaces and that (X, dX ) is complete. Suppose there exists a bijective function
f : X → Y such that f is continuous and f −1 is uniformly continuous. Then Y is
complete.
Proof: Suppose han i is a Cauchy sequence in Y . Then f −1 han i is Cauchy in X by and so

converges in X because X is complete. Therefore han i, being the image of f −1 han i under the
continuous function f , also converges. Since han i is an arbitrary Cauchy sequence in Y , Y is
complete.
(i) When the closed interval [1, ∞] is endowed with its usual metric, it is a complete metric
space. Now endow it with the inverse metric (a, b) 7→ |a−1 − b−1 | ; this space is not
complete because the sequence hni is Cauchy but does not converge. Notice, however,
that the identity map from [1, ∞] with the Euclidean metric to [1, ∞] with the inverse
metric is continuous and has continuous inverse.
(ii) Let d denote the usual metric on R+ . and m denote the exponential metric (a, b) 7→ |ea −eb |.
(R+ , d) is complete, being a closed subspace of R with its usual metric. The identity map
from (R+ , d) to (R+ , m) is bijective; it is easily verified that continuity of the exponential
function implies continuity of this identity map, and the identity function from (R+ , m)
to (R+ , d) is a Lipschitz map because |a − b| ≤ |ea − eb for all a, b ∈ R+ . So R+ with the
exponential metric is complete.
3.1 Complete Subsets

We will notice that the property of universal closure, unlike that of closure, is independent of
any particular enveloping metric space. So whether or not a metric space is being considered as
a space in its own right or as a subspace of some larger space is irrelevant when we are talking
about completeness.
Definition 7 Suppose (X, d) is a metric space and Y is a subset of X. We say that Y is a

complete subset of X if, and only if, the metric subspace (Y, d) of (X, d) is a complete metric
space. For example,
(i) Because R is complete, the complete subsets of R are its closed subsets; that is, precisely
those that have the nearest-point property. In particular, every closed interval of R is
complete, and, despite its obvious fragmentation, the Cantor set C is also complete.
(ii) An open subset of a complete metric space is not complete unless it is also closed. However,
open subsets can be made into complete spaces by judiciously altering the metric.
Theorem 16 Let (X, d) be a metric space, and let Y be a subset of X. Then Y ,

considered as a metric space, is complete if and only if Y is closed.
Proof: Let Y be a complete subspace of X, we are to show that it is closed in X. Let y ∈ Y

be a limit point of Y , then for any positive integer n, S(y, 1/n) must contain a point yn of Y .
Then hyn i is a Cauchy sequence.
Since Y is complete, the hyn i converges to some element of Y . But lim yn = y and so y ∈ Y .
n→∞
Thus we see that Y contains all the accumulation points and consequently Y is closed.
Conversely, let Y be closed and we shall show that Y is complete. Let hyn i be any Cauchy
sequence of elements of Y . Since Y ⊆ X, hyn i is also a Cauchy sequence in X, and so converge
to a point y0 ∈ X. We now show that y0 ∈ Y , for this we consider the following two cases :
(i) If the range set of hyn i consists of finite number of distinct points, then yn = y0 for infinitely
many values of n. Since hyn i is in Y , it follows that y0 ∈ Y .
(ii) If the range set of hyn i has infinitely many distinct points, then y0 is a limit point of the
range set of hyn i and so y0 is also a limit point of Y . Since Y is closed, y0 ∈ Y .
Thus we have shown that every Cauchy sequence in Y converges to a point in Y . Hence Y is
complete.
Deduction 3.1 Let y be a limit point of Y . Then ∃ a sequence hyn i in Y where yn 6= y ∀ n

such that yn → y as n → ∞.
∴ hyn i is a convergent sequence in (X, d).
Since every convergent sequence is Cauchy, hyn i is a Cauchy sequence in (X, d). Also yn ∈ Y
∀n, hyn i is a Cauchy sequence in (Y, dy ).Thus y ∈ Y . Hence Y is a closed set.
Therefore, if (Y, dy ) be a sub-space of a metric space (X, d) and if Y is complete then Y is a
closed sub-set of X.
Theorem 17 (Unions of complete subsets) Let (X, d) is a complete metric space.

Prove that if E1 and E2 are complete sub-sets of (X, d) then E1 ∪ E2 is also com-
plete.
Proof: Let hxn i be a Cauchy sequence in E1 ∪E2 . Then at least one of E1 , E2 contains infinitely
many members of hxn i. Without loss of generality suppose E1 contains infinitely many members
of hxn i. Then ∃ a sub-sequence hxnk i of hxn i such that xnk ∈ E1 ∀ k ∈ N. Since hxn i is Cauchy
sequence, hxnk i is a Cauchy sequence in E1 . Since E1 is complete, hxnk i converges to some point
x ∈ E1 .
Thus hxn i is a Cauchy sequence and a sub-sequence hxnk i of hxn i converges to x in E1 ∪ E2 .
This implies hxn i converges to x in E1 ∪ E2 . Hence E1 ∪ E2 is complete.
Note: Union of a finite number of complete subsets of a metric space is complete.
Theorem 18 (Intersections of complete subsets) Intersection of any number of com-

plete sub-sets of a metric space is complete.
T space and {Fα : α ∈ A} be an arbitrary collection of complete

Proof: Let (X, d) be a metric
sub-sets of X. Suppose F = Fα . Since every complete sub-set is closed, for each α ∈ A, Fα
α∈A
is a closed sub-set of X and so F is a closed sub-set of X (as intersection of arbitrary collection
of closed set is a closed set).
For a fixed α0 ∈ A, Fα is a complete metric space and F is a closed sub-set of Fα0 . Since a
closed sub-set of a complete metric space in complete, F is complete.
Theorem 19 Let (X, d) be a complete metric space and (Y, dY ) be a sub-space of

(X, d). Then Y is closed implies Y is complete.
Proof: Let hxn i be a Cauchy sequence in Y . Then hxn i is a Cauchy sequence in (X, d) (since
Y ⊆ X). But (X, d) being complete, hxn i converges to some point x in X. Let A be the range
of the sequence hxn i.
Case I: If A is finite then xn = x for infinitely many values of n. Since xn ∈ Y ∀ n, x ∈ Y .
Case II: If A is an infinite set then x is a limit point of A. Also since A ⊆ Y, s a limit point
of Y . Since Y is closed, x ∈ Y .
∴ From the both cases we have hxn i converges to x in Y . Hence (Y, dY ) is a complete metric
space.
Theorem 20 Completeness is preserved under isometries.
Proof: Let (X, d) and (Y, ρ) be metric spaces. Let T : X → Y be an isometry from (X, d) onto
(Y, ρ) and also let (X, d) be complete. Since every isometry is a one-one onto mapping, hence
for each n ∈ N, ∃xn ∈ X such that f (xn ) = yn . Since f is isometry, we obtain
d(xn , xm ) = ρ(T (xn ), T (xm )) = ρ(yn , ym ); ∀n, m ∈ N. (15)
Since < yn > is a cauchy sequence, hence for giving ε > 0, ∃ n0 ∈ N such that
ρ(yn , ym ) < ε ; m, n ≥ n0
⇒ d(xn , xm ) < ε ; m, n ≥ n0 .
⇒ hxn i is a Cauchy Sequence in X.
Since (X, d) is complete and hxn i is a Cauchy Sequence, then hxn i is a convergent sequence.
Hence there exists x ∈ X such that xn → x. Hence corresponding to ε > 0, ∃ n1 ∈ N such that d(xn , x) <
ε, ∀n ≥ n1 . Since T is an isometry, we have
d(xn , x) = ρ(T (xn ), T (x)) = ρ(yn , f (x)) < ε for n ≥ n1 .

⇒ yn → f (x) : n → ∞.
Thus every Cauchy Sequence in y is a convergent sequence and hence (y, ρ) is complete.
Result 7 We know the real line is complete and space X = (0, 1) with usual metric is not
complete. Now, the homeomorphic image of R in X. But R is complete whereas X is incomplete.
Thus homeomorphism need not presence completeness.
4 Completeness and Contraction Mapping

Strong contractions on a metric space, when iterated, tend to pull all the points of the space
together into a single point. Banach’s Theorem, also called the Banach Contraction Principle,
is that such a fixed point must exist if the space is complete and can, in that case, be computed
by iteration. This theorem is invaluable for developing algorithmic procedures and generally for
computing solutions to equations.
Definition 8 Suppose X is a non-empty set and f : X → X. A point x ∈ X is called a fixed
point for f if and only if f (x) = x. For example,
(i) Let X = R be a nonempty set and T : X → X be a mapping defined by T (x) = x + a, for

some fixed number a 6= 0. Then the translation T has no fixed point.
(ii) A rotation of the plane has a single fixed point. Indeed, the centre of rotation is the only
fixed point.
(iii) The mapping T : X → X defined by T (x) = x/2. Then x = 0 is the only fixed point of T .
(iv) The mapping T : X → X defined by T (x) = x2 . Then x = 0 and x = 1 are two fixed
points of T .
(v) The mapping T : X → X defined by T (x) = x. Then T has infinitely many fixed points.
In fact, every point of X is a fixed point.
(vi) The projection (x, y) → x of R2 onto the x-axis has infinitely many fixed points. In fact,
all points of the x-axis are fixed points.
(vii) Every continuous function from [0, 1] to [0, 1] has at least one fixed point, though it may
have many. At least, the following assertion is intuitively true: suppose f : [0, 1] → [0, 1]
is continuous, f (0) = α ∈ (0, 1] and f (1) = β ∈ [0, 1). Then the graph of f joins (0, α)
continuously to (1, β) and must cross the line {x ∈ R2 : x1 = x2 } somewhere along the
way (Fig. 1).
x2
6 x1 = x2
r
(0, α) q (1, β)x1

-
O
Figure 1: The graph x1 = x2 .
The examples show that a mapping may not have any fixed point, it may have unique fixed
point, it may have more than one or even infinitely many fixed points.
Definition 9 Let (X, d) be a complete metric space and T : X → X be a mapping ( or called

an operator). Then T is said to be
(i) non-expansive, if
d(T (x), T (y)) ≤ d(x, y), ∀x, y ∈ X. (16)
(ii) contractive or weak contraction mapping or contractible or contractive or shrinking map

if
d(T (x), T (y)) ≤ d(x, y), ∀x, y ∈ X, x 6= y. (17)
(iii) a contraction mapping or contraction on X if there exists a real number α with 0 ≤ α < 1
such that

d T (x), T (y) ≤ α d(x, y) < d(x, y); ∀x, y ∈ X, x 6= y. (18)
The number α is usually referred as Lipschitz constant of T .
It is stressed that α in this definition (18) is independent of x, y in X. Thus in a contraction

mapping, the distance between the images of any two points is less than the distance between
the points. Hence the application of T to each of two points ‘contracts’ the distance between
them. Every contraction mapping is uniformly continuous.
Theorem 21 (Banach fixed point theorem) Every contraction mapping on a com-

plete metric space has a unique fixed point.
Proof: Existence: Let (X, d) be a complete metric space and x, y be any two points in X.
Let T : X → X is a contraction on X, there exists a α ∈ R with 0 ≤ α < 1 such that

d T (x), T (y) ≤ α d(x, y); ∀x, y ∈ X

∴ d T 2 (x), T 2 (y) ≤ d T (x), T (y) ≤ α2 d(x, y); ∀x, y ∈ X
.. ..
. .
n n n−1
d T (x), T (y) ≤ d T (x), T n−1 (y) ≤ αn d(x, y); ∀x, y ∈ X. (19)
Now let x0 be any point of X and inductively construct the sequence hxn i of points in X as:
x1 = T (x0 ), x2 = T (x1 ) = T 2 (x0 ), · · · , xn T (xn−1 ) = · · · = T n (x0 ), · · · .
Clearly, hxn i is the sequence of images of x0 under repeated application of T . We claim that hxn i
is a Cauchy sequence. For, let m, n be any positive integers such that m > n. Write m = n + p,
where p is any positive integers greater than or equal to 1. Using triangle inequality, we have
d(xn , xm ) = d(xn , xn+p )

≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) + · · · + d(xn+p−1 , xn+p )

≤ d T n (x0 ), T n (x1 ) + d T n+1 (x0 ), T n+1 (x1 )

+ · · · + d T n+p−1 (x0 ), T n+p−1 (x1 ) ; Eq. (19)
≤ αn d(x0 , x1 ) + αn+1 d(x0 , x1 ) + · · · + αn+p−1 d(x0 , x1 )
h i αn
≤ αn d(x0 , x1 ) 1 + α + α2 + · · · + αp−1 = d(x0 , x1 ). (20)
1−α
Since 0 ≤ α < 1, so lim αn = 0, it follows that d(xn , xm ) can be made less than any pre-assigned
n→∞
positive number ε by taking n (and hence m) sufficiently large. Thus hxn i is a Cauchy sequence.
But by hypothesis X is complete, so that there exists x ∈ X such that xn → x as n → ∞. It
follows that
lim T (xn ) = T (x). (21)

n→∞
Now T (xn ) = xn+1 so that hT (xn )i is a subsequence of hxn i and consequently hT (xn )i must also
converge to x, i.e., lim T (xn ) = x. Using Eq. (21), we have
n→∞
lim T (xn ) = T (x) = x. (22)

n→∞
This shows that x is a fixed point.

Uniqueness : All that remain to show is that if y ∈ X, y 6= x, then y cannot be a fixed point.
Suppose, if possible, y is a fixed point. Then T (y) = y. Since, T (x) = x, we have
d(x, y) = d(T (x), T (y)) ≤ α d(x, y)

⇒ α d(x, y) ≥ d(x, y) > 0; as d(x, y) 6= 0
⇒ α≥1
which is a contradiction. Hence T (y) 6= y and the proof is complete.
(i) This theorem is often stated as ‘ T has precisely one fixed point’.
(ii) Let T : (X, d) → itself be a contraction, and ε > 0 be given. Let us take a δ > 0 to satisfy
δ < ε/2α. Now, if d(x, y) < δ, we have
d(T (x), T (y)) ≤ α d(x, y) < αδ < ε/2 < ε.
Hence T is uniformly continuous over (X, d).
(iii) If the condition of completeness in Banach fixed point theorem be dropped, then T may
not have a fixed point. For example, let X = (0, 1) and the mapping T : X → X defined
by T (x) = x/2, then T (0) = 0 6∈ X. Here X is not a complete metric space with the usual
metric and T does not have any fixed point.
(iv) Let X = {x ∈ R : x ≥ 1} be a complete metric space with usual metric of reals and the
x 1
function T : X → X be defined as, T (x) = + for x ∈ X. Let x, y ∈ X, then
2 x
x 1 y 1 1 1 1
|T (x) − T (y)| = + − + = (x − y) − −

2 x 2 y 2 y x

1 x − y 1 1
= (x − y) − = (x − y) −

2 xy 2 xy

1 1
= |x − y| − .

2 xy
Since x ≥ 1 and y ≥ 1, we have 1/xy ≤ 1. So 1/xy lies either between 0 and 1/2 or
between 1/2 and 1 and in each of these cases
1 1 1 1
− ≤ ⇒ |T (x) − T (y)| ≤ |x − y|.

2 xy 2 2
So T is a contraction and Banach contraction√principle says that T has unique fixed point
∗
x∗ given by x∗ = x2 + x1∗ or x∗ = 2 or x∗ = 2.
(v) If T is not contraction in Banach fixed point theorem, then it may not have a fixed point.
For example, consider the metric space X = [1, ∞) with the usual metric and the mapping
T : X → X given by T (x) = x + x1 . Now, X is a complete metric space but T is not a
contraction mapping. In fact
h 1 i
|T (x) − T (y)| = |x − y| 1 − < |x − y|; ∀x, y ∈ X
xy
and so T is contractive. Of course, T does not have any fixed point.
(vi) Consider the complete metric space X = [0, ∞) equipped with the metric of absolute value
and consider the mapping T : X → X given by T (x) = 1/(1 + x2 ). Then
(a) the mapping T satisfies d(T (x), T (y)) < d(x, y) and hence T is a contractive map,
while T is not a contraction.
(b) T has no fixed point.
(vii) The mapping T : (0, 13 ] → (0, 13 ] defined by T (x) = x2 is a contraction, but it has no fixed
point in (0, 31 ].
(viii) Let X = R be the real number space and for x ∈ R, let T : X → X be defined as
T (x) = x + π2 − tan−1 x, then T is not a contraction mapping.
Theorem 22 Let (X, d) be a compact metric space and T : X → X a contractive map
(not necessarily contraction). Then, T has a unique fixed point in X.
Proof: Existance : Define a mapping f : X → R by f (x) = d(x, T (x)). We first show that f
is continuous. Let ε > 0 be chosen arbitrary. Then

|f (x) − f (y)| = d(x, T (x)) − d(y, T (y))

≤ d(x, T (x)) − d(T (x), y) + d(T (x), y) − d(y, T (y))

( by triangle inequality )
≤ d(x, y) + d(T (x), T (y)) ≤ 2d(x, y), ∵ T is contractive
< ε, provided d(x, y) < δ < ε/2.
Since f is continuous and X is compact, ∃x ∈ X such that f (x) ≤ f (y), ∀y ∈ X, i.e., f attains
minimum at x. We, now, show that x is a fixed point of T .
Let, if possible, x be not a fixed point of T . Then T x 6= x. Since f (x) ≤ f (y), ∀y ∈ X, taking
y = T (x), we have
f (x) ≤ f (T (x))
⇒ d(x, T (x)) ≤ d(T (x), T (T (x))) < d(x, T (x)), ∵ T is contractive and x 6= T (x).
which is impossible. Hence x is a fixed point of T .

Uniqueness : Let, if possible, x and y be two fixed points of T in X. Then T (x) = x and
T (y) = y. Now,
d(x, y) = d(T (x), T (y)) ≤ α d(x, y)

⇒ d(x, y) = 0 ⇒ x = y.
This completes the proof of this theorem.
Theorem 23 Let (X, d) be a complete metric space and T : X → X be a mapping.

If T m is a contraction on X for some positive integer m, then T has a unique fixed
point.
Proof: Let S = T m . Under the hypothesis, S is a contraction. By Banach Contraction

Theorem, the mapping S has unique fixed point, say x∗ . Thus
S(x∗ ) = x∗ ⇒ S n (x∗ ) = x∗ .
Again, in view of Banach Contraction Theorem, we have
lim S n (x∗ ) = x∗ , ∀x ∈ X.
n→∞
In particular, taking x = T (x), we get
x∗ = lim S n (x∗ ) = lim S n (T (x∗ ))

n→∞ n→∞
= lim T (S n (x∗ )) = lim T (x∗ ) = T (x∗ ).
n→∞ n→∞
This shows that x∗ is a fixed point of T . Also, since every fixed point of T is a fixed point of
S = T m , it follows that T cannot have more than one fixed point and uniqueness follows.
Banach fixed point theorem has many applications in analysis.
(i) Suppose f : R → R is a differentiable function and there exists k ∈ [0, 1) such that
|f 0 (x)| ≤ α for all x ∈ R. Then, for each x, y ∈ R with x < y, it follows from the Mean
Value Theorem that there exists c ∈ (x, y) with f (y) − f (x) = (y − x)f 0 (c). From this we
get |f (y) − f (x)| ≤ α|y − x|, so that f is a strong contraction. Then Banach’s Theorem
tells us that f has a unique fixed point.
(ii) The graph of the cosine function clearly crosses the line y = x somewhere between x = 0
and x = π/2, (Fig. 2) and it does so exactly once. In other words, the cosine function
restricted to [0, π/2] has a unique fixed point. It turns out that we can apply Banach’s
Theorem to discover it despite the fact that the cosine function is not a strong contraction
on [0, π/2].
y
6
y=x
r
y = cos x
O x
× -
(0, π/2)
Figure 2: The graph y = x and y = cos x.
(iii) Let [a, b] be a closed interval, which is a complete metric space with usual metric of reals
and g : [a, b] → [a, b] be a function with g 0 (x) satisfying |g 0 (x)| ≤ α < 1 in a ≤ x ≤ b. If
x1 , x2 ∈ [a, b] with x1 < x2 we have by Mean-value theorem of differential calculus
g(x2 ) − g(x1 ) = (x2 − x1 ) g 0 (ξ); x1 < ξ < x2
⇒ |g(x2 ) − g(x1 )| = |(x2 − x1 )| |g 0 (ξ)| ≤ α|x2 − x1 |; 0 < α < 1.
So g is a contraction and Banach Contraction Principle says that there is exactly one
member x∗ ∈ [a, b] satisfying g(x∗ ) = x∗ or g(x∗ ) − x∗ = 0 or g(x) − x = 0 has exactly one
root in [a, b].
(iv) The Cantor set is an example of a fractal. On the Fig. 3 of fractal, Sierpi Lnski’s triangle,
@
@
@
@
@
@
Figure 3: Sierpi Lnski’s triangle.

this time in R2 rather than R. The picture was obtained by iterating the function
n o
A → (.5a1 + c1 , .5a2 + c2 ) : a ∈ A, c ∈ {(1, 1), (1, 50), (50, 50)}
on the space of non-empty closed bounded subsets of R2 starting at a single point.
4.1 Applications of Banach’s Fixed Point Theorem

Banach’s fixed point theorem has wide and important applications in diversified fields of study.
(i) The most interesting applications of fixed point theorem arise when the underlying metric
space is a function space. Here we discuss the existence and uniqueness of the Volterra
and Fredehelom integral equations.
(ii) We give an application of Banach fixed point theorem to prove a theorem of Picard on the
existence and uniqueness of solutions of a certain class of first order ordinary differential
equations.
Theorem 24 Let φ be a function in C[a, b], and K be a continuous real-valued mapping
defined on the square [a, b] × [a, b] with subspace metric of R2 . Then there exists a
real number λ such that the Volterra equation
Z x
f (x) = φ(x) + λ K(x, y) f (y) dy; for all x ∈ [a, b]. (23)
a
has a unique solution in C[a, b].
Proof: Let K be a continuous function on [a, b] × [a, b] and let φ be a continuous function on
[a, b]. For each parameter λ ∈ R, consider the Volterra equation (23). Let X = C[a, b] be the
set of all continuous real-valued functions defined on [a, b] with the uniform metric. Since K is
continuous, ∃ a constant M > 0 such that
|K(x, y)| ≤ M ; for all x, y ∈ [a, b].
Define the transformation T : f → T (f ) on X by

Z x
T (f (x)) = φ(x) + λ K(x, y) f (y) dy. (24)
a
For all f, g ∈ X, we have

Z x
|T (f (x)) − T (g(x))| = λ K(x, y) |f (y) − g(y)| dy

a
≤ |λ| M (x − a) d(f, g); ∀x ∈ [a, b].
Since T 2 (f ) − T 2 (g) = T (T (f ) − T (g)), we have

Z x
2 2
|T (f (x)) − T (g(x))| = λ K(x, y) |T (f (y)) − T (g(y))| dy

Za x
≤ |λ| |K(x, y)| |λ| M (y − a)d(f, g) dy
a
Z x
≤ |λ|2 k 2 (y − a)dy d(f, g)
a
|λ|2 M 2 (x − a)2
≤ d(f, g).
2
Continuing this iterative process, we obtain
|λ|n M n (x − a)n
|T n (f (x)) − T n (g(x))| ≤ d(f, g); ∀x ∈ [a, b].
n!
[|λ|M (b − a)]n
∴ |T n (f ) − T n (g)| ≤ d(f, g).
n!
rn
Recalling that → 0 as n → ∞ for any r ∈ R, we conclude that there exists n such that T n is
n! n
a contraction mapping. Taking n sufficiently large to have [|λ|M (b−a)]
n! < 1. Hence there exists
an unique solution f ∈ X satisfying T (f ) = f. Obviously, if T (f ) = f, then f solves Eq.(23).
Theorem 25 (Picard’s theorem) Let D be a nonempty open subset of the Euclidean

∂f
plane R2 , f : D → R, be continuous maps which satisfy the Lipschitz condition in
∂y
the second variable
|f (x, y1 ) − f (x, y2 )| ≤ α |y1 − y2 |, (25)
for all (x, y1 ), (x, y2 ) ∈ D and for some α > 0, and let (x0 , y0 ) ∈ D. Then the differential
equation
dy
= f (x, y) (26)
dx
has an unique solution y = g(x) which passes through the point (x0 , y0 ).
∂f
Proof: Since f (x, y) and are continuous in D and D is a compact subset of R2 ; f (x, y) and
∂y
∂f
are bounded, and so there exists real numbers M and N such that
∂y
∂f
|f (x, y)| ≤ M and ≤ N ; ∀(x, y) ∈ D. (27)

∂y
Now, if (x, y1 ) and (x, y2 ) be any points in D, then by Lagrange’s mean value theorem of calculus
∂f
|f (x, y1 ) − f (x, y2 )| = |y1 − y2 | (x, y1 + (y1 − y2 )θ); 0 < θ < 1.

∂y
⇒ |f (x, y1 ) − f (x, y2 )| ≤ |y1 − y2 | · N ; (x, y1 ), (x, y2 ) ∈ D. (28)
Now, if y = g(x) is the solution of the differential equation (26), then
g 0 (t) = f (t, g(t)); and g(x0 ) = y0 . (29)
Then integrating Eq.(29) yields

Z x0
g(x) − g(x0 ) = f (t, g(t)) dt
x
Z x0
⇒ g(x) = g(x0 ) + f (t, g(t)) dt. (30)
x
Conversely, if y = g(x) satisfies Eq. (30), then it follows that g(x0 ) = y0 . Thus y = g(x)
such that y0 = g(x0 ) is a solution of Eq. (26) if and only if it satisfies the integral equation, Eq.
(30). Therefore, it is sufficient to prove that Eq. (30) has a unique solution. Choose a positive
constant δ such that N δ < 1 and the closed rectangle
n o
D0 = (x, y) ∈ R2 : |x − x0 | ≤ δ and |y − y0 | ≤ M δ ⊂ D.
Let X be the set of all real-valued functions y = g(x) which is defined and continuous in
|x − x0 | ≤ δ, such that
d(g(x), y0 ) = |g(x) − y0 | ≤ M δ.
Let set X is a closed subspace of the complete metric space of continuous functions C[x0 −δ, x0 +δ]
with the sup metric d. Let ψ in C[x0 − δ, x0 + δ] be a limit point of X, then ∃ a sequence hgn i
in X which converges in C[x0 − δ, x0 + δ] to φ. So corresponding to a real number ε > 0, ∃k ∈ N
such that
d(gn , ψ) = |gn (x) − ψ(x)| < ε; ∀n ≥ k and x ∈ [x0 − δ, x0 + δ].
Using Triangle inequality, we get
|ψ(x) − y0 | ≤ |φ(x) − gn (x)| + |gn (x) − y0 | < ε + M δ.
Since ε > 0 is arbitrary, we have
|ψ(x) − y0 | ≤ M δ; ∀x ∈ [x0 − δ, x0 + δ].
Thus ψ ∈ X and ψ is a closed set. Thus X is complete. Consider a mapping T : X → X defined

by
Z x0
T (g) = h; where h(x) = y0 + f (t, g(t)) dt.
x
Now h(x) is a continuous function on [x0 − δ, x0 + δ] and

Z x0
d(h(x), y0 ) = sup f (t, g(t)) dt ≤ M (x − x0 ) ≤ M · δ,

x
shows that h(x) ∈ X and T is well-defined. Now

Z x0 h i
d(T (g), T (g1 )) = d(h, h1 ) = sup f (t, g(t)) − f (t, g1 (t)) dt

x
Z x0
≤ sup f (t, g(t)) − f (t, g1 (t)) dt

x
Z x0
≤ α |g(t) − g1 (t)| dt
x
≤ α · d(g, g1 )(x − x0 ) ≤ αδd(g, g1 )
∴ d(T (g), T (g1 )) ≤ βd(g, g1 ); 0 ≤ β = αδ < .
Thus is contraction mapping on X into itself. Thus by Banach contraction theorem, the con-
traction mapping T has a unique fixed point g ∈ X, i.e., T (g) = g. This means that there is a
unique solution g ∈ X as given in Eq. (30).
Theorem 26 (Implicit Function) Let D be an open set in R2 containing a point

(x0 , y0 ) and f : D → R be a mapping such that
(i) f (x, y) is continuous in D,

∂f ∂f
(ii) and are both continuous in D, and
∂x ∂y
∂f
(iii) f (x0 , y0 ) = 0 and 6= 0.
∂y (x0 ,y0 )

n o
Then ∃ a rectangle R = (x, y) : |x − x0 | < h, |y − y0 | < k ⊂ D and a function
g : [x0 − h, x0 + h] → [y0 − k, y0 + k] such that
(i) f (x, g(x)) = 0; ∀x ∈ [x0 − h, x0 + h], and

∂f . ∂f
(ii) g is differentiable in [x0 − h, x0 + h] and g 0 (x) = − .

∂x ∂y (x,g(x))

∂f ∂f
Proof: Let = p and = q. Define a mapping F : D → R by
∂x (x0 ,y0 ) ∂y (x0 ,y0 )
1
F (x, y) = y − f (x, y); ∀(x, y) ∈ D.
q
∂F ∂F
Then F (x, y) is continuous in D; F (x0 , y0 ) = y0 and = 0. Since is continuous in
∂y (x0 ,y0 ) ∂y
D, ∃h, k ∈ R+ such that
∂F 1
< ; ∀(x, y) ∈ R.
∂y 2
Since F (x, y) is continuous and F (x0 , y0 ) = y0 , we can take smaller h if necessary such that
1
F (x, y0 ) − y0 | < k; ∀x ∈ [x0 − h, x0 + h].
2
Now, let X be the set of all continuous real valued functions
g : [x0 − h, x0 + h] → [y0 − k, y0 + k]; with g(x0 ) = y0 .
X is obviously non-empty. Let d : X × X → R be a mapping defined by
n o
d(g1 , g2 ) = sup |g1 (x) − g2 (x)| : ∀x ∈ [x0 − h, x0 + h] ,
then (X, d) is a complete metric space. Now, we define a mapping T : X → X by

o
T (g) = ψ, where ψ(x) = f (x, g(x)); ∀x ∈ [x0 − h, x0 + h] .
Then ψ is continuous in [x0 −h, x0 +h], φ(x0 ) = f (x0 , g(x0 )) = y0 and using Lagrange mean-value
theorem
|ψ(x) − y0 | = |f (x, g(x)) − y0 |
≤ |f (x, g(x)) − f (x, y0 )| + |f (x, y0 ) − y0 |
∂f 1
≤ |g(x) − y0 | + k;

∂y (x,ξ) 2
for some ξ between g(x) and y0 . Therefore |h(x) − y0 | ≤ k. Thus T is induced a mapping from
X to itself. Also,

|ψ1 (x) − ψ2 (x)| = f (x, g1 (x)) − f (x, g2 (x))

∂f 1
≤ |g1 (x) − g2 (x)| ≤ |g1 (x) − g2 (x)|.

∂y (x,η) 2
This shows that T is a contraction marring. Therefore, T has a unique fixed point g in X so
that T (g) = g, that is, g(x) = F (x, g(x)). Thus f (x, g(x)) = 0; ∀x ∈ [x0 − h, x0 + h].
Now it remains to prove that g is differentiable. Let x and x + 4x be any two points in
[x0 − h, x0 + h]. Then f (x, g(x)) = 0 = f (x + 4x, g(x + 4x)), and
0 = f (x + 4x, g(x + 4x)) − f (x, g(x))
= f (x + 4x, g(x + 4x)) − f (x + 4x, g(x)) + f (x + 4x, g(x)) − f (x, g(x))
∂f ∂f
= [g(x + 4x) − g(x)] + 4x
∂y (x+4x,ξ0 ) ∂x (η0 ,g(x))

for some ξ 0 between g(x + 4x) and g(x), and η 0 between x and x + 4x. Therefore,
g(x + 4x) − g(x) ∂f . ∂f
=−
4x ∂x (η0 ,g(x)) ∂y (x+4x,ξ0 )
g(x + 4x) − g(x) ∂f . ∂f
∴ g 0 (x) = lim =− .
4x→0 4x ∂x ∂y (x,g(x))
Example 18 Let X = Rn being the set of n-rowed column vectors of real numbers,
is a metric space with the metric d∞ (x, y) = max |xi − yi |. Find the condition for the
1≤i≤n
system of equations
 
a11 a12 ··· a1n
 a21 a22 ··· a2n 
y = Ax + b; where, A =  . (31)
 
.. ..
 . ··· ··· . 
an1 an2 ··· ann
and x = (x1 x2 · · · xn )T ; b = (b1 b2 · · · bn )T to have precisely one solution.
Proof: Let us consider the space Rn , with the metric is given by d∞ (x, y) = max |xi − yi |.
1≤i≤n
On Rn , let us define T : Rn → Rn given by y = T (x) = Ax + b. Let u = (u1 u2 · · · un )T and
v = (v1 v2 · · · vn )T be any two elements of X. Then

d∞ (T (u), T (v)) = max aij (ui − vj )

1≤i≤n

≤ max |ui − vj | max aij

1≤i≤n 1≤i≤n

≤ d∞ (u, v) α; α = max aij .

1≤i≤n

Assume that α = max aij < 1, then T is a contraction mapping, and so Banach Contraction

1≤i≤n
Theorem, there is a unique fixed point x ∈ X of T . Then the system of linear equations
x = Ax + b and the system of linear equations (31) has an unique solution x.
Consider the following system of n linear equations with n unknowns:

a11 x1 + a12 x2 + · · · + a1n xn = b1 

a21 x1 + a22 x2 + · · · + a2n xn = b2 

.. .. .. (32)
. ··· . . 


an1 x1 + an2 x2 + · · · + ann xn = bn

Taking, x = (x1 x2 · · · xn )T and b = (b1 b2 · · · bn )T , then in matrix notation, Eq.(32) can be

written as Ax = b, where A is given in Eq. (31). This system (32) can be written as

x1 = (1 − a11 )x1 − a12 x2 − · · · − a1n xn + b1 

x2 = −a21 x1 + (1 − a22 )x2 − · · · − a2n xn + b2 

.. .. .. (33)
. ··· . . 


xn = −an1 x1 − an2 x2 − · · · (1 − ann )xn + bn

5 Incomplete Metric Space

A metric space (X, d) is called incomplete if it is not complete. If (X, d) is an incomplete
metric space, then we say that d is an incomplete metric on X. To justify the motivation and
meaningfulness of the above definition, we must cite examples of incomplete metric spaces:
(i) Let the P[a, b] be the set of all polynomials defined on [a, b] ( a subspace of the metric
space C[a, b]) with the uniform metric d∞ defined in it as
d∞ (f, g) = max |f (t) − g(t)|; f, g ∈ P[a, b].

t∈[a,b]
Let us take a = 0 and b = 1. Consider the following sequence

n
X t k t t2 tn
fn (t) = =1+ + 2 + · · · + n ; ∀t ∈ [0, 1].
2 2 2 2
k=0
Clearly, fn (t) ∈ P[0, 1] for each n ∈ N. Taking m < n and observe that
d∞ (fn , fm ) = max |fn (t) − fm (t)|

t∈[0,1]
n m n
X t k X t k X t k
= max − = max

2 2 2

t∈[0,1] t∈[0,1]
k=0 k=0 k=m+1
n 1 k
X 1 1
≤ max = m − n.

t∈[0,1] 2 2 2
k=m+1
This difference is arbitrarily small for large enough m and n, which implies that hfn i is
a Cauchy sequence in P[0, 1]. However, this sequence does not converge in (P[0, 1], d∞ ),
because
2
lim fn (t) = , for all t ∈ [0, 1]
n→∞ 2−t
is not a polynomial. Thus (P[0, 1], d∞ ) is not complete. Also, in view of the Weierstrass
polynomial approximation theorem, the uniform limit of a sequence of polynomials need
not be a polynomial and so (P[a, b], d∞ ) is not a complete metric space.
(ii) Consider the sequence hxn i of rational numbers in the usual metric space Qu as follows
x1 = 1.7, x2 = 1.73, x3 = 1.732, x4 = 1.73205 · · ·

√
The sequence hxn i converges to 3. Hence the sequence hxn i is a Cauchy sequence.
However, it does not converge to any point of Qu so Q is not complete.
(iii) Consider the space X = C[0, 1], with the metric

Z 1
d1 (x, y) = |x(t) − y(t)| dt; x, y ∈ [0, 1]
0
then we know (X, d1 ) is a metric space. Consider hxn i in X, where
0 ≤ t ≤ 21 − n1

 0;
1 1 1 1
xn (t) = nt − 2 n + 1; 2 − n <t≤ 2
1
1; 2 <t≤1

which can be represented graphically as in Fig. 4. Now we are to show that hxn i is a
xn (t)
6 1/n -
1 p
p -
O 1
− 1 1/2 1 t
2 n
Figure 4: Graph of xn (t)
Cauchys sequence.
Z 1
d1 (xm , xn ) = |xm (t) − xn (t)| dt
0
Z Z 1/2
1 1 1
≤ xm (t) dt + xn (t) dt = +
1/2−1/m 1/2−1/n 2 m n
< ε; ∀m, n > n0 , where n0 = [2/ε] + 1 ∈ N.
This shows that hxn i is a Cauchy sequence. Geometrically, the measure d1 (xm , xn ) repre-
xn (t) 6
1/n1/m
-
-
1 qB
xn xm
Aq qC -
O 1 1 1/2 1 t
2 − n1 12 − m
Figure 5:
sents the area of the triangle as shown in the Fig. 5. Suppose that there is a function x(t)
such that d1 (xn , x) → 0. But
Z 1/2−1/n Z 1/2 Z 1
d1 (xn , x) = |x(t)|dt + |xn (t) − x(t)|dt + |1 − x(t)|dt. (34)
0 1/2−1/n 1/2
Since the integrands are non-negative, so is the each integral on the right hand side of Eq.
(34). Since d1 (xn , x) → 0 as n → ∞, we have
Z 1/2−1/n Z 1
lim |x(t)|dt = 0 and |1 − x(t)|dt = 0. (35)
n→∞ 0 1/2
Since f is continuous, we have

0; if 0 ≤ t < 1/2
x(t) = (36)
1; if 1/2 ≤ t ≤ 1
Therefore, x is not continuous which contradicts to our supposition that x is a continuous

function. Hence C[0, 1] is not complete.

(iv) Consider the space of all natural numbers N with the metric d(x, y) = x1 − y1 , for all

x, y ∈ N. Let hnin≥1 be a sequence in N. Let ε > 0 and n0 be the least integer greater
than 1/ε. If m, n > n0 , then
1 1 n 1 1o 1
d(m, n) = − ≤ max , < < ε.

m n m n n0
Thus, the sequence hnin≥1 is Cauchy. Suppose contrary that the sequence hnin≥1 converges
to some point p ∈ N. Let n1 be any integer greater than 2p. Then n ≥ n1 implies that
1 1 1 1 1 1 1 1 1
d(p, n) = − = − ≥ − > − = .

p n p n p n1 p 2p 2p
This shows that the sequence hnin≥1 cannot converge to p, a contradiction to our suppo-
sition. Hence (N, d) is not a complete metric space.
Example 19 Let X = R and d : X × X → R be defined by

|x − y|
d(x, y) = √ p , for all x, y ∈ X.
1 + x2 1 + y 2
Show that (X, d) is a metric space but not complete.
Solution:
(i) By the definition of absolute value of a real number, it follows that
|x − y|
d(x, y) = √ p ≥ 0; ∀x, y ∈ R.
1 + x2 1 + y 2
Thus d is non-negative.
(ii) Now d(x, y) = 0 ⇒ |x − y| = 0. A necessary and sufficient condition for |x − y| = 0 is that
x = y. Thus
d(x, y) = 0 ⇔ x = y; ∀x, y ∈ R.
(iii) Since |p| = | − p|, it follows that

|x − y| | − (y − x)|
d(x, y) = √ p =√ p
1 + x2 1 + y 2 1 + x2 1 + y 2
|y − x|
= √ p = d(y, x); ∀x, y ∈ R.
1 + x2 1 + y 2
Thus d is symmetric.
(iv) Finally, ∀x, y, z ∈ R, we have
|x − y| |(x − z) + (z − y)|
d(x, y) = √ p = √ p
1+x 2 1+y 2 1 + x2 1 + y 2
|x − z| |z − y|
≤ √ √ +√ p
1+x 2 1+z 2 1 + z2 1 + y2
= d(x, z) + d(z, y).
So, d satisfies the triangle inequality.
Combining the above conditions, (R, d) is a metric space. Consider the sequence hnin≥1 of
natural numbers. Observe that
|n − m| | n1 − m
1
|
d(n, m) = √ √ = q q
1+n 2 1+m 2
1 + n12 1 + m12
1 1 1 1
≤ − ≤ + → 0 as n, m → ∞.

n m m n
Thus hnin≥1 is a Cauchy sequence in (X, d). Suppose contrary that the sequence hnin≥1 con-
verges to some point p ∈ R. Then
|n − p| |1 − np |
d(n, p) = √ p =q
1 + n2 1 + p2
p
1 + n12 1 + p2
1
→ p 6= 0 as n → ∞.
1 + p2
This shows that the sequence hnin≥1 does not converge to p ∈ R, a contradiction. Hence (X, d)
is not a complete metric space.
6 Separable Metric Space

It is well known that the rational numbers or irrational numbers are densely packed along the
real line. This phenomenon can be characterized interms of distance by saying that every point
of R is zero distance from the set of rational numbers Q or from the set of irrational numbers
R/Q. The concept of closure embled us to write this density property by the formula Q̄ = R or
¯ = R. In this section, we extend the idea of density to an arbitrary metric space.
R/Q
Definition 10 A metric space (X, d) is said to be separable if there exists a countable, every-
where dense set in X. Thus, ∃ a subset A of X such that
1. A is countable.
2. A = X.
In other words, X is said to be separable if there exists in X a sequence hx1 , x2 , · · · i such that
for every x ∈ X, some sequence in the range of hx1 , x2 , · · · i converges to x.
If the space is not separable, it is called inseparable. For example
(i) The real line usual metric space Ru is a separable metric space as Q ⊂ R is a countable
dense subset of R. The set of rationals Q is a dense subset of R (usual metric) and so is
the set of irrationals. Note that the former is countable whereas the latter is not.
(ii) A finite metric space is separable. Any countable metric space is separable.
(iii) The usual metric space C is separable, since the set A = {a + ib ∈ C : a, b ∈ Q} is dense
in C.
(iv) The Euclidean space Rn is separable since the set Qn = Q × Q × · · · × Q is countable and
dense in Rn .
(v) The metric space C[a, b] is separable since the set P[a, b] of all polynomials defined on [a, b]
with rational coefficients is countable and dense in C[a, b].
(vi) Let (X, d) be a discrete metric space. Since every subset is closed, the only dense subset
is X itself, so, X is countable and as such the space is separable.
(vii) Let X be the uncountable set and d be the discrete metric on X then (X, d) is not separable
as no proper subset of X in the discrete metric space (X, d) can be dense subset of X.
(viii) Consider the set A of elements x = {x1 , x2 , · · · } of X for which each xi is either 0 or 1.
Let E be any countable subset of A, then the elements of E can be arranged in a sequence
s1 , s2 , · · · . We construct a sequence s as follows: If the mth element of sm is 1, then the
mth element of s is 0, and vice versa. Then the element s of X differs from each sm in
the mth place and is therefore equal to none of them. So, s 6∈ E although s ∈ A. This
shows that any countable subset of A must be a proper subset of A. It follows that A is
uncountable, for if it were to be countable, then it would have to be a proper subset of
itself, which is absurd. We proceed to use the uncountability of the subset A to argue that
X must be inseparable.
(ix) The distance between two distinct elements x = {x1 , x2 , · · · } and y = {y1 , y2 , · · · } of A is
d(x, y) = sup{|xi − yi | : i = 1, 2, 3, · · · } = 1.
Suppose, if possible, that E0 is a countable, everywhere dense subset of X. Consider the

balls of radii 1/3 whose centres are the points of E0 . Their union is the entire space X,
because E0 is everywhere dense, and in particular contains A. Since the balls are countable
in number while A is not, in at least one ball there must be two distinct elements x and
y of A. Let x0 denote the centre of such a ball. Then
1 1
1 = d(x, y) ≤ d(x, x0 ) + d(x0 , y) < + < 1,
3 3
which is, however, impossible. Consequently, (X, d) cannot be separable.
(x) Consider l∞ space. Consider a subset A = {x = hxn i ⊆ l∞ : xn = 0 or 1} of l∞ . With x

we associate a real number x∗ such that the binary expansion of x∗ is
x∗ = 0.x1 x2 x3 · · · ∈ [0, 1].
Since each real number in [0, 1] has a unique binary expansion, distinct real numbers in
[0, 1] give rise to distinct sequences of zeros and ones. Since [0, 1] is uncountable, the set of
sequences consisting of zeros and ones in l∞ is uncountable. Also, the metric of l∞ shows
that the distance between two distinct elements x and y of A is d(x, y) = sup |xn −yn | =
1≤n<∞
1.. If we let each of these sequences to be the centre x0 of an open sphere of radius 1/3,
then
1 1
1 = d(x, y) ≤ d(x, x0 ) + d(x0 , y) < + <1
3 3
and there will be uncountable non-intersecting open spheres. If E be the dense set in l∞ ,
then it will be intersecting with each of these open spheres. Therefore, E is uncountable.
Since E is arbitrary set, l∞ can not be separable.
Theorem 27 Let (X, d) be a metric space and Y ⊂ X. If X is separable, then Y with

the induced metric is separable, too.
Proof: Let E = {xi : i = 1, 2, · · · } be a countable dense subset of X. If E is contained

in Y , then there is nothing to prove. Otherwise, we construct a countable dense subset of
Y whose points are arbitrarily close to those of E. For n, m ∈ N, let Sn,m = S(xn , 1/m)
and choose yn,m ∈ Sn,m ∩ Y whenever this set is nonempty. We show that the countable set
{yn,m : n and m positive integers} of Y is dense in Y.
For this purpose, let y ∈ Y and ε > 0. Let m be so large that 1/m < ε/2 and find xn ∈
S(y, 1/m). Then y ∈ Sn,m ∩ Y and
1 1 ε ε
d(y, yn,m ) ≤ d(x, xn ) + d(xn , yn,m ) < + < + = ε.
m m 2 2
Thus, yn,m ∈ S(y, ε). Since y ∈ Y and ε > 0 are arbitrary, the assertion is proved.
Theorem 28 A metric space (X, d) is separable if and only if it is second countable.
Proof: Suppose (X, d) is separable. Then X has a countable dense sub-set A (say). Let
A = {an : n ∈ N}, so A =n X. o
Let us construct B = S 1 (ai ) : ai ∈ A, n ∈ N . Here B is a countable collection of countable
n
set. So B is countable. We now prove that B is a base of (X, d).
Let G be any open set in (X, d) and x ∈ G. Since G is open, x is an interior point of G. So,
∃ r > 0 such that Sr (x) ⊆ G. Now by Archimidian property ∃ a natural number m such that
1 r
< . Since
n 2
x ∈ X = A, S 1 ∩ A 6= φ
m
1 1
so ∃ ai ∈ A such that ai ∈ S 1 (x), i.e., d(x, ai ) < m. Let y ∈ S 1 (ai ). Then d(y, ai ) < m. Now
m m
1 1 2
d(x, y) ≤ d(x, ai ) + d(y, ai ) < + =
m m m
1 r
< r; as <
m 2
i.e., y ∈ S 1 (ai ) ⇒ y ∈ Sr (x).
m
So, S 1 (ai ) ⊆ Sr (x) ⊆ G, i.e., x ∈ S 1 (ai ) ⊆ G. So, B is a base of (X, d). Hence (X, d) is second
m m
countable.
Conversely suppose (X, d) is second countable and B = {Bn : n ∈ N}, be a countable base of
(X, d). We take a point bn ∈ Bn for each n ∈ N and construct A = {bn : n ∈ N}. So, A is a
countable sub-set of X. We shall prove that A = X.
Let x ∈ X and r > 0. Since B is a base, ∃ Bk ∈ B such that, x ∈ Bk ⊆ Sr (x). Now
bk ∈ Bk ⊆ Sr (x) and bk ∈ A
so bk ∈ Sr (x) ∩ A ⇒ Sr (x) ∩ A 6= φ
Since r > 0 is arbitrary, Sr (x) ∩ A 6= φ for each r > 0 ⇒ x ∈ A. So x ⊆ A consequently A = X.

Hence (X, d) is separable.
Example 20 Prove that the space lp , 1 ≤ p < ∞ is separable.
Solution: Let lp (p ≥ 1) consists of all those sequences x = hxn i of real or complex numbers,
∞
|xn |p is convergent. We define a function d : lp × lp −→ R given by
P
such that
n=1
∞
hX p i1/p
d(x, y) = x − y ; x = hxn i, y = hyn i ∈ lp .

n n
n=1
Let us construct a set A containing those elements of lp , the first n components of which are
rational numbers and the rest are all zeros, i.e.,
n o
A = (β1 , β2 , · · · βn , 0, 0, · · · ); βi ∈ Q, 1 ≤ i ≤ n .
S∞
Then A = k=1 Ak , where for k ∈ N,
n o
Ak = (β1 , β2 , · · · βk , 0, 0, · · · ); βi ∈ Q, 1 ≤ i ≤ k .
Each Ak is countable. For (β1 , β2 , · · · βn ) ∈ Qn , the mapping
(β1 , β2 , · · · βn ) → (β1 , β2 , · · · βn , 0, 0, · · · )
is a one-to-one correspondence between Qn and An , and Qn is a countable set. Thus, A, being

a countable union of countable sets, is a countable set.
Now, we show that A is dense in lp . Let x = hαi i be an arbitrary element of lp . Then
αn p is convergent, and so corresponding to an ε > 0, ∃m ∈ N, such that
P∞
n=1
∞
X p εp
αi < .
2
i=m+1
Since Q is dense in R, for each real number αi (i = 1, 2, · · · , m) ∃ a natural number βi such that
m p
αi − βi p < ε .
X
2
i=1
Let y = {β1 , β2 , · · · βn , 0, 0, · · · } then y ∈ A. Now

(∞ )1
X p p
d(x, y) = αi − βi
i=1
(m ∞
)1
X p X p p
= αi − βi + αi
i=1 i=m+1
εp εp
< + =ε
2 2
So, y ∈ Sε (x). Since ε > 0 is arbitrary, Sε (x) ∩ A 6= φ for each ε > 0.
∴ x ∈ A. So lp ⊆ A ⇒ A = lp . Hence lp is separable.
Example 21 Let (X, d) be a metric space and Y be a subset of X. If Y is separable

and Ȳ = X, then prove that X is separable.
Solution: Let A be a countable dense subset of Y . Let x ∈ X and ε > 0 be given arbitrary.
Since Ȳ = X, there exists y ∈ Y such that d(x, y) < ε/2. Also, since A is dense in Y , there
exists z ∈ A such that d(y, z) < ε/2. By the triangle inequality, we have
ε ε
d(x, z) ≤ d(x, y) + d(y, z) < + = ε.
2 2
It follows that A is dense in X and hence X is separable.

Complete Metric Space: 1 Sequence

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Complete Metric Space

1.1 Convergent Sequence

Definition 1 [Convergence of a Sequence] Let (X, d) be a metric space and A be a non-

xn ∈ S(a, ε); ∀n ≥ k. (1)

Result 1 (Criteria for Convergence): Suppose X is a metric space, a ∈ X and hxn i is a

Theorem 1 In a metric space, every convergent sequence has an unique limit.

d(xn , a) < ε/2, for n ≥ n1 and d(xn , b) < ε/2, for n ≥ n2 .

Using the triangle inequality, we have

d(xn , xn ) ≤ d(xn , a) + d(a, b) + d(xn , b)

Also, interchanging xn , a and xn , b in Eq.(2), we obtain

d(a, b) − d(xn , xn ) ≤ d(xn , b) + d(xn , a). (3)

Combining (2) and (3), we get

⇒ 0 = d(xn , xn ) → d(a, b).

∴ d(a, b) = 0 and so a = b. Thus the limit of the sequence hxn i is unique.

1.2 Convergence of Subsequences

Definition 2 Suppose X is a metric space, x ∈ X and hxn i is a sequence in X that converges

Since hnk i is a strictly monotonically sequence of natural number so ∃ a k1 ∈ N such that

d(xnk , a) ≤ d(xnk , xn ) + d(xn , a) < ε, ∀ k ≥ k1 .

∴ The sub-sequence hxnk i converges to a. Thus every subsequence of a convergent sequence is

1.3 Bounded Sequence

Theorem 3 In a matric space every convergent sequence is bounded.

d(xm , xn ) ≤ d(xm , x) + d(x, xn )

d∞ (x, y) = sup |xn (t) − 0| = sup |e−nt − 0|

Theorem 4 (Characterization of limit points of a set in terms of convergent sequence)

Proof: Let (X, d) be a metric space, A ⊆ X.

Then xn 6= a; ∀n ∈ N. We now prove that hxn i converges to a. Let ε > 0 be given by

Thus {xn } converges to a.

d(xk , a) < r ⇒ xk ∈ S(a, r).

Since xk 6= a and xk ∈ A,, so

Since r > 0 is arbitrary, a is a limit point of A.

Conversely, assume that every convergent sequence of points of A converges to a point of

Theorem 5 Let hxn i be a sequence in a matric space (X, d) which converges to x in

(i) If A is a finite set then xn = x for infinitely many n.

(ii) If A is an infinite set then x is a limit point of A.

Suppose r1 = min{d(y1 , yi ); 2 ≤ i ≤ m}. So, yi 6∈ S(y1 , r1 ) for i = 1, 2, 3, · · · , m; d(y1 , yi ) >

d(xn , a) > r ⇒ xn ∈ S(a, r), ∀n ≥ k.

Since A is an infinite set, ∃ a natural number m ≥ k such that xm 6= a. So

Since r > 0 is arbitrary, so x is a limit point of A.

d(xn , a) < ε/2; ∀ n ≥ k1 ; and d(yn , b) < ε/2; ∀ n ≥ k2 .

Let k = max{k1 , k2 } ∈ N, then

1.4 Convergence in Product Spaces

Proof: Consider the product metric d : X × Y → R given by

Clearly, (X × Y, d) is a metric space. Assume that first xn → a in X and yn → b in Y . Then,

1.5 Cauchy Sequence

A sequence hxn i in K (R or C) is a Cauchy sequence in the sense familiar from elementary

Example 2 Let X = C[0, 1] be a metric space with the metric d∞ defined by

d∞ (f, g) = sup |f (t) − g(t)|, for all f, g ∈ X.

Theorem 8 In a metric space every convergent sequence is a Cauchy sequence.

d(xm , xn ) ≤ d(xm , a) + d(a, xm )

x1 = 1.4, x2 = 1.41, x3 = 1.414, x4 = 1.4142, · · · .

Proof: If hxn i is a convergent then, by Theorem 2 every sub-sequence of hxn i is convergent.

d(xn , a) ≤ d(xn , xnk2 ) + d(xnk2 , a)

Hence the sequence hxn i is a convergent.

Theorem 10 Let (X, d) and (Y, ρ) be metric spaces and f : X → Y be a uniformly

Proof: Since f is uniformly continuous, corresponding to an ε > 0, ∃ a δ > 0 such that

Since hxn i is a Cauchy sequence in X, given δ > 0, ∃n0 ∈ N, such that

d(xm , xn ) < δ, ∀m, n ≥ n0 . (6)

Therefore, from Eqs. (5) and (6), it follows that