Unit - 4
Contraction mappings and applications
The concept of contraction mapping is very useful in classical analysis.
Definition-
Let (X, d) be a metric space. A mapping T of X into itself is said to be a contraction (or contraction mapping) if there exists a real number a, 0 < a < 1, such that d(Tx, Ty)
d(x, y) for all x, y 
X.
Note- a contraction mapping is uniformly continuous
A point x 
X is called a fixed point of the mapping T : X 
X if Tx = x.
Theorem- Let T: X 
X be a contraction of the complete metric space (X, d). Then T has a unique fixed point.
Proof:
Let 
X and let 
be the sequence defined iteratively by 
= T
for n = 0, 1, 2, . . . . We shall prove that 
is a Cauchy sequence. For p = 1, 2, . . . , we have
………..(1)
Where 0 <
< 1 is such that
For all x, y
X.
Repeated application of the inequality (1) gives
Now, let m, n be positive integers with m > n. By the triangle inequality,
But 
it follows 
is a Cauchy sequence in (X, d), which is complete. Let y = 
Since T is a contraction, it is continuous. It follows that Ty = T(
= 
. Thus, y is a fixed point of T.
Moreover, it can be shown to be unique: If y 
z are such that Ty = y and Tz = z, then d(y,z) = d(Ty,Tz) 
d(y,z)<d(y, z). This implies d(y,z) = 0, i.e., y = z.
Proposition: Suppose 
is a sequence in any metric space (X, d) and k a positive integer such that each of the k subsequences
Converges to the same limit x. Then 
converges to x.
Proof:
Given 
> 0, there exist positive integers 
,
. . . ,
such that
Take N = max {
,
. . . ,
}. Consider any n 
Nk. By the division algorithm, N = mk + j, where 0
j
k - 1. Since m
N -1 would imply that N 
(N - 1)k+ j
(N - 1)k+ (k- 1) = Nk - 1, we must have m>N - 1, so that m
N
for j = 0, 1, 2, . . . ,k- 1. It follows that d(
,x)<
, i.e., d(
,x)<
.
This has been proved for any n
Nk. Therefore, 
converges to x.
Proposition- Let X consist of continuous mappings from [a, b] into [l,m], so that X is a subspace of the space C[a, b] of all continuous real-valued functions defined on [a, b]. Then X is a complete metric space.
Proof:
Since X 
[a, b] and the latter is complete, it is enough to show that X is a closed subset of it. Accordingly, we assume that f 
[a, b] is a limit point of X and proceed to show that f 
X. There exists a sequence 
in X that converges to f with respect to the metric of C[a, b]. For each x 2 [a, b], we have
And hence 
But for each x 
[a, b], we have
And so,
Which shows that f 
X. This completes the proof.
Picard’s existence theorem-
Consider the domain D: = {(x,y) 
: |x-
| < a, |y-
| < b } . Assume that f: D 
R is a continuous function satisfying the Lipschitz condition in the y-variable uniformly in the x-variable, that is,
Then there exists a 
> 0 and g:[ 
]
which is a solution of the initial value problem-
g’(x) = f (x, g(x)) satisfying the initial condition g(
) = 
Proof:
By an application of the fundamental theorem of calculus, equivalent to the integral equation
Let M > 0 be such that |f (x, y)| 
Choose 
> 0 such that
Define
Then Y is closed in 
and hence is a complete metric space. The map
Is a contraction on Y.
Definition
A (metric) space X is said to be connected if the only sets which are both open and closed in X are 
and the full space X, when X is a metric space.
A subset A of a metric space X is said to be connected if A is a connected space when considered as a (metric) space with the induced (or subspace) topology. More explicitly, this amounts to saying that (A, 
) is connected, where .5 is the restriction of the metric d on X to A.
Definition:
A metric space (X, d) is said to be disconnected if there exist two nonempty subsets A and B of X such that
(i) X = A 
B;
(ii) A 
=
and 
B = 
That is, the subsets must be nonempty, together they must constitute the whole space and neither may contain a point of the closure of the other. If no such subsets exist, then (X, d) is said to be connected.
Theorem- Let (R, d) be the space of real numbers with the usual metric. A subset I 
R is connected if and only if I is an interval, i.e., I is of one of the following forms:
Proof:
Let I be a connected subset of real numbers and suppose, if possible, that I is not an interval. Then there exist real numbers x, y, z with x < z < y and x, y 
I but z 
I . Then I may expressed as I = A 
B, where
Since x 
A and y 
B, therefore, A and B are nonempty; also, they are clearly disjoint and open in I. Thus, I is disconnected
To prove the converse, suppose I is an interval but is not connected. Then there are nonempty subsets A and B such that
Pick x 
A and y 
B and assume that x < y. Observe that [x, y] 
I , for I is an interval. Define
The supremum exists since A 
[x, y] is bounded above by y and it is nonempty, as x is in it. Then z 
. (We shall show that if z 
A, then z is a limit point of A. Let 
> 0 be arbitrary. By the definition of supremum, there exists some element a 
A such that z - 
< a 
z, i.e., every neighbourhood of z contains a point of A.) Hence, z 
B, for 
; in particular, x 
z < y.
If z
A, then x < z < y and z 
I. This contradicts the fact that [x, y] 
I . If z 
A, then for
So there exists a d > 0 such that 
This implies that there exists 
satisfying the inequality z < z1 < y. Then x 
z < z1 < y and z1 
I, for z1 being greater than sup (A 
[x, y]) is not in A. This contradicts the fact that [x, y] 
I.
Theorem-
Let (X, 
) be a connected metric space and f: (X, 
) 
(Y, 
) be a continuous mapping. Then the space f (X) with the metric induced from Y is connected.
Proof:
The map f : X 
f (X) is continuous. If f (X) were not connected, then there would be, a continuous mapping, g say, of f (X) onto the discrete two element space (
, 
). Then g o f :X 
would also be a continuous mapping of X onto 
, contradicting the connectedness of X.
Theorem-
Let I = [_ 1, 1] and let f: I 
I be continuous. Then there exists a point c 
I such that f (c) = c.
Proof:
If f (- 1) = -1 or f (1) = 1, the required conclusion follows; hence, we can assume that f (-1) > -1 and f (1) < 1. Define
g(x) = f (x) - x, x 
I :
Note that g is continuous, being the difference of continuous functions, and that it satisfies the inequalities g(-1) = f (-1) + 1 > 0 and g(1) = f (1) - 1 < 0. Hence, by the Weierstrass intermediate value theorem, there exists c 
(- 1, 1) such that g(c) = 0, that is, f (c) = c.
Theorem-
If Y is a connected set in a metric space (X, d) then any set Z such that Y 
Z 
is connected.
Proof:
Suppose A and B are two nonempty open sets in Z such that A 
B = Z and A 
B = 
1; as Y is dense in Z, Y 
A and Y 
B are nonempty open sets in Y and we have
a contradiction.
Local Connectedness-
A metric space (X, d) is said to be locally connected if, for x 
X, there is a base of connected neighbourhoods of x. Thus, X is locally connected if and only if the family of all open connected sets is a base for the open subsets of X
Theorem- Let (X, d) be a metric space. X is locally connected if and only if the connected components of the open sets in X are open in X. (For any subset A of X, the connected components of the points of the subspace A are called the connected components of A).
Proof:
Let G be an open subset of X, and C be a component of G, and {
: 
L} be a basis consisting of open connected sets for the open sets of X.
Let x 
C. Since x 
G, there is a 
such that x 
G; but since C is the component of x and 
is connected, x 
C.We have thus shown that C is a union of open subsets of X and is, therefore, open.
On the other hand, if O is any open set containing a point x 
X, the connected component of x in the subspace O is a connected neighbourhood of x contained in O; hence, X is locally connected.
Theorem- Let O be a nonempty open subset of (R, d). Then O is a union of an
At most countable family of open intervals, no two of which have common points.
Proof:
The connected components of O are intervals and that they are open subsets of O and, therefore, of R as well. Thus, they are open intervals. Note that O 
Q, where Q denotes the set of rationals in R, is countable and each component of O contains a point of O 
Q. The mapping r 
C(r) of O 
Q into the set of connected components of O is onto and, therefore, the set of connected components of O is countable.
Path-wise Connectedness
Definition-
Let (X, d) be a metric space, Y 
X and I = [0, 1]. A path in Y is a continuous mapping f : I 
Y . If f : I 
Y is a path in Y, we call f (0) 
Y its initial point, f (1) 
Y its final point and say that f joins f (0) and f (1) or that f runs from f (0) to f (1). A subset Y of X is said to be path-wise connected if, for any two points in Y, there exists a path running from one to the other, i.e. the points can be ‘‘joined by’’ a path in Y.
Note- Path-wise connectedness is also known as arc-wise connectedness.
Bounded sets-
A metric space (X, d) is said to be compact if every open covering G of X has a finite subcovering, that is, there is a finite subcollection {G1,G2, . . . , Gn} 
G such that
A nonempty subset Y of X is said to be compact if it is a compact metric space with the metric induced on it by d.
If G is an open covering of X, then the collection F of complements of sets in G is a collection of closed sets whose intersection is empty, and, conversely.
Thus, a metric space (X,d) is compact if and only if every collection of closed sets with empty intersection has a finite subcollection with empty intersection.
Note- A collection F of sets in X is said to have the finite intersection property if every finite subcollection of F has a nonempty intersection.
Proposition-
(i) (X, d) is compact;
(ii) every collection of closed sets in (X, d) with empty intersection has a finite subcollection with empty intersection;
(iii) every collection of closed sets in (X, d) with the finite intersection property has nonempty intersection.
Definition-
Let (X, d) be a metric space and e be an arbitrary positive number.
Then a subset A 
X is said to be an «-net for X if, given any x 
X, there exists a point y 
A such that d(x, y) < 
. In other words, A is an 
-net for X if
A finite subset of X that is an e-net for X is called a finite
-net for X.
Definition-
The metric space (X, d) is said to be totally bounded if, for any 
> 0, there exists a finite 
-net for (X, d). A nonempty subset Y of X is said to be totally bounded if the subspace Y is totally bounded.
Proposition- A totally bounded metric space is bounded.
Proof:
Let (X, d) be totally bounded and suppose 
> 0 has been given. Then there exists a finite 
-net for X, say A. Since A is a finite set of points, d(A) = sup{d(y, z): y, z 
A} < 
. Now, let x1 and x2 be any two points of X. There exist points y and z in A such tha
It follows, using the triangle inequality, that
So that
And, hence, X is bounded.
Proposition-
Let (X, d) be a compact metric space. Then (X, d) is totally bounded.
Proof:
For any given 
> 0, the collection of all balls S(x, 
) for x 
X is an open cover of X. The compactness of X implies that this open cover contains a finite subcover. Hence, for 
> 0, X is covered by a finite number of open balls of radius 
, i.e., the centres of the balls in the finite subcover form a finite 
-net for X. So, X is totally bounded.
Note- A metric space is compact if and only if it is complete and totally bounded.
Proposition-
Any continuous function f from a compact (metric) space (X, d) to another metric space (Y, d) is bounded, that is, f (X) is a bounded subset of Y.
Proof:
Fix any y 
Y. For each n 
N, the set B(y, n) is an open set in Y and hence
Is open. Note also that U 
. The collection {
: n 
N} is an open cover of the compact space X. If {
: 1 
i 
m} is a finite subcover and if N := max{
}, then X = 
, that is, f (X) 
B(y, N).
Hence f (X) is bounded.
Theorem:
Any continuous function from a compact (metric) space to R attains its bounds.
Or
Let X be compact and f: X —> R be continuous. Let M := M := sup{ f (x) : x 
X} and m := inf{f (x) : x 
X } . Then there
Exist 
, 
X such that f (
) = M and f (
) = m.
Proof:
There exists an 
R such that |f(x)| 
for all x 
X. Hence m and M exist as real numbers.
We shall prove the existence of 
by contradiction. If there is no p 
X such that
f (p) = M , then the set
Form an open cover of X.
(For, if x 
X, then f(x) <M and hence by Archimedean property of R , there exists n 
N such that n > 1/M-f(x) hence f(x) < M - 1/n or x 
.) Note that 
Un+ 1 for all n. As in the last proof, we conclude that X = 
for some N. But this leads to the contradiction: sup{ f (x) : x 
X} 
M — 1/n !
Theorem- Any continuous function from a compact metric space to any other metric space is uniformly continuous.
Proof:
Let f: (X, d) 
(Y, d) be continuous. Assume that X is compact.
We need to prove that f is uniformly continuous on X.
A naive attempt would runs as follows. For a given 
> 0, for each x, there exists 
by continuity of f at x. Since X is compact, the open cover {B(x,
): x 
X} admits a finite subcover, say, {B(
, 
) : 1 
j 
n} where 
= 
. One may be tempted to believe that if we set 
= min{ 
: 1 
i 
n, it might work. See where the problem is. Once you arrive at a complete proof, you may refer to the proof below. We now modify the argument and complete the proof. Given 
> 0, by the continuity of f at x, there exists an 
> 0 such that
Instead of the open cover {B(x, 
) : x 
X}, we consider the open
Cover {B(x, 
/2) : X 
X and apply compactness. Let {B(x„ 
/2) : 1 
i 
n} be a finite subcover. (Here 
= 
„ 1 
j 
n.) Let
:= min{
/2 : 1 
j 
n}. Let x, y 
X be such that d(x, y) < 
. Now
B (
,
/2) for some i. Since d(x, y) < 
, we see that
Thus y 
B (
,
). It follows that
By our choice of 
. Thus f is uniformly continuous.
Note- Let f be a homeomorphism of a metric space (X, d) onto a metric space (Y, d). Then X is compact if and only if Y is compact.
References:
- S. Kumaresan, Topology of Metric Spaces, Narosa Publishing House, Second Edition 2011
- Topology of metric spaces, S. Kumarsen, Alpha science international Ltd.
- Metric spaces, Satish shirali and Harkrishan L. Vasudeva, Springer.
