Unit - 3
Continuity
For a real-valued function f with domain A 
R, a rough and rather inaccurate description of continuity at a point a 
A is the statement ‘‘f(x) is close to f(a) when x is close to a’’. The measure of ‘‘closeness’’ of two numbers, or distance between them, is the absolute value of the difference of the numbers. In terms of the standard metric d on R, continuity involves a relationship between d(x, a) and d(f(x),f(a) ). This observation makes it possible to extend the concept of continuity to functions with domain and range in metric spaces.
Definition- Let (X, 
) and (Y, 
) be metric spaces and A 
X. A function f: A 
Y is said to be continuous at a 
A, if for every 
> 0, there exists some
> 0 such that
If f is continuous at every point of A, then it is said to be continuous on A.
Note-
- If one positive number d satisfies this condition, then every positive number
< 
also satisfies it. This is obvious because whenever x 
A and 
(x, a) < 
, it is also true that x 
A and 
(x, a) < 
. Therefore, such a number 
is far from being unique. - In the definition of continuity, we have placed no restriction whatever on the nature of the domain A of the function. It may happen that a is an isolated point of A, i.e., there is a neighbourhood of a that contains no point of A other than a. In this case, the function f is continuous at a irrespective of how it is defined at other points of the set A. However, if a is a limit point of A and {xn} is a sequence of points of A such that xn
a, it follows from the continuity of f at a that f (xn) 
f (a).
Proposition: A mapping f for a metric space (X, 
) into metric space (Y, 
) is continuous at a point a 
X iff for every 
> 0, there exists 
> 0 such that
Where S(x, r) denotes the open ball of radius r with centre x.
Proof-
The mapping f : X 
Y is continuous at a 
if and only if for every 
> 0 there exists 
> 0 such that
Which means
Or
This is equivalent to the condition
Theorem- A mapping f: X 
Y is continuous on X if and only if 
is closed in X for all closed subsets F of Y.
Proof-
Let F be a closed subset of Y. Then Y\F is open in Y so that f
is open
In X.
[Note-A mapping f: X 
Y is continuous on X if and only if 
is open in X for all open subsets G of Y.]
But
Is open in X. Since every open subset of Y is a set of the type Y\F, where F is a suitable closed set, it follows by using (note) that f is continuous.
The characterisation of continuity in terms of open sets leads to an elegant and brief proof of the fact that a composition of continuous maps is continuous.
Note- Let (X, 
) and (Y, 
) be metric spaces and let
f : A 
Y then the following statements are equivalent-
(i) f is continuous on X;
(ii) 
for all subsets B of Y;
(iii) 
for all subsets A of X.
Definition of extension and restriction-
Let X and Y be abstract sets and let A be a proper subset of X. If f is a mapping of A into Y, then a mapping g : X 
Y is called an extension of f if g(x) = f (x) for each x 
A; the function f is then called the restriction of g to A.
Theorem- Let (X, 
) and (Y, 
) be a metric space and let f:X, g : X 
Y be continuous maps. Then the set 
X:f(x) = g(x)} is a closed subset of X.
Proof: Let F = 
X:f(x) = g(x)}. Then X\F =
X:f(x) 
g(x)}. We shall show that X\F is open. If X\F = 
, then there is nothing to prove. So let X\F 
and let a 
X\F. Then f (a)
g(a). Let r > 0 be the distance dY (f (a), g(a)). For 
= r/3, there exists a 
> 0 such that
By the triangle inequality, we have
Implies
For all x satisfying 
(x, a) < 
. Thus, for each x
S(a, 
), 
(f (x), g(x)) > 0, i.e., f (x) 
g(x). So,
Hence, X\F is open and thus F is closed.
Proposition: Composite of continuous functions is continuous: Let X ,Y, Z be metric spaces. Let f: X 
Y be continuous at x 
X and g: Y 
Z be continuous at y = f (x). Then the composite map gof: X 
Z is continuous at x 
X
Proof:
Let 
x. Then 
:= f (
) 
y = f (x) by the continuity of f at
x. Since g is continuous at y, it follows that g(
) ---* g(y) or what is the
Same, g ( f (
)) 
g(f (x)).
Definition
Let X be a nonempty set. Given mappings f and g of X into C and a 
C, we define the mappings f + g, 
f , fg and |f| into C as follows
For all t 
X. Further, if f (t) 
0 for all t 
X, we define the mapping 1/f of X into C by
Uniform Continuity-
Let (X,
) and (Y, 
) be two metric spaces and let f be a function continuous at each point 
of X. In the definition of continuity, when 
and 
are specified, we make a definite choice of 
so that
Definition
Let (X,
) and (Y, 
) be two metric spaces. A function f : X 
Y is said to be uniformly continuous on X if, for every 
> 0, there exists a 
> 0 (depending on 
alone) such that
For all 
, 
X.
Every function f: X 
Y which is uniformly continuous on X is necessarily continuous on X. However, the converse may not be true.
Proposition-
Let (X, d) be a metric space and let x 
X and A 
X be nonempty. Then x 
if and only if d(x, A) = 0.
Proof:
Suppose d(x, A) = 0. There are two possibilities: x 
A or x 
A. If x 
A, then x 
. We shall next show that if x 
A, then x is a limit point of A. Let e > 0 be given. By the definition of d(x, A), there exists a y 
A such that d(x, y) < 
, i.e., y 
S(x, 
). Thus, every ball with centre x and radius e contains a point of A distinct from x; so x 
. Conversely, suppose x 
.. If x 
A, then obviously d(x, A) = 0. We shall next show that if x is a limit point of A, then d(x, A) = 0. By the definition of limit point, every ball S(x, 
) with centre x and radius 
> 0 contains a point y 
A distinct from x.
Consequently, d(x, A) < 
, i.e., d(x, A) = 0.
Theorem-
Let A and B be disjoint closed subsets of a metric space (X, d). Then there is a continuous real-valued function f on X such that f (x) = 0 for all x 
A, f (x) = 1 for all x 
B and 0
f (x)
1 for all x 
X.
Proof:
The mappings x 
d(x, A) and
x 
d(x, B) are continuous on X. Since A and B are closed and A 
B = 
,
d(x, A) + d(x, B) > 0 for all x 
X. Indeed, if d(x, A) þ d(x, B) = 0 for some x 
X, then d(x, A) = d(x, B) = 0; so x 
= A and x 
= B, and hence x 
A 
B, a contradiction
Now define a mapping f : X 
R by
Then f is continuous on X. Moreover,
And 0
f (x)
1:
Theorem-
If f and g are two uniformly continuous mappings of metric spaces (X, 
) to (Y, 
), and (Y, 
) to (Z, 
), respectively, then g O f is a uniformly continuous mapping of (X, 
) to (Z, 
).
Proof:
Since g is uniformly continuous, for each 
> 0, there exists a 
> 0 such that
For all x, y 
X.
Thus, for each 
> 0, there exists an 
> 0 such that
For all x, y 
X and so g o f is uniformly continuous on X.
Let (X, 
) and (Y, 
) be two metric spaces. A function f: X 
Y is said to be uniformly continuous on X if, for every 
> 0, there exists a 
> 0 (depending on e alone) such that
For all x1, x2 
X.
Every function f : X 
Y which is uniformly continuous on X is necessarily continuous on X. However, the converse may not be true.
Proposition-
Let (X, d) be a metric space and let x 
X and A 
X be nonempty. Then x 
if and only if d(x, A) = 0.
Proof:
Suppose d(x, A) = 0. There are two possibilities: x
A or x 
A. If x 
A, then x 
. We shall next show that if x 
A, then x is a limit point of A. Let e > 0 be given. By the definition of d(x, A), there exists a y 
A such that d(x, y) < e, i.e., y 
S(x, e). Thus, every ball with centre x and radius e contains a point of A distinct from x; so x 
. Conversely, suppose x 
. If x 
A, then obviously d(x, A) = 0.
We shall next show that if x is a limit point of A, then d(x, A) = 0. By the definition of limit point, every ball S(x, e) with centre x and radius e > 0 contains a point y 
A distinct from x. Consequenly, d(x, A) < e, i.e., d(x, A) = 0.
Theorem-
If f and g are two uniformly continuous mappings of metric spaces (X, 
) to (Y, 
), and (Y, 
) to (Z, 
), respectively, then g o f is a uniformly continuous mapping of (X, 
) to (Z, 
).
Proof:
Since g is uniformly continuous, for each e > 0, there exists a 
> 0 such that 
(f (x), f (y)) < 
implies 
( (g o f )(x), (g o f )(y)) < e for all f (x), f (y) 
Y.
As f is uniformly continuous, corresponding to 
> 0, there exists an h > 0 such that
For all x, y 
X.
Thus, for each 
> 0, there exists an 
> 0 such that
For all x, y 
X and so g o f is uniformly continuous on X.
Theorem-
Let (X, 
) and (Y, 
) be two metric spaces and f : X 
Y be uniformly continuous. If 
is a Cauchy sequence in X, then so is 
in Y.
Proof:
Since f is uniformly continuous, for every 
> 0, there exists a 
> 0 such
That
For all x, y 
X.
Because the sequence 
is Cauchy, corresponding to d > 0, there exists n0 such that
From above result,
And so 
is a Cauchy sequence in Y.
Homomorphism-
Let (X, 
) and (Y, 
) be any two metric spaces. A function f : X 
Y which is both one-to-one and onto is said to be a homeomorphism if and only if the mappings f and inverse of f are continuous on X and Y, respectively. Two metric spaces X and Y are said to be homeomorphic if and only if there exists a homeomorphism of X onto Y, and in this case, Y is called a homeomorphic image of X.
Let d1 and d2 be metrics on a nonempty set X such that, for every sequence 
in X and x 
X,
Or
A sequence converges to x in (X, d1) if and only if it converges to x in (X, d2).We then say that d1 and d2 are equivalent metrics on X and that (X, d1) and (X, d2) are equivalent metric spaces.
Theorem-
Two metrics d1 and d2 on a nonempty set X are equivalent if there exists a constant K such that
For all x, y 
X.
Proof:
Let xn 
x in (X, d1). Then we show that xn 
x in (X, d2). This follows from the inequality
If xn 
x in (X, d2), then xn 
x in (X, d1) in view of the inequality
Which is the complete proof.
Isometry-
Let (X,d) and (
) be two metric spaces. A mapping f of X into 
is an isometry if
(f (x), f (y)) = d(x, y) for all x, y 
X. The mapping f is also called an isometric embedding of X into 
If, however, the mapping is onto, the spaces X and 
themselves, between which there exists an isometric mapping, are said to be isometric. It may be noted that an isometry is always one-to-one.
Definition-
A mapping f of X into Y is an isometry if
For all x, y 
X. It is obvious that an isometry is one-to-one and uniformly continuous.
Definition
Let f : X 
Y and fn : X 
Y, n = 1, 2, . . . Be given. We say that 
converges pointwise to the function f if and only if
A sequence 
converges pointwise to f on X if, for a given 
> 0 and given x 
X, there exists an integer n0 such that
Definition
Let 
be a sequence of mappings of (X, dX) into (Y, dY ). We say that the sequence 
converges uniformly on X to a mapping f : X 
Y if, for every 
> 0, there exists an n0 (depending on 
only) such that
For all n
n0 and all x 
X; which means-
Note- It is clear that uniform convergence implies pointwise convergence; the converse is not true.
Theorem-
Let (X, dX) and (Y, dY ) be metric spaces, 
a sequence of functions, each defined on X with values in Y, and let f : X 
Y. Suppose that fn 
f uniformly over X and that each fn is continuous over X. Then f is continuous over X. Briefly put, a uniform limit of continuous functions is continuous.
Proof:
Let x0 
X be arbitrary and let 
> 0 be given. Since fn 
f uniformly over X, there exists n0 (depending on 
only) such that for each x 
X,
……….(1)
…………..(2)
Using (1) and (2)-
And therefore, f is continuous at x0.
Cauchy’s criterion-
Let 
a sequence of functions defined
On a metric space (X, 
) with values in a complete metric space (Y, 
) ). Then there exists a function f : X 
Y such that
Fn 
f uniformly on X
If and only if the following condition is satisfied: For every 
> 0, there exists an integer n0 such that
For every x 
X.
Definition: Let f1, f2, . . . Be a sequence of real-valued functions defined on a set X. We say that 
converges on X to a function f : X 
R if the sequence of functions {Sn}n$1 converges to f on X, where Sn = f1 + f2 + . . . + fn. In this case, we write
Or
Definition-
Let f1, f2, . . . Be a sequence of real-valued functions defined on a set X. We say that 
converges uniformly on X to a function f : X 
R if the sequence of functions 
converges uniformly to f on X, where Sn = f1 + f2 + . . . + fn. In this case, we write
Or
Weirstrass M-test-
Let f1, f2, . . . Be a sequence of real-valued functions defined on a set X and suppose that-
For all x belongs to X and all n = 1, 2, . . . . If 
converges, then 
converges uniformly.
Proof:
If
converges, then for arbitrary 
> 0,
For all x provided that m and n are sufficiently large.
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.
