So, today we will discuss concept of Cauchy
sequence and completeness; the convergence
concept we already discussed. However, we
will repeat what we have done in last time.
We know the sequence of the real or complex
number, it plays a vital role in calculation
and the metric define on it that is the usual
metrics enables us to define the concept of
the convergence.
When we say a sequence of the real or a complex
number n goes to a, it means the difference
between a n minus a; this tends to 0 as n
tends to infinity. So, this basically define
is nothing but the distance between a n and
a and this concept on the real line, we wanted
to enhance over an arbitrary metric space
X, d where, X is the set of point and d is
the notion of the distance defined on it.
So, we say a sequence X n, a sequence convergence.
First, a sequence X n in a metric space 
X d is said to converge 
or to be convergent, if there is an X, belonging
to capital X such that, the distance between
x n X, when n is sufficiently large is 0 tending
to 0 or limit of this is 0, there, X is called
the limit of of the sequence x n and we write
of course, write it as X n goes to X under
the limit, when n tends to infinity on the
limit of this 1.
Now, here, if we look, this what is d of x
n x? It is basically a real number, it is
basically a real number a sequence of the
real number. So, the concept of the sequence
of that real number, the same concept is applied
here. We are converting the sequence a points
in x n in x which goes to x in the form of
the real number and then applying the same
definition of the convergence of a real or
complex numbers, real numbers.
Now, here, when we say the x n converges to
x, this means, here we say when x n converges
to x means for given epsilon greater than
0, there exist an integer n which depends
on epsilon such that, all x n with n greater
than capital N lie in the neighborhood, in
the epsilon neighborhood of x, that in in
epsilon neighborhood B x epsilon of x, that
is the meaning of this. So, in terms of the
neighborhood, we can say sequent x n convergent.
ok
Now, the point x must be a point in x, that
if x does not belong to capital x, then we
cannot say the limit converges. For example,
if we take a sequence x n say, 1 by n and
our capital X is a semi close, say, interval
0 to 1 and the metric d, if I take as x minus
y, then, though the sequence x n goes to 0
when tends to infinity. In the sense of real
over a real line, but the 0 is a point which
is not available in this plus x. So, we say
the x n does not converge to 0 in the metric
space x d because, the limiting point is not
available here ok.
So, the important part, area that limiting
point, must be a point of the space x. That
is why the convergence of a sequence is not
an intrinsic property, it basically depends
on the metric on the space which you are choosing.
Here, if I take this closed interval, then
the same sequence will behave as a convergent
sequence in this replacement.
So, this much we have already discussed earlier.
Now based on this or using this concept of
the convergence, we cannot define the concept
of the boundedness and the relation between
the convergence and the bounded sequence.
So, boundedness, bounded a non-empty subset
M of x is said to be bounded, to be a bounded
set, if it is diameter delta M, which is equal
to supremum d of x y x y are the points of
M is finite. So, a set M, which is a subset
of a metric space x d is said to be bounded
if the diameter of the set is finite. That
is, if we pick up any 2.0 here, find out the
distances and like this if you continue, then
supremum of this, if it comes out to be finite,
then we say the set m is bounded.
And similarly, we say a sequence x n in a
metric space x d is bounded, is said to be
bounded if the corresponding point point set
that is x 1 comma x 2 and so on is a bounded
set, is a bounded set a bounded subset or
set subset of x just like.
So, basically what we if. So, clearly, if
a set m is bounded, then m can be content
inside a open ball centered at x 0 with a
suitable radius r where, x 0 is a point in
capital X and any point in capital X and r
is sufficiently large large real number, if
we have the set x d a metric space and this
is our set M.
If the set is bounded, it means if I pick
up any arbitrary point x 0 in capital X need
not be then corresponding to this point, we
can always find an open ball centered at this
point, with a suitable radius, say r, such
that all the points of the set M or the points
of this ball, when we say, the set m is a
bounded set or bounded subset of x. So, there
is a relation between the convergence and
bounded sequence.
So, we go for some relation in the form theorem
or let it be, let x which is a metric space
x d be a metric space, then the following
thing of course, a convergence sequence 
in x is bounded and its limit is unique, this
is the first session we made it.
Second point which is true if sequence x n
converges to x and y n converges to y in the
metric space x d, then the corresponding distances
x n y n, this will go to d of x y this. So,
basically, what this result says, every convergence
sequence is a bounded sequence. We are not
talking about the converse part; converse
basic is basically not true always. In general,
the sequence need not be a convergent always
ok.
Second part says, if there are 2 sequences
which goes to the limit x and y, then the
difference if I take the x n y n and find
out the distances, then that distance will
also go to the distance between x and y. So,
that is what is. So, let us see the proof
of this.
Part a what is given a a convergent sequence
is given, we wanted to show the convergent
sequence is bounded. So, suppose sequence
x n converges to x, this is given so, by definition
of the convergent. So, for given epsilon,
say suppose 1, we can find we can find an
n an integer n such that, distance between
x n and x can be made less than epsilon, that
is, 1 after certain integer capital N, so,
all n, greater then capital N.
So, let us consider d of x n comma x. We claim
that this is less than 1 plus a, where a stands
for maximum of d x 1 comma x d x 2 comma x
d x n comma x. So, if this is true, then obviously,
a point set x 1 x 2 x n will be a bounded
set because, d of f x n x is less than a.
So, supremum was taken its finite, therefore,
the convergence sequence will be a bounded
sequence ok.
Now, how does it follow this? By the triangle
inequality, we can write d of x n x, when
n is say, n is less than or equal to n, we
can write this thing as this will be less
than equal to d of x n comma x n minus 1 plus
d of x n minus 1 comma x n minus 2 and lastly,
it goes to d of x 1 comma x like this. So,
here when n is there, so, we can write this
as capital N also. So, we can say capital
N, that is, it goes a d of x comma x comma
x x n minus 1 and d of x n minus 1 x n minus
2 and like this, continues till d of x 1 x.
Now, the maximum value of this, say a, then
we can write, this is less than equal to some
n times a. So, let it be another constant,
say, a. So, we can say d of x n x n is less
than a 1 for all n 1 2 up to n minus 1 and
for n greater then this 1 n its already less
than 1. So, basically, when you take the both
these simultaneously, we can write the d of
x n x is less than 1 plus a for all n. This
this is true for all n, for all n for every
n. So, it is. So, we get this, shows that
sequence x n is bounded ok.
Now, second part of this is, the limit is
unique. So, for the second part, suppose x
n has 2 limit, suppose x n converges to x
and also x n converges to z, we wanted x and
z are identical, equal. So, now, consider
d of x comma z since the distance cannot be
negative, it will be always be greater than
equal to 0 apply the triangle inequality.
So, we get d of x comma x n plus d of x n
comma z.
Now, as n tends to infinity, x n converges
to x x as well as x n converges to z; it is
our assumption. So, the right hand side will
go to 0, therefore, this d of x z is less
then equal to 0, but basically it is always
greater than 0. So, this implies d of x z
is 0 that implies x equal to 0 by definition
of the metric. So, limit will always be unique,
if a sequence converges, it will converge
to the same limit. So, this will
The second part of it proof that we wanted
to show that, if x n converges to x y n converges
to y, then d of x n y n will go to d of x
y. So, consider d of x n y; start with this
and then apply that triangle inequality d
of x n comma x plus d of x comma y plus d
of y comma y n ok.
Now, transfer this d of x y this side. So,
we get d of x n y n minus d of x y is less
than equal to d of x n comma x plus d of y
n comma y. Because, it is symmetric in nature
now, as n tends to infinity, the right hand
side tends to 0; right hand side will go to
0 because, this goes to 0, this goes to, so,
this will go to 0.
Now, interchange the role of x n x n and x
that is x n is interchange with x y n interchange
with y, then here you are getting d of x y
minus d of x n y n and this can be further
written because, if I interchange x n x, there
will be no change. So, this will also go to
0 as n tends to infinity. So, what we get
it from here d of x n y n minus d x y tends
to 0, while in this case d of x y minus d
of x n y n goes to 0. This implies the modulus
of d x n y n minus d of x y this modulus will
go to 0, as n tends to infinity and that is
what we wanted to prove clear.
Now, so far, we have considered the convergence
of the sequence and the corresponding sequence
said to be bounded set and the relation between
the convergent bounded the converse of this
converse of part a need not we need not be
true or be true, that is every bounded sequence
need not be convergent need not be a convergent
sequence.
For example, if I take a sequence x n to be
minus 1 to the power n, by say n now, obviously,
the sequence when or just we take minus 1
to the power n, where this is convergent,
we can say minus 1 to the power n. Take the
sequence x n to be minus 1 to the power n.
Now, this sequence is not convergent sequence,
minus 1 to the power n is not convergent as
n tends to n because, when n is say, even,
it goes to plus 1, when it is odd it goes
to minus 1. So, it is not a convergent, but
it is a bounded sequence 
of real numbers. So, every bonded sequence
need not be convergent that we.
Now, another concept which we wanted to introduce
is the completeness. The completeness concept
is not dry, with the help of those four properties,
which the metric joined, that is, the early
metric space d of x y is greater than equal
to 0 real non negative and positive real finite
numbers. It is 0 when it is x equal to y and
vice-versa d of x y equal to d y x and d of
x y is less then equal to d x z plus d of
z.
So, these properties are satisfying, then
only we say the metric is the d is a metric
on x, but, it does not give any information.
It does not say whether the metric is complete
or not. So, before going to the definition
of the complete, and let us see what is the
meaning of these completeness in a general
metric space and how can we show, how can
we prove the metric is a complete metric space.
But this is a different property and this
is an extra property which the metric is joined.
Some of the metric spaces are complete, where
there are examples, where the metrics are
also incomplete metrics space.
The concept of the completeness is taken basically
form the sequence of the real number or sequence
of the complex number. We know that sequence
of the real number or complex number, if it
is convergent, it will satisfy the Cauchy
convergence criteria. What is that Cauchy
convergence criteria?
So, let us say, a sequence x n. We know a
sequence x n of real or complex number converges
on the real line R or in the complex plane
C 
if and only if if and only if, for every given
epsilon, there is an N which depends on epsilon
such that mod of x n minus x n is less than
epsilon for all n and M greater than N.
Now, this mod of x n minus x m is less than
epsilon, this is equivalent to basically the
distance between x n and x m in a real line
or complex plane, because, distance, now,
some of distance on real or complex plane
is defined as the usual way mod of x n minus
x m. So, when we say a sequence x n of real
or complex number is convergent, if and only
if, for any epsilon, they greater than 0,
there exist in such that the distance between
x and m is less then epsilon and this criteria
is known as the Cauchy convergence criteria
ok.
Or, we say if a sequence satisfies this condition,
then we say it is a Cauchy sequence or we
say if a sequence x n of real or complex number
satisfies satisfy the condition of Cauchy
cauchy criteria, then such a sequence, we
call such sequence x n as a Cauchy sequence.
So, we can also say real sequence of a sequence
of real or complex number converges.
So, we can say. So, we can say, a sequence
of real or complex number number converge,
numbers converges on R or C if and only if
it is a Cauchy sequence, it is a Cauchy sequence.
So, this is the criteria, is well known and
we know all the sequences of real or complex
number, if convergent, it will be Cauchy and
vice-versa and that is why, we say the real
or complex number is a complete metric space.
But, in general metric space, this criterion
may not be hold may not hold good. A sequence
x n in x may be a convergent sequence or may
be a Cauchy sequence, but need not be convergent.
For example, if we take the earlier sequence
x n to be 1 by n which we have discussed and
metric, if I take the usual metric and x to
be 0 1 where, 0 is a metric, then, we have
see the sequence is not convergent. However,
that can be shown to be a Cauchy sequence.
So, such a space, we cannot say it is a complete.
So, we define the completeness related to
this, a sequence which is Cauchy or not Cauchy
sequence. So, before going, this let us see
the definition of the Cauchy sequence in a
general metric space.
A sequence in a metric space x d is said to
be Cauchy or we also say, fundamental sequence.
If for every if for every epsilon greater
than 0, there is an n depends on epsilon such
that, distance d of x m comma x n is less
than epsilon for every m n greater than m
and the space x is said to be complete definition
of a complete metric space. A metric space
x d is said to be complete if every Cauchy
sequence 
in x converges, that is the limit point, that
is has a limit point has a limit, which is
an element of x 
of x, which is an element of x.
So, a metric space is said to be complete
if every Cauchy sequence in x convergent,
that it has a limit, which is an element of
x. So, obviously, with this definition, 1
can say the set of real number or set of complex
number under the usual metric is a complete
metric space, because, a Cauchy convergence
criteria says every sequence of the real or
complex number is Cauchy; of if a sequence
is Cauchy, it must be convergent, vice-versa
for them clear. So, that is clear.
But in general, metric space, whether it is
true or not, let us see. So, example is, if
I take a metric space x to be 0 1 and d of
x y as mod of x minus y and take a sequence
x n to be 1 by n of real number, then, clearly
sequence x n is a Cauchy sequence, because,
the difference between because difference
between two terms of the sequence can be made
less than epsilon after certain integer n
m greater than for n n m greater than equal
to n, it can be shown ok.
1 by n minus 1, when n m is sufficiently large,
it basically reducing to 0. So, we can identify
the capital n such that, after certain stage,
the difference between these 2 terms can be
made as small as we pleased. So, it is a Cauchy
sequence, but the limit of the sequence x
n 1 by n, this tends to 0 as n goes to infinity,
which is not which is not in x because, x
is the semi closed interval 0 and one.
Therefore, this Cauchy sequence is not convergent.
So, this space x under d is incomplete metric
space incomplete. In fact, those space, where
all the Cauchy sequences are not convergent,
the space will be incomplete, that is, there
exist even a single Cauchy sequence which
is incomplete, which is not convergent, then
it will be a incomplete metric space ok.
There are many examples, others also for incomplete
metric space. If I take the space x to be
r minus a and the notion of the distance d
x y, if I choose to be mod of x minus y, then,
we say, we see that this space is an incomplete
metric space, because, we can choose a sequence
x n which is of the form, say, a plus 1 by
n, then, this sequence is a cauchy sequence
which converges to a as n tends to infinity.
While a is not a point, is a does not belongs
to the spaces, where a does not belongs to
this. So, it is incomplete metric space.
Then, another example we can choose from the
set of real number. If I remove, take out
all the rational point, then the set of rational
number is an incomplete metric space, even
the open interval under this metric d x y
as mod x minus y with a induced topology will
be an incomplete metric space. So, here, I
can choose the sequence x n as a plus 1 by
n or x n, we can choose to be b minus 1 by
n. This type of sequence if I take, then,
what all Cauchy sequence says, but but does
not or do not converge converge. So, that
is why, it is a not.
Now, what is the relation between convergence
and the Cauchy sequence? Every convergent
sequence is a Cauchy sequence, that we will
show and Cauchy sequence need not be convergent.
We have already take seen by many examples,
where the sequence are Cauchy, but not convergent.
So, we have result every convergent sequence
in a metric space x d is a Cauchy sequence,
is a Cauchy sequence the proof is very simple;
suppose we have a sequence x n which converges
to x.
So, by a definition, for given epsilon greater
than 0, there exist an n depending on epsilon
such that, such that the distance between
x n comma x can be made less than epsilon
or epsilon by two, for all n greater than
equal to, say, capital n 1, then, similarly,
we can say a sequence say x n converges to
x. We can say for the given epsilon greater
than 0, there exist say here n one. So, I
am taking here n 1, here n two, depending
on epsilon such that, d of x m comma x is
less than epsilon by 2 for all n greater than
equal to.
Now, if I picked up N to be maximum of N 1
comma N 2, then both these conditions hold
for all n m greater than equal to capital
n. This is also less than epsilon by 2, this
is also less than. So, consider d of x n comma
x m, now this will be less than equal to d
of x n comma x plus d of x comma x m.
Now, this is less then epsilon by 2, this
is less than epsilon by 2 for all n m greater
than or equal to capital N. Therefore, this
whole thing will be less than epsilon, hence,
we say the sequence x n is a Cauchy sequence
Cauchy. So, this completes the proof converse
we have already discussed.
Second result which gives the relation between
the closer and the convergence, the result
is, let m be a nonempty subset of a metric
space x d 
and m bar its closure, closure then, the following
result hold x belongs to m bar if and only
if there is a sequence x n in m such that,
x n goes to x yes b part m is closed if and
only if the situation x n belongs to capital
N x n converges to x implies implies that
x is a point of m, the proof is like this.
So, if x is an m closure, m closure means,
set of all the points m together with its
limit points, m closure means means the set
of all points of m together with its limits
points all limits points, then this is the
m closure.
So, we will prove the first part that is,
part a. What we want is, x belongs to m bar,
then, there is a sequence x n n m such that.
So, let us suppose x is a point in the closure
of this ok.
So, the possibility is either x will be a
point in M or x may be limit point. So, if
x is in m, then there is a sequence of the
type x comma x etcetera in m which converges
to x. So, our result is follow. So, result
follows, that is what we wanted to show there
will be a sequence x n n m, which goes to
x. There will be a sequence x n n m which
converges to where x belongs to m so, obviously,
this x will be in. So, this follows.
Now, if as x is not in m, then it will be
a limit point. Then, it is a point of accumulation
of m. So, if it is a point of accumulation,
then, for each n for each n say, 1 2 3 and
so on, the ball b centered at x with a radius
say, 1 by n contains n and x n belongs to
m and this x n sequence will go to x because,
1 by n is tending to 0, as n tends to infinity.
So, again, the session is complete. So, if
x m, is then, there will be a sequence n,
such that x n converges to x. So, that will.
So, this part, the part conversely be, now,
I have this result suppose, a sequence x n
in m that goes to x, then, we prove that x
belongs to m closure. So, conversely, if the
sequence x n is in m and x n goes to x, then
there are 2 cases. Either x will be a point
of m or every neighborhood of x or every neighborhood
of x contains points x n other than x, because,
this is a sequence goes to x. So, either x
will be a point in m, because, x n all in
m or if it is not, then, every neighborhood
of x will contain the point x n, which is
different from x.
So, this is this shows that x must be an accumulation
point. Hence, x is a point of accumulation.
So, it must be a point in m bar, hence, x
belongs to m closure m bar, by definition,
closure of M. So, this comes the proof for
the part b is very simple, it follows form
part a.
We know that m is closed if and only if if
and only if m is equal to m closure, this
we know now if m is equal to m closure. So,
let us see the part n. Now, when m is closed,
it means m is equal to m closure. So, if m
is m closure, then according to the part a,
there will be a sequence x n which goes to
x. So, we get situation x n belongs to x,
but the point x will be in m, because, m is
closed. So, it will be the point of m bar,
means, it will be in m and vice versa. If
I take this part, suppose x n is the x n converges
so that x belongs to m, then, the x, either
it will be limit point or will be a point
of the m itself. So, it must be in m closure.
So, m is called to m closure and m this 1.
So, all the limits points lies in m, it means,
m is equal to m closure, it means m will be
closed. So, this completes the proof of this,
nothing to prove. So, it proof you can say,
proof follows from part a, that is complete
ok.
Now, there is another result, a subspace space
M of a complete metric space X is itself complete
if and only if the set m is closed in X. So,
what is the meaning of this? A subspace M
of a complete metric space X is itself complete
if and only if the set m closed. We know by
definition of the completeness, if x d is
a metric space and here is the set m, and
say, this is under the induced metric is a
metric space, then, we say this set m is a
complete metric space, if every Cauchy sequence
is a convergent 1.
Now, instead of proving this part, we can
simply say, if a subset subspace m of a complete
metric space, this is a complete given is
closed. That is, if I simply prove m is equal
to m closure, then, automatically it will
give you the conclusion that m will be a complete
subspace of X.
So, you need not to consider the Cauchy sequence.
So, that simply says, prove all the limits
points of the m lies in that time itself.
So, that is very good proof, let m be complete
suppose. So, what we wanted to show is that,
m is closed. If it is complete, means m is
closed, this is our proof. Now, to show m
is closed, then, what we show here is that,
it contains all of its limit point.
So, let x belong to the closure of this. Now,
if x belongs to m, then, automatically it
will be a closed set, this proof. So, suppose
x be a point in that, then, because x is in
m bar, by the previous result if x belongs
to m bar, then, there is a sequence x n in
m. This result is, if x in it, there is a
sequence in it such that x n converges to.
So, by the previous. So, for every. So, for
every x for every x belongs to m bar for every
x belongs to m bar, there is a sequence x
n in m, which converges to x.
But every convergent sequence is Cauchy. So,
x n is Cauchy is a Cauchy sequence and m is
complete. So, every Cauchy sequence is convergent.
So, this implies that x n converges to x is
a convergent sequence, because, it is Cauchy
and the limit point is unique. So, it belongs
to m, as limit point is unique. So, this proves
m is closed.
Now, conversely, if we take m to be closed,
closed and x n be a Cauchy sequence in m,
then, because, Cauchy sequence x n is a Cauchy
sequence. So, we say x n converges to, and
this sequence converges to x n, which is a
point in x. Suppose, now we wanted to x belongs
to m bar, now this follows x belongs to m
bar. By this property again, fourteen a if
m is closed, if m is closed then, the situation
x n converges to x implies x belong. So, this
belongs to m bar. So, this shows x belongs
to m, but m is equal to m bar. So, we get
because, it is closed set. So, it belongs
to m bar.
Hence, by assumption, arbitrary Cauchy sequence
converges hence arbitrary Cauchy sequence
converges in m. So, m is complete, this completes
the proof, thank you, thanks.
