WEBVTT
Kind: captions
Language: en
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Alright so now what we are going to do is
we will continue with the proof of the Arzela-Ascoli
00:00:41.650 --> 00:00:43.080
Theorem, okay.
00:00:43.080 --> 00:01:00.390
So so conversely suppose that A in C X, R
suppose A in C X, R or ofcourse you could
00:01:00.390 --> 00:01:05.650
have taken C X, C okay instead of real valued
functions you should have taken complex valued
00:01:05.650 --> 00:01:32.140
functions is is a closed subset such that
A is bounded and equicontinuous, okay we will
00:01:32.140 --> 00:01:45.680
show show A is compact and how we will show
that A is compact we will show A is compact
00:01:45.680 --> 00:01:51.700
by showing that A is sequentially compact,
okay because you know compactness is equivalent
00:01:51.700 --> 00:02:01.900
to sequential compactness for a metric space.
So we will show A is compact by demonstrating
00:02:01.900 --> 00:02:13.880
that it is sequentially compact compact, okay.
00:02:13.880 --> 00:02:24.760
Now so so what you have to do is that you
know we will pick a sequence from A, okay
00:02:24.760 --> 00:02:32.750
this is sequence of functions in A and we
will assume and then we will try to extract
00:02:32.750 --> 00:02:37.670
the subsequence which is convergent that is
what sequential compactness means, sequential
00:02:37.670 --> 00:02:44.730
compactness means that every sequence admits
a convergent subsequence, okay. So for this
00:02:44.730 --> 00:03:12.630
we pick any sequence f 1, f 2, etc f n and
let me call it as S of A and show that there
00:03:12.630 --> 00:03:35.840
is show that there is a convergent subsequence,
okay.
00:03:35.840 --> 00:03:41.810
So we start with any given sequence in A and
we try to show that there is a convergent
00:03:41.810 --> 00:03:49.449
subsequence but now notice that C X, R okay
the space of continuous bounded real valued
00:03:49.449 --> 00:03:57.280
functions on X mind you is is a Banach algebra
it is complete as a it is a complete norm
00:03:57.280 --> 00:04:03.290
linear space with the norm given by the sup
norm, okay the supremum norm. So any close
00:04:03.290 --> 00:04:06.690
subset of a complete space is also complete,
okay.
00:04:06.690 --> 00:04:11.660
Therefore since A is closed A is also complete
this means that every Cauchy sequence converges
00:04:11.660 --> 00:04:17.420
ofcourse you know that convergent sequences
are Cauchy, okay. So in order to show that
00:04:17.420 --> 00:04:21.300
there is a convergent subsequence it is enough
to show that there is only a Cauchy subsequence.
00:04:21.300 --> 00:04:26.449
So finally what we will do is we will use
the closeness of A and we use the completeness
00:04:26.449 --> 00:04:34.860
of the bigger space the Banach algebra to
reduce to showing just that we will be able
00:04:34.860 --> 00:04:41.169
to extract a Cauchy subsequence, okay so we
have to just check that there is a subsequence
00:04:41.169 --> 00:04:43.310
which satisfies the Cauchy property, okay.
00:04:43.310 --> 00:05:05.550
So let me write that down since A is closed
and C X, R is complete A is complete
00:05:05.550 --> 00:05:22.830
so you know we just need to show that S the
sequence S which consist of f 1, f 2, etc
00:05:22.830 --> 00:05:40.650
f n admits a Cauchy subsequence, okay. So
so this is what we have to show, alright and
00:05:40.650 --> 00:05:51.620
what is it that we are given we are given
we are given that the family A, the collection
00:05:51.620 --> 00:05:56.290
of functions A, the subset of functions A
is equicontinuous and it is bounded, okay.
00:05:56.290 --> 00:06:01.620
Now let us let us analyse what these condition
mean, okay and ofcourse we are also given
00:06:01.620 --> 00:06:05.789
the condition that X is compact mind you X
is a compact metric space so that also needs
00:06:05.789 --> 00:06:12.000
to be used, okay and mind you because X is
a compact metric space automatically all the
00:06:12.000 --> 00:06:18.810
continuous functions on X are automatically
bounded, okay normally if you take a non-compact
00:06:18.810 --> 00:06:25.780
space if you look at continuous functions
but if you want to define the supremum norm
00:06:25.780 --> 00:06:31.639
you will have to restrict to bounded functions,
okay so that you get a finite norm, alright
00:06:31.639 --> 00:06:32.810
for every function.
00:06:32.810 --> 00:06:37.610
But then in this case you do not have to restrict
to bounded functions among continuous function
00:06:37.610 --> 00:06:41.280
because every continuous function is bounded
and that is because X is already compact,
00:06:41.280 --> 00:06:42.280
alright.
00:06:42.280 --> 00:06:51.620
Now so what I am going to do is that let us
analyse the condition that that A is bounded,
00:06:51.620 --> 00:07:06.319
okay we are given A is bounded what does this
mean? This means that there exist an M greater
00:07:06.319 --> 00:07:22.990
than 0 such that the norm of f is less than
M for all f in A, okay. So basically A is
00:07:22.990 --> 00:07:31.979
bounded means that A is bounded as a subset
of a the of C X, R as a bounded as a metric
00:07:31.979 --> 00:07:41.920
space, okay bounded as a metric space means
it is contained in a finite open disk I mean
00:07:41.920 --> 00:07:46.940
an open disk of finite radius centred at some
point, okay but you can choose the point to
00:07:46.940 --> 00:07:52.409
be the origin because this is a norm it is
not just a metric space it is a norm linear
00:07:52.409 --> 00:07:54.320
space it is a actually a vector space.
00:07:54.320 --> 00:08:00.539
So there is a 0 element and every vector every
element there AB vector has a norm and it
00:08:00.539 --> 00:08:06.240
is the norm that induces the metric because
it is a norm linear space. So saying that
00:08:06.240 --> 00:08:11.300
a subset is bounded is a same as saying that
all the norms the lengths of the vectors in
00:08:11.300 --> 00:08:17.740
the subset they are all bounded by a positive
value that is what I have written, okay. So
00:08:17.740 --> 00:08:25.140
mind you what does this mean but what is norm
f but norm f is what it is supremum over all
00:08:25.140 --> 00:08:34.599
small x belonging to capital X mod of x this
is what norm f is this is the supremum norm
00:08:34.599 --> 00:08:42.380
and this is less than M and this is true for
all f in A, okay and what does this mean this
00:08:42.380 --> 00:08:52.820
implies that for all f in A and for all x
in A for all x in X okay mod f x is less than
00:08:52.820 --> 00:08:58.820
or equal to M sorry is less than M, I could
have used less than or equal to but probably
00:08:58.820 --> 00:09:05.970
it does not matter let me stick to less than
or equal to if you want does not matter, okay.
00:09:05.970 --> 00:09:11.740
Now you see look at this condition if you
look at this condition now I have rewritten
00:09:11.740 --> 00:09:21.529
this condition with with the points of the
the domain x of the functions. So if you look
00:09:21.529 --> 00:09:26.870
at it like this you see that what you are
seeing is that for all the functions in A
00:09:26.870 --> 00:09:36.970
okay M is uniform bound for the function values
and that is why this boundedness as a subspace
00:09:36.970 --> 00:09:41.290
of C X, R is actually uniform boundedness
that is why this Arzela-Ascoli Theorem is
00:09:41.290 --> 00:09:47.139
called the uniform boundedness principle namely
it says that uniform boundedness plus equicontinuity
00:09:47.139 --> 00:09:53.920
will imply compactness, okay that is the that
is the other way in which this theorem is
00:09:53.920 --> 00:09:56.900
often stated so the uniform boundedness comes
from this.
00:09:56.900 --> 00:10:09.779
So what this tells you is that A is uniformly
bounded uniformly bounded and this uniform
00:10:09.779 --> 00:10:17.279
this uniformness is a kind of double uniformness
because it is uniform bound for all the functions
00:10:17.279 --> 00:10:24.100
in A and for each each of those functions
it is uniform bound for all points of x so
00:10:24.100 --> 00:10:29.980
it is a double uniformness it is uniform in
the f, it is uniform in the x, okay in that
00:10:29.980 --> 00:10:33.040
sense it is uniform bound, okay.
00:10:33.040 --> 00:10:40.440
So that is one thing that is given to us than
the other thing is the other thing that is
00:10:40.440 --> 00:10:48.310
given to us is that this family or this collection
A subset A of function is equicontinuous and
00:10:48.310 --> 00:11:03.750
what is that so let me write that down we
are also given that A is equicontinuous i.e,
00:11:03.750 --> 00:11:18.889
given epsilon positive there exist delta positive
such that d distance between x and x prime
00:11:18.889 --> 00:11:32.310
lesser than delta will imply that the distance
between f x f x and f x prime which is just
00:11:32.310 --> 00:11:36.910
the distance function is just the modulus
of the difference of the function values since
00:11:36.910 --> 00:11:46.940
these are real values less than epsilon for
all f in A so this is the equicontinuity,
00:11:46.940 --> 00:11:55.459
the equicontinuity is that it is just the
same the equicontinuity for a family is just
00:11:55.459 --> 00:12:00.240
the same as the epsilon delta definition for
a single function but the only thing is that
00:12:00.240 --> 00:12:07.029
the for a given epsilon the delta works for
all functions, okay at any given point, okay.
00:12:07.029 --> 00:12:13.899
And usually the delta will also change as
you change the point but here you can even
00:12:13.899 --> 00:12:19.820
make the delta independent of the point also
because the functions are all continuous functions
00:12:19.820 --> 00:12:22.820
on a compact space and therefore they are
uniformly continuous, okay.
00:12:22.820 --> 00:12:29.579
So so you have this so this is so these are
things that we have we had given and we have
00:12:29.579 --> 00:12:35.880
to use, okay. So the first step of the proof
will be to so the so let me tell you the idea
00:12:35.880 --> 00:12:40.380
of the proof the idea of the proof is that
the first thing we will do is that we will
00:12:40.380 --> 00:12:46.230
show that see basically we have a sequence,
okay we are given a sequence of functions
00:12:46.230 --> 00:12:52.850
from A, we have to pick out a Cauchy subsequence,
okay. So mind you saying that this Cauchy
00:12:52.850 --> 00:13:03.940
that you have a Cauchy subsequence is means
this is Cauchy in the metric for the space
00:13:03.940 --> 00:13:10.779
of functions but that means uniformly Cauchy
with respect to points of x, okay.
00:13:10.779 --> 00:13:16.230
See what you must understand is that whatever
you say for functions on with respect to the
00:13:16.230 --> 00:13:23.370
metric on C X, R okay that will translate
if you because if the sup norm if you translate
00:13:23.370 --> 00:13:30.720
it to an equivalent statement concerning points
of x it will become a uniform statement, okay
00:13:30.720 --> 00:13:36.660
it will be a property which does not depend
on the particular points of x so it will be
00:13:36.660 --> 00:13:38.319
uniform on x, okay.
00:13:38.319 --> 00:13:44.290
So for example convergence in the space of
functions will correspond to uniform convergence
00:13:44.290 --> 00:13:50.260
on x, boundedness in the space of functions
as we just now saw corresponds to uniform
00:13:50.260 --> 00:13:54.220
boundedness, okay. So whatever you say in
the space of functions that will correspond
00:13:54.220 --> 00:13:58.781
to a property which is uniform with respect
to points of x that is what we have to remember,
00:13:58.781 --> 00:14:04.620
when you translate from function theoretic
property to the space theoretic property the
00:14:04.620 --> 00:14:08.170
property you will get a uniform version it
is a uniform version because it does not depend
00:14:08.170 --> 00:14:11.959
on the points of x value of verifying it,
okay.
00:14:11.959 --> 00:14:17.560
So basically what we have to do is from the
given sequence we have to pick a Cauchy subsequence
00:14:17.560 --> 00:14:22.949
but this Cauchy this is the Cauchy subsequence
in the in the with respect to the metric on
00:14:22.949 --> 00:14:28.770
the space of functions but that will transform
that will translate to a uniformly Cauchy
00:14:28.770 --> 00:14:34.140
sequence of functions with respect to x, okay
that is what we want, alright. Now the trick
00:14:34.140 --> 00:14:38.230
is that what we will do is that we will use
we will make use of the fact that because
00:14:38.230 --> 00:14:44.699
x is compact it is a compact metric space
x is actually separable that is it has a countable
00:14:44.699 --> 00:14:46.139
dense subset, okay.
00:14:46.139 --> 00:14:51.700
Then what we will do is that we will be using
this countable dense subset we will find we
00:14:51.700 --> 00:14:57.899
will extract a subsequence, okay using a diagonal
argument from the original sequence of functions
00:14:57.899 --> 00:15:03.250
such that this subsequence will converge at
all points of this countable dense subset
00:15:03.250 --> 00:15:07.970
and then we will use the equicontinuity to
show that that is good enough to say that
00:15:07.970 --> 00:15:15.270
the subsequence is Cauchy on all of x, okay
because because it is already uniformly Cauchy
00:15:15.270 --> 00:15:19.370
on you know a dense subset, okay so this is
the trick.
00:15:19.370 --> 00:15:24.089
So the first step that we need is a compact
metric space is separable we need that fact,
00:15:24.089 --> 00:15:40.930
okay. So let me say state that here step 1
since X is a compact metric space, hence X
00:15:40.930 --> 00:15:57.640
is separable
that is X has a countable dense subset and
00:15:57.640 --> 00:16:04.740
how does one prove this? This is just an exercise
saying that a compact metric space is separable
00:16:04.740 --> 00:16:14.120
the method is very very simple so what you
do is basically for each for each n for each
00:16:14.120 --> 00:16:21.009
n greater than 1 actually you could have taken
any sequence of numbers going to 0 which is
00:16:21.009 --> 00:16:24.800
any sequence of rational numbers going to
0 but I will choose 1 by n for each n greater
00:16:24.800 --> 00:16:44.689
than or equal to 1, look at look at all the
open balls centred at various points of x
00:16:44.689 --> 00:16:57.399
of x of sorry of capital X of radius 1 by
n, okay.
00:16:57.399 --> 00:17:04.430
So you take n equal to 1, 2, 3 and so on and
for each fixed value of n, look at all the
00:17:04.430 --> 00:17:13.760
open balls centred at various points of X,
radius equal to that n 1 by n reciprocal of
00:17:13.760 --> 00:17:21.600
that n, right. So in the first case I will
look at all balls open balls at various points
00:17:21.600 --> 00:17:29.170
of X centred at various points of X with radius
1, then I look at all open balls of X centred
00:17:29.170 --> 00:17:33.750
at various points of X with radius half and
then I will next look at radius 1 by 3 and
00:17:33.750 --> 00:17:34.750
so on, okay.
00:17:34.750 --> 00:17:39.750
Now each of these for any fixed n each of
this is an open cover and X being compact
00:17:39.750 --> 00:17:46.690
admits a finite sub cover, okay and from that
finite sub cover you pick the centres that
00:17:46.690 --> 00:17:50.810
will give you a finite set of points. So for
every n you get a finite set of points and
00:17:50.810 --> 00:17:55.540
you put together all this all the sets of
points as n goes to infinity you will get
00:17:55.540 --> 00:18:00.200
a countable union of countable sets and that
is that will serve as the countable dense
00:18:00.200 --> 00:18:02.170
subset of X that you are looking for, okay.
00:18:02.170 --> 00:18:15.600
So let me write that down
this set this is an open cover cover of X,
00:18:15.600 --> 00:18:38.560
hence admits a finite sub cover
and we are we label the finitely many elements
00:18:38.560 --> 00:19:08.180
the finitely many points of X which are centres
of balls in that finite cover and we label
00:19:08.180 --> 00:19:21.630
the set of let me put set of as C sub n, okay.
So C n is those finitely many points of X
00:19:21.630 --> 00:19:28.310
such that the open balls centred at those
points, radius 1 by n covers X, okay and you
00:19:28.310 --> 00:19:32.040
do this for every n each C n is a finite set,
okay.
00:19:32.040 --> 00:19:47.880
Then union n equal to 1 to infinity, C n is
a countable dense subset ofcourse it is countable
00:19:47.880 --> 00:19:54.010
because it is a countable union of finite
collection of sets so it is countable, okay
00:19:54.010 --> 00:19:59.550
and why is it dense? It is dense because give
me any point of X and give me any epsilon
00:19:59.550 --> 00:20:04.890
greater than 0, I can find an n large enough
so that 1 by n is less than epsilon, okay
00:20:04.890 --> 00:20:14.210
and then if you that point has to lie in the
open cover consisting of balls of radius 1
00:20:14.210 --> 00:20:26.250
by n so it has to be within distance of 1
by n from one of the points of C n, okay and
00:20:26.250 --> 00:20:31.710
that means that the points of your set the
union of all the C n’s they come in infinitely
00:20:31.710 --> 00:20:36.681
close (()) close to every point of X that
means that they are closure this all of X
00:20:36.681 --> 00:20:40.720
that is the reason why X is this subset is
dense, okay.
00:20:40.720 --> 00:20:43.770
Now what we will do is that since it is a
countable subset we will again re label the
00:20:43.770 --> 00:20:59.820
points x 1, x 2, x 3 because it is countable,
okay. So let us label the points of the set
00:20:59.820 --> 00:21:14.290
of this set as x 1, x 2 and so on, okay alright.
So so the step 1 is to get this you can think
00:21:14.290 --> 00:21:25.810
of this as some you know test set of points
where you want to test convergence, okay so
00:21:25.810 --> 00:21:29.460
this test set is just it is countable dense
subset, okay.
00:21:29.460 --> 00:21:34.620
And you know why you need the countability
because you want to be able to extract a subsequence
00:21:34.620 --> 00:21:40.530
and extracting a subsequence means you need
some algorithm for getting hold of a countable
00:21:40.530 --> 00:21:47.710
subset, okay so you need countability, right
okay and now you see now what we will do is
00:21:47.710 --> 00:21:55.500
so this is step 1 we have got this this countable
subset of points which is dense. Step 2 is
00:21:55.500 --> 00:22:01.270
to show that you can extract a subsequence
from your original sequence of functions which
00:22:01.270 --> 00:22:08.050
converges at all of these points converges
point wise at each of these points, okay.
00:22:08.050 --> 00:22:13.270
Now that is done by a standard trick which
is called a diagonal argument, okay so idea
00:22:13.270 --> 00:22:18.910
is very very simple the idea is see I have
already a sequence of functions f 1, f 2,
00:22:18.910 --> 00:22:24.080
f 3 which I call as S now what you do and
I have this sequence of functions on the one
00:22:24.080 --> 00:22:28.460
hand, on the other hand I have this sequence
of points I have these sequence of points
00:22:28.460 --> 00:22:34.300
x 1, x 2, etc which is a countable dense subset
of your metric space compact metric space,
00:22:34.300 --> 00:22:35.300
okay.
00:22:35.300 --> 00:22:41.870
What I do is I test point by point I first
take this point x 1 I substitute in these
00:22:41.870 --> 00:22:49.410
functions so I get f 1 of x 1, (f 2 of x 2)
f 2 of x 1, f 3 of x 1 I get this I substitute
00:22:49.410 --> 00:22:53.340
the point in the function in the set of functions
and what I am going to get I am going to get
00:22:53.340 --> 00:22:59.640
a bounded sequence of real numbers, why is
it bounded? Because I know all these functions
00:22:59.640 --> 00:23:04.460
are uniformly bounded so bounded sequence
of real numbers and you know a bounded sequence
00:23:04.460 --> 00:23:09.010
of real numbers has to admit a convergent
subsequence.
00:23:09.010 --> 00:23:16.200
So there is a sudden convergent subsequence
of functions there is a subsequence of functions
00:23:16.200 --> 00:23:21.960
which is converges at x 1 I will extract that
sequence, then what I will do I will repeat
00:23:21.960 --> 00:23:28.920
the process with x 2 for that subsequence
of functions. So from that subsequence of
00:23:28.920 --> 00:23:33.880
functions I will get another subsequence of
functions which will converge at x 2, okay
00:23:33.880 --> 00:23:39.670
and then I will repeat this process ad infinitum.
So what will happen is at the nth stage I
00:23:39.670 --> 00:23:46.400
would have extracted nth at the nth level
I would have extracted a subsequence of subsequence
00:23:46.400 --> 00:23:50.680
of subsequence at the nth level, okay which
converges at x n.
00:23:50.680 --> 00:23:57.300
And if I do this ad infinitum okay the point
is that you take you write the functions on
00:23:57.300 --> 00:24:05.290
each stage in the form of a row, okay and
you have a you write it as rows and then you
00:24:05.290 --> 00:24:10.920
get countably many rows and then you take
all the functions in the diagonal the point
00:24:10.920 --> 00:24:14.580
with all the functions you are taking the
sequence of functions in the diagonal is a
00:24:14.580 --> 00:24:22.630
ofcourse you will get a countable subsequence
but the point is that this sequence will converge
00:24:22.630 --> 00:24:30.880
at all points of the dense subset because
at some stage if you take all the functions
00:24:30.880 --> 00:24:40.750
in the diagonal the ith functions will converge
at it will converge at x 1, x 2, x 3, etc
00:24:40.750 --> 00:24:47.390
upto x i okay and then what will happen is
that if you if you keep letting i become larger
00:24:47.390 --> 00:24:53.830
and larger you will see that all the functions
converge at all points of this countable dense
00:24:53.830 --> 00:24:58.170
subset, okay and that is the critical step.
00:24:58.170 --> 00:25:11.760
So let me write that down, so step 2 we extract
a countable countable means a subsequence
00:25:11.760 --> 00:25:38.660
S subsequence of S which converges point wise
at each x i, okay that is it converges point
00:25:38.660 --> 00:25:44.260
wise on that countable dense subset, how do
you do that? So let me do it figuratively
00:25:44.260 --> 00:25:59.080
so you have S is equal to f 1, f 2 and so
on, okay and what you do is well substitute
00:25:59.080 --> 00:26:11.360
x 1 what you will get is you will get f 1
x 1, f 2 x 1, f 3 x 1 and so on, okay and
00:26:11.360 --> 00:26:20.260
note that mod f j x 1 is less than or equal
to M because this M is uniform bound for all
00:26:20.260 --> 00:26:24.890
functions of A and after all the sequence
S has been taken from A, okay.
00:26:24.890 --> 00:26:32.630
So but this is a this bounded sequence of
real numbers okay and you know if we were
00:26:32.630 --> 00:26:35.960
working with complex functions you would have
got a bounded sequence of complex numbers
00:26:35.960 --> 00:26:42.390
any bounded sequence of real numbers are complex
numbers does admit a convergent subsequence,
00:26:42.390 --> 00:27:03.630
okay. So let me write this here this admits
a convergent subsequence so what you will
00:27:03.630 --> 00:27:12.220
get is you will write you see you get a convergent
subsequence of real numbers, okay or complex
00:27:12.220 --> 00:27:15.990
numbers whatever it is, okay depending on
other we are now looking at real valued functions
00:27:15.990 --> 00:27:18.990
if it work on complex valued functions you
will get a convergent subsequence of complex
00:27:18.990 --> 00:27:19.990
numbers.
00:27:19.990 --> 00:27:24.970
But then forget now the numbers and look at
the functions that have come up, okay forget
00:27:24.970 --> 00:27:30.410
the x 1 and look at the f so what you let
us call that as f 11 because I want to use
00:27:30.410 --> 00:27:39.150
this I am relabeling that so f 12, f 13 and
so on, okay mind you this is a subsequence
00:27:39.150 --> 00:27:53.510
of S okay and I am relabeling in principle
I should write as f if you want well you know
00:27:53.510 --> 00:27:56.920
I could use different methods of relabeling
but let me use this okay.
00:27:56.920 --> 00:28:03.910
So thee f i j’s f 1 j’s are just these
the first subsequence I picked out from this
00:28:03.910 --> 00:28:11.740
set, okay. Now what I do is I repeat the same
process for this S 1with x 2 okay and and
00:28:11.740 --> 00:28:20.640
I proceed so let me write that once more.
So though this substitute x 2 then what you
00:28:20.640 --> 00:28:33.730
will get is you will get this f 11 x 2, f
12 x 2 and so on and again you will have mod
00:28:33.730 --> 00:28:41.620
f 1j x 2 is certainly less than or equal to
M because it is a uniform bound it works for
00:28:41.620 --> 00:28:43.430
all functions and for all points.
00:28:43.430 --> 00:28:58.020
Again this admits the convergent subsequence
this admits a convergent subsequence, okay
00:28:58.020 --> 00:29:06.310
and again now what you do is relabel the subsequences
f 21, f 22 so I will get S 2 which is now
00:29:06.310 --> 00:29:15.800
f 21, f 22, f 23 and so on. Now this is a
subsequence of S 1 which intern is subsequence
00:29:15.800 --> 00:29:25.970
of S, okay but the point is that this S 1
this converges at x 1 when you plug in x 1
00:29:25.970 --> 00:29:32.300
this converges what about this? This converges
both at x 2 and x 1 because it is already
00:29:32.300 --> 00:29:39.670
a subsequence of S 1. So this converges at
x 1, comma x 2, okay and then now you proceed
00:29:39.670 --> 00:29:48.330
by induction what you get is you will get
S n which will consist of functions f n1,
00:29:48.330 --> 00:29:55.340
f n2, etc this is a subsequence of S n minus
1 which is which intern is subsequence of
00:29:55.340 --> 00:30:06.350
S and this will converge at x 1 dot dot dot
upto x n, okay and you can do this for every
00:30:06.350 --> 00:30:07.660
n, alright.
00:30:07.660 --> 00:30:12.940
Now what you do is you write down all these
f ij’s as an array and pick the diagonal
00:30:12.940 --> 00:30:17.070
sequence that is the sequence that we are
looking for that is the sequence that will
00:30:17.070 --> 00:30:22.170
converge at all points at all these points
of this countable dense subset in that sequence
00:30:22.170 --> 00:30:25.370
that is going to be a Cauchy sequence as we
will see, okay.
00:30:25.370 --> 00:30:44.450
So write all the functions all the S i in
rows so here is S 1 this is f 11 f 12 f 13
00:30:44.450 --> 00:31:03.650
f 14 and so on, there is S 2 that is f 21
f 22 f 23 f 24 and so on
00:31:03.650 --> 00:31:12.360
and it goes on like this and if you write
S n I am going to get f n1 going to get f
00:31:12.360 --> 00:31:24.480
nn, okay and ofcourse and well that is how
it is and there are commas in between if you
00:31:24.480 --> 00:31:36.290
want but any way, okay. Now what we are worried
about is the diagonal subsequence so so we
00:31:36.290 --> 00:31:50.860
are worried about this guy pick this put g
1 is equal to f 11, g 2 is equal to f 22 and
00:31:50.860 --> 00:31:56.670
so on g n equal to f nn and so on, okay.
00:31:56.670 --> 00:32:03.120
Then the claim is that ofcourse this g 1,
g 2 that is a subsequence of the original
00:32:03.120 --> 00:32:10.180
sequence S mind you this S 2 is contained
in S 1, S 3 is contained in S 2, S n is contained
00:32:10.180 --> 00:32:15.550
in S n minus 1 and all this is contained in
S which is the original sequence which consisted
00:32:15.550 --> 00:32:25.230
of you know f 1, etc f n and so on, okay.
So what you get is you get this sequence g
00:32:25.230 --> 00:32:31.220
of g j’s these g j's are all subsequence
of the original S but the beautiful thing
00:32:31.220 --> 00:32:36.890
with this g j is that these g j's will all
converge point wise at every point of x, okay.
00:32:36.890 --> 00:32:42.440
So in fact at every point of the countable
dense subset, okay.
00:32:42.440 --> 00:33:12.130
So then so let me write it here then g j is
a subsequence of S, g j converges at each
00:33:12.130 --> 00:33:24.620
x i and why is that true because if you give
me an x i then from g i onwards okay all g
00:33:24.620 --> 00:33:38.530
i, g i plus 1 etc they will all converge at
x i okay alright because g i x i g i plus
00:33:38.530 --> 00:33:52.150
1 x i converges, okay. See for a sequence
to converge at a point it is enough if beyond
00:33:52.150 --> 00:33:57.270
a certain stage all the members of the sequence
converges.
00:33:57.270 --> 00:34:05.010
So if I want to check the g i if I want to
check g i when I substitute when I take the
00:34:05.010 --> 00:34:10.240
sequence g j and substitute x i, if I want
to say g j of x i converges it is enough to
00:34:10.240 --> 00:34:19.210
show that g j of x i converges beyond a certain
j and what is that j? that j is from i onwards
00:34:19.210 --> 00:34:23.839
okay that is what I am saying okay so this
is the so you see this is the diagonal trick
00:34:23.839 --> 00:34:41.440
which is very cleverly used, okay so we have
this now the claim is that g j converges
00:34:41.440 --> 00:34:59.339
in fact I should say I should let me say g
j is the required Cauchy subsequence of S,
00:34:59.339 --> 00:35:07.809
okay so this is this is claim this is the
claim, okay.
00:35:07.809 --> 00:35:13.289
So finally we are able to extract a Cauchy
subsequence and that is good enough, right.
00:35:13.289 --> 00:35:17.359
Now how do you check it is Cauchy, how to
check it is uniformly Cauchy okay because
00:35:17.359 --> 00:35:22.119
it is Cauchy in the space of functions means
that is it uniformly Cauchy with respect to
00:35:22.119 --> 00:35:28.260
points of x, okay so you have to bring in
points of x somehow alright and the point
00:35:28.260 --> 00:35:33.420
is that we will have to use what we have not
used so far we have already used the closeness
00:35:33.420 --> 00:35:39.210
of A because we have reduced from finding
a convergent subsequence to a Cauchy subsequence,
00:35:39.210 --> 00:35:45.190
okay and we have used the closeness to used
the fact that A is already complete, okay
00:35:45.190 --> 00:35:48.630
close subspace of complete subspace is complete,
we have not used the equivicontinuity we have
00:35:48.630 --> 00:35:50.420
to use the equivicontinuity, okay.
00:35:50.420 --> 00:35:54.390
And what you will have to do is that you will
have to use the equivicontinuity of this family
00:35:54.390 --> 00:36:00.860
A in particular it also applies to the functions
in the subsequence g j because after all it
00:36:00.860 --> 00:36:06.549
is they are all functions from the same family
A, okay this collection A, we have to use
00:36:06.549 --> 00:36:13.369
the equivicontinuity and but then the thing
is that we will also have to use the see we
00:36:13.369 --> 00:36:16.890
have to we will have to use the compactness
of A again.
00:36:16.890 --> 00:36:21.630
See this is because first of all what we have
proved is that we have got a because of the
00:36:21.630 --> 00:36:27.039
diagonal argument we have got an hold of a
subsequence that converges at all points of
00:36:27.039 --> 00:36:31.330
a countable dense subset, okay but form that
you want to conclude that it converges at
00:36:31.330 --> 00:36:38.430
all points, okay first of all you have to
conclude that it converges at all points,
00:36:38.430 --> 00:36:43.210
you have to conclude that using the information
that it converges only on points of a countable
00:36:43.210 --> 00:36:49.349
dense subset, okay and for that how do you
reach an arbitrary point from from this given
00:36:49.349 --> 00:36:55.430
dense point there you will have to use compactness,
okay and there you will have to use equivicontinuity.
00:36:55.430 --> 00:37:05.049
So what we will do is that so let me write
out so this is the final step more or less
00:37:05.049 --> 00:37:12.680
so this is well let me put this as a so what
was I what I had called it as step 2 may be.
00:37:12.680 --> 00:37:19.019
So this is step 3 actually I have to show
that this is the required Cauchy subsequence.
00:37:19.019 --> 00:37:29.250
So what you will do is that so let epsilon
greater than 0 be given, okay. What do we
00:37:29.250 --> 00:37:43.859
have to show? We need to show there exist
an N such that m, comma n greater than or
00:37:43.859 --> 00:37:58.569
equal to N implies mod g j sorry g m so this
is the Cauchy condition so I should say distance
00:37:58.569 --> 00:38:04.670
so let me write down so norm g n minus g m
can be made less than epsilon this is the
00:38:04.670 --> 00:38:09.950
Cauchy condition mind you this is the norm
g n minus g m is the distance between g n
00:38:09.950 --> 00:38:13.630
and g m in the metric induced by the norm
in the space of functions, okay.
00:38:13.630 --> 00:38:22.650
I have to show that this can be made less
than epsilon for if the subscript of the sequence
00:38:22.650 --> 00:38:27.720
goes beyond a certain stage, okay. Let points
of the sequence come to within an epsilon
00:38:27.720 --> 00:38:31.130
beyond certain stage that is the Cauchy condition,
okay but ofcourse this norm is the sup norm
00:38:31.130 --> 00:38:39.050
what this means is that that is for all x
belonging to X you must show that mod g n
00:38:39.050 --> 00:38:52.900
of x minus g m of x can be made less than
epsilon, okay and you see the important thing
00:38:52.900 --> 00:38:59.570
is that here you see this this is because
the norm is a sup norm, right. So we will
00:38:59.570 --> 00:39:01.119
have to show this.
00:39:01.119 --> 00:39:07.990
So how do you do it? Mind you the x is an
arbitrary point of X, so you have to tackle
00:39:07.990 --> 00:39:13.970
an arbitrary point of X so what you do is
so at some points you know you have to reach
00:39:13.970 --> 00:39:25.000
an arbitrary point of X from one of the points
in the countable set, okay but then this is
00:39:25.000 --> 00:39:27.880
but there are too many points in the countable
set there are countable meaning but you have
00:39:27.880 --> 00:39:32.359
to reduce to finitely meaning so this is where
you again use compactness of X, okay.
00:39:32.359 --> 00:39:41.980
So what you do is you look at a cover of X
by open balls of radius delta, okay where
00:39:41.980 --> 00:39:47.670
you see where delta is this delta that we
are already we already have for the equivicontinuity.
00:39:47.670 --> 00:39:55.700
So where is it you see here so there is this
delta here if you go back A is equivicontinuous
00:39:55.700 --> 00:40:00.529
so given epsilon greater than 0 there is a
delta such that whenever the distance between
00:40:00.529 --> 00:40:06.480
x and x prime is less than delta then mod
f x minus mod f x prime is less than epsilon,
00:40:06.480 --> 00:40:08.650
okay we have to use that delta.
00:40:08.650 --> 00:40:16.319
But then you see since we are going to interpolate
with points from finite subset of points the
00:40:16.319 --> 00:40:20.279
distance in the triangle inequality will be
broken in three terms so we will take this
00:40:20.279 --> 00:40:23.190
delta corresponding to epsilon by 3, okay.
00:40:23.190 --> 00:40:30.430
So what we will do is so let me write that
down let delta greater than 0 be such that
00:40:30.430 --> 00:40:44.269
whenever the distance between x and x prime
is less than delta, the distance between f
00:40:44.269 --> 00:40:55.460
x and f x prime is less than epsilon by 3
for all f in A this is where we are using
00:40:55.460 --> 00:41:02.640
the equivicontinuity, okay by equivicontinuity.
00:41:02.640 --> 00:41:07.430
So you started with an epsilon but you consider
epsilon by 3 and then for that epsilon by
00:41:07.430 --> 00:41:12.779
3 you pick a delta now use this delta to form
an look at all look at this open cover which
00:41:12.779 --> 00:41:18.720
is formed by open balls of radius delta that
will have a finite sub cover because (A) because
00:41:18.720 --> 00:41:25.170
x is compact, okay and take those points those
finitely many points, okay and use those points
00:41:25.170 --> 00:41:31.480
to interpolate between f x and f x prime using
the g j's and you will get what you want,
00:41:31.480 --> 00:41:32.600
okay so that is the idea.
00:41:32.600 --> 00:41:57.930
So look at the open cover of X centred at
various points of X, in fact in fact but I
00:41:57.930 --> 00:42:03.269
would like those points to be one of the points
in the countable collection so instead of
00:42:03.269 --> 00:42:08.440
taking all points of X I leave and take the
points from this countable dense subset even
00:42:08.440 --> 00:42:12.559
that will form a open cover because the subset
is dense, okay. So look at the open cover
00:42:12.559 --> 00:42:28.590
of X centred at at points so not all points
at the points x 1, x 2, etc of X, okay.
00:42:28.590 --> 00:42:33.380
So I could have taken all points of X but
I am purposely taking the points of the countable
00:42:33.380 --> 00:42:37.339
dense subset, okay that will also form an
open cover because their closure is all of
00:42:37.339 --> 00:42:55.809
X, okay. Since X is compact this admits a
finite sub cover
00:42:55.809 --> 00:43:07.740
so say with centres x i1 so you will get a
subsequence of their x i's so you will get
00:43:07.740 --> 00:43:29.260
which I am calling as x i1, x i2, etc x im,
okay my i looks like j, okay.
00:43:29.260 --> 00:43:34.180
Mind you these x i's are finitely among those
points which form a countable dense subset,
00:43:34.180 --> 00:43:42.260
okay and now what you do is that you see now
the trick is that you know I will I will have
00:43:42.260 --> 00:43:51.849
to interpolate mod f x minus f x prime with
these x ij’s and the g j's okay and get
00:43:51.849 --> 00:43:54.410
what I want alright.