 So, we were discussing the properties of closed sets in the last class. Let us just recall a few properties which we discussed. Let us say that x d is a metric space. Then the first thing that we saw was that empty set and full space x are closed sets. And then another thing was that if we take any arbitrary family of closed sets, then its intersection is also closed. So, if f alpha, say alpha in sub indexing set lambda is a family of closed sets, then intersection f alpha, alpha in lambda is closed. And lastly, if I mean this is not true of unions. In case of unions, you have to take only a finite family. So, if f 1, f 2, f n are closed sets, then union f j, j going from 1 to n is closed. And we have seen that as far as proofs of these are concerned, I mean one can proceed in several ways, but one way will be that use the fact that any set A is closed if and only if its complement is open. And well using that we can say that since each f alpha is closed, complement of f alpha is open. That is x minus f alpha is open. And we have seen that in case of open sets, union of any arbitrary family of open sets is open. So, union of these over alpha in lambda, this will be open. And this union is nothing but x minus intersection f alpha. And this must be open. And this is open means this must be closed. Similarly, you can prove about the intersections also. But of course, that is not the only way. You can prove this directly also. For example, how does one prove that any set is closed? You just take any point in the closure and show that that point is in the set. So, for example, suppose you want to do it that way, take a point in the closure of this, x is a closure of intersection of alpha. What does it mean that if you take any open ball with let us say x is such a point, if you take any open ball with centre at x, then it must intersect this set which is intersection. But if it is something intersects intersection, then it must intersect each of the set which is same as saying that if it is the closure of f alpha for each f alpha. But f alpha is closed. So, x belongs to f alpha. Now, if it happens for each alpha, it belongs to the intersection. So, whichever way either you use the complements or prove directly, the proofs are fairly straight forward. Then next thing that we will also see is that see for example, in the discrete metric space, we have seen that every subset is open. It also means that every subset is closed because your complement will be open. Now, in case of the real line, we have seen that every open set, every non-empty open set is a union of a countable family of disjoint open intervals. Now, does it say that how to get closed sets in the real line? It does because all that you do is that you take any countable family of open intervals and drop that family. So, whatever remains must be a closed set. So, that gives again several examples of closed sets. We have also seen that closed ball is a closed set. Let me just say couple of things about this. Is it clear that in any metric space singleton x is always a closed set? Suppose you take a singleton set x, is it clear that it must be closed? Is it obvious that its complement must be an open set? Because you take the complement x minus singleton x. Can you easily find a ball contained at which disjoint from this set singleton x? So, complement is open. So, this must be closed. Again it means it will, if you combine this with this last observation here, it will be in that every finite set is closed. Because every finite set is a finite union of singleton sets. So, it is closed. Of course, again as we have seen in the case of open sets, we cannot replace this by an arbitrary family. You cannot say that arbitrary family of close, the union over the arbitrary family of closed sets is closed. Because what we can do is that you can take any arbitrary set, then that will be a union of singleton sets. So, if that were true, it will be in that every set is closed, which is not the case. So, this last and if otherwise also you can easily find counter examples to show that you cannot replace the finiteness. You cannot dispense with the assumption of finiteness here. But this whole process of removing countable family of open intervals from real line can lead to fairly complicated sets. We shall see one example shortly, but before that let me also make one more comment. So, suppose A is a subset of a real line. And suppose A is non-empty and bounded above. Suppose A is non-empty and bounded above. Then we know that by LUB action, if a subset is non-empty and bounded above, it must have a least upper bound. So, let us say alpha is LUB of A. Then what I want to say that alpha must be in the closure of A. Suppose alpha is supremum of A, then alpha must be in the closure of A. Then alpha belongs to A closure. How does this follow? Suppose you take any open ball with center at alpha. Now open ball in the real line is nothing but the open interval. It is a form alpha minus r to alpha plus r. Or we can say in of the form alpha minus epsilon to alpha plus epsilon. Now is it clear that such an interval must contain a point from A? We have already seen that if you take any positive epsilon, alpha minus epsilon is not an upper bound. So, there must exist x in A such that x is strictly bigger than alpha minus epsilon. And it is obviously less than or equal to alpha because alpha is an upper bound. So, such an x belongs to this interval alpha minus epsilon alpha plus epsilon. So, intersection of this with A will be always non-empty for every epsilon bigger than c. That is same as saying that alpha belongs to A closure. And in particular it means that if a set is closed, then it contains its least upper bound. Can we say a similar thing about the greatest lower bound also? That is clear. We can give a similar proof to show that if a set is bounded below, then it must, then its greatest lower bound also belongs to the closure of A. So, similarly we can say that if A is bounded below, is bounded below and beta is the let us say infimum of A, then this beta belongs to A closure. So, supremum of A and infimum of A they are always in the closure of A. And so in particular if A is a closed set, it will contain its supremum as well as infimum. Now, let us go to the example which I was saying that dropping this countable family of open sets can lead to a fairly complicated examples of closed sets. And one of the most famous example of this type is what is called canter set. To begin with, we take the interval, closed interval let us say suppose I call that set f 1. I will take f 1 as the interval 0 to 1. It is an example of a closed ball. So, it is a closed set. So, we already know f 1 is close and we can also say that f 1 is obtained from real line. So, this is let us say closed interval 0 to 1. So, it is obtained from the real line by dropping the interval minus infinity to 0 and 1 to infinity. These two are open intervals. So, if you drop those two intervals from the real line what remains is the set f 1. So, that is a closed set. Now, what we do in the next stage is we drop the middle one third of this interval that is you take this interval 1 by 3 to 2 by 3, take this interval 1 by 3 to 2 by 3 and drop this from this drop that also. So, what will remain is it will be closed interval 0 to 1 by 3 and then closed interval 2 by 3 to 1. So, what will remain is this. So, let us take the union of these two. So, suppose I call that as f 2, suppose I call that as f 2 then that is also closed set that is also closed set. So, it is obtained from the real line by dropping these two and plus this one that dropping 3 open intervals. Then we continue this process further. Now, what we do is that. So, what is f 2 f 2 is 0 to 1 by 3 and then 2 by 3 to 1 0 to 1 by 3 and 2 by 3 to 1. Then from each of these two intervals I again drop middle one third open interval. So, what is that in this case it will mean 1 by it is 0 to 1 by 3. So, you can I can say I can write this. So, it will be 1 by 9 to 2 by 9 and from here also it will be something like. So, this is 3 by it is 4 by 9 to 5 by 9. So, suppose I drop that. So, what will remain is this f 3 will be say 0 to 1 by 9 union 2 by 9 to 3 by 9 that is nothing but 1 by 3. So, 2 by 9 to 1 by 3 and then from here it is 2 by 3 that is 2 by 3 is 4 by right. So, 2 by 3 to 5 by 9 union say 8 by 9 2 9 by 9 that is 1 2 by 3 same as 4 by 9. So, 4 by 9 to 5 by 9. So, 6 by 9 to 7 by 9 is drop. So, remaining is 8 by 9 to 1. So, 4 unit of 4 close sets or you can say the same thing as dropping something like 5 open intervals from or. So, f 3 is also closed and then continue this way. So, at each stage whatever are the see each. So, continue this way. So, suppose I call the nth set like that is f n. What is f n? f n is each f n is a finite union of close intervals. Each f n is a finite union of close intervals and each f n is obtained from f n minus 1. f n minus 1 will contain a certain finite number of close intervals. So, from each of those close intervals you drop the middle one third open intervals. So, it will contain. So, whatever remains will be f n. So, each f n is close also f 1 contains f 2 f 2 contains f 3. So, in general f n contains f n plus 1. So, what we get is each f n is closed each f n is close and also f n or I will say f n plus 1 is contained in f n. Such a family of sets is called a decreasing family when f n plus 1 is contained in f n it is called a decreasing family of sets. So, it is a decreasing family of close sets and what is the cantor set C? Cantor set C is intersection of all such f n. So, let me just write here suppose I call this cantor set C it is intersection of since each f n is closed and we already seen that intersection of any family of close sets is closed C itself is a closed set C is an example of closed set. Now, the question is whether the cantor set contains any points at all because we have dropped so many intervals, but you can see that at no stage 0 or 1 gets dropped. So, cantor set obviously contains 0 as well as 1 similarly it will contains all these n points 1 by 3 2 by 3 these n points of the intervals are not getting dropped. So, there will be several such points this 1 by 9 2 by 9 etcetera. So, so many sets or so many points are there I will do an exercise show that 1 by 4 belongs to C. So, that 1 by 4 belongs it will require some work, but this cantor set has several interesting properties and you will come across this set again and again. There is one way of deciding whether a particular number belongs to a cantor set or not and to do that we will look at what is called ternary expansion. Ternary expansion is similar to decimal expansion or binary expansion. Decimal expansion is what it is nothing, but you write each real number as a decimal in its decimal expansion where the digits occurring are 0 to 9, digits occurring are 0. In binary expansion you take only 0 and 1. So, similarly in ternary expansion you take 0 1 and 2. So, each number since the number are lying between 0 to 1 there will be no integer part. So, every number will be of this form 0 point say x 1 x 2 etcetera where each of this x n will be either 0 1 or 2 that is ternary expansion. So, every number in the interval 0 to 1 you can write using its ternary expansion. So, what I want to say is that suppose such a number is x, what I want to say is that x belongs to C. If none of these x n's are 1 if the ternary if the no digit in the ternary expansion is equal to 1 then that number is in C. So, x belongs to C if and only if that is x, x is equal to this point x which is a ternary expansion if x n if and only if x n is not equal to 1 for all n or in simple language it means its ternary expansion does not contain the digit 1 it will contain only 0 or 2. So, show this once you show this that will be easy to show that 1 by 4 belongs to C will be easy all right. Secondly, once you show this you will also be able to show the following that C is uncountable how will this follow that C is uncountable you have to show that set of all points whose ternary expansion contains only 0 and 2 that set is uncountable the proof will be again similar the so called diagonal procedure so called diagonal procedure of cantor using that you can see is uncountable. So, this cantor set is uncountable it contains uncountably many points, but another interesting fact is suppose you look at the length of the intervals which are dropped from this interval 0 to 1. So, in the first stage we had dropped the intervals 1 by 3 to 2 by 3 so that length of that interval is 1 by 3 in the next next stage which intervals we have dropped we have dropped 2 intervals and the length of each of those intervals is 1 by 9. So, the total length that it dropped is 2 by 9 of course, remember those intervals are disjoint from this 1 by 3 to 2 by 3 because they are coming from here and length of each of that is 1 by 9 there are 2 such intervals in the next stage what will happen you will drop 4 intervals and the length of each of them will be 1 by 27. So, the total length drop will be plus 4 by 27 etcetera. So, if you want to look at the sub of the all the intervals dropped some of the lengths of all the intervals dropped that will be given by this infinite series that will be given by this infinite series. Now, the obvious question is whether this series converges now what is the answer it is a geometric series with the common ratio 2 by 3 and the first term is 1 by 3. So, obviously it converges and what is it sum the first term is 1 by 3 and divided by 1 divided by 1 minus 2 by 3 that will be the sum now what is that it is nothing but 1 that means the sum of the length of the intervals dropped that is 1 that is 1 or which is same as saying that if you look if you had some notation or some notion of the length of any set then the length of the counter set is 0 because you are from the interval 0 to 1 you have dropped the intervals whose total length is 1. So, what remains is of length 0 of course, this can be made more precise when you have a when you define what is called a major of a set which you will do in the next semester when you learn a course on whenever you learn a course on major on integration you will make these ideas more precise. So, counter set is an example of a set whose measure is 0 but it is an uncountable set. I think for the time being we will conclude this discussion about the closed sets and properties of closed sets and let us go to the next topic and suppose we take a matrix space x d and what we want to now say is what is meant by a sequence in x. Of course, we already know what is meant by sequence sequence is a function whose domain is the set of all natural numbers. So, sequence in x is nothing but a function from n to x function from n to x. So, what we want to say now is that what is meant by saying that a sequence in x and converges what a sequence in x and converges to a point in x. So, let us say x n is a sequence in x where now x is an arbitrary matrix space x n is a sequence in x and suppose you take a point x in x. So, what is the meaning of saying that x n converges to x that is what we want to define. So, x n converges to x to x in this big x. So, by definition this means the following. Of course, one can define in several ways and all those definitions are equivalent and different definition will be useful in different context. So, we shall just see all those things shortly, but the first is this. Given any sequence x n like this and a point x in x, we can form this d x n x. This will be a sequence of real numbers, distance between x n and x that is a real number and it is a sequence of non-negative real numbers. So, we can always talk of what is meant by saying that this sequence of real numbers converges to something that is something that we already defined. So, if this sequence converges to 0, we say that x n converges to x. So, let me again repeat x n converges to x it is same as saying that the sequence d x n x converges to 0 and sequence d x n x is a sequence of real numbers and that is something that we have already defined. So, this is the definition of a convergent of a sequence. Now, but only thing is that this is a sequence in real numbers whereas, this is a sequence in this metric space. So, now we have a definition of what is meant by saying that a sequence in any metric space converges to some point in that metric space. Now, let us write this in a little more elaborate form and so that way we will be able to prove the various properties of convergent sequences. Now, what is the meaning of this? Suppose I want suppose I write this definition in the full form it means that given any epsilon positive, I can find some natural number n 0 etcetera. Let us let us write that in the full form. So, this means the following for every epsilon bigger than 0 there exists n 0 in n such that n bigger than or equal to n 0 n bigger than or equal to n 0 should imply what distance between x n x and 0 that is what d x n x minus 0 that should be less than epsilon, but d x n x is already non negative. So, what d x n x minus 0 is nothing but d x n x. So, this means distance between x n and x is less than epsilon. So, what it means is that given any epsilon bigger than 0 this will exist some n 0 such that for all n bigger than or equal to n 0 distance between x n and x is less than epsilon. Now, what I will do is that I shall just rewrite this last thing distance between x n and x is less than epsilon. Is it same as saying that x n belongs to an open ball with center at x n radius epsilon. So, that means this last thing is same as saying that x n belongs to open ball with center at x n radius epsilon. So, what does it mean that sequence x n converges to x means every open ball with center at x whatever be the radius every open ball with center at x should contain all points of the sequence after n bigger after some stage n 0. Let us again go back to our terminology of what we have said eventually we say that something some property of a sequence of the different elements of a sequence holds eventually if there exist some n 0 such that for n bigger than or equal to n 0 that property is true. So, what we can say is that given any open ball with center at x the sequence x n lies eventually in that open ball sequence x n lies eventually in that open ball. Can I also replace open ball by any open set can I also set every open set containing x will contain the sequence eventually. So, x n converges to x means every open set containing x will contain the elements of the sequence eventually that is another requirement formulation. Does it also mean that every neighborhood of x contains the sequence x n eventually. So, let me just write that last thing again that is this is if and only if every neighborhood of x every neighborhood of x contains x n eventually. We also describe this same thing saying that x n converges to x by also the same symbol limit of x n as n tends to infinity is equal to x this is same as this is just another notation for this same thing x n converges to x and also this notation x n converges to x x n tends to x sometimes we say x n tends to x as n tends to infinity. Sometimes we drop this also and simply say x n tends to x all this is meaning is the same the sequence x n converges to x. So, now we have a notion of what is made by saying that a sequence in any arbitrary matrix space x converges. Now, there is one obvious thing to follow from here that if a sequence of course a general sequence may or may not converge, but whenever it converges it has to converge to a unique point it cannot converge to two different elements in x is that clear. Because suppose you are given any two points let us say x and y in x with x not equal to y what we have seen is that if you take any two different I think I had given that to you as an exercise you can always find a number r such that open ball with center at x and radius r and open ball with center at y and radius r are disjoint. So, it is see if x is not equal to y then we can always find let us say suppose I call instead of calling it r suppose I call it epsilon there exists epsilon bigger than 0 such that that is u x epsilon and u y epsilon these two balls are disjoint. Now, suppose it so happens that x n converges to x also and x n converges to y also then all points of x n must lie in this ball eventually and they also must lie in that ball eventually obviously that cannot happen or other words to make this whole argument precise we can say that there will exist some n 1 such that when n is bigger than or equal to n 1 x n is in u x epsilon similarly when n is bigger than or equal to n 2 x n is in u y epsilon and so you can take the maximum of n 1 n. So, whenever n is bigger than or equal to both then x n must lie in u x y also sorry u x epsilon also and u y epsilon also and that cannot happen because these two are disjoint open balls. So, in other words if a sequence cannot converge to two distinct points. So, let us so we say that limit of a sequence limit of a convergent sequence is unique limit of a convergent sequence is unique. Now, we have already seen several examples of the sequences that converge in real number with the usual metric. We can also see a few examples in other metric spaces before going to some more general concepts. So, let us take say x d suppose is a discrete metric space then what will be what will be the convergent sequences in this discrete metric space of course constant sequence will be converged and it will converge to the same constant. But are there any other examples right that is sequence did not be constant right from the beginning, but it has to be eventually constant because why because if you can take this suppose is epsilon is less than 1 then this ball with center at x n radius epsilon is nothing but singleton x and that means x n has to be equal to x for n bigger than or equal to n 0 which is same as saying that after n bigger than or equal to n this is the terms of the sequence become constant. So, in discrete metric space the only convergent sequences are the ones which are eventually constant. So, we can say that then we can say that x n is convergent convergent if and only if x n is eventually constant. Let me just write it this way eventually constant that means the first few terms may be different, but after some stage it must become a constant sequence. So, these are the only sequences which converge these are the only sequences which converge all right. Let us take some other space in case of r we already know in case of r we already know what are the convergent sequences etcetera we have done a complete thorough discussion on that. So, let us take some other space. So, let me take the space r 2. Now, what to discuss the sequences. So, till now we have been using this suppose I take a point x and denote that point x as x 1 x 2 right. Now, I will slightly change this notation because this notation is not very convenient when I want to discuss the sequences in r 2 because then I will have to take say suppose this is x n then that will confuse with this x 1 and x 2. So, instead of this I shall now use this notation x 1 and x 2 x 1 and x 2 that is the first coordinate is x of 1 second coordinate is x of 2. What is the advantage if I take a sequence x n now if I take a sequence x n now I can say that this x n that will be x n 1 and x n 2. So, in other words every sequence x n in r 2 will give rise to 2 sequences of real numbers say x n 1 and x n 2 x n 1 and x n 2 all right. Of course, you talk about the convergence in r 2 we have to start with some metric in r 2. Let us take this so called one metric given by the one norm. What is the one matrix? That is norm of x is mod x 1 plus mod x 2 mod x 1 plus mod x 2 mod x 2 mod x 2 mod x 1 plus mod x 2. Then what I want to say is the following this sequence x n converges to x in r 2 converges to x in r 2 if and only if if and only if what should happen that is both the sequence x n 1 and x n 2 both should converge and converge to what x n 1 should converge to x 1 and x n 2 should converge to x 2 x n 1 should converge to x 1. So, if and only if I will say x n 1 converges to x 1 and x n 2 converges to x 2 I mean everywhere you can write as n tends to if you want to. Now, how does that follow? How do we prove this? I mean basically we have to look at the way in which this is what we had called norm suffix 1 this what we had called norm suffix 1. So, basically we have to look at the way in which this norm suffix 1 is defined. So, suppose we look at say distance bit see what is distance between x n and x what we have to show is that the sequence x n converges to x means given any epsilon you can find some n 0 such that whenever n is bigger than or equal to n 0 this is less than epsilon. But what is the distance between x n and x n this case it is nothing but norm of x n minus x suffix 1 norm of x n minus x suffix 1 and what is this by that definition it is nothing but mod x n 1 minus x 1 plus mod x n 2 minus x 2 that is the definition. Is it also correct to say that this will be always bigger than or equal to mod x n 1 minus x 1 will it also be bigger than or equal to this mod x n 2 minus x 2. Now, does it follow from here that if this is less than epsilon these two also will be less than epsilon. So, this way it is clear if x n converges to x then x n 1 must converge to x 1 and x n 2 must converge to x 2. What about the converse suppose you know that x n 1 converges to x 1 and x n 2 converges to x 2 how will x n converges to x again the usual that epsilon by 2 proofs we can say that given any epsilon you find n 1 such that this part becomes less than epsilon by 2 there will exist some n 2 because. So, that whenever n is bigger than or equal to n 2 this becomes less than epsilon by 2 and then this whole thing will become less than epsilon for n bigger than or equal to both n 1 and n 2 nothing new idea usual way of proof. So, what it means is that if you take a sequence x n in r 2 then that is that converges to some point x in r 2 if and only if x n 1 converges to x 1 and x n 2 converges to x 2. Is there anything particular about this number 2 here can I replace that by say r r so what can we say that the same thing will be very even if I take r 3 r 4 r 5 anything arguments will be similar. So, we can say that in general x n converges to x in r k r k of course, r k with this norm r k with this norm what will the norm there it will be mod x 1 plus mod x 2 is up to mod x k. So, instead of this I will say x n j converges to x j for j equal to 1 to k what will be the obvious change in argument obviously, if x n converges to x each of this number mod x n j minus x j each of them are going to be less than or equal to distance between x n and x. So, this way it is trivial and then for the reverse argument you will have to take something like epsilon by k and then take n 1 n 2 n k etcetera for each of those coordinates that is that is straightforward. In other words what we can say is that in this space r k a sequence x n converges to x if every sequence x n will give rise to k sequences formed by this different coordinates and if all those k sequences converges then x converges or vice versa. So, let us take a slightly different question instead of this norm suppose I use some other norm suppose I will look at what let us say norm suffix infinity what is norm suffix infinity it is it is the maximum of mod x 1 mod x 2 etcetera will the same similar not the same but will the similar proof work this part is true because this is the maximum. So, each of this for x n j minus x j will be less than or equal to this here you will have to devise some other argument. So, what is important is that it does not matter even you take some other norm this thing will be still true whatever norm you take here it will be still be true that if x n converges to x if I do not give x n j converges to x n j for each and a similar thing you can say about this space also c k because there are also norms are defined in a similar manner and again not only for this norm 1 or norm suffix infinity but even those. So, many norm suffix p that we have defined for that also for those norms also this will still be true we shall not repeat the argument because the arguments are essentially similar. Now, let us go back to again a general metric space and we consider a similar concept of what is known as a let me write here in case of real numbers we have defined what is meant by a Cauchy sequence we can we can take the same similar concept in arbitrary metric space. So, suppose x d is a metric space and x n is a sequence if x sequence in x then we will say that x n is said to be a Cauchy sequence in x if Cauchy sequence if what should happen for every epsilon bigger than 0 every epsilon bigger than 0 there exists n 0 in n there exists n 0 in n such that if you take any 2 n and m bigger than or equal to n 0 then the distance between x n and x n should be less than epsilon. So, what we say is n and m bigger than or equal to n 0 this implies distance between x n and x m is less than epsilon these are the illustrations of the concepts which we generalize from the real numbers to any metric space what we have done is that we have just replaced the definition of a usual Cauchy sequence by see in the usual Cauchy sequence of real numbers this would have been mod x n minus x n which is nothing but distance between x n and x m in the real numbers. So, what we have done is that the kind of things that we define in the case of real numbers we just generalize that to any metric space. So, now we can talk of what is meant by Cauchy sequence in any metric space all right. Now, what do we know about the relation between a Cauchy sequence and convergent sequence in real numbers? We know that every convergent sequence is Cauchy and we also know that every Cauchy sequence is convergent all right. Now, let us see which of these things are true in arbitrary metric space all right. So, first of all it is fairly easy to show that in any metric space every convergent sequence is Cauchy. So, that is we can say that is a theorem every convergent sequence is Cauchy every convergent sequence is Cauchy all right. Now, before going to the proof of that theorem let me again make a comment on these spaces r 2, r 3, r k etcetera. Now, we have seen some description of convergent sequences in each of these spaces. Suppose I want a similar description about the Cauchy sequences instead of saying that x n converges to x in r k suppose I say that suppose I want to say that x n is a Cauchy sequence in r k x n is a Cauchy sequence in r k what will this be replaced by if and only if what can you say Cauchy sequence can we say if and only if each of this x n j we can say that this will be replaced by say that x n j is a Cauchy sequence of is a Cauchy sequence for j equal to 1 to k. So, what we are saying is that in each of these spaces r k if you take any sequence it leads to k sequences of real numbers by taking coordinate wise. So, if the sequence x n is convergent each of those k sequences should be convergent similarly if the sequence x n is Cauchy each of those k sequences also should be Cauchy and wise first. Let us also look at some other spaces we will come to this theorem little later. For example, let us take the space l 1 what is one of the space l 1 we have said it is the space of all sequences such that sigma mod x n is convergent. So, this is the space of all sequence x such that x is from n to r such that sigma mod instead now in a using a similar instead of calling it x n I shall call it x x of let us say x of j j going from 1 to infinity this is a convergent series that is description of x n. And we have seen that this is a metric space in fact it is a norm linear space with the norm given by this norm suffix 1 of x is sigma mod x j going from 1 to infinity now I can ask similar questions in this space itself. So, suppose now I take a sequence x n in l 1 sequence x n in l 1 in spaces like r k and c k what was happening that if you take one sequence in r k that lead led to k sequences of real numbers. In this case what will happen? . There will be infinitely many sequences if x n is a sequence in l 1 each of this x n j this will be a sequence in r each of this x n j will be for each j. So, this is for each j in for each j in n and then we can ask a similar questions here also suppose x n tends to x in l 1 suppose x n converges to x in l 1 then can be say that each of this x n j converges to x j as a sequence of real number I think I will stop with this we shall consider proof of this theorem tomorrow.