 Hello, welcome to NPTEL NOC on Introductory Core from Point Setupology, Part 2. Today we will do module 16 and a new concept, total boundedness. In a metric space, there is another important concept which is a consequence of compactness namely boundedness. However, boundedness itself is too fragile a condition because any metric space can be equivalently changed, the metric can be changed into bounded metric without changing the topology. So, we are looking for some version of boundedness property which is not so fragile. Once again, we go back to the link between abstract topology and metric spaces namely the fundamental open subsets, open balls. So, here is a definition, let x t be a metric space, let epsilon be positive real number, n in m. By an epsilon net in x, we mean a finite subset x 1, x 2, x n of x such that the entire space x is contained inside the finitely many open balls centered at x i and radius epsilon. So, these epsilon balls are enough to cover the whole thing and of course, if you take all the points is always possible, but finitely many points only. So, we say x d is totally bounded if for every epsilon there is an epsilon net. So, obviously this is much stronger if we if we have taken epsilon balls that will admit a finite sub cover that is what it means, not all open subsets are admitting finite. So, this almost comes to very close to compactness. So, it has something to do with compactness already. So, let us see what. So, total boundedness is some kind of restricted compactness that is the first thing to know. Every open cover of x by balls of a fixed radius, they admit a finite sub cover. It is clear that any compact matrix space is totally bounded because it can take balls of fixed radius positivistic that will be that will allow you a finite sub cover because it is compact. In general it is not clear why a total bounded matrix space should be compact or compact. So, observe that total boundedness implies boundedness automatically because all that you have to take is the x 1, x 2, x n and look at the diameter of this finite set x 1, x 2, x n. Add a twice epsilon that is all. Every point will be inside that much distance between any two points will be less than that that is what you can see. So, total boundedness implies boundedness. In the Euclidean space is even the converse is true. Thus our intuition may easily mislead us. So, it is total boundedness is much more stronger than boundedness but in the Euclidean space you do not have this problem. So, indeed it is clear that the bounded subset need not be totally bounded since the matrix can be simply changed to a bounded matrix all the time. Then it would that was the place then everything would have been. So, this is just a you know gas work but we will see the examples unless you see an example you will not be satisfied. Recall that if a Cauchy sequence admits a subsequence which is convergent then the sequence itself is convergent. Thus in a sequentially compact matrix space sequentially compact means remember what it is every sequence is a subsequence which is convergent. Every Cauchy sequence will be convergent because it will admit a subsequence which is convergent because it is Cauchy that Cauchy sequence is convergent. This latter property is known as completeness. Every Cauchy sequence convergent is completeness. So, somehow when studying the sequential compactness etc. we are forced to think about completeness also. Then there is another important property all the time used in analysis of matrix spaces namely compact matrix spaces namely the Lebesgue property. So, I will just recall it what is the meaning of Lebesgue. There is a theorem of Lebesgue and then whatever it satisfies Lebesgue property we have made for each open cover u of x you must find a positive delta such that the family of balls of radius delta form a refinement of u. If you take any ball of radius delta anywhere in the space such a ball will be contained in one of the members of u that is the meaning of refinement. So, that is the Lebesgue property. So, it is also worth recalling that if x satisfies Lebesgue property then every continuous function from x to any space y is uniformly continuous. Continuity implies uniformly continuity. You have been studying that on a compact space. It is compact you know intervals close intervals or a close joint bounded subset of r in and so on that is analysis. But now you see that all that unity is the Lebesgue property not compactness as such. So, how far can we go? How far are they right actually from compactness? Our next aim is to obtain a characterization of compact matrix spaces in terms of these properties. Lebesgue property sequential compactness, total boundedness and so on. So, that is the theorem here. There are six criteria here the first one is x compact. So, you can say five different criteria for a compact matrix space. You start with a matrix space then the following conditions are equivalent. X is compact, x is comparably compact, x is limit point compact, x is sequential compact, x is totally bounded and has Lebesgue property, x is totally bounded and complete. So, these first three are the ones which we have studied last time. These two new things are there now which are our old things except this total boundedness is new. Lebesgue property and completeness is old friends right. So, let us go through the proof of this. Equal 1 implies 2 implies 3 implies 4 implies 5 implies 6 implies 1 that would have been ideal thing the easiest way. Somehow I am not able to arrange it in that way. So, I will just follow a slightly different approach here. This is 1 implies 2 implies 3, 4, 5 and directly 1 and 4 and 6 are equivalent separately. So, observe that a matrix space is both T 1 and first countable. This is what I have already told you last time. Thus, we have already seen 1 implies 2 implies 3 implies 4 right. Compactness implies countable compactness, sequential compactness which is the same order. Countable compactness, limit point compactness and sequential compactness ok. These things we have already seen alright. We have seen more than that but right now the concentration is up till here we have already proved. So, these two implications and these two implications are left out right now ok. So, let us prove 4 implies 5 namely this is sequential compactness, sequential compact matrix space. So, we want to show that it is totally bounded and satisfies Lebesgue property that is what we have to show right. To prove total boundedness suppose there exist epsilon positive such that no finite number of pulse of epsilon radius cover x. That means there is no epsilon net ok that is that negation of total boundedness. Choose a point x naught in x any point does not matter. After that x 1, x 2, x 3 inductively how do you choose? x n will be chosen in x minus b epsilon of x k union k rings 1, 2, n minus 1. So, x 2 will be chosen x minus b 1 and b epsilon of x 1. Then take b epsilon of x 1, b of x 2 after choosing x 2 choose x 3 in the complement of that and so on. By sequential compactness a subsequence x n k will convert to some point a. So, there exist some k naught such that k is bigger than k naught will imply all these x n k points of the subsequence. They are in the side b epsilon by 2 ball of a. I can take epsilon by 3 epsilon by 4 order epsilon by 2 seems to be sufficient here ok. Hence the distance between x n k and x n k plus 1 n plus k plus 1 or anything after that will be less than epsilon right. On the other hand n plus k n of k plus 1 is bigger than n k I mean it is coming later on. Therefore, this x n plus k n k plus 1 is not in the previous ball epsilon ball that is absurd here you see. This says that this point is in epsilon ball of this point, but it says it is not all right. So, that is already a contradiction. So, we have proved total boundedness. So, no problem is things are things are straight forward. If this is not true what happens just ok. Second part we have proved what is that total boundedness Lebesgue property. Lebesgue property is also very close here. Let now u i be any open cover assume that this cover has no Lebesgue number. What is the meaning of that? This means that you know the Lebesgue number there exist some delta positive. So, for every delta positive something should happen that is true all right. Then for every n so now instead of every delta I am choosing n then 1 by n I am applying to 1 by n ok. There exist x n instead of x such that b 1 by n x n is not contained in any u i you can go to b 1 by 2 ok there will be 1 point b 1 itself b 1 of x 1 will not be contained in any u i next b 1 by 2 x 2 will not be contained in so on. So, you get a sequence let y k be a subsequence of x n which converges to y. Let y belonging to u i say ok. So, y must be in one of the open subsets here because u i is an open cover choose m such that b 1 by m y is contained inside. Now there exist a k naught such that for all k bigger than k naught all the y k must be in b 1 by twice m ok half of b 1 by m ball of radius 1 by 2 m around y ok. Now if you take y k equal to x n k is such that this n k is bigger than 2 m right it is a subsequence. So, you can take this n k as large as you want n k bigger than 2 m then what happen b 1 by n k at x n k will be contained inside b 1 by m y right. So, this entire ball will be inside this one because this point will be there and it is in the distance between any point in the ball and this point is less than 1 by 2 m 1 by 1. So, together it will be less than 1 by m. So, that will be less than. So, it will be contained in the b 1 by m y. But b 1 by m y is contained inside u i is by choice and that is a contradiction ok. You see the proof of Lebesgue property as well as this one is more or less similar here ok using the sequential compare. This is all because we can use the metric here that is the otherwise there is no problem. So, now I will prove 5 implies 1 that namely what is 5 total boundedness and satisfy Lebesgue property ok that will give you compare. Just like we observed that countably compare and Lindelof implies compare. So, if the proof is not that simple almost you know. So, 5 implies 1 is also not difficult. So, let us go through the proof. Start with any open cover let r positive be a Lebesgue number. Now, by total boundedness there exist finitely many balls b r x i which cover x. Since each of these balls is contained in some u i we get if b i is cover it u i corresponding u i's will cover ok. So, that is all right. So, we have left with 4 implies 6 and 6 implies 2 6 implies 4 4 is what 4 is sequentially compare 6 is total boundedness and completeness ok. So, these are also not difficult given an open cover u i i belong to x ok this we have finished already. Here 4 implies 6 the first part of this has been proved in the implication 4 implies 5 ok. So, total 4 total boundedness 4 is a sequential comparison I have to say total boundedness plus completeness, but we already prove total boundedness and Lebesgue property. So, there we have already proved that one completeness have been observed already in the remark 3 above right. Start the Cauchy sequence as a sequence it has a subsequence it is convergent therefore it is convergent. So, 4 implies 6 is easy. So, that is why I have taken that one all right. Now finally, I have to show that 6 implies 4 let x n be any sequence in x the idea is to show that it has subsequence which is Cauchy by completeness the Cauchy sequence will converge and that is that will over instead of so that is all you know you want to show that there is a sequence which is convergent you have to just show it is Cauchy here ok. Now from total boundedness for each k greater than equal to 1 let us first get a finite subset a k such that x is covered by b 1 k 1 by k balls around points of this a k what is a k a k is a finite set ok that is total boundedness for each k there will be a 1 by k is epsilon here. So, this epsilon net for epsilon equal to 1 by k. Now by the pigeon hole principle there exist an infinite subset n 1 of n and a 1 belong in to a capital n capital 1 such that all n belong in to n 1 we have x n is in b 1 of a 1 ok a 1 a 2 a k are there for each k right. So, look at a 1 ok there are only finitely many balls here and there is infinite subset infinite sequence. So, you will have pigeon hole principle is infinite to finite that is all exist in infinite subset n 1 such that n 1 a 1 is a point of a 1 such that all n inside n 1 ok this infinite subset all the x n are in b 1 of a 1 ok in a 1 there are finitely many one of them has to have infinite many one of the balls have infinite many. So, I am calling that as b 1 of a 1 similarly you can choose in infinite subset n k of previously chosen n k minus 1 because each is infinite subset such that this a k there will be a k inside a capital k some point such that the ball of radius 1 by k contains all this n k all this infinite subset ok. So, once you have chosen this nested sequence of infinite subsets of starting with infinite that here these are indexing sets by the way. So, that there are x n corresponding sequence will come out ok let n k be the least number in n capital k ok capital n k which is bigger than n k minus 1 choose n 1 to be anything inside n 1 then n 2 you have to choose bigger, but the smallest one you have to give what choice. So, I am telling you least number ok then we claim that this sequence x n k is a course is equal ok. So, this choice was not all that obvious how to do this, but we have done this kind of even more complicated choices here anyway. So, why x n k is a course here even epsilon positive a k naught such that 1 by k naught is less than epsilon by 2 then if k and l are greater than k naught x n k and x n l will be inside 1 b of 1 by k naught a k naught right both of them will be here what is the meaning of that distance between them is at most epsilon by you know twice 1 by k naught twice 1 by k naught is less than epsilon. So, given every epsilon I have chosen I have given you some k naught such that bigger than that distance is less than epsilon means that this sequence is course here they are all coming nearer and nearer to some finite even point each time. So, that is why so this becomes a course is sequence course is sequence convergence ok. So, that completes proof of almost big theorem and looking big right. So, we have this these are the implication that we have grown now alright. So, here is my exercise I will go through that you can you are welcome to solve them and get answers from a checked and so on. Easy application of theorem 4.10 that is what you know take x to r we continuous real world function where x is countably complex then show that f is bounded and attains its extremum ok. So, this is similar to what we remark namely the big property implies uniform continuity. So, compact under compactness this is vice trust theorem any continuous function on a compact set to real numbers is bounded and attains its extremum both maximum and minimum right no need for compactness countably compactness is enough. So, try your x p a metric space for each r positive show that there exist a subset a r of x which is maximum with respect to the property that for any two points x y not equal to x not equal to y inside a r the distance between x and y is bigger than r you get the point you have for each r you have to subset each any two points are at a distance bigger than r. For example, if you have just two points r is one the distance between these two points must be at least one then third point all the distance between the one one between one and two two and three all of them must be bigger than one and so on ok. Choose a maximal such subset a r ok. So, that is there exist a maximal subset that is what you have to show further x is limit point compact then show that any such a r is finite in a compact metric space you have proved such thing now you have to prove it for limit point compact space limit point compact metric space show that every metric space which is limit point compact is separable and hence second countable in particular conclude that every compact or countably compact metric space is second countable which you might have proved in a different way elsewhere ok. So, next time we will prove the very important functional result namely a scholes theorem. Thank you.