 So, let us try to give some more applications of this compactness. It says here is an important thing. If A is a non-empty compact set, then we know it is bounded because it is closed and bounded. So, in particular it is bounded. So, by the completeness property of real numbers, its greatest lower bound and least upper bound must exist. That is only by boundedness, but compact as implies that those points least upper bound and greatest lower bound will be elements of that set if A is compact. It is a very important thing. So, let us try to approve that. That greatest lower bound and least upper bound both are inside. So, let A be compact. When I do not write A, it could be an Rn also. A in R or Rn, any subset. A is compact. So, let us write A non-empty because otherwise for empty set everything is okay. Let alpha be equal to greatest lower bound of A. Alpha exists by completeness property. There is a slight logical issue because what is completeness for Rn? I have not discussed actually. Let us keep it on the real line for time being. I can define it, but I think I do not see much point of view knowing that one can define what is called completeness property for Rn. Let us not go into that. So, let us keep this A is compact, A is subset of real line. Let us do it only for the real line so that we do not have to bother about much. So, claim alpha belongs to A. We want to claim that alpha must belong to A. Now, this is very simple and it just depends on what is the definition of A is my alpha, which is the greatest lower bound. That means what? All the elements of the set A are on the right side of alpha. Alpha is the greatest lower bound. It is a lower bound and the greatest of them. So, everything else is on the right side. Now, definition of it says that if I for every epsilon bigger than 0, if I look at alpha plus epsilon, that will be a point on the right side. Can that be an upper bound for A? Can alpha plus epsilon be an upper bound for A? Alpha is the least upper bound. Alpha is the greatest lower bound. So, can something bigger than alpha be A lower bound? No, other than definition of greatest lower bound is contradictory. That means what? That means between alpha and alpha plus epsilon, there must be a point of the set A, because I want alpha plus epsilon should not be a lower bound. So, there must be a point on the left side of it. So, there must be a point. So, implies there exists some point. So, let us call it x epsilon such that alpha less than x epsilon less than alpha plus epsilon. So, there must be a point here, x epsilon and that x epsilon should belong to the set A. There must be a point of that. Otherwise, it cannot contradict. This is the crucial thing. This is the definition of it. See, at one point I had said that if you want to understand what is true, then you should also understand what is false. So, it is the negation or the definition, whichever way you like, alpha plus epsilon cannot be least upper bound, greatest lower bound. So, there must come something inside. So, whenever you want to understand what is true, you should also look at what is negation of that statement, what is false to understand that. So, this is the definition and what is given to me? A is compact and how is compactness defined in terms of sequences and convergent sub sequences. So, from here, I should try to produce some sequence somehow or the other and then go to a sub sequence and convergent inside the set using compactness. So, how can we produce a sequence by using this fact? And I want it to be convergent. So, the obvious thing is, this is true for every epsilon. So, specialize epsilon equal to 1 over n. So, in particular, we write in particular for epsilon equal to 1 by n, there exist some xn such that xn belongs to a and alpha less than xn less than alpha plus 1 by n for every n. Because something is happening for every epsilon. So, I have specialized it to epsilon equal to 1 over n. Is it okay? What is the sequence now xn? And that sequence xn is between alpha and alpha plus 1 over n. So, can I say sequence is convergent? Yes, convergent where? To alpha. So, no, xn belongs to a, xn converges to alpha. What does compact? Till now, we have not used any compactness or anything. We have only used the definition of alpha. Now, compactness says that this must have a sub sequence converging in the set. But the limit is alpha and the sequence itself is convergent. That implies alpha must be inside a. Because sequence has a sub sequence which is convergent, but that sub sequence will converge only to the same limit as a sequence converging and that is alpha. So, by compactness alpha must belong to a. So, by compactness or if you like xn is a sequence converging to alpha, a is closed. So, limit must be inside. You can shortcut using the fact that compactness means closed and bounded. Alpha is alpha belongs to a, because a compact equivalent to closed plus bounded, if you like. Either way you can write. So, what we have done? Till now, till this point, what I have done is, if alpha is greatest lower bound of a set, then there is a sequence converging to it. Till this point, we have not used anything, only definition. So, if a number alpha is greatest lower bound, there must be a sequence converging to of the set, elements of that set, which you are taking the greatest lower bound, converging to alpha. In fact, something more I can do, I can make this sequence xn a monotonically decreasing sequence. I can specialize it, because I can always choose something on the other side. If it does not work, then go to the next number and go to the next one, because if once I have chosen x1 between x1 and alpha, there must be an element of the set, because x1 cannot be, so there may be x2 bigger than x1. Similarly, x3 and so on, you can go on doing such kind of things. So, you can improve this, but this is very nice and important thing that if, in terms of sequences, that if alpha is greatest over bound, then there is a sequence of elements of the set converging to that alpha. Compactness says this must belong to the set. So, alpha belongs to the set. So, this proves that if alpha is greatest over bound, then it belongs to the set. Same property true for least upper bound. Why? It will be on the other side, only you will be, instead of waving your, there on the right side everything, now you will be saying everything on the left side. So, if beta is least upper bound, then beta minus epsilon cannot be least upper bound. So, there must be something inside. So, take epsilon equal to 1 over n again. So, alpha minus 1 over n will be doing the same analysis you will do. So, this proves the theorem that if it is a compact set, then greatest over bound and least upper bound are elements of the set. This will be useful later on when you want to maximize and minimize functions. In many situations, your functions will have domain which is compact and the functions will be continuous. I have not really defined, but all if we know continuity and so on, then we will get that the range is compact and hence upper bound and function must have a, the range must have a least upper bound and greatest over bound and that must belong. So, that says that will give you a theorem for continuous functions. Every continuous function must attain is bounded and must attain is maxima and minima. So, that will come in calculus later on. So, compact sets are defined as sets. In terms of sequences, we are defining. Every sequence in that set has a subsequence which is convergent in the set that is compactness. Using that, we said every compact set must be closed and bounded. That is one. Second, if a set is compact, then very interesting and application of that is that it is bounded. So, greatest over bound and least upper bound exist as real line. So, they must be attained. That means they are in the set itself. For open interval 01, the greatest lower bound is 0, but that is not part of the set. Similarly, least upper bound of open interval 01 is 1, but that is not inside the set and that is because open interval 01 is not compact. We know that here. So, let us look at some more properties of this. So, here is another way of defining compactness which finds why we should be doing this at all. Say there is something you should be doing that because it is a part of slavers. But there are some things which why it is a part of slavers because there is some other reason for that. The reason for looking at something called a cover of a set is that in this way, we will define what is called an open cover of a set and define some properties of compact sets in terms of this. Then that will make it independent of sequences and such things. There are spaces where you can define compactness where there are no notion of a sequence, but compactness becomes important. So, I will probably indicate this later on. So, let us just look at for the time being that it is mandatory for us to know what is a covering and how compactness is defined in terms of it. It is important. So, what does the English word cover mean? It should cover. What else it could be? So, we want to say, given a set, there is something that covers. So, if A is a subset of B, then you will say B covers A. But in general, one set may not cover another one, may not be subset, but there will be a collection of sets. There are union covers that are given set. Then we say this collection is called a cover for the given set. So, a set theory purely, a collection of say here is G alpha of subsets of Rn is called a cover for A if A is inside the union. It covers. If each G alpha is open, then we say it is an open cover for A. We are specializing cover by sets. That is a cover. And if I can say each G alpha is open, then we say it is an open cover for A. Let us look at some examples of this. You can manufacture many. For example, if I take the open interval 0 1 and look at 0 comma 1 minus 1 over n, 0 to half minus 0 minus 1 by 3. So, that is increasing. I am making that point slowly increasing and increasing. There union will be the open interval 0 1. Is that okay? Union of open interval 0 to 0 comma 1 minus 1 over n forms a cover of the interval 0 1. Because it is starting at 0 ending at 1 minus 1 over n. So, slowly it is becoming going to the right and right. So, every point of 0 1 sometime or the other will be inside on the left side of 1 minus 1 over n. Is that okay? Is it clear? Because 1 over n goes to 0. So, 1 minus 1 over n goes to 1. So, it is increasing to the right side. So, this will be a cover and each is an open set is an open interval. So, this is an open cover of 0 1. Some other examples are given. So, look at 1 to infinity. Then you can look at 0 to infinity is an open set which covers it. Is an open interval. So, open cover of it and so on. There are many examples you can construct to yourself many more. I am just giving you some to get an idea of open. So, here is a very important property is called the Heine-Bohrel theorem. It is for the real line. We are looking at it says in a closed bounded interval a, b is given and you are given an open cover of this by open sets j i. i is an interval that is a, b and there is a covering of i by open sets j. j belonging to some collection. Some collection often open sets cover it. Claim that we do not need a full cover. There is a finite number of them. So, that means there exists j 1, j 2, j n such that i is contained in this finite union only. So, from an arbitrary collection you have come down to a finite sub collection which covers it. It is a very important thing. You understand what I was saying? Given any arbitrary open cover of the closed bounded interval a, b, there is a finite sub cover of it. There is a finite sub collection of the given collection which covers it. This goes by the name Heine-Bohrel theorem. So, let us try to prove this theorem. The idea of the proof is rather simple. So, let me try to give you the idea and then prove it. So, what we are given? Given i equal to a, b is inside union of j, j belonging to some collection s, j open. Each j is open. It implies there exists j 1, j 2, j n such that i is contained in union of j i i equal to 1 to n. Given any open cover, cover of a closed bounded interval by open sets, there is a finite sub cover. There is a finite sub collection which is enough to cover it. You do not have to do all. So, let us write the proof. So, here is my a and here is my b. So, the idea of the proof is I start with the point a and go to a nearby point say x and look at this interval a, x. Look at a part of the interval a, b starting from a. Close interval a to x. Possibility is a to x is covered by finitely many members of that given collection. This interval which is a sub interval part of a, b has the property that a to x has a finite sub cover or it may not have. It may or it may not. So, the idea is try to collect such points and show that they have a largest element and that is b. So, a to b will have a finite sub cover. So, the idea is let us construct let a be the set of all x belonging to a, b such that a to x has finite sub cover. What is our aim to show? b is an element of this set. A means to show that b belongs to it because if b belongs a to b will have a finite sub cover will be through. So, we are reformulating the problem in some way so that we can get a hold of it. This set a, so note this set a is non-empty set. Why it is non-empty? That means what I have to show? There is at least one element of x between a and b which belongs to this set a. I can take x equal to a itself for example, a belongs to a. a, a, x is equal to a. So, it is a closed interval a, a. It is part of a, b. It is only a single point. So, I can pick up any element in the cover, open cover. One element is enough to say that a, a is covered by a open set j. Is that okay? Only one is enough. Why finite? Only one is enough for the single term. Is that okay? Because a, the point a belongs to the interval a, b because a belongs to a, b is contained in union of j, j belonging to s. So, what does it imply? a belongs to j for some j because it belongs to this. This is inside the union. So, a has to belong to one of them. Pick up any one you like. So, implies a, a is inside j. So, interval a, a has a finite sub cover. So, a belongs to the set a. So, it is non-empty. So, that is a first observation. A is bounded above. It is bounded above. Why it is bounded above? Where are elements x? You are picking up elements of the interval a, b. So, x cannot be bigger than b. They are all inside a, b. So, they are bounded above by b. It is bounded above by b. So, we have got a non-empty set, which is bounded above implies l, u, b of a, call it alpha, exists. l, u, b exists by the completeness property of real numbers. It is a non-empty set, which is bounded above. So, l, u, b must exist. Note, alpha is between a and b. So, alpha is between a and b, because b is an upper bound. b is an upper bound. So, least upper bound cannot be bigger than b. It has to be less than or equal to b, and all elements are bigger than a. So, this is between a and b. Alpha is between a and b. Claim that alpha is equal to b. That will prove the theorem. This will prove. Will that prove also? Because if a is equal to b, then a, b will have finite sub-cover. So, I have to only show that this b is upper bound. So, look at the picture. That is what I said that a to x has got a finite sub-cover. Pick up all these x's and try to take the largest possible x with that property. We try to show the largest possible x has to be b. It cannot be something smaller. So, a to b will have a finite sub-cover. That is the only thing to show that it has. Let us prove that. Most often, then not such proofs go by contradiction. Suppose alpha is strictly less than b. So, here is the picture. a, b, and here is my alpha. Here is my alpha. Claim that alpha is less than b. That is the assumption. We will try to prove a contradiction that it cannot be strictly less. It has to be equal to b. Now, if it is strictly less than b, then it is in the interval a, b. Then it is in the interval a, b. And a, b is covered by those open sets j. So, alpha will belong to one of the j's. So, let me just draw a picture. So, here is my something. This is my j. This yellow one is j. That is a open set. Because a, b is covered by j's, open sets j's. And alpha is inside a, b. So, alpha will belong to one of them. If it belongs to one of them, it is an open set. So, there will be open interval which will include alpha. So, call it as something c to d. So, what I am saying is there is an open interval c to d, which includes alpha. Now, if this open interval includes it, then there must be a some point beta on the right side of alpha. There are many. So, alpha is on the left side, but there is at least one. And on the left side, c to alpha, alpha is least upper bound of all those x's. So, there must be an x belonging to a, which is on the left side of alpha. Because alpha is the least upper bound of those x's. Now, look at the interval a to x. x belongs to a. So, finite number of j's will cover a to x. And this beta is covered by c to d, which is inside j. This c to d is inside j. j was an open set. So, one more j, if I include, that covers beta also. So, that means from a to beta is covered by finite number of j's. Because a to x is covered, because x is in a. So, by the definition of a, a to x is covered. And beta belongs to c to d, which is inside a j. So, if I take that j also earlier finite plus one more, they cover alpha to a to beta. But beta is bigger than alpha. Beta is bigger than alpha. And we said alpha is the least upper bound of those, which are covered by finitely many. And we have found another something bigger than alpha, which is beta, which has the same property. That means beta must belong to a. That is not possible, because alpha is the least upper bound of a. That is a contradiction. Simply nothing more. So, probably I will repeat it next time, but it is very easy to understand. Alpha is the least upper bound. So, assume alpha is less than b. So, this alpha must belong to one of the j's. And alpha is an open set. So, you must have an open interval, including that point inside. So, c to d is the open interval, including alpha. On the right side, pick up any point beta. On the left, there has to be a point of a, because alpha is least upper bound. So, a to x is covered. x to beta is covered by this. So, this finitely many cover a to beta. That is a contradiction to the fact that alpha is least upper bound of this. And that proves the theorem. So, probably I will repeat it next time again. The proof very simple, nice proof. So, what we are saying is that close bounded interval a to b has the property every open cover has got a finite sub cover. And next lecture we will prove the converse is also true. In the sense that for a compact set, this is one of the equivalent of saying every open cover has a finite sub cover.