 Let A contained in R. So, the following statements are equivalent. Then, the following statements are equivalent. So, what are the statements? 1, A is compact, 2, A is closed and bounded and 3rd one says A has property. And what is that property? We say every open cover of A has a finite sub cover. So, recall what was our definition of compactness? That was in terms of sequences. We said A set is called compact if every sequence has got a sub sequence which is convergent in the set. The second property closed and bounded. We defined closeness of a set again in terms of sequences. We said A set is closed whenever a sequence A n of elements of A converges somewhere, then the limit must be inside the set A. That was closeness. Bounded. That is interval of intervals. There is some interval big enough which contains the set A. That is boundedness. So, closeness and boundedness are mixture of some property of sequences and of property of intervals. Now, saying A has high number of property, it is just purely in terms of open sets. But recall what was an open set? A set is open if every point is an interior point. That is same as saying every point given any point in the set, there is an open ball around it which is inside the set. That was open set. Again in terms of sets, open balls, every open cover has got a finite sub cover. So, the important thing of this theorem is that later when there is no notion of sequences in higher courses in mathematics, there are courses where sequences are not not possible. That is called, if you happen to study what is called a course in topological spaces. There one defines only what are open sets. One does not go to sequences or anything. Defining is dealt in terms of declaring some sets to be open and then defining a set to be closed if its complement is open and then doing whatever we are doing. So, there the notion of compactness is defined in terms of Heine-Borel property. Because saying that a set is closed or bounded, closeness you can define in terms of openness, but boundedness requires notion of distance, boundedness. So, space is where this is not possible to define notion of distance, but we want to have a notion of compactness. So, there the definition is the third one that every open cover of that set has got a finite. So, that is taken as a definition in topological spaces of compactness. So, I am just giving a snapshot of something which you may come across. So, let us write a proof of this. One is equivalent to two that we have already proved that A is compact if and only if it is closed and bounded already proved. Let us prove two if and only if three. So, that we want to prove. So, let us look at suppose A has, I will just sort an Heine-Borel property as HB property. Given any covering has got a finite sub cover. We want to show A is closed and bounded. We want to show it as it is closed and bounded. So, let us first prove A is closed. It is bounded either way. It is easier to show bounded. So, let me just show that first. It does not matter actually because both have to be shown. So, A is bounded. Now, to show that A is bounded, A is a subset of, everything is in real line. By the way, A is a subset of real line. We are in the real line. A is bounded. That means it lies between bounds. Now, given any set A and I have to use somehow my tool that every covering has got a finite sub cover. So, keep these two things always in mind what you want to show and what is given. So, to show a bounded, I have to somehow manufacture some covering of the set A and use a fact that it has got a finite sub cover and conclude as a consequence that it is bounded. Now, to show it is bounded, if you can show A is between some minus n to plus n. Every A is a subset of the interval minus n to plus n that will be bounded. Now, once I want to show it is between minus 1 to n to n, can I sort of take that interval and make it a covering of the set A? How do I get a covering of, so that one of these elements is minus n to plus n? First of all, to apply Ionimbral property, I should have open interval. So, let us take the open interval minus n to plus n because if A is contained in minus n to plus n, then A is bounded obviously. How can you manufacture a covering so that minus n to plus n is one of the elements in that covering? I can let very n minus 1 to plus 1 minus 2 to plus 2 and go on expanding it. So, that will be a covering of the whole real line actually. In particular, it will also be a covering of the set A because A is subset of, so let me write, so note, so this is how one thinks of a proof. A is contained in union minus n to plus n, n equal to 1 to infinity which is contained in R, which is actually A is subset of R and this is actually equal to Y subset. This is equal to R. So, what you have done is we have given a covering of R and hence it is a covering of A also. But now this being a covering of, this is a covering of A and all elements are open intervals, so open sets. So, it is an open covering of A. So, by high-neighboral property, it must have a finite sub cover. So, let us apply that. So, implies, implies there exists some m such that A is a subset of n equal to 1 to m minus n to 1 m, n plus m and that is nothing but minus m to plus m implies A is bounded. How do we think of the proof? We want to show A is bounded, lies between some limits. So, and that should be something like an interval and we should go to a covering. So, we manufacture a covering, so that this is a particular element and use any particular property. So, A is bounded. Let us show it is also closed. So, A is closed. Now, here lies a very nice idea as far as the notion of distance is concerned. See, we want to show A is closed. I can just give the proof directly and be happy about it. But I just want to give you a feeling for how does one think of a proof. If A is closed, how do I show something is closed? There are two ways. One, take a sequence and say it is converging somewhere and try to show that limit is inside. But how is that the convergence of a sequence will relate it to open coverings? My tool available is every open cover has got a finite sub cover. Somehow I have to bring in those things. To show closeness, I should bring in coverings somehow. Now, instead of showing A is closed, it is equivalent to showing A complement is open. So, here it is the idea. Why A complement is open is useful because openness is in terms of neighborhoods. Every point in A complement, if I want to show is open, then I have to show given any point in A complement, there is open interval around it. Somehow I am coming to some kind of a covering kind of a thing. So, let us use that definition. To show A is closed, I will try to show A complement is open. So, for that, let us take a point x belonging to A complement. To show, there exists some epsilon bigger than 0 such that x minus epsilon to x plus epsilon is contained inside A. So, that is what is to be shown. Sorry, A complement Now, how do I manufacture coverings from here? Every point I want to show this. Now, here is something which is very useful. So, here is a node. How to get open sets? So, look at x is a point. Look at all y belonging to real line. Say that the distance between x and y, x is a point. If x is less than or equal to, then it is a closed set. It is a closed ball. If I make it bigger than epsilon, what kind of a set is this? In the real line, x is fixed. I am looking at all y such that the distance of x and y is bigger. Obviously, it is a complement of the closed ball. So, it is an open set. You can look at that way. So, this set is equal to r minus y belonging to r. Say that x minus y is less than or equal to epsilon. That is a closed set. So, that means, so implies that this y belonging to r says that x minus y bigger than epsilon is a general fact. It is an open subset. You can do it also in Rn. In Rn also, it is a look at all the distance between x and y bigger than epsilon in Rn. That also is an open subset of Rn because only the notion of distance is required. Now, I have already got an open set. Now, I want to manufacture a covering. So, how do I go to a covering? This is happening for every epsilon. This is happening for every epsilon. So, how do I go to a covering? That is the question. For every epsilon, I want a family. How can I generate a family by varying epsilon? 1 over n. In particular, for every n, y belonging to r says that x minus y bigger than 1 over n is an open set. x is fixed. Is it okay? This is an open set. Also, what is the relation of this open set with a? Where was the point x? x was in a complement. So, what is the relation of this with a complement? Here is a complement. Can you say a complement? Here are the sets y belonging to r such that mod of x y minus x is bigger than 1 over n. So, somehow, what is my aim? I want to get a covering of, see, what is given to me? Given to me is, a is, I want to show a is close and given to me is, a is, a has a high borrel property. So, that means every, a has high borrel property. So, every covering of a has got a finite sub cover. So, what is the relation between, so, let me also, let us look at this family. These are sets where x belongs to a complement and n belongs to n. So, am I getting some kind of a covering of a from these things or not? Or a of a complement something? Are you able to see something or not? I have got a set a. I am looking at a complement. So, if point is, so, if x is in a complement, then all things which are bigger, they are open sets and x is in a complement. So, from here, I want a covering, so we want a covering of a from these things. So, what could be a family which will give me a cover of a? Getting something, some idea anybody has? See the, if I can show that the ball, see this is what is this? This is a ball, this is open set, open interval around x of radius of distance 1 over n. So, now, if I am able to show, so, if we are able to show that x belongs, see x is in a complement that is given to me. If x belongs to x minus 1 over n, x plus 1 over n, if x belongs to this, which is inside a, then I am trying to show that this kind of thing comes inside. So, I am trying to work backward now to indicate how is the proof going on, go on. Now, look at this x minus 1 over n to x plus 1 over n, this interval. If m is bigger than n, what is happening? If what is the relation between this for m bigger than or equal to n? What is the relation between these two intervals? Which one is contained? Because if m is bigger than n, then 1 over m is less than or equal to 1 over n. Both are centered at the point x. So, this is smaller than this one. So, let us say x minus 1 over m to x plus 1 over m is inside x minus 1 over n to x plus 1 over n, for every n bigger than or equal to n. That means the union of, if I take this union, m equal to 1 to n, x minus 1 over m, x plus 1 over m, actually that is equal to, I am just playing with intervals. What I want to show? I want to show this kind of a thing. I am just working out an argument which will work. So, I want to show this is contained in A. So, to show, this is what I want to show. If I revert it, that means what? A complement, I should be inside. So, where should be A complement? If these unions are inside A, where should be a complement? In the union, this is inside this. So, x minus 1 over m, x plus 1 over m complement. If something is inside A, complement is inside the complement of that. So, inside intersection of m equal to 1 to n, x minus 1 over m to x plus 1 over n, complement of that set. So, let me write complement of this interval. And what is this thing? So, this is equal to intersection m equal to 1 to n of, what is this thing? The same as y, such that x minus y is bigger than or equal to 1 over, 1 over, this is 1 over m. So, you do those, the sets coming again here. So, let us now go back. I want finite number of them should contain it. A complement should be inside finite number of them. So, let us go back. So, these were all open sets and here is, here is that. So, why this is an open set? And x belongs to it, singleton x belongs to it. So, A should be a finite union of, so we want to show it is open. So, every point is a neighborhood. If I put it bigger than or equal to, then it will be a closed set. So, for every x belonging to A complement, we are looking at the sets y belonging to R, such that y is bigger than or equal to, now mod of y minus x is bigger than or equal to, y is bigger than or equal to n, n bigger than or equal to 1. So, if x belongs to A complement and if I call these sets, so this is the ball centered at x of radius 1 over n and the closed thing. So, implies x belongs to this ball. I think I will have to postpone the proof because, see the basic idea is I want to show A is closed. That means it should be, can I use that it is open? It is open. I have to go to a covering of A. So, how to go to a covering of A? A minus, why do I vary? x belongs to A complement. I think x belongs to A complement that implies R minus a singleton x. A is a subset of, is it okay? If x belongs to A complement, A must be a subset of the complement of the singleton. Is it okay? Because x does not belong to A. So, even if I remove it, it is still a subset and what is this? This is equal to union. I think this seems to be okay. y minus x is bigger than or, why bigger than or equal to? Let us just write bigger than. I think this is what I was trying to reach 1 over n, n equal to 1 to infinity. Let us look at this statement. x belongs to A complement. So, A is a subset of R minus x. If I take any point in R minus x, then y is not equal to x. If I take any point in y in this, then that point is not equal to x. So, distance between x and y, absolute value of x and y should be bigger than something. So, for sufficiently large n, it will be bigger than 1 over n. So, I am saying these two are equal. Is that okay? This is equality. Is this equality okay for everybody? Yes. Because right hand side is a union of subsets of R. So, right hand side is a sub and x does not belong to it. So, that is inside this. Conversely, if I take a point y in R minus x, then this distance is something and that distance will be bigger than or equal to or bigger than 1 over n for some n. So, the left hand side is a subset of right hand side, right hand side is a subset of left hand side. Is that okay? Yes. So, a, now that is what I wanted to say that for a, I have got a covering now. A is compact and I have got, these are all open sets. We have already shown these are open sets. They are the complements of the closed intervals. So, there must be a finite sub cover. There must be a finite sub cover. So, implies by LUB property. I was just trying to motivate that this should be. So, that means a is a subset of R minus x. But R minus x now it is gone. We are directly using a is a subset of this union. So, that implies a is n equal to 1 to m, y mod y minus x bigger than 1 over m. Is that okay? Because not LUB, that Heine-Berl property. Sorry, what is given is Heine-Berl property. A is inside R minus x and R minus x is union of this open sets. We have complements of closed intervals. So, a must be covered by finitely many of them by Heine-Berl property. Compactness of, so if you see what I this is what I was trying to motivate you that a should be inside finite number a should be finite number of them anyway. So, this is I think sometimes trying to explain too much complicates the things. So, it is simplest is this is very simple. So, this implies what? Now, this bigger than 1 over m, bigger than 1 over n, n equal to 1 to m. But what is this right hand side? So, that implies a is a subset of what is this union? We just now observed for m bigger than n, 1 over this is a smaller one. So, what is the largest one? So, this is all y such that y minus x is bigger than 1 over n. 1 over n equal to 1 to m. So, this is m is 1 over m is the largest length possible and that implies now go to complement. What is the complement of this? That means, so y minus x less than or equal to 1 over m that set all y such that is a subset of a complement that implies the interval x minus 1 over m to x plus 1 over m, x plus 1 over m is inside a complement and x belongs to it. So, the crucial thing in this is only to observe that these are all mod y minus x bigger than epsilon for x fixed, epsilon fixed is an open set for every. So, that proves that a is also, so implies a is sorry not in a complement less than or equal to, so implies a complement is open. So, that proves. So, what we have proved is that if a has ain Borel property then it is closed and bounded. So, 3 implies 2, hence 3 implies 2. So, let us look at converse. What I want to prove? 2 implies 3. So, that means given is always good idea to write what is given. What is 2? They are given a is closed and bounded, a is closed and bounded to show a has ain Borel property. So, where do we start? To show it as ain Borel property I should start with an open covering of a and get a finite sub cover. So, let say u alpha, alpha belonging to i, b and open cover. So, I have to construct a finite sub cover. We only know a is closed and bounded. If a were a interval that we have already proved. If a were an interval we just now proved in the beginning that if a is a closed bounded interval then it has ain Borel property. But a may not be a closed bounded interval, but it is a closed bounded set. So, that means what? If it is a bounded set at least it will be inside a closed bounded interval. So, a bounded implies there exists n along the natural number such that a is inside minus n 2 plus n. Now, the aim should be from a covering of a, I should try to go to a covering of minus n to n and then use the theorem that for minus n to n I want a finite sub cover. So, from a covering of a I want to go to a covering of minus n to n. For a I already have a covering. For a I already have a covering. So, note a is inside union u alpha, alpha belonging to i. So, what I want? a is inside this, a is already covered and I also know a is. So, what is outside a? That is not covered. What is not outside? That is a complement. What can you say about a complement? What can you say about a complement? It is open. So, what is not covered by a? It is covered by a complement which is open. So, covering of a and a complement will give me a covering of everything in particular of minus n to plus n. So, that is the idea. A complement is open implies u alpha, alpha belonging to i. These sets, union with a complement is an open cover of, is open cover of r and hence of minus n to n, because minus n to n is a subset. Now, we use that fact, but close bounded intervals have high number of property since minus n to n has h b property. This is implies, I have got a covering. That means there is a finite sub cover. That finite sub cover of minus n to n may include a complement, may not include a complement, but positively it will have some finite sub collection of u alpha. So, implies u alpha 1, u alpha 2. So, there exists u alpha m may be a complement cover minus n to n. From this collection, how do you get a finite sub cover? Keep a complement and pick up probably some finite number of elements from the given cover. So, that is what we have written. Now, if I delete from this, if I delete a complement, what will be that cover? This covers minus n to n, a is a subset of it. So, that implies what? The remaining one should cover a, because this is only covering a complement. So, implies u alpha 1, u alpha 2, u alpha m cover a. So, that proves that given a covering of a, we have got a finite. So, the idea is, boundedness puts it inside a interval and comp, collusion as gives me covering of complement. Put together, I get a covering. Select a finite sub cover of that interval minus n to n, remove a complement from that. Because we do not need, we need a covering of only a. So, remaining thing should cover a. So, that proves. So, hence we have proved that closed bounded implies high in Borel property. So, we have proved an important theorem namely for a subset of the real line. In this proof, 1 implies 2 and 2 implies 1. We did not use the fact that we are real, in the real line. That is anywhere it is to, R n is also true. Only proving 2 implies 3 and 3 implies 2, we heavily use the notion of distance that we are on the real line and the distance is given by the absolute value. So, this proof actually works for R n also. Because in R n, you can define notion of a sequence, closeness, openness, covering, open covering. But the proofs become slightly more involved. So, we will not be proving those theorems for R n. But with a remark that these theorems also, this theorem also remains true for R n. In fact, we will define later on what is called a metric space. Just we will define and is a generalization of R n, basically notion of distance and this theorem remains true even for metric spaces. But the proofs are slightly involved because there is no order, there is no way of saying bigger than or less than.