 So, intervals. Now, here is a property of intervals, which is very important. So, let us take a, I think he said do not use red, because it looks bad in the, so let us take a green one. So, this is a line, real line. So, let us take a point on the line. So, this is a point P. I want to locate that point P. Where is that point? P is a point on the geometric object, namely the line. I want to pin point. Where is that point? Imagine as if there is a big road, going from one end to another, in a desert. There is no sign of anything anywhere around and there is somebody stranded on the road and he sends a message that, locate me, find me, help me. So, how will you locate that point? So, here is a possibility. Let us draw some point on the line, some object, some city. So, try to see whether the person is on the left. So, person is able to answer you, yes or no. Are you on the left side of that city or the right side of it? Then our, if it is not on this side, then our search becomes only on that part. So, I can go on doing that. A better way would be, let us take an interval, are you between this city and that city? Have you crossed that city? Have you reached that city? So, no. So, you are in between that city. Now, I want to narrow down the search. So, I will make it smaller and make it smaller. So, search, search window smaller and smaller. So, eventually I should be able to pin point that point. Where is that person? Is that okay? So, this we want to formalize mathematically. So, let us write down the statement and then try to prove it. So, here is, let us take a sequence i n of intervals. So, keep that picture in mind now. At the nth stage, you have got this interval a n to b n. And we say it is nested. If the next one is inside this, we want to make it shorter. The next interval is inside, locating. So, with the property that i n is a subset of i n plus 1. Is that okay? This is saying i n is inside. i n is bigger, and i n plus 1 is, is it right or wrong? Which way? Have I written it rightly or wrongly? i n plus 1 should be smaller. So, this is written wrong. There is a typo in this. So, this one should be, i n plus 1 should be inside i n. I want to say that. That picture, i n, i n plus 1, smaller part of it. So, this is wrong here. And what should happen? In that picture, I may want to make it smaller and smaller. So, how do you say that really they are becoming smaller and smaller? Eventually, they should capture that point. That means, the sequence a n, that is the right end points, left end points, b n, the difference of the two should go to 0. The length of the interval should go to 0. What is the length of the interval i n, b n minus a n? That should go to 0. So, I want to prove. So, let us write here. This is called nested interval property. So, i n is a sequence of closed bounded intervals. i n equal to a n comma b n for every n bigger than or equal to such that 1, i n plus 1 is a subset of i n for every n. Second, limit b n minus a n, n going to infinity is equal to 0. The intervals are becoming smaller and smaller. That is captured by saying this is 0. So, this implies, eventually what do you expect? I should capture that point and only that point. So, it says intersection of i n 1 to infinity is a singleton point x for some x belonging to R. It is a nested sequence. It means the next one is inside the previous one. That is why we say it is a nested interval property. It is a sequence of intervals which is nested. The next one is inside the previous one and they are becoming smaller and smaller. So, saying that the length is going to 0 and one more property is very important that each i n is a closed bounded interval. Otherwise, you cannot say b n minus a n is going to 0 anyway. If any one of the i n is unbounded, then you cannot say that length will be infinity. It will be bigger than anything. So, it cannot go. So, either I should prove or let us prove this first and then say why all these conditions are required. See, whenever you prove a theorem under some conditions, you try to see whether any one of the conditions can be dropped or not. You can remove that condition from the theorem or not. Because if you can remove it, then you have got a much stronger result at your hands without that condition also. So, i n is a sequence. So, let us try to understand the proof. So, this is a n and this is b n. So, where is a n plus 1? So, here will be a n plus 1 and possibly here will be b n plus 1 because it is nested. It is a nested sequence. So, now, look at the left end points a n. It is a sequence of points. What can you say about that sequence? It is monotonically increasing. What about sequence b n? Are both bounded? So, they must converge. So, first observation, the left end points must converge. The right end points must converge. So, limit a n, n going to infinity equal to a exists. That is the first step because a n is a monotonically increasing sequence which is bounded above and limit n going to infinity b n equal to b exists because of the property every monotonically sequence which is bounded must converge. What is the relation between a and b? Claim a is less than or equal to b. Is that okay? a is less than or equal to b. Is that okay for everybody? Because a n is less than b n. So, or look at each a n is less than b n and b n is decreasing so b n are upper bounds for the sequence n. So, the limit of that that is a, so a must be less than or equal to b. Now, the question is can I say is it possible a is strictly less than b? Is it possible that a is strictly less than b? See, saying that limit of a n exists, limit of b n exists. We have used the property that is sequences nested. That property has been used once nested. If you want to say that if a is strictly less than b, that means what? If a is strictly less than b, then what? All that does not help actually. That means the limit of a n and limit of b n has got a length in between. Is that possible? Because length of b n minus a n, so limit of b n minus limit of a n, what is that limit? Limit of a n is a, limit of b n is b and that is going to 0. The property says limit of, so this property says a cannot be strictly less than b. So, a is equal to b, so that much is because of 2. So, implies, so where is, so a implies a is equal to b. Now, can I say this number, so claim, can I say this number, call it, if you want to call it as x now, a is equal to b, call it as x or keep it a and b. That this number x belongs to i n for every n. It belongs to each one of the intervals, because where is x? It is less than, it is bigger than a n, less than b n. Is it okay? Because it is a limit of a n and limit of b n, so this must be inside a i n, a less than, is it okay? x is a n less than x less than b n for every n. So, this implies, is it okay? I am just writing a weaker statement because that is true. So, implies x belongs to, x belongs to i n. Can I say x belongs to the intersection i n? It belongs to each i n, so it must belong to intersection. So, implies singleton x is a subset of the intersection i n. Now, question is, can I say it is equal to? I want to say it is equal to. How can I say that this is equal to intersection i n? If not, the simplest way of doing this, if not, suppose there are two points, x and y both in, then what will happen? So, length of i n will be bigger than y minus x, right? Both are inside i n. So, length cannot go to 0 then. If x and y, x not equal to y are inside i n, then what will be the length of i n? It will be bigger than or equal to y minus x, which cannot go to 0. It is number. So, that is not possible again because length goes to 0. So, this is okay. Now, my question is, we have used nested property. We have used the property that, what else we have used? Nested property we have used and then property to length going to 0 we have used. Where have you used the fact that they are closed intervals? Where did we use the fact that each i n is a closed interval? In this proof, can you spot somewhere some line which I should have justified? I did not. There must have been somewhere. Otherwise, I can remove that condition that i ns are closed intervals. I could just say i ns are nested sequence length going to 0 should be happy. a n, b, r, what are a n, a n, b, r? So, you intend to say that here, right? But I use property to here. Namely, the length is going to 0. So, in this proof, where I have used the fact that closed, what you are trying to say is, I can give you a counter example where for open it will not be true. But I am saying in this proof, see that is how we should try to understand in what line of this arguments did I use the fact that i ns are closed intervals. Till now, saying a ns are increasing, that is nested property. b ns are decreasing, that is nested property. Bounded intervals are bounded. So, both sequences are bounded. So convergent, no problem. And because a n is less than or equal to b n, so limit a will be less than or equal to b. That is property of limits. So, we are saying, if possible a is less than b, that is because if a is less than b, then what happens? We said x will be between a and b n, that is because a ns are increasing, b ns are decreasing, so x will be inside. But does it imply that x has to be in the interval i n b n, closed interval? If it is not closed, would you say that x has to be in the interval a n b n? So, something to think about. So, try to go through the proof yourself and try to understand. This is how you will understand and then remember the conditions, what are the conditions what are required. Where in the proof, we use the fact that i ns are closed intervals. So, let me give you some example and then probably that will help you to understand. So, what we are saying is, this nested interval property is no longer true if i ns are open intervals. So, let us look at an example of that type. So, let us look at i n to be equal to, so where is the first one? 0 and 1, where is the next one? 1 by 2, 1 by 3. So, what is the length of i n? Length of i n is 1 over n, that goes to 0. So, this is a nested sequence of intervals, i ns, length going to, what is the intersection of i ns? What do you think is the intersection of i ns? If you say it is 5, that means I have got a counter example indicating that in the theorem of nested interval property, you see when you are in the proof, when you are taking limit a n is a, limit b n is b, then a n b need not be part of the, that is what is happening here of the intersection, a n b need not be part of the intersection, that is what is happening. So, to say that a n b, when you want to say x is in a i n for every n, that is the limit of a n and b and you want to say that belongs to i n, then you have to have close intervals, otherwise this may not be true. So, that is the place they reuse. So, we have proved what is called the nested interval property, that means, and there is a consequence of basically LUB property you can say, because LUB property implies every monotonically increasing bounded sequences convergent and we did not prove that, but one can prove that actually that is equivalent to LUB property. So, LUB property, monoton increasing and bounded convergent, quotient, all are equivalent properties. If you assume one, you can prove the other, but we are doing only LUB implies monotonically increasing and bounded is convergent. Monotonically and LUB property implies Cauchy and we also proved Cauchy implies LUB property. In the way that we did not prove Cauchy implies LUB, we only proved that every Cauchy sequence, a sequence is convergent if and only if it is Cauchy. One can prove all these are equivalent, we will not go on. We also proved as a consequence of this what is called nested interval property, locating a point on the line. One can prove that this is also equivalent to LUB property, converse also. If you assume this property on a field, then it has to have LUB property. These are all slightly deep theorems in construction of real numbers, we will not go into that. But as a consequence of LUB property, these properties are true. So, what we have done today is nothing much actually. What we have done is, we have defined one, the Cauchyness is equivalent to convergent, one. Second, we defined what is an interval, we try to characterize what are intervals left, open, right, close in the familiar geometric form and then we proved in the nested interval property. So, let us stop. Thank you.