 first place of A1. So, what will happen? This number x, whatever be the decimals after the first place, it is not going to be equal to A1. x1 will be different from A1. Is that okay? x1 will be different from A1 because at the first decimal place, they are different. It differs from the first decimal place of A1. So, x1 is not, x is not going to be equal to A1. Now, I have a method of doing. Look at A2. Look at the second place of A2. Pick up some x2, which is different from the second place, digit whatever appears in the decimal representation of A2. It will be 0, 1 to 9, right? Something. So, pick up some, whatever come, pick up some different one. So, what will happen? This will also be different from A2. And I can go on doing it. xn. So, what will be xn? xn will be the number, the natural, natural, the non-negative number between 0 and 9, which is different from the nth place of An. And go on doing it. So, I have an inductive way of writing knowing the nth place. I have constructed the next place, right? Is it okay? So, what can you say about this number x? x is a real number. Is x a real number? This is a decimal representation we have given you. I have given a decimal representation of x. It is a real number. x is not equal to An for every n, because at the nth place, it differs from the nth place of An, whatever digit comes. So, it cannot be equal. Two decimals are equal, if all are equal, right? Simply. So, I have produced an x in R, which is not equal to, that is a contradiction, because every real number must be one of the An's. And we have produced a real number, which is not equal to any one of the An's. So, that solves the problem. Yes, okay? Yeah, what? No, no, no, no. I want to construct, I want to construct a real number, which is different from each one of An's, right? So, the method is the following. Let us write a decimal representation of that number. Writing a number is same as giving a decimal expression for that number, right? Is that okay? How is the decimal representation constructed? And I keep in mind that two real numbers are different if at some place in the decimal representation, those differ, right? So, keeping that in mind, I am constructing x. At the first place, it is different from A1. At the second place, it is different from A2 and so on. That is all, nothing more than that. That way, x will be different from each one of the An's. And this 0 dot x1, x2, xn will be a real number, right? So, I am saying this is the decimal representation of the x I am constructing. So, it must be one of the An's, which is not true, because it is not equal. It has to be also one of them. So, that is a contradiction, right? That is the way anything about infinity is proved. Still not? Okay. Recording shuru kardiya? Okay, okay. Anyway, just at this stage, probably let me just... So, here is 1, 2, n. Let us write something. It is one of the most number lines. I am there. Probably, I should not give a picture of that. Let me just write 1 less than 2, less than 3, less than n. The number of elements in natural numbers, let us denote it by something. It is called Lf0. So, this is called Lf0. Lf is the Greek letter. So, this is called Lf0. Lf is the Greek letter, like n and 0 is... Why 0? You will soon see. And this thing, there is no one-to-one correspondence between these and real numbers. How many real numbers are there? You say, uncountable. Let us give it a name. Let us... This is divided by small c. So, that is small c. So, this is n and that is r. That is small c. So, that is the cardinality of continuum. That is c. There is a continuous of points kind of thing. And in some sense, this is less than this, because there is no one-to-one map, but n is sitting inside it. So, is it reasonable to say that n0 is strictly less than c? It does not make sense, because both are infinities. But we can intuitively say that this is the kind of relation between them. Question in mathematics, a very deep one. Is there anything in between n and c? Natural numbers, one infinity, real numbers, another infinity. Is there any other infinity in between them or not? One question. Are there infinities beyond c? Are there infinities beyond c? That means, is there a set whose number of elements is much more than c? But r is sitting inside it, like n is sitting inside r. r should be inside something which is much bigger. So, I think I will stop here by saying, look at the power set of real numbers. How many elements do you think power set of real numbers has? So, keep thinking. Let us come back to our lecture for today. So, a consequence is of LUB property. So, one consequence we said LUB implies what we called as a sequential completeness. That is saying that every monotonically increasing sequence that is bounded above is convergent. So, this is what we had proved last time. So, second, to prove the second thing, first observation. So, observe, we started looking at convergent implies it is Cauchy. That we have seen last time. That if a sequence is convergent, it must be Cauchy. That means, elements of the sequence must be coming close and close to each other. So, we need a page more. Observe, let me give it some numbers. 1, 2, Cauchy implies it is bounded. We want to claim that if a sequence is Cauchy, elements are coming closer. It cannot grow. It cannot become very, very large or very, very small. So, intuitively every Cauchy sequence must be bounded. So, let us prove that. Proof is quite so bounded. So, proof. So, let us take epsilon greater than 0 given. There exist some n naught such that mod of a n minus a m is less than epsilon for every n bigger than or equal to n naught. That is Cauchy-ness. So, let us start a n naught. Here is a n naught. n naught is now known. And that means, for all n bigger than n naught, everything is close to each other. That means, if I take a n naught minus epsilon and a n naught plus epsilon, then everything must be inside, right, after the stage n naught. Is that okay? Because they are close to each other. So, how many will be outside? May be a 1, may be a 2, may be a n naught minus 1. Again, finitely many. So, I can take the, like we have shown, very convergent is bounded. Same proof repeated, essentially saying that, if it is Cauchy, then it must be bounded. The minimum of a 1, a 2, a n naught minus 1 and a n naught minus epsilon, call it alpha, maximum of a 1, a 2, a n naught minus 1 and a n naught plus epsilon as beta, then everything will be inside. So, let me, so continue. Is that okay? Same proof, basically same idea. Okay. So, Cauchy implies bounded. And here is the third observation. Every sequence has a monotone subsequence. Given any sequence, it should have a subsequence. We defined the notion of a subsequence, right? Picking up elements of the sequence, but going ahead and ahead. So, the claim is given a sequence, there must be a subsequence which is monotonically increasing or monotonically decreasing. It should be a monotonous sequence. So, I try to give you a visualization of this. Let us imagine the sequence looks like this. This is the first one is a 1. So, this is a 1, a 2 and so on. What do you observe in this picture? That after this stage, I am able to see the height of every building, top of every building. Nothing is hidden from me. So, let us look at those numbers n, those indices n, say that m is bigger than n, then a m is bigger than a n. For example, here, except for this, everything else is okay. So, the set, let us look at the set. Let us call it as c to be the set of all n, such that x m, what is the sequence given to us? We have written. So, let x n be the given sequence. Let x n be the sequence which is given to me. So, look at the indices n, say that x m is bigger than x n if m is bigger than n. So, what we are saying is at any place you are standing, look at the buildings. After that, you are able to see all of them. Is it okay? So, this set c is a well-defined set. Now, what are the possibilities? It may be an empty set. So, case 1, c is empty. That means what? That means this statement is not true. So, whenever m is bigger than n, x m is less than or equal to x n. Is that okay? That means if I take n and n plus 1, then what is the relation between x n and x m? n plus 1 is bigger than n. So, x n plus 1 should be less than or equal to x n for every n. What does that mean? Sequence is monotonically decreasing. If this is so, this implies x n is monotonically decreasing. Sequence itself is monotonically decreasing. What is the second possibility? It is a finite set. c is non-empty, but is finite. c is not equal to empty, but finite. If it is a finite set, it is a set of subset of natural numbers. If it is finite, there is some largest element in it, perisopermal property. It must have a largest element. What happens after that largest element? This property should not be true. This property x m bigger than x n for m bigger than or bigger than n is not true. That means what? After finite number of steps, finite number of elements, after that the sequence starts decreasing. So, is that okay? If everybody implies that sequence is monotonically decreasing after finite number of elements. Understand what I am saying? If it is a finite number set, c is a finite set, it has a largest element. Call it n naught something. Then after n naught what happens? After n naught, whenever m is bigger than n, x m will be less than or equal to or strictly less than x n. Is that okay? Because here we have to be bigger than. Negation of that. We have collected all those for which this is true and there is a finite set. So, pick up the largest number of them. That is n naught. After that, none of the indices n can be inside the set c because we have taken the largest of c. That means what? That means after that stage n naught, if m is bigger than n, then x m is less than x n. That means after that stage, the sequence is monotonically decreasing. So, that is what I have written. So, after that stage it is. So, case three, that c is not empty, but infinite. So, that means what? It is a infinite set. So, what does infinite mean? Whenever given any element in c, it is not bounded. There is something beyond that also possible. So, given any n in c, there is some stage k n which is also in c. Is that okay? Because it is infinite. So, if you like, let us write it as, so let c be equal to, it is an infinite set. So, let me write k 1. So, I can write it as k 1 less than k 2 less than k 3. So, this is a set. I should not be writing then less than that because it does not make sense of saying. So, let me write, let c be equal to k 1 k 2 k n and so on. It is an infinite set such that k 1 is strictly less than k 2 strictly less than k 2 is bigger than k 1. So, x k 2 should be bigger than x n. So, this is a monotonically increasing sequence is monotonically increasing. Is that okay? Because all the elements of c, k n plus 1 is bigger than k n. So, x k n plus 1 should be bigger than or equal to x k n because they are in c. So, I have got an infinite number of elements of c of the sequence. That is a subsequence which is monotonically increasing. Is that okay? Everybody clear about it? So, that is something saying, if you want to look at the pictures, see given this, I am able to see one. But next one I am not able to see, but I go here, I am able to see the next one. So, this is going to be my k 1 and this is going to be my k 2 and such kind of things. Is it okay? So, every sequence has got a subsequence which is monotonically decreasing. So, let us write a consequence of that. If every sequence that is bounded has a convergent subsequence. So, given a sequence which is bounded by the previous property, it has a monotonically increasing or decreasing subsequence. So, that is monotonically increasing or decreasing and bounded that subsequence. So, it must converge by our earlier property, sequential completeness. So, we have to assume that every sequence that is bounded has a convergent subsequence and this property goes by the name Bolzano Vistras property. This goes by the name of Bolzano Vistras property. Now, let us look at the fifth consequence of what we are doing. If X n is Cauchy, then it has a convergent subsequence. Why? Now, we said Cauchy sequence is bounded. So, and just now we said every bounded sequence must have a convergent subsequence. So, if a sequence is Cauchy, it has got a convergent subsequence and 6. If a Cauchy sequence has a convergent subsequence, then the sequence itself is convergent. If a sequence is Cauchy, elements are coming close to each other and if it has a subsequence which is convergent, is it clear that the sequence itself must converge to that limit? Not clear? So, let us try to understand. So, let us try to write a proof of this. So, let us write a sequence X n be Cauchy. Let X k n be a convergent subsequence. So, let us write a convergent subsequence. I should write better be a convergent subsequence. So, let us say X k n converges to L. So, that is every Cauchy sequence. Then it has a convergent subsequence because Cauchy is bounded. Every Cauchy sequence is bounded and every sequence has got a monotonically increasing or decreasing subsequence. So, every Cauchy sequence, if a sequence is bounded, it must have a convergent subsequence. In particular, Cauchy is bounded. So, it will have a convergent subsequence because every sequence has got a monotone subsequence. But that monotone subsequence may not be bounded. But if a sequence is Cauchy, then it is bounded. So, the subsequence will be monotone and bounded and hence convergent. So, a sequence is Cauchy. Then it has a convergent subsequence because it is bounded. Now, what we are saying is if a Cauchy sequence, we already know that it has a convergent subsequence actually. But what we are saying is that subsequence converges to L implies the sequence itself must converge to L. That is what we are trying to say. So, if a sequence X n is Cauchy, let us assume it has a convergent subsequence. X n k converges to L. So, claim X n itself converges to L. Why is that? So, what we want to show? So, let epsilon greater than 0 be given. Here is L, here is L minus epsilon and here is L plus epsilon. What we have to show is that X n is convergent. After some stage, all the X n's must come inside it. But what we know, elements of the subsequence come inside. So, given there exist some n naught such that what happens? X k n belongs to L minus epsilon to L plus epsilon for every n bigger than n naught because the subsequence is convergent. So, here is my X k n naught. But the sequence is Cauchy. The sequence is Cauchy. That means what? Elements are coming closer to each other. So, after some stage, elements will be closer to these elements of the subsequence also. Because whole elements are coming. So, I can say without loss of generality, my X n is also close to X k n naught for n bigger than. It will be some other stage, but I can take the maximum of the two stages if you want. Cauchy says X n and X m are close by distance epsilon for n bigger than some stage n 1. So, if I take this stage bigger than n 1 and n 0, that tail, then what will happen? X k n will be inside L minus epsilon to L plus epsilon and X n also will be inside for n bigger than n naught. What does that mean? That means sequence is convergent. X k n is inside and X n is closer to X k n. So, both have to be inside that. That is all we are saying. Nothing more than that. So, let us write that also. So, let us say without loss of generality by Cauchy-ness X n belongs to L minus epsilon, X n and X k n, X n and X n, X k n both belong to L minus epsilon to L plus epsilon for every n bigger than n naught. Is that okay? That is the Cauchy-ness. But that is same as saying implies X n converges to L because X n is coming inside L minus epsilon and L plus epsilon for n bigger than n naught. Is it okay? So, that proves that this fact that if a Cauchy sequence has a convergent subsequence, then the sequence itself must converge. So, what we are saying now? Take a Cauchy sequence. Cauchy is bounded. Cauchy has got a monodont subsequence. So, that must converge because it is bounded. So, Cauchy sequence also converges. Earlier we proved every convergent is Cauchy and now we are proving the converse that every, we proved convergent is Cauchy. Now we are proving every Cauchy is also convergent. So, all this together. So, here is L minus epsilon, here is L plus epsilon by convergence of subsequence. This must have X k n inside it for n bigger than n naught. But for n bigger than n naught, if you like, or some other stage, X n must be closer to X k n by Cauchy-ness. There is a stage where the tail of the sequence elements are closer. I can assume that stage is same as this one or you can take any other tail bigger than that. So, after some stage, k n's are close to L. And X n's are close to k n. Where is X n's? X n's are close to L. That is all. Nothing more than that. So, finally, let us write seventh Cauchy implies convergent. So, every Cauchy sequence in the reals is also convergent. This is a very important property of real numbers because this property is not true for rational numbers. There are sequences of rational numbers which are Cauchy, but are not convergent to irrational. They will of course converge because they are converged to a real number. So, let us try to see an example of that. So, all these steps are, let us just revise what we have done. If it is convergent, then it is Cauchy. If a sequence is convergent, then it is Cauchy. We proved the first thing. Every sequence has a subsequence which is either increasing or decreasing that building heights. Every bounded, if a sequence is bounded, it has a convergent subsequence because if a sequence is bounded by the previous one, it has got a monotonically increasing or decreasing subsequence and hence it will converge. Cauchy sequence is always bounded. We proved that. And if a sequence has a convergent subsequence converging to say, then I think there is a typo here. If it is a Cauchy sequence which has a convergent subsequence converging to alpha, then this itself must converge to, because a sequence can have many different sequences convergent. So, this one there is a step missing. If a sequence alpha n is Cauchy and has a convergent subsequence, then the sequence itself must converge. That is what we proved. So, every Cauchy sequence is convergent. So, l u v property is also equivalent to saying every convergent sequence is Cauchy. A sequence is convergent if and only if it is Cauchy. So, this is called the Cauchy completeness of real numbers and that is equivalent to l u v property, because as a consequence of that we are showing.