 Welcome to modulating. I have boldly named it as the metric trinity. What we are going to do today to discuss three most important theorems according to me in matrix space theory. So one of them is Cantor's intersection theorem. The second one is Banach's contraction mapping principle. The third one is Baer's category theorem. You can call all of them as principles. They have from theorems to they have become principles anyway. High status of a theorem is to become principle. The CIT, CIT I mean contraction mapping, okay. Sorry the Cantor's intersection theorem. This is so fundamental that even in the proof of Baer's category theorem we will be using it. Banach's contraction mapping theorem is slightly of different flavor. It doesn't use CIT directly, okay. The method of the proofs of all these three, they are themselves quite agitative. So often you may have to employ that in your own research work if you want to do, you know, deep analysis or topology. So I would like you to pay attention to not only the statement here, how things have been arrived. The proofs also should be learned properly. So let us begin with freezing this symbol X, D or just X for a metric space once for all in this section. The first theorem Cantor's intersection theorem to start with a complete metric space. Take a nested sequence of non-empty closed subsets of X. Nested means one contained in the other, but the other way around. Fn contained in containing Fn plus 1 and so on. So here they are decreasing here. Okay nested could be the other way around also. So here they are decreasing but that comes automatically after the second condition here. So delta Fn denotes the diameter of each of them. So this sequence of real number converges to 0. So this is also condition. So each Fn is closed, which contains the next one and the diameter goes to 0. Then the statement is that intersection of all these Fn's consists precisely one element. This is stronger than saying that it is non-empty. This is precisely one element. The proof is surprisingly very simple. Surprisingly very simple. You will see that. All that I do is Fn's are non-empty. So pick up Xn belong to Fn for each n. Then look at this sequence Xn plus m. Okay it is less than delta Fn because both of them are inside Fn. The delta is what? Delta is the diameter is supremum of all such numbers where X and Y range over Fn. Therefore the distance between Xn and Xn plus 1 becomes less than delta Fn. But delta Fn itself tends to 0. Therefore this can be made less than epsilon etc. So as soon as this is a Cauchy sequence, this will be also Cauchy sequence. That is the condition. So we have got a Cauchy sequence. But now I use that X is complete. Therefore Xn is convergent. So let us take X to be the limit. Okay now we also know that if you have sequence Xn, you see the entire sequence Xn is not in any one of the sets. But suppose you choose some Fk here, then k onward that sequence is there. The limit of that portion of the sequence is also X. Right? The first k terms does not matter here. Therefore each of these sequences can be thought of as a sequence inside Fn. Okay? And the limit therefore must be inside this closed set because Fn's are closed. So it follows that this limit point is inside all the Fn's. So here we have used that Fn's are closed and they are monotonically, you know, decreasing one content in the other. That is also understood. That just means that intersection is non-empty. The second part that it must be a singleton comes very easily because of its delta Fn goes to zero. You see that this, the diameter of this one will also become zero. So diameter is zero, the set can be only singleton. Anyway, if Y is not equal to X, you know, then DXY will be positive. Therefore it will be bigger than delta Fn because delta Fn is converging to zero. Once it is bigger than delta Fn, both X and Y cannot be inside Fn. Right? If it is not in one Fn, it cannot be in the intersection. So any point which is not equal to X is not inside the intersection. So there is only one point. So that is the proof. Now I have to make a definition because there is some term in the Benach contraction mapping that in the name itself. So it will come in the statement of the theorem also. So take a function from one metric space to itself or to some other metric space also. This will work. But right now I need only this definition. We say F is a contraction mapping or just a contraction. If there exists a positive constant strictly less than one. This strictly less than one is important such that the distance between FX and FY is less than or equal to C times the distance between X and Y. For example, if C is half, each time you apply F, okay, so two points at distance one will go to distance less than half. Next time they will go next 10, 1 4 and so on. Okay, that is the meaning of contraction mapping. Okay? If you take the layman's language, a map of a say your campus or a country or just a state, if the map is up to scale, definitely it will be a contraction mapping. Okay? From the actual object of which it is a map to the map, you can think of holding the map in your hand sitting, you know, you are standing in the country. So the map is inside, so the function F is from the country into the country, but it is a contraction mapping. Now what does this theorem say? Let us say, okay. So for example, contraction mapping is always continuous because if I can, I can control, I want to control distance between epsilon, this one less than epsilon. I can make this one less than epsilon by C, that is all. So it is always less than epsilon. So continuity is obvious. By the way, this condition constant C on both sides, we are familiar with that is equivalence of matrix, right? Similarity of matrix. So this is not a strange condition at all. This is quite a nice condition. It implies continuity. Contraction mapping can also be defined from one matrix space to another matrix space, which I already told you. So the theorem says that if X is a complete matrix space and you have a contraction mapping F from X to X, then F exists a point. Fixed is a point means there is some X such that F X equal to X. Moreover, such an X is unique. So then there exists a unique X belong to X such that F X equal to X. That is the statement. Once again, proof is this time slightly, little longer than the proof for Cantor's Intersection Theorem, alright? So let us first take care of uniqueness. Suppose there are two points X and Y such that F X equal to X and F Y equal to Y. Then you apply, you know, D of F X, F Y, okay? D of F X, F Y. F X is X, okay? F Y is Y. So it is DXY. DXY equal to D of F X, F Y. But because it is a contraction mapping, it is less than would C times DXY. But C is less than 1. So this is observed unless DXY is 0, which is same thing as unless X is equal to Y. So the proof of that uniqueness theorem, uniqueness part is as easy as in the first part. Now for the existence. So one may say that, you know, this Benach's contraction mapping, where he got the idea? Maybe he got the idea from Newton. People who are familiar with some miracle analysis and so on, they will know this kind of iteration method, you know, which was again used by Picard and so on. Start with any point, okay? Apply F. If it is different, apply F again. If it different, apply F until you get, you know, whatever you have F X equal to X, that is the whole idea. But it may not happen at all. So that is where the ingenuity comes. So don't give up the method yet. So look into deeper into it. That is the point. So given, if you start with X1 belong to X, inductively define Xn to be the image of F of Xn minus 1. So apply F to that previous one and take that as Xn. So X1 is any point, I do not know. X2 will be F of X1. X3 will be F of X2 and so on. The claim is that this Xn is a Cauchy sequence. So this time, it is not all that obvious. You have to do little more work, that is all. Otherwise, you know, in the case of CIT, it was easy to see that Xn is a Cauchy sequence. Once it is a Cauchy sequence, completeness comes into picture. There will be a limit point, the limit point of this sequence. The beauty is that that limit point cannot go anywhere. It has to be that F X is equal to X. So that is the point we are seeking. So let us see how the proofs of these claims are coming. So put r equal to distance between X1 and X2. We are assuming that X2 is not equal to X1. If X1 is r equal to X2, then there is nothing to bother about going further at all because then what is X2? X2 is F of X1. If it is equal to X1, we are home already. So we are not going to do that. Put r equal to D of X1, X2. Then look at distance between X2 and X3. That is by definition, distance between Fx1, Fx2 because X2 is Fx1, X3 is Fx2. But then this is by contraction mapping part the C times distance between X1 and X2. This we have denoted by r. So it is a C times r. So distance between X2 and X3 will be less than or equal to C times r. What happens to distance between X3 and X4? One more C will come. One more C will come and so on. So what happens is inductively distance between Xn and Xn plus 1 will be less than or equal to C power n times r. So X2 and X3 is the r. Xn and Xn plus 1 will be just the index will be 1 and less here. So maybe I should write it Xn minus 1 times r which I had written earlier. Anyway, it does not matter. Then it follows that distance between Xn plus 1 and Xn plus 2 is less than or equal to C power n plus 1 times r. Once you have proved this one. Therefore, if you look at distance between Xn and Xn plus m, see we are trying to prove that this is a Cauchy sequence. So distance between Xn and Xn plus 1 is less than or equal to I range from 0 to m minus 1. Distance between I start with Xn go to Xn plus 1, Xn plus 1, go to Xn plus 2, Xn plus 2, go to Xn plus 3. I use triangle inequality and then I get this summation. The distance between Xn plus I to Xn plus I plus 1. So this is a summation. But each of them is r times C raise to n plus I, 0 to m minus 1. So you can rewrite it as r will come out. I range from n to n plus m. Do not worry about n plus I. Now it is C I. It starts with I equal to instead of 0, I equal to n. This is just a rewriting this. What you know is C is less than 1. Therefore, this is a geometric series. Mother of all series. So this is a convergent series. We know how to conclude the limit also anyway. So geometric series because C is less than or equal to 1. C is less than 1. And hence the partial sums say 1 to n C I, they are cos sequences. So this is just difference between n plus 1 partial term to nth term. Only n to n plus 1 I have taken. So if this partial sum is a cosy sequence, this number can be made less than epsilon, any epsilon bar r. So r times epsilon will be less than epsilon for sufficiently large n. That means this is a cosy sequence. So we have got starting with a point x and repeatedly applying f, f, f and so on, we have got a cosy sequence. That sequence converges, take the limit. Now we apply, if we say continuous function, we apply f 1 x, f of x, what is x? x is the limit of all these x n's. But x n I can write to the f of x n minus 1. So I can take this x n plus 1 and write to the f of x n because f is continuous. Applying f of x n, the same thing as limit of x n plus 1. But limit of x n plus 1 or x n plus 2 or whatever it is, add tends to infinity is same limit which is x. So x is a fixed point of f, uniqueness you have already proved. So the third theorem takes a little more time. Therefore let us first consolidate these two theorems today and we will do the third theorem next time. So let us look at some easy comments here on this. Look at the Cantor's Intersection theorem. There are, what are conditions? Delta f n goes to 0, f n or a n whatever. a n's are closed, x a is complete. And a n's are one contains the other. The point is, if you violate any of those conditions, something will go wrong. So conclusion will not be as strong as you like. Conclusion may not be there at all. For example, delta a n does not tend to 0 but because they are smaller and smaller it may tend to something fine but positive. Then you will not get the uniqueness. You may be able to prove the existence that it is namely intersection is non-empty. There is some point but you will not be able to prove that it is a unique point. Similarly existence itself is violated if you do not put the hypothesis nested. If you take arbitrary subsets f1, f2, fn without any relation between them maybe half of it is contained in there or they are disjoint and so on. There is no intersections can be easily empty. That is not very surprising. The third comment I make is here. It is easy to construct a nested sequence a n of close of sets of 0 to infinity. By the way, 0 to infinity is complete metric space such that all the diameters are infinity yet the intersection is empty. So it is not going, not only it is going to, it is not going to converge into 0, they are all infinity. Intersection is empty. Can you think of a set here? What should be a n for each n? Remember it must be close of sets and a n should be containing a n plus 1, containing a n plus 2 and so on, smaller and smaller. Can you think of that as a sequence? Yes sir, if I take n to infinity close to infinity. Put a n equal to n to infinity very good. So intersection is empty. It satisfies all other things but now delta n is not going to converge into 0. All the time it is infinity. Yet the intersection can be empty. So here another question is there. Can you construct a nested sequence of subspaces of a n with finite diameter such that intersection of a n is empty? You can do it in R itself actually but this time I am not putting you see nested sequence I am putting diameters also finite. I am not saying diameters are converging to 0 but what I am not saying is that they are closed. Right? There is no word closed here. So is it possible now think of some subsets of open interval 0, 1. Open interval 0 to 1 by n. n or instead of going up n to infinity 0 to 1 by n. Right? Intersection will be empty. Now can you do the same thing as you did a but this time a n's must be all closed. Closed subsets nested that's all. Diameters I am not putting. Diameters can into 0. Is it possible? This is this is a little more harder right. Think about this. The next one is a little more harder also. Can you do the same thing as B but in the usual metric space in the R with the usual metric? So this is a hint for this one. Maybe you can change the metric okay and then try to do that not in the usual metric. So C is much harder to think about them we will answer them in the due time. Maybe as an assignment you can work it out then we will explain if after checking your answers we will explain you the result okay. So that is all for today. So let us meet next time. Thank you.