 Last time we introduced compact spaces and lindelow spaces. Today module 39, let us discuss compact metric spaces. In fact, last time we did not have any examples. Why? Because now we will have plenty of examples naturally, without spending any more time. So let us come to compact metric spaces. Earlier, so whatever we have done, we never used the word compactness, right? Now we will bring it and get familiar with the metric spaces themselves. In a metric space, every compact subset is closed and bounded. You see, I could not use closed and bounded words in an arbitrary topological space, right? As soon as I have metric space, I can use them and suddenly a compact subset is closed and bounded. Let us go through this. Many of these things you must have seen at least for R and R and R and so on. But now, for any compact metric space, the proof will be similar to almost the same as. So I have to show boundedness as well as closeness, right? Fix a point x0 belonging to x and look at all balls, open balls bn of x0 of radius n and center at x0. Every point, every point in x is at a finite distance from x0, right? Because d of xn x0 is some finite number. Therefore, you can always choose n large enough so that that xm whatever x1 you have taken that will be also there in the open ball containing x0, center at x0 and radius n. This means that x is contained in the union of all these balls. So x is, so this is an open cover now, okay? Therefore, it is a cover for our subset A also, which is compact. That means what? There is a finite sub cover for A. So bn1 x1, bn2 x2, sorry x, x0 is fixed, bn1 x0, bn2 x0 and so on, right? For some n, what does that mean now? If you take the n away large enough, bigger than all the n1, n2, nk, okay? That bn x0 will contain all those other balls, smaller balls of smaller radius because all of them are center at x0. So a subset is contained inside a ball means it is bounded already. That is why boundedness is the exactest one. That shows A is bounded, okay? Now we have to show that A complement is open. So take a point set in the complement of A, okay? For each x inside A put epsilon x is some number I am going to put. What is it? It is d of x, comma z by 3, okay? This x0 I can put x equal to x0. For each x inside A I am putting d of x, comma equal to z by 3. Z is in the complement, x is variable, sorry here, not x0. Z is in the complement, that is what we said. For each x, look at the d of xz. That is a positive number because z and x are different. One is in the complement and in the A. Take one-third of that. That is your epsilon x. Now you take epsilon x bar around x, okay? Very x. What you get? You get an open cover for A, right? A is compact. So you get a finite sub cover. That will be B epsilon x1 x1 epsilon xk xk. Union of these finite many balls will cover A. Now you take the minimum of all these epsilon x1 epsilon xk. There are finitely many positive number. Take the minimum, okay? Now look at B epsilon z. That was in the complement. Claiming B epsilon z is contained inside A complement, okay? To work out this one, all that you have to do is take a pencil and a paper and just plot your points here. One is some A, some other point and there is some finite cover and so on, right? You have to do that for understanding this thing. I am not going to do that kind of doobling here, okay? I want to finally get the truth out of this one just by logic. No pictures, okay? Only that way you will learn, you know, your learning of point set topology will be strong. So all that you have to do is use triangle inequality to show that B epsilon z is contained inside a complement. In other words, if you take some y, y distance between y and z is less than epsilon, you should show that y cannot be inside A. Cannot be inside A comes to this one. Cannot be in one of these balls, then it cannot be inside A. So at that level, I will leave it to you. You verify that. The importance of compactness stems from the so-called Heineborel theorem which states that a subset of the Euclidean space is compact if Fendon leaves closed and bounded. We proved that every compact space inside a metric space is closed and bounded. Converse holds for Euclidean spaces, okay? So that is the classical result which goes under the name Heineborel theorem, okay? So you might have learnt it in your analysis course, but let us do it here if you haven't learned that. We begin with R. Inside R, we want to say that the closed interval AB is compact. This also you must have learnt, but I will redo this one here. So closed interval is compact, is what I mentioned. Take an open cover where all these UIs are open subsets of R. Nothing more is assumed, okay? AB is a closed interval. These are open subsets of R. Union is, Union covers AB. Union is, you know, contains AB, okay? Now I define a subset A of this interval. All points inside AB such that the closed interval A to T admits a finite sub cover from U. You see, A to T is also covered. So make these hypotheses and then put all those T which satisfy these hypotheses. A to T must be admitting a finite sub cover. Put that T inside this A, okay? Each element in AB is in one of the UIs. In particular, A is inside one of the UIs, right? Let us say A is inside U1. It follows that there is some epsilon positive such that A, A plus epsilon is contained inside U1 by the definition of opens of sets in R, okay? Actually A minus epsilon to A plus epsilon will be there, but I do not need A minus epsilon part here because I am working in the closed interval AB. So this part is contained inside U1. Once this is there, okay? Everything up to epsilon satisfies this property. Therefore, this entire, you know, half open interval is contained inside A. In particular, A is non-empty, right? Right? Up to A, A plus epsilon, some positive part here, not just A, will be already inside A. Now put S equal to supremum of A. Since A is non-empty, this is a finite number, alright? Least upper bound. It has to be inside AB anyway. It is enough to show that this supremum belongs to A and it is equal to B. Understand this statement. What I am going to prove? AB admits a finite cover. If that happens, what happens? This entire AB will be equal to A. All the points will be there. B itself is in A is enough. I want to show that this B is in capital A. That is the same thing as saying that AB admits a finite cover, okay? So what I am trying to say? S is inside A and S is equal to B. So I am proving it in two stages. Finally, I want to prove S equal to B, right? So first I put S is inside A and then S is B. So that is the end of the proof, alright? So let us prove this. First of all, since I up to A plus epsilon is already A, supremum will have to be bigger than A plus epsilon. So S is bigger than A plus epsilon. And being the supremum of a subset of A to B, it will be in A bar, okay? Since it is A bar, okay? Therefore, A bar is contained as AB because AB is a closed interval, okay? It is a closed subset. Therefore, there is some member say U naught belonging to U in the same family such that this S is also inside one of the members. That is all I am telling. I am calling it as U naught I should say, okay? Now to U1. See U naught I said here, okay? So S belongs to U naught. Choose 0 less than epsilon prime, less than S minus A such that S minus epsilon prime, S plus epsilon prime is contained inside U naught. So that is again the property just like this one here. Some open interval around S must be inside U naught because U naught is an open set and S is inside U naught, alright? Nothing very great I have done so far, okay? Once supremum is inside AB, AB is covered. So I am taking one of the members here to which it belongs. Now, nice observation start. By property of the supremum, if you take anything smaller than supremum, it will be inside the set, okay? There must be element here, okay? So it follows that S minus epsilon prime by 2 must be inside A. If this is not inside A, S cannot, could not have been the supremum. So this is definitely inside A, okay? Where epsilon prime is some positive number, it has been chosen such that this open interval is inside U naught. So once it is inside A, what does it mean? There will be U1, U2, UK, okay? U1 I have chosen for this one. I can include that number all the way because it contains little A here, U1, U2, etc. UK I am calling this the finite sub cover from this family UI for the set A to S minus epsilon by 2, epsilon prime by 2. That this belongs to A means there is a finite cover like this, okay? Now all that you have to do is put the extra member U naught also. Remember U naught covers this portion. So this portion overlaps with this one up till here and it goes up to S plus epsilon prime. So A comma S plus epsilon prime, okay? Intersection with A, B. Now I cannot say this is contained here, unless I intersect with A, B, okay? So that is contained inside U1, U2, UK, okay? You can just put epsilon, epsilon prime by 2 also if you want. No problem, okay? If T is the maximum of S plus epsilon prime and B, like there are only two elements, this implies that T itself is inside A, up to epsilon prime it is there, okay? If you see these two are both intervals starting from A. So intersection I am taking, right? So minimum of the two will be the intersection, okay? Look at the maximum, T is maximum of this one. I have taken S plus epsilon prime and A intersection B. Maybe I should take minimum. Then T will be inside A definitely, okay? So this will imply that S is inside A because up to S plus S prime is there, okay? Or it will be all the way up to B. If it is B, anyway smaller than it is going to be. So in either case, S will be inside A. So if S is less than B, then it must be this number, right? S plus S prime. So S will be always less than T. This T, T is larger, one of, it goes above that one, that is the whole thing which contradicts S is supremum of A. No number bigger than S will be inside A because S is supremum. But this shows that up to, up to here or up to AB it will be here, okay? One of them is contained in such this one, which just means that there is a number, there is a sum element which is bigger than S and it is inside that A and that is a contradiction, okay? Therefore S must be equal to B. No, that completes the proof. So the next thing is high-neighboral theorem that a subset of R n is compared if filled only if it is closed and bounded. So this is what we wanted to prove, okay? So we are going to use the earlier theorem here, namely closed intervals are compact. Then we have also proved yesterday that product of compacts at least compact, finite product. Therefore you can take product of finitely many closed intervals, take the closed boxes inside R n, they are all compact. So that is the thing that I am going to use now. See now you have a lot of compact subsets, right? Suddenly once you have closed and bounded subsets of R n, all of them are compact. So many examples you have now. A subset of R n is compact if filled only if it is closed and bounded, okay? From 3.67 the only part follows. Once it is compact if it is closed and bounded over. Now suppose it is just closed and bounded, okay? That is the way Weistra started with hypothesis because he was working only inside R n anyway. So and he called by the way he called these sets limited sets. So there are so many different words by different authors and so many, you know dozens of people around the same time were working to develop the topology. So then there exists delta positive such that A is contained inside minus delta to plus delta power n, a large square cube whatever any cube we are taking inside R n, okay? So this is another way of looking at what is the meaning of bounded set. You can take a ball also centered at origin but balls are always contained inside the squares and squares are contained inside the ball larger and larger or smaller and smaller. That is what we have seen in the picture, right? So A is contained inside some minus delta plus delta raised to n, right? So it is bounded, that is the meaning. Now A is closed. This is compact. So A is compact. The proof is so. Life was not so easy for people who started these concepts but now for us these things look so easy. Thus we have plenty of examples of compact spaces as well as those which are not. All that you have to take is a non-close set. All that you have to take is a non-bounded set, plenty of non-compact spaces and plenty of compact spaces. So you can vary whatever way you like everything which is in R n now also. So outside R n of course outside other metric spaces and so on you have to take. All metric spaces this will do. Any non-close subset of a metric space cannot be compact. Any non-bounded thing cannot be compact, okay? An important consequence of Heineborg theorem is that every continuous real-world function on a compact metric space attains its supremum and infimum. So this is known as Weisstrass theorem. So this could have been actually the motivation for Borel to come up with this thing. Heine was independently working on his own and he had the correct ideas and Borel expanded on them and came up with all this. So we shall prove here slightly more general result instead of just let us see what it is. Every continuous function f from x to r on a compact subspace x attains its supremum and infimum. See Weisstrass theorem was inside R n and closed and bounded. We do not need that. Now we use the word compact and then we can go inside any space. You see I am mixing now metric spaces general spaces. I have told you that I want to study both of them simultaneously. You do not know any metric now. X is compact. R is a metric space. Usual infimum. Any continuous function from x to r where x is compact attains its supremum and infimum. To make sense supremum infimum you have to come to R in some order topology you have to have. And of course least upper bound, greatest lower bound such things should be there. So supremum is attained is the same thing as some people call it a maximum it is a maximum word is used only after the supremum is attained. Attained means what it is actually a value. So supremum means thus the upper bound, least upper bound of all the values it need not be a value. So here it is attained means it becomes maximum one of the maximum points. Similarly infimum when it is attained it becomes minimum. So put s equal to sup of all fx, x belong to x. R equal to enough of all fx, x belong to x. The supremum infimum is defined for any set of points inside R including infinity. This may be infinity. This may be plus infinity. This may be minus infinity. That is also allowed here. Since x is compared, fx is compared by Heineborel theorem it is bounded. Therefore s and r are finite numbers. They are the closure points of fx. Every supremum is a closure point of the corresponding set. So this is a part property of real numbers. What is the meaning of supremum? Therefore, but fx is also closed because it is a compact thing. Therefore both s and r are in fx. That is precisely the meaning that they are values. s is equal to f of some x0. It becomes a maximum. R equal to fx, y. So f of some y0. So it becomes minimum. Heineborel theorem has plenty of applications. So here is one illustration. I can't go on doing everything. On a finite dimensional vector space, any two norms are similar. This was one of the theorems that I promised you that I will prove. So now we can prove it. Remember on Rn, we had lots and lots of norms. L1 norm, L2 norm, Lp norm in general where p is any number bigger than or equal to 1 and then L infinity norm also. So we had seen that they are similar. But there may be many other norms. There are in fact, lots and lots of norms. This theorem says that on a finite dimensional vector space, any two norms are similar. So they will have all geometric properties, similarity. By fixing a basis, we can see that any finite dimensional vector space over R is linearly isomorphic to Rn. This is linear algebra. Therefore, the statement of the theorem is equivalent to the following. Now instead of arbitrary vector space, I can just assume we are working in Rn. Any two norms on Rn are similar. So do not worry about arbitrary vector space and so on. On Rn, you can use coordinates, everything you can use now. We shall show that any norm I am just denoting by just no suffixes here is similar to the L1 norm. If everything is similar to L1 norm, any two of them will be also similar to each other because similarity is an equivalence relation. All right. So let us prove that this arbitrary norm on Rn is similar to the L1 norm. We shall show that any norm on Rn is similar to L1 norm. That is, we shall find constant c1 and c2 positive such that c1 times norm x1 is less than norm x is less than c2 times norm x1 for every x in Rn. So start with a basis even e to en vector space basis for Rn. Now for any x in Rn, we can write x as a linear combination of these standard basic vectors. So let x equal to iRn into 1 to n aiEi put c2 equal to the maximum of the norms of e1, e2, en with respect to the new norm that we are going to estimate. So there are these n elements are there, you take the maximum of them, none of them is 0, so maximum is some positive number that is going to be our c2. Now norm of x, the norm of xi is written as summation iRn to 1 to n, x is summation aiEi. Norm of that is less than equal to iRn from 1 to n summation modulus of ai times norm of ai. Each of these norm ai, I will replace it by the maximum number c2 here. c2 times summation iRn to 1 to n ai, modulus of ai, that is nothing but the l1 norm of x. So we get the equation that every x or the extra norm, the new norm is always less than equal to the c2 times norm of x1. So one side inequality of already got it. On the other hand, as we have already observed that the unit sphere with respect to l1 norm which we have denoted by s1 that is compact and this inequality already implies that this norm is continuous with respect to the l1 norm. Therefore, we can apply Weiss-Truss theorem. So I repeat because of this equation 24, inequality 24, what we get is that norm from Rn, l1 to R is continuous. Therefore Weiss-Truss theorem, whatever we have proved, this norm attains its infimum also on s1. In any case, the norm will never be 0 on a non-zero set of vectors that is s1. So this infimum will have to be strictly positive. In other words, what we have is norm of x is bigger than or equal to some c1 positive for every x in this unit sphere with respect to the l1 norm s1. Therefore, now we take any x0 equal to 0 in Rn. We can write norm x equal to, if we divide by norm x1, x is not 0 and multiply by norm x1 outside. So this is a constant, this is the norm x1. What I am taking is the given norm, norm of x is norm of x2 by norm x1 into norm x1 outside. Now the inside thing is inside s1. Therefore, I can apply this inequality. So that means that this is bigger than or equal to times c1. So this is norm x1 times c1. So combining 24 and 25, we get whatever you want, namely 23. So what is the corollary now? There is a corollary here. Any finite dimension in the vector space m of any norm in your space n is a closed subspace and complete, actually complete and closed subspace of n. You start with n, you say a norm in your space. Take a finite dimension subspace m, that will be automatically complete and a closed subspace. See finite dimension vector space is not compact. Vector spaces are never compact. But now they become complete and closed. How? The first part follows from the previous theorem combined with the fact that similarity preserves completeness and L2 norm on Rn. Any norm, it is a norm, you restrict it to m. But m is finite dimensional. Therefore, the norm coming from this n is equivalent to say let us say L2 norm. But L2 norm on a finite dimension vector space, what is finite? Some Rn. So it is complete. Right? So it is completeness follows because similarity preserves completeness. Similarity preserves quotientness also, all these we have seen. The second part follows was general principle, namely if something is complete then it must be closed in any metric space. Complete means what? You take the closure point. There is a sequence converging to that. But every sequence converges, it is a Cauchy sequence. But the Cauchy sequence is convergent in the subspace itself because subspace is complete. You cannot have two different limits of a sequence inside a metric space. You start with a closure point. The closure point may be in the larger space. Right? But there is a sequence converging to that point inside the smaller space. That sequence will be Cauchy sequence. So it is complete means what now? The Cauchy sequence must converge inside the space. So that limit must be inside. So it just means that closure is equal to that n itself, as m itself. So this is a general, it is nothing to do with this finite dimensional vector spaces and so on. Finite dimensional vector spaces are similar to Rn. Therefore they are complete. That part you needed is theorem. I never had a theorem or whatever. All right? So we shall complete, we shall continue a little bit of study of compact metric spaces and then come back again to just compact a couple of like spaces. Okay? Next time. Thank you.