 Welcome to module 40 of point set apology part 1. Today, we will continue the study of compactness. I begin with one of the most important result about compact metric spaces as important as the three big theorem that we have proved for complete metric spaces if not less. Okay, this may be even more important. That is called Lebesgue Covering Lemma. Start with a compact metric space given any open cover UI. There exist a real number R positive such that every ball of radius R in Hydex is contained in one of the members of UI. So that is the statement. Several applications of this even at your level at the right in the beginning while doing Riemann integrations and so on especially for the closed interval 0, 1 function defined 0, 1 directly or indirectly you must have used this theorem. So let us have a proof of this one which is not at all difficult. As soon as you have an open covering because X is compact, there is a finite covering. X is contained inside U1 cup, Un etc. where UIs are coming from the given open covering UI. If one of the UIs is the whole of X, then there is nothing to prove. You can take R to be any number. Every ball of radius R will be contained inside X which is one of the UIs. So that is nothing very great. Okay, so we may implicitly and explicitly assume that the complement of UIs at which I will denote by FI is non-empty for i equal to 1, 2, 3 up to n. If the only X there is nothing to prove. So we are assuming that FIs are non-empty. Now consider the distance function from FI. Let us call it FI of X. We call what is FI of X? It is the infimum of all DXY where Y range is over FI. So that is called a distance function and we know that the distance function is continuous. Okay, distance of X or X here is over all the facts. From a given set, this set is close of set which we will use soon. Right now any set will do that will be distance well function will be still continuous. Therefore, you take all F1, F2, Fn and take the maximum that will be also continuous. Okay, so let us call it as Fx. Now I use the fact that FIs are closed. Therefore, FI of X is 0. The distance is 0 if and only if X belongs to F1. So this is where FIs are closed as the industry. But U1, U2 even cover the whole of X. Therefore, if you take intersection of FI by De Morgan law that must be empty. So if all the FIs are 0, that would have been X would have been inside the intersection. But intersection is empty. Therefore, even any X at least one of the FI is not 0. Therefore, this Fx which is maximum will be always positive. Okay, it will never be 0. Okay, Fx itself is the maximum of all the FI effects as a function. Now this function is continuous will attain its minimum on X. Why? Because X is compact. So compactness is used twice here. Okay, that minimum will be a positive real number because F is never 0. This number will do the job namely take any ball of radius R, any open ball of radius R wherever you take the center. That ball will be contained in one of the numbers U1, U2, Un. Okay, so that is the claim. One of the numbers to contain. Okay, so proof is very easy. If this is not true, what does that mean? A point belonging to this one is not here means it will be inside FI. FI is a complement of UI. Right? So Y will be in FI. As soon as Y is in FI, the distance between X and Y will be bigger than FI of X because FI of X is distance of X and FI which is a minimum of infimum of all dx Y. Okay, so dx Y is bigger than FI of X. But FI of X are all equal to, you know, I can take, I have taken VR of X contained inside UI. One of the UI. If FI of X is equal to FX, Y is equal to FX is maximum of one of them. Right? So one of the FI is equal to, that is bigger than equal to R. Right? Some FI, I am trying to say this is not true. So FX or each X will be equal to one of the FI of X. So I am putting equality here. But this is true overall, whatever, whenever Y is there, this is true. Which implies that Y is not inside VR of X. You see, because dx Y is bigger than equal to R. To be inside VR of X, the radius must be, the distance must be less than the radius. Okay? Given actually what we have shown here is given any X, look at that index I for which FX is equal to FI of X. Because maximum is always for finite things, you will be put one of them. Right? The same UI will do the job. This is what it means. That is not very important to notice. But this is how we have proved it. That is all. Look at the way the argument goes. First you have some kind of an infimum here. Then you have a supremum here, maximum. Then again you have an infimum. So the proof of this already gives you another principle which can be used in several mathematical concepts, which is called Minimax principle. I have no time to explain that one. Also it is out of our way. But this is used in complex, is used in all sort of analysis always. Okay? The proof itself is of importance here. So we can make a definition to be happy with this kind of concept. So that we are able to recall this concept very easily. Any number, positive number such that this every ball of radius r is contained in one of the UIs is called the Lebesgue number of the cover UI. Start with any cover. If there is an R like this, that R will be called a Lebesgue number. It is very easy to see that if S is less than R, of course I have to take R always positive. So 0 less than S less than R. Then R is a Lebesgue number implies S is also a Lebesgue number. So you can take the supremum of all such values and call that Lebesgue number that is not necessary. So the practice is to get any number and then call that Lebesgue number. Okay? A metric space Xd which has this property that every open cover has a Lebesgue number. Okay? That is to satisfy Lebesgue property. If this happens for every open cover in a metric space, you can call that X as Lebesgue property. So in this terminology, what we have proved is that every compact metric space has a Lebesgue property. So there may be other spaces and other metrics. Okay? They may have this property. We do not know whether that will imply that this is compact. So that is another aspect. So I am not going to touch that one here. But this is the definition. So we can just work out with this one. Lebesgue cover is a very important result in real analysis. Let us derive one immediate consequence from this one named another important concept in metric space theory named uniform continuity. Okay? In real analysis, you have seen that continuous functions on closed intervals are uniformly continuous. Okay? So that can be extended. By the way, that is very much used in Riemann integration theory. So uniform continuity itself is an important thing. Take any function on a compact metric space to another topological space and a continuous one. Okay? The assumption is X is compact metric space. Then given any open cover V i of Y, there exists R positive such that for every X inside X, F of the ball V R of X is a subset of one of the V i's. Okay? Why I call this one? By the way, the statement is directly straightforward because all that you have to do is when you have taken a covering V i, take F inverse of V i, those things that will be covered for X, then choose this R to be the big number and then you have this property. So that is that is an easy consequence. But why this is called uniform continuity, I will explain. Okay? So if both X and Y are metric spaces, then how do you define uniform continuity given epsilon positive, there exists a delta such that X Y distance is less than delta should imply F X, F Y distance is less than epsilon. That given epsilon, there is no metric on Y. So I have to convert that one and that is converted into open covering V i here. If you have metric space, when you have epsilon here, you have all epsilon balls. That is a cover, open cover. So that has been replaced by this V i now. These V i's play the epsilon. Okay? All over whole thing you have to talk about one single code. Okay? Then this R is actually plays the role of delta, there is no problem. So this R is independent of X, you see. If it is dependent upon X, that is ordinary continuity. Okay? So this is the way the uniform continuity is attained. So we shall take a break from compact metric spaces now and go back to the study of compact spaces in general again. Now I come to one of the important results, what is called as Stublemann or Valais theorem. But what I am going to do is I am going to combine it with another important result which is not so central and do a bit of circus. We give you a characterization of compact spaces. See our definition of compact spaces is what every open cover has a finite sub cover. So whereas many other examples we have seen that there are several definitions. Right? So such an important thing you should have different ways of looking at it. Okay? So here is two other way of looking at a compactness. Compactness as a property. That is my two characterizations I am going to give. Okay? Before that I will recall that you might have, by now you might have seen such a thing but let us look at this one. Give an example to show that in general projection map is not closed. So I still put it as an exercise but since I want to illustrate the coming theme here, so let me tell you that in general projection maps from X cross Y to X or Y projection map they are not closed maps. They are open maps remember that. Okay? While studying the product space we have seen all the time we have used it also. So what is the simplest example? There are many simplest example is you know from R cross R to R itself. Look at a hyperbola given by X Y equal to 1. Okay? Its projection on the X axis just misses the point 0, nothing else. Therefore it is not closed. Right? So having said that now we come to the characterization of compact spaces. No metric now. So I would like to call it the whole thing as Valais theorem. In classical book standard books Valais theorem is only one-third or even one-sixth of whatever we are doing here. Okay? I will tell you what it is exactly. Let X be any topological space. Then the following three conditions are equivalent. So I have put deliberately the third one is X is compact. The other two are going to be equivalent to that. That is my aim. The first thing is every topological space Y for every topological space Y, look at the projection to the Y coordinate from X cross Y. That is a closed mapping. Okay? Not just some Y. For every topological space Y this should happen. Okay? Second thing is X satisfies the following condition which is somewhat longer. So I have put this one carefully very much visible. So you should know this one. This is the very important thing which goes under the name tube lemma. So for every topological space Y again pick up a point inside Y and an open subset V of X cross Y such that the copy of X at Y namely X cross little Y contained inside this open set. So V is a neighborhood, open neighborhood of X cross little Y. Suppose this is the situation then there exists an open neighborhood N which may depend upon Y inside Y of little Y such that the entire X cross N is contained inside V. Okay? So this is also very much used in analysis. So this is a condition I want to say which will be equivalent to X being compound. So third condition is X is compound. Okay? The proof will go through not 1 implies 2 implies 3 implies 1 as usual but I will prove 1 and 2 are equivalent and 2 and 3 are equivalent. So that is why I have put this one in the center. So 1 implies 2. What is the meaning of that? Start with this condition that projection maps to every Y is continuous, is closed. Then I must prove this condition. Okay? So start with any Y inside Y and a neighborhood V etc. of X cross Y this part as well as the condition 1 then I must produce a capital N with this property. Right? So take F to be the complement of V. See V is a subset of X cross Y. So take its complement that will be a closed subset of X cross Y. Okay? Now use this property 1 and come to Y. Pi Y of F is a closed subset. Right? So its complement in Y will be an open subset. So where were you going? X cross little y was completely contained inside V. Therefore X cross little y intersection with F is the complement of V will be empty set. That just means that the complement of the projection of the complement is just does not contain Y. That is an open subset. This is what I have. Start with F equal to we see Pi Y of F is closed. Okay? Now take the complement of that Pi Y of F inside Y. That is an open set. X cross little y intersection F is empty. Right? Because X cross Y was contained inside V. Therefore this Y will be inside this N. Y cannot be inside Pi of F. That is otherwise this would be a non-empty. Okay? Projection X comma Y if it is there, then projection would be inside F. Right? So this is empty means now this Y is inside N. Okay? So Y is a neighborhood of, sorry N is a neighborhood of Y. Right? That is what we are trying to prove here. See N is a neighborhood of Y. And what I want to prove? X cross N is contained inside V. Right? So that also comes this just freely here. Look at any point X cross N X comma Y prime. Okay? Suppose it belongs to this one. It just means that Y prime is in N. Okay? Just take into coordinate. Projection of this one cannot be inside F because because you have taken N is a complement. So Y prime is in N. Okay? So Y prime, sorry Y prime is in N is direct X comma Y prime is in X cross Y. Y prime is in N implies Y prime is not in Pi Y of F. Okay? That means X comma Y prime is in the complement of that. That is V. Otherwise it will be in F. So the entire X cross N is contained inside V. So that was easy proof. Right? So just set theoretically you have to chase it. Alright? So one implies this condition. Now we will revert to this one. Namely that condition implies one. Namely take any space Y, X cross Y to Y, you have to show it is a closed mapping. So start with a closed subset. Then Pi Y of F is closed is what you have to show. Take the complement. You must show that this complement is open inside Y. Same way I am trying to go backwards there. Okay? Now take Y belonging to U. Again these are all set theoretically. They are reversible. This just means that X cross little of Y intersection F is empty. Now you take V is X cross Y minus F. Okay? To begin with Y was F was closed. So X cross Y minus F into, okay? To get once you have this an open subset. Right? The property 2 gives you a neighborhood of Y, okay? N such that Y is in N and N is open and X cross N is inside V. So it follows that as soon as this is possible X cross N intersection F is empty. Because what by definition V is X cross Y minus F. Okay? Therefore N is contained inside U because U is a complement. Okay? So more or less, you know, if you choose correctly the notation, every arrow can be reversed here from 1 and 2. Because it is somewhat complicated statement, so it is better to write independent proof that is what I have written. Okay? So 1 implies 2 and 2 implies 1 is over. Now come to 2 implies 3 and 3 implies 1, 3 implies 2. So I first prove 3 implies 2. I save the last thing, okay? Namely 2 implies 3 for the last thing. That is the theorem which is called Valais theorem. Sorry, this is the one which is called Valais theorem or tube lemma. 3 implies 2. 3 is compactness, right? Yeah, 3 is compactness. Compactness implies this one is a standard result which is called Valais theorem or tube lemma. Okay? So what I have done is I have made it into if and only if along with another condition here. Okay? Let us prove this one. Let X be compact. We have to prove X satisfies this W, this property Valais property I have denoted by W, W for Valais. Okay? So the entire property we have to prove. So what is the meaning of property proving? I have to start with the hypothesis of the property. Y inside Y, X cross little y contained inside V, V open. This much you have to start with. We have to find N as required, namely N open, Y inside N, the whole N is contained inside V. So whole X cross N is contained inside V. That is what you have to prove. For each X comma Y in X cross Y, we can find N open subset WX, GX in because either Y is fixed here, X is the variable here. But X comma Y is inside X cross Y. Y is fixed. This is given. So WX, GX in X and GX is open in Y. Okay? Open subset such that X, Y is in the product. So this is the definition of the product typology. Right? These are the basic opens of sets. WX cross GX. Since V is open, there is a basic open set. The basic open set looks like a product. That is all I have used. That is the more the more. That V is an open subset containing X cross little Y. Okay? Here only X varies. Y is fixed. As X varies, these WXs will cover capital X. That is an open cover. Now use the property that X is compact. Get a finite cover. WX1, WXK. Okay? All that you do is take the intersection. Your finite WX1, take N equal to intersection of all these WXIs. Intersections, all these I want to I want open subset of Y. So GXIs. Okay? The corresponding GXIs. WX1, WXNs are covering for X. So GX1, GH2, GXN are all neighbourhoods of Y. Take the intersection. That will be neighbourhood of little Y. And that will do the job because now take any point X comma Y prime here. Okay? Y prime is N means Y prime is in all the GXIs. Right? X is in one of the WXIs which you do not know. Suppose it is in WX1. Then X comma Y will be in WX1 cross GX1. That is inside the O. So that is why you have to take the intersection here. Okay? So the proof of Wallace theorem as such, classically is just that much only. Okay? 3 implies 2. Now finally I come to 2 implies 3. Pay attention to this. Okay? Because many expositions, many books do not have this theorem. Definitely not this proof. Alright? Now I have to prove that the space X is compact by using a condition 2. 2 implies 3 means R. Okay? So start with any open cover for X. How to use 2 so that this is going to give me a sub cover. The 2 says for every Y something is true. So I must cook up some, you know, some nice space Y. Nice means what? You know, friendly which will give you something. Okay? Space Y such that when I apply this condition, this condition to that Y, this condition. This is true for every Y. But how to use that? You have to cook up some Y so that when you use this condition that it should give you X is compact. Okay? So that having said that much, I will just go ahead with the proof. How I got this one? Start with an open cover for the given topological space X. Now go to the power set of X. Now my Y is going to the power set of X and I must give you topology on this power set of X. Okay? So the open subsets of power set of X will be what? There will be collections of subsets of X. Okay? It is a subset of the power set of X, sub collection of power set of X. So you have to be careful here what is going on. We shall construct a topology and then take Y as this space in the hypothesis to arrive at a conclusion that you admit say finite cover. So this Y is going to be depending upon the covering U. The underlying space is always P X, the power set of X. But the topology I choose will depend upon the covering U. Okay? That will say that this covering has finite sub covers, finite sub family that covers. Okay? So that is the trick. For any A inside X, let us make a notation here. A plus means all the supersets of A, collection of all supersets, everything which contains A and of course subsets of X. All A is contained inside B. That is a notation A plus. For instance, if A is empty, what will be A plus? It will be the whole of all the subsets, power set of X which will come. Okay? Every subset is contains empty set. Right? So empty plus C, the power set itself. Similarly, if A is X, then what will be A plus? It will consist of only one element, namely X itself. It is a singleton X. You remember that. It is not X. It is a singleton X. Whereas A is empty, it is all the one entire, all the subsets of X, the power set of X. It is not X. Okay? So that is the convention here. Okay? Let B be the family of all U pluses where U range is over this open cover. Okay? This collection B, I do not know what topology, what property it has. It does not matter. Take this as a subspace, sub-base for a topology, tau hat. Any collection of subsets of a given set, this time, the given set itself is Px. Okay? That will generate a topology. The smallest topology generated by this as a sub-base. So tau hat I am denoting. It is nothing but tau of B that we have earlier used denotation. So this is a topology on Px. That is what I am interested in now. Clearly, it depends upon U. Okay? An important point to note is that the singleton X is in every non-empty member of tau hat. Take an open subset of tau hat. Singleton X will be always there. Why? Because you get U, U plus. Of course, empty set does not have. Empty set is also there in every topology. Take any non-empty set. Look at this U plus, all supersets. So in every U plus, it X is there. So when you take sub-base, remember, you have to take finitely many open sets in the sub-base and then intersect. Intersection will also contain the singleton X. Then the union, of course, will contain singleton X. So singleton X is in every member, okay, other than empty set. Okay? So I have taken Y as Px tau hat. All right? So now this is Y. So what is your B open subset? It is equal to all those which look like U cross U plus. U is a member of this one. Okay? That is a subset of X cross U plus which is a subset of Px. So this is going to be a subset of X cross Y. Okay? Then clearly B is an open subset of X cross Y. Now singleton X is always in the second factor here U plus for all X, for all U. Therefore, whatever you take here, singleton X is there, but when you rearrange U, this curly U covers X. Therefore, what happens? This entire X cross singleton X is contained inside V. So this is the situation of the condition W, valence condition. Given this little X is the point of Y, remember that. Okay? And X cross little Y contained inside V, V open. What does it give you? It gives you a neighborhood N of the little X here. Okay? Capital X here, singleton X here, such that X cross little X singleton X is contained inside X cross and contained inside V. So that is the conclusion of all right? So after cooking up this Y, which is a topology on the power set, I have used that one in condition 2 to get this neighborhood. Now I want to say that this will give you a finite cover out of U. Can you see how finite cover comes? That is a trick here. That is a very strange thing. Even now, you may not see why the finite cover comes. Okay? The point is the B here, this is a sub base. What is the meaning of sub base for an open subset in a topology? You have to take, given any point X, belongs to some open set, then there will be finitely many members from B such that their intersection contains the point as well as contained in the given open set. A basic open set out of this one is finite intersection of members of this one. So that is what I am going to use here. Little X cross singleton X is X cross in N. Now I will replace this little X belong to N. I will replace this N by a basic open set, members of B and then finitely many of them intersect with that. N is not equal to this but in this statement I can replace it. In 25, this statement I can replace it because this singleton X will be inside this one and this is contained inside capital N. So X cross N will be also. So instead of this one, I can write this and that is all. I can write the intersection area to 1 to U plus to make it simple. In other words, I am getting another equation here namely X cross, the singleton X is contained in the X cross, this intersection contain inside B. Instead of calling it, you can call this as N prime as a result. So 25 can be read with N equal to intersection of U i plus where I range from 1 to N. Now you verify another property of this plus, the supersets of A. A plus intersection B plus, what are they? Something belongs to this one means it contains both A and B. This is subset of, members of this one are subsets of X. If it is contains A, it is here, it is contains B, it is here. If it contains A and B, it contains A union B. So it is here and conversely, any set contains A and B, A union B will contain both A and B. So A plus intersection B plus is the same thing as A union B plus. So this happens for every member. In particular it happens for the members of U also. And hence put G equal to union of U i. See this U 1, U 2, U N, U have got finite. Now you take G equal to i range to 1 to N, U i. It follows that this W is equal to G plus. What was W? I forgot. There is no W here. W is G plus here. W equal to G plus. But what is W? G is this one. G plus, G plus is what? G plus is union of this, apply this one inductively. So what do you get? This is for two of them. You can do it for N of them. So G plus is intersection of U i plus. That is N. So this W is N here. That is all. So this is a typo. This must be N. That is all I want. By this notation, if you look at the intersection of U plus is the same thing as take union of this and then take the plus. So this N is G plus. So it remains to show that I want to show G itself is X. I want, what I want is not this one but there is a on the way. But I want to show that this union i range to 1 to N U i covers X. That will prove that this open sub, open covering has a finite sub covering. And since U is an arbitrary open covering, that will prove X is compact. So that is the whole idea. So I have to show that these U i's, these are U i plus, these U i's they cover X. So look at X comma singleton G. This is a member of X cross G plus. Remember what was G plus, whatever G is. G plus contains all those members which contain G. In particular G is there. So this is a member here in G plus. So what is the meaning of this X comma G belongs this one. X comma G will be inside V because the whole thing was this X cross N, G plus is X cross N. That is what I have used here, N here. And X cross N is contained inside V by 25. So you write it as N that is all. So now come here. So this implies X comma singleton G is inside V which is same thing as now X comma G belongs to U cross U plus for some U inside U. Just go ahead. What was V? V is the union of all this. So it must be inside one of the U cross U plus. For some U inside that, what is the meaning of this? This is an ordered pair. X is inside U and G is inside U plus means what? G is inside U plus means G contains U. So X inside U inside G for some U inside U. Therefore remember all the U will cover X that is the starting point. Therefore all of X is contained inside G. Start with any X here. Then you go through this one. It shows that it is inside G. Therefore this X is contained inside G. But G is a subset of X after all. So X is G. In some sense this has the magic of Furstenberg's proof of natural, you know the primes, infinity to your primes. The key here, passage from arbitrary to finite comes only because the topology has the property that such families generate topologies by taking finite intersections first and then going to the arbitrary union. That is all. There is no other way I can explain this one. Okay. Suddenly you get finite. From where? See compactness in the definition we have put the finiteness ourselves. Now we have to produce it. Right? To understand what is going on. So you have to choose a proper, you know, properly thought open cover here. The starting with open cover, properly thought topology. Okay. Yeah. So having said that, let me again repeat a few things which I have told already. Usually the three implies two of this theorem is known as Tubler-Marvara's theorem. This itself is a very subtle result and it is very useful also. For example, we can derive the theorem 3.64 from it. So this I will leave it as an exercise to you. The uniform continuity etc. It is useful in many other situations as well. Okay. There are proofs of one implies three. See now we have all, all the three are equivalent. Right? So this, all the three are equivalent. How, how I have proved it? I have proved three implies two and two implies three. One implies two implies three. I always tell you, one implies three. So some people, you know, classically have proved that not very classically. This is a modern approach now. One implies three using ideas of nets and filters. These things you can find in many books. Okay. Proof of one implies three by using nets or filters. The, what are called as ultra filters really produce this thing in a magic just the way this proof has produced it. Okay. The nets also give you, they pretend to explain it, but it is still a some kind of mystery only how the proof comes. So that is one thing I wanted to tell you. Whereas the proof that I have given namely two implies three. Okay. Compactness implies three is, yeah, two implies three. Namely from valence condition producing the compactness. This seems to be new. I have not seen it anywhere. Okay. Now in part two of this course, I plan to give a not so difficult proof of Tyknoff's theorem namely arbitrary product of compact species compact using one implies three of this theorem along with principle of transfinite induction. So this transfinite induction takes some time that is why I have put it in part two. Otherwise I could have done it right now also. Okay. So that is the thing I want to tell you. So next time onwards we will take up some other properties namely countability, separability and so on. Thank you.