 Welcome to module 38 of point set biology part 1. So far we have studied connectivity, path connectivity, local connectivity, local path connectivity and so on. So it is time to take up another very important concept, compactness. As a tagging along, the lindelofness can also be studied, which is not all that important for us and quite often just like you study connectivity and path connectivity to the compactness and lindelofness, go hand in hand. However, from the point of view of point set biology compactness is so important, if you try to tell the story of compactness by which I mean how the concept developed and so on, it will be half of the story of point set biology. So, so important in the compactness concept. I will not be able to do historical any remarks and so on because quite often they can be done only in the perspective, only after knowing what we wanted to do, what has been done and so on. Then why this one, why this one and so on, then you can go back and dig up history, historical development motivations and so on. So, let us start studying compactness in its earnest. Start with the topological space X. A subset of X, so let us say A will be called compact if the following happens. Suppose you cover A by open subsets of X, some arbitrary cover, then it must admit a finite sub cover. So, let me define all these things carefully, sub cover, open cover, what is the meaning of finite sub cover and so on. So, a family UI of subsets of X is called a cover. This I think we have defined earlier, but let us recall. If A is contained inside the union of this family, it is called an open cover. If every member of this family, this all these UIs are open. So, one can have closer coverings also, they seem to be not so important. By a sub cover of a family, for a given set, first of all it must be a cover, then you take a sub family of UIs, which will be also a cover for the same set A, such a thing will be called sub cover. If this sub cover happens to be a finite or countable, then we will say that is a countable sub cover or finite sub cover accordingly. So, once we have these definitions, a subset is called compact if every open cover of A has a finite sub cover for it. Likewise, we say A is lindle of, if every open cover of A has a countable sub cover for it. Deliberately I have not put compact space and compact set and so on. Whatever you want to call it, if you want to think of this as a subspace, you are welcome. If you are think of this as set, we are welcome, compact set or compact subspace. What I have defined is a compactness here. Similarly, lindle ofness here. Okay. However, you may wonder why the subsets? Suppose this A is the whole of X, then it makes sense. All these definitions make sense. Then we may say that X is a compact you know, space by itself. No subspace, no subset. A compact space. So, its own topology take an open cover, any open cover, you can extract a finite sub cover. And you may not be able to extract. The definition gives you a finite sub cover out of it. That is the whole point. So, that is what I have repeated here. Namely, you can call it a topological space compact or lindle of space if it is compact lindle of as a subset of itself. Okay. It is immediate that a subset A contained in that X is compact or lindle of if and only if it is compact or lindle of as a subspace. You take subspace topology, then as a space, it must be compact and lindle of these two are equivalent because you have an open cover, open subsets coming from the larger space, intersect it with the given set, they will cover and those are the open subsets into subspace. So, these two notions are same. There is no difference between them. In particular, you will see that once a space is compact, it does not matter where it is contained inside as a subspace. Suppose X is a compact space. Suppose X is contained inside Y as a subspace or Z as a subspace, both the case is it will be still compact subspace. Now, if you take a closed subspace of a compact space, automatically it will be compact. Similarly, if you take a closed space of a lindle of space, automatically it will be lindle of. So, how does it get it? Just add one extra open set namely the complement of the closed set. That will become a cover for the whole space. From that, you can get a countable or finite sub cover. Now, throw away the extra thing, you do not need it to cover it. So, you have still countable sub cover or a finite sub cover. So, that is all the trick here. By extending the given cover for the subspace which is a closed subspace by putting one extra element namely the complement of that set. And then you can come back. Let X be a topological space and B be a base for X. Then X is compact respectively lindle of. See quite often whatever I do for compactness, I can put it inside a bracket lindle of, but not always. When it is not, I will mention that. If and only if every cover for X admits a finite, but every cover of it from members of B admits a finite cover, sub cover. You see, in the definition of compactness, you want to have every open cover should admit a finite sub cover. But here is a restricted thing here. Members of B cover it. They do not give all open covers. But if it is satisfied, condition is satisfied for members of B which is a subclass of open covers. Suppose you take a covering of X with only members of B and they admit a finite sub cover. That is enough is what this theorem says, this lemma says. So, this is the role of base to cut down our tile. You do not have to check it for all open covers. You have to just check it for covers coming from members of B. This is the gist of this lemma here. The proof is very easy. One way is clear. Namely, if you take a cover out of B, that will be also an open cover in the general case. So, it must admit a finite sub cover. Now, suppose this happens for only members of open covers coming from members of B, then why it should be true for any general covering? That is what you have to start. So, let the condition hold and Uj be an open cover for X, not necessarily from members of B. Then for each X in X, we have X must be inside one of the open sets here, Uj X depends upon X. But then, B is a base means what? There must be an element BX inside B, such that X is inside BX contained inside Uj X. Now, if you will vary all the X, that will cover, this BX is will cover X. Now, by this condition, there will be a finite cover out of these BX. Let us call them as BX1, BX2, BXN. But each BXI is contained inside the corresponding Uj XI. Therefore, X is contained inside union BXI, contained inside union of JXI. So, these were the members from the original cover. If you write and replace n by infinity here, you will get lindelofness. So, proof of, for the lindelof property is also the same thing here. The next immediate thing is the following. Under a continuous map, compactness and lindelofness are preserved. What is the meaning of this? F from X to Y is a continuous map. X is compact, implies FX is compact. X is lindelof, implies FX is lindelof. Just like connectivity, local or path connectivity, etc., not local connectivity. Connectivity and path connectivity we are seeing. What does that mean? Immediately it means that under homeomorphisms, compactness and lindelofness are preserved. In other words, they are topological properties now, in our formal definition. Take A contained inside X, B contained inside Y, where X tau and Y tau prime are topological spaces. Suppose A is compact respectively, lindelof and we have a surjective function F from A tau A to B tau prime B. This tau restricted to A, tau prime restricted to B, they are subspace topologies. And this is a continuous function here. You do not need F to be defined on the whole of X and Y, whole of X to Y. We have to show that B is compact, if A is compact, respectively lindelof. So, how do you do that? Very easy. Start with an open cover for B. F inverse of UI intersection B. See, UIs are now open subsets in tau prime. And all that you have is B is contained in the union of UIs. But if you intersect UI intersection B, this is an open subset of the subspace topology. F is continuous. F inverse will be open inside A. And since this is a covering, F inverse of all this union is the whole of A, which is same thing as union of all F inverse of UI intersection B. So, I have got an open cover for A. You can go back to topology X here just for fun. What do you have to do? What are the, what are the meaning of these are open subsets? Each open subset F inverse of UI intersection B is some UI intersection A where UI is open in X. In any case, these VIs will now cover A because they are obviously larger than the original space here. A itself is contained here. So, A will be contained here also. Therefore, there is a finite respectively countable subset I contained inside J. Subset A is contained inside the union of VI, I over I. This is by compactness or little delofeness. But now you can come back because these things are larger than F inverse of UI intersection B. So, B will be contained inside the corresponding UI's which is either finite cover or a countable. So, that is the consequence of this elementary result namely under continuous functions compactness is preserved. Therefore, it is a topological property. In particular, it follows that compactness of a subset A of topological spaces does not depend on how A is sitting inside X. I am repeating it again. Once A is homomorphic to another A prime, A and A prime are sitting wherever they like, okay, inside the subspaces. If one is compact, the other one is compact, over. This is the same thing with little delofeness, connectivity, path connectivity, etc., that we have studied. They are all topological property, that is the whole idea. Now there is a partial converse here. Converse means what? Compactness from the domain to core domain we have seen. Now we want to count the other way around. Suppose you have a surjective open mapping. In particular, it will be quotient also, but not just a quotient, surjective open mapping. If Y is compact and each fiber is compact, then X is compact. So, this is similar to connectivity wherein we had just a surjective quotient map only, not necessarily open map. Open surjective map is a quotient, but not the converse. So, weaker condition, we had similar result for connectivity under weaker condition. But now for compactness we have put a stronger condition, surjective open mapping. You will see why this is true in the proof, why we needed open set here, open mapping here, okay. Right now don't worry about why it doesn't work for quotient, just quotients and so on. So, let us work out this one. Take an open cover for X, we have to extract a finite sub cover. For each Y in Y, choose a sub cover U, Y1, U, Yk, Y for Q inverse of Y. The Q inverse of Y is given to be compact, a compact subset of X and U is an open cover for the whole of X. So, it will be a cover for Q inverse Y also, okay. So, now I can take a cover for finite cover for Q inverse Y, okay. For each fiber you get a cover, number of elements here will depend upon Y, okay. I have just listed them as QY1, QYky, okay, which is not a very good notation. I should have perhaps QY1 and so that is not correct because Y is fixed here. So, indexing has to be only changed, okay. So, put WI equal to intersection of all these, this Q of UI1, UI1, UI2, etc., but take Q of that. I want this one to be an open set. This is a finite intersection. If each of them is open, this will be open. So, what is the guarantee that these are open? These sets are open. Q of that will be open provided Q is an open map. That is why I have put that condition. I want this one to be an open map, open set. This open set definitely contains Y, okay. So, call this as WI, all right. For each point Y, I have constructed an open subset in a very peculiar way, okay. That is very important. Now, you vary the point Y in the capital Y. What do you get? You will get a cover for Y, that Y is compact. Therefore, what you get is a finite sub cover here Y, Y, Y belong to Y. For Y, say Y is contained inside WIJs, J varying from 1 to N. There will be finitely many points. Neighborhoods of those things which I have chosen already, they will cover the whole of Y, okay. On each UI1, there will be N sets sitting above, some KY sets sitting above. So, for each of them you take those sets here, U, Y, J, I. I varying from 1 to KYJ, J itself running over Y1, Y2, YN. So, this is a finite family. For each fixed J, there are finitely many members and J itself runs over a finite. These will cover now the entire of X, okay. That is the whole idea. Take any point X, okay. Q of X must have, must be inside one of these, right. Then Q inverse of that will be covered by one of these elements. That is because of this one. Q inverse of that part will cover by that, okay. So, we have proved one very important result here, namely under subjective open mappings, compactness can be obtained backwards, namely Y is compact and fibres are compact, then X is compact. So, you can cover the reverse way now, okay. This does, this may not hold for lindel ofness, right. Yeah, why? Because you see, you cannot take, you know, you cannot take count, you may have to take countable intersection here because here you may get countable things, right, right. But there are different way you can generalize this one. Think about that. How can I generalize, how far I can generalize, okay. Think about that. There is no time for going into all the details. Directly lindel ofness is not possible here, okay. This statement in general is not, not true also. Not only the proof does not work, in general it is not true. Now another important landmark result, finite product of compact spaces is compact. Converse is easy. Why? Because if product is compact, you can take the projection maps, they are open surjective. Therefore, each factor Xi is compact if the product is compact. It is the converse that we want to attack. But the converse will all come just by this theorem itself inductively. Once I prove it for 2, then you can use inductively, right. So, for 2 what do I do? Look at pi y from x cross y to y which is the projection to the y coordinate. You can, you can check x coordinate also if you want. One of them. This is an open surjective mapping. Y is compact, x is compact. So, x cross, singleton y, all of them are what? They are all homomorphic x. Okay. And they are the fibers here. Pi y inverse of singleton y is x cross singleton y. Okay. So, you can directly apply the theorem to say that x cross y is compact. Now, induct. x cross y cross z will be compact because z is compact and x cross y is compact and so on. So, finite product there is no problem, all right. The theorem is true for infinite products also. And that goes, celebrated theorem of Ticknauf. But for that you will have to wait a little bit. Now, another interesting diversion here suddenly. The following result has a flavor of Cantor's Intersection theorem for metric spaces, complete metric spaces. Here there is no metric, no completion, no deltas and so on. Something funny happens, but you have to put, what you have to put? Start with a compact apological space. Okay. So, let x be a compact apological space. F1 contains F2 contains F3, etc. Sequence of non-empty closed sets. Okay. Non-empty is obviously necessary whatever I am trying to say. They are decreasing sequences they must hear. Okay. And they are closed subsets of the compact set. Then the entire intersection is non-empty. You see in the Cantor's Intersection theorem, finally you had a unique point there. But non-emptiness was very important. So, the same kind of conclusion can be got out of compactness instead of complete metric space and so on. In the complete metric space you needed more stringent conditions. Here is almost less condition. Apply De Morgan law. It is a one line proof that too. Okay. If this is empty, what does it mean? If you take the complement, if the whole space, the complement of the intersection is the union of the complements. What are the complements? They will be increasing, you increasing, you know, subsets each of them open and they cover the whole space X. That means what? By compactness, at some final stage, it must be equal to the whole space. Right? Some Fn will be equal to the whole space because it is a finite cover now. Whatever you take, the biggest one will cover the whole thing. Right? But then what happens? If you go back to the De Morgan law, the corresponding Un is whole space means Fn is empty and that is a contradiction. Fy is a non-empty sorting. So just apply De Morgan law, you get the proof. Okay. So like this, we can go on taking some glimpses of, you know, compact spaces and so on. By the way, there is no such result for Lindelow spaces. So let us do, take, look at metric spaces again and get some more hints for what kind of things we can do with compact spaces. Okay. So that is next time. Thank you.