 Welcome to NPTEL NOC course on introductory course on point set topology part 2. Today we shall do compactly generated spaces further, last time I already told you that it is not easy to find examples of spaces which are not compactly generated. So, now we will see a lot of examples of compactly generated spaces, they come from wide spectrum quite unexpectedly. So, let us go through these things carefully. Let x be any host dust project, we are restricting ourselves to host dust spaces, if we do not then we get many more but for practical point of view they are not much important. So, let us restrict ourselves to host dust space. Then x is in cg, if x satisfies the following limit point test condition. I am calling it a limit point test condition, you will see why. So, I will keep calling it LpT. So, what is this condition? Given any subset f of x and a point x inside x, x is a limit point which I remote it by Lf. Remember we have introduced this notation for the set of limit points of f, it is derived set of f, x is a limit point if and only if there exists a compact subset k of x such that x is a limit point of k intersection f. So, pay attention, this is there exists a compact subset, not for every compact subset that will be totally incorrect. Given any subset f and a point x, the point is inside a left. If there exists some k such that x is inside L of k, if part is cg, only if part puts an extra condition here. So, that you know if and only if it is important. So, if this condition is satisfied then x is cg that is the lemma. This lemma therefore gives you a gives you a nice you know sufficient condition, but this is not a criteria that is what I, it does not give you a criteria. This condition itself is not different only. So, here is if x satisfies, it is not only if. However, I do not know any example of a cg, it does not satisfy this condition. I am sure that there are such examples, otherwise it will not float like this, it would have been stated as a criteria. So, such example may be quite complicated. So, that is the story. So, let us go through this example. Those in this lemma, let f be a subset such that f intersection k is closed for every compact subset k of x. Then we have to show that f is closed in x, that is what x equal to kx means. So, to show that f is closed in x, we take a point x inside lf. If I show that this x itself is inside f, then we are through y. We know that for if f is closed then lf must be inside f, that is cg, but lf inside f also implies f is closed because what are the closure points of f? Closure points of f are either points of f or they are limit points of f. Therefore, limit points of f is contained is lf means f bar is contained inside f, therefore f bar is equal to f. So, I start with a point x inside lf, then I want to show that it is inside f. Now condition lp t comes into picture. x is in lf means there is a compact subset k of x such that x is inside l intersection. But if as soon as I take compact subset, this f satisfies this condition and f intersection k is closed. So, x is a limit point of x intersection k, but f intersection is closed in k. Therefore, it is closed in x also. We started with a compact subset of a house door space. This is why house door system is important here. So, that is a closed subset. Something is closed in k, therefore it is closed in x. Therefore, if x is inside this one in this subset, it is closed inside x itself will be inside k intersection f. But that is f. That is contained inside f. So, what do I have to show? x is inside f and that is all you wanted. So, this condition of house door space is important here. As a corollary, we have two classes of compactly generated. What are they? Any first countable space is compactly generated. Any locally compact space that is what we are being studying, it is also compactly generated. When I say that, I have in my back of my mind, I have that they are house door also. A is a house door space and that will be in CG if these two conditions happen. That means house door first countable house door plus locally compact. See, completely diverse field. We have also seen something like this happening. Any metric space and locally compact space, quite different things, both of them are bare spaces. Now, here is another example where in first countability of metric spaces, local compactness are coming together in a different way. So, this is just for your observation. See, every metric space is first countable. Now, you do not have metric space, just first countability. So, that and local compactness, completely different things, but both of locally along with house door space, they will give you compactly generated. Why? So, I am putting this corollary, all that you have to do is, you have to see that both of them satisfy LPT and then apply this lemma. So, how do you see that they are satisfying LPT? As soon as you have house door first countable set, a point is a limit point if and only if you have a sequence converging to that point. Along with that sequence, take that point also. Suppose Xn converges to Xn, then take the set Xn union X, that is always a compact set. So, X will be a limit point of that compact. So, that means that X satisfies what? LPT. Now, for local compactness, it is even more further easier because take a point, X belong to limit point of some set, there will be a neighborhood around that point which is compact. You do not have to go outside that neighborhood at all to see whether it is a limit point. You can restrict yourself to that compact set already. So, local compactness immediately gives you that LPT, very easy. The first countability also gives you because of this, a sequence converging to a point, union with the limit point is always a compact. Not all spaces of compactly generated spaces are compactly generated. By this I mean that compactly generatedness is not hereditary. To see such an example, we have to wait a little bit. It is not coming so easily. Right now, on the positive side, we have the following. Take X belong to CG. Every closed subspace is compactly generated. That is, it is weakly hereditary. In one sense only, not closely weakly hereditary, not open. X belong to CG, every closed subspace is compactly generated. The proof is not veritable. A be a closed subset of X. B be a subset of A such that B meets each compact set L of A inside a closed subset of L. You want to show that B is actually closed in X from which it follows that B is closed in A. See, X is compactly generated. A is a closed subspace. I want to show that A is compactly generated. So what should I do? Start with the set B which has its property. B means each compact subset L of A inside A in a closed subset of L. From this I have to show that this B is closed in A but I will actually show that B is closed in X. So let K be a compact subset of X. Since A is closed, start with A is closed subset. L which is equal to K intersection A is compact subset of A. A is closed. So K intersection A will be a closed subset of the whole space, all of them. So it is a closed subset of a compact subset. So it is a compact subset of A. Therefore K intersection B which is K intersection A intersection B. See because B is already a subspace of A. It is a closed in A, K intersection A and hence closed inside K. So this is true for all K. It follows that B is closed in X. So is K intersection A is L. L intersection B is closed. It is the property with which I start with. B means every compact subset L of A in a closed subset. So this K intersection A put L. L intersection B is closed in L. So this whole thing is K intersection B is closed in K intersection A. Therefore it is closed inside K itself because A intersection K is closed inside K. So see it is true for all K. It follows that B is closed in X because X is in C H. That is all. We say a subset U of X is a regular subset. If for every X inside U, there exists an open set V in X such that X belongs to V contained in the closure V, closure V contained inside U. To begin with though we have used the usual notation for open set, there is no assumption that U should be open. But however this condition for every X inside U is an open set V in X. So X contained V in content inside U automatically says that any regular subset U has to be open. Also if U is a closed open set that is closed and open then it is automatically regular because we can then take V equal to U for all X. X belongs to U. U contained inside U bar because U is closed one and U bar is U so U bar is contained in U. So that will be satisfied. So where is the regularity coming intact? You have to be very careful here. Note that being a regular subset neither implies nor implied by the condition that if you take the subspace topology on this subset that subset U is a regular space. So these two motions are quite different for you can take U to be any singleton set in X which is not an open set. There are plenty of subspaces. Then U as a singleton is already a regular space but not a regular subset because it is not open inside X. Similarly you can start with any non regular space U even if you can allow it to be host or false so you have seen such examples and then take X to be disjoint union of two copies of U. Then this U will be open in X as well as closed therefore it will be regular but as a subspace it will not be regular because we started with a non regular space. So that these two examples should convince you that this is just a ad hoc definition. This is going towards what to ensure that subspace of a compactly generated space is again compactly generated. So this is the theorem. Let U be a regular subset of a space X then X is in CG implies U is in CG. Let B be a subset of U such that B means each compact subset of U in a closed subset. In order to show that B is closed in U let X belong to the closure of B. The closure point of B inside U. We want to show that X is inside B. So let V be an open subset of X such that X belongs to V contained inside V power contained inside U. This is by regularity of the regular subset U. But if this is the case once we have an open subset of the whole space X will be inside the closure of V intersection B itself. Any closure point we need to say B for this open subset of it must be a closure point. That means it is contained inside the closure of V bar intersection B because it is a larger set. Therefore it suffices to show that V bar intersection B is closed in U so that X will be inside V bar intersection B but V bar intersection B is contained inside B. Our aim is to show that X is inside B. Therefore we have come here namely show that V bar intersection B is closed inside U. Everything is inside U. So take a compact subset of X. Then K intersection V bar is a compact subset of U. Compact subset it is and it is contained inside U therefore it is compact subset of U. And hence K intersection V bar intersection B is a closed subset of K intersection V bar K intersection V bar is compact subset. So K intersection V bar intersection B will be a closed subset of K intersection V bar. This is the hypothesis we have been starting with. So B has that property. So it is closed in K intersection V bar which is closed inside X itself. So this is a closed subset of X. But X is CG. So we have verified that for arbitrary K K intersection V bar K intersection V bar intersection B is closed for every K. Therefore V bar intersection B is closed in X. So once this closure of this, its closure inside U will be itself whole thing. Therefore X will be inside. Before ending up today's lecture, let me give you one very important application of this one in compression analysis. Why I say compression analysis? I do not know personally any application of this in other analysis. But that does not mean that it is not applicable. So as far as compression analysis is concerned, I have used it that I know. And I have taken this theorem from a very famous book by Raghavan Narasimhan's complex analysis one way or another. So what does it say? It is a peculiar statement here. This kind of study we will do in later on in the course in a different context altogether. Right now take a locally compact half-dwarfs pair and a connected component of X which is compact. Given any open set U in X containing K, there exists a set N such that K is inside N contained inside U and N is both open and closed in X. Think of this K as single point. Single point could be a component also. And single point is automatically compact. What does it mean? This means that for every open subset of that point or every neighborhood of that point, we have another open set which both open and closed. Open and closed. It is not open and its closure is closed. That kind of regularity is stronger than that. N is both open and closed inside X. So this is happening to every compact component of X. So I suspect that this will be useful in all these complex dynamics also. Not just complex analysis of one variable. Where we have to study Uliacets and such things. Let us see the proof. Proof is not all that easy at all. Indeed I as a practice when I read a thing like this, I try to prove it myself. This one I could not prove it myself. So first case is when X is compact. So we are going step by step here. Assume X itself is compact. That does not mean that, you know, it is connected. We have to take a component actually compact point. So connected component are compact. For each compact component something is happening. That is what we want to show. Okay. Let F be the family of all N contained inside X. Such that K is contained inside N. N is both open and closed. We do not know whether this F is non-empty one. We have just take F to prove this one. Okay. Fine. Look at L which is intersection of all members of N. All members of this family. Okay. Since each member is a closed set, intersection is closed. Therefore L is closed. Clearly all of them contain K. Therefore this K, this L will contain K. Okay. L will contain K. What are all N contains such that K is contained inside N and N is both open and closed inside N. There is no condition on N to be connected. K is a component. Okay. First we shall prove the claim for L in the place of K. So we are trying to prove something for K. But we have taken a slightly larger thing. Maybe very large. I do not know. But it contains K. So we shall prove it for this L. Okay. L itself is not assumed to be connected or anything. We do not know that. But what we will show is that even any open subset of L, there X is take closed and open subset like N containing L inside that open subset. So let you be an open subset of X containing L. Then look at X minus U. Okay. That is contained in X minus L, which is the union of X minus N. De Morgan law. Because L is the intersection of all N's. Okay. Now X minus U is compared. U is open. X is compared. So that is why this first case X is compared to L. X minus U is compared. It follows that you have finitely many N1, N2, N2 cases that X minus U is contained inside the complement of X minus N i, i contain finitely many of them. Okay. Now take N2 intersection of N i, finite intersection. Finite intersection of open sets is open. Finite intersection of closed sets is anywhere close. So this contains L clearly, right? Because L is intersection of all of them. I have taken on the finite family here. Contained inside U because they are all, you know, it is even smaller. X minus U itself is union of X minus N i. And it is both open and closed. So we have already proved this for L. Okay. Now in order to prove the property for K, we shall actually prove that K is L. Okay. So intersection of all such neighborhoods is actually K is what we shall prove. This will follow if we show that L itself is connected. Anything larger than a connected component has to coincide with it. Right? K is a component of X. So if we show that L is connected, okay, K must be equal to L. All right? If possible, let L be disjoint union of closed N, you know, two closed subsets, non-empty closed subsets. Then we will get a contradiction. K is connected and contained inside the union. Therefore it must be contained in one of them. Only one of them. So K is contained inside A, let us say. So we shall actually prove that B is empty. That is the meaning of saying that there is no such separation which just means that L is connected. So we have assumed that A is disjoint union B is L. We want to show that B is empty under the assumption that K is inside A. All right? Now, use X is half star. Okay? Since A and B are disjoint compact subsets, there exists disjoint open subsets to end B in X, S, A is inside U and B is inside V. Remember this theorem? A half star space can have two compact subsets. They can be separated by open subsets. Two disjoint compact subsets can be separated by. So this is what we have to write. For first people point and a closed subset, then we improve for two different closed subsets. So that I am using here. Okay? A and B are disjoint closed subsets of X. Therefore they are compact. So again I am using X as compact here. All right? So find two open subsets larger than A and B. A and B. Right? And they are themselves disjoint. But now L is contained inside U union V. There exists N which is both open and closed in X and such that L is contained inside U. Because this is an open subset. Okay? Right? Clearly N intersection U is open. Because N intersection U is open. Also N intersection U is N intersection X minus V. Right? Hence it is closed also because N is closed and X minus V, V is open. So X minus V is closed. So N intersection I see is closed. N is closed also. Since K is contained inside A and A is contained inside U, right? We have K is contained inside N intersection U. Therefore N intersection U which is both open and closed and contains K is in a member of F. It is inside F. This means L is contained inside N intersection which is contained inside U. But N is A union B. Therefore B must be empty. Thus we have shown that K itself is L and L has satisfied its property. So K satisfies its property under the assumption that X is compact. So that completes the case one. Okay? The general case is much simpler now. Okay? X is locally compact and K is compact. Therefore there exists a compact neighborhood X0 of K in X. Right? Each point has a compact neighborhood. Right? Each point of K has a compact neighborhood. It takes a union that will be a covering. That will you can extract a finite covering that will give you a compact neighborhood. Compact neighborhood means what? An open subset disclosure is compact. That much you can say. Okay? So there exists a compact neighborhood X0 of K inside X. K0 with a connected component of X0 containing K. K is connected. Remember that. Okay? Now K is a subset of K X0. Okay? K is a connected. So K is a neighborhood of X0. So take K0, a connected component of X0 in containing K. Every connected subset is contained inside a connected component. Then K0 will be connected in X as well. Okay? In a subspace, something may not be connected. But if it is already connected in subspace, larger space will be connected. Okay? So K0 is a connected component. So K0 is, sorry, K0 will be connected in X as well. Since K is a component of X, this implies K is K0. So what I am saying is from X to X0, X0 is a smaller space. Okay? Advantage is now X0 is compact. But is K a component? This is what you have to see, right? So I say that K is a component of, component in X0 itself. And K is equal to K0. That is all. Therefore, K is a component of X0 as well. Now, let U be an open subset of X such that K is contained inside U. Since X0 is a neighborhood of K. Okay? Right? There is an open subset V of X such that K is contained inside, we can say X0 intersection U. I can actually take this X0 intersection U itself. Right? So by case one now, applied to X0, because X0 is compact. There exist n such that K is contained inside n, contained inside V. And n is both open and closed inside X0. Everything is happening in X0. Since X0 is closed in X, you see, X0 is compact, right? X0 is closed in X, n is closed in X. Because n is closed in X0. Since n is open in X0, n is also open in V which is open in X. See, you cannot go via X0, you cannot make it open. But via V, you can see that this n must be open inside X. Therefore, we have got already n is contained inside U. The proof of theorem is complete because we have found that both open and closed ends of X which is contained inside U and containing K of course. Alright? Later on, when we are studying one point compactification, I will give you some relevance of this one there, okay? For compactifications of Rn. In complex analysis, R equal to 2, you have the complex plane. Compactification of that is nothing but the extended complex plane. And that is how it is important in complex analysis. Thank you.