 Welcome to module 46 of point set topology part 1. Last time we stated and proved Tyknoff's theorem, of course we assume Alexander's Sub-Base theorem. So in order to complete the proof of Tyknoff's theorem, we should now prove Alexander's Sub-Base theorem. We have already made some set theoretic preparation for that also last time. So let us start with the proof of Alexander's Sub-Base theorem, which I restate here that X be at a level space and S be a subspace of its topology. X is compact if and only if every cover of X by sub-family of F admits a finite sub-cover. The necessity of this condition for compactness is obvious because X is compact every open cover has to admit a finite sub-cover. After all members of S are open, so some sub-family of S covers means there is an open cover. So that must admit a finite sub-cover, that condition is necessary. The crux of the matter is that we have to prove a converse. So we expected that this proof will be sufficiently complicated. So we have to be ready for that. So fix a sub-base S for X and let us have notation that B is the base generated by S, namely elements of B are those which are finite intersection of members of S. So in what follows, I shall use the word cover to mean a cover for X, so that that much shortness of notation is short time, words, okay, a little bit of small word set. What we have is, let us just recall what is the meaning of whatever you are talking about. Every cover U of open subsets means the content is at tau means U is a sub-family of tau means they are open subsets admits a finite sub-cover. This is the compactness, right. The second statement is every cover F contained inside B, that means what? Only members of this particular B are allowed. This is a smaller family than tau after that admits a finite sub-cover. The third one is even shorter. Every cover epsilon contained in the sub-base admits a finite sub-cover. Obviously A implies B implies C, okay. We have also seen that B implies A, okay. Now what we want to prove is C implies B. So this is the gist of Alexander sub-base theorem. So indeed what we shall do is C implies the contrapositive of B, namely if F is a family of B which has no finite sub-cover then it is not a cover which is the same thing as if it is a cover then it is a finite sub-cover. So it is in this form I am going to prove C implies B, okay. So this is all just to clarify the ground situation. Now the plan of attack is as for fix a family contained inside B which admits no finite sub-cover. Construct F's of family, another one like this, namely these are all now families of families, okay. Theta is H contained inside B such that F piece inside H is F piece one such family which admits no finite sub-cover. Finally we want to show that it is not a cover for X, okay. So F contained inside H and H has no finite sub-cover, okay. So look at all those H which also have this property no finite sub-cover but they are larger than F. So that is my family of, so families of tau here. The first claim is that you seem to show that under the set theoretic inclusion the partial order set theta which is containment. So it is a partial order has at least one maximal element. From one arbitrary F we want to have something which is maximal, okay with respect to this property that it has no finite sub-cover, okay. We have to assume that is one F then only this theta will be non-empty we are unsure. Then we have the maximal element for that we have to apply John's lemma which means we have to prove something there, okay this is a plan of that. Second step will be take such a maximal element H in theta. Now put this epsilon equal to H intersection H that means those members of H which are in the sub-base, those members of H which are in the sub-base, the sub-base is fixed. Okay, so that is a smaller family than H, right? Okay, this does not admit any finite sub-cover because H does not admit any finite sub-cover. So how can epsilon admit finite sub-cover? If this does not admit finite sub-cover from the hypothesis in seed follows that this sub-family epsilon is not a cover for X, okay. So up till here we have arrived by using John's lemma and our assumption that there is a F such that which is now does not admit finite sub-cover, okay. Now second claim is look at the union of all members of this H and so that is clearly, so this is a claim, sorry, look at all the members of H that is not a cover, that is what we know, it does not admit finite sub-cover but now take the entire set, we would show that this is contained inside unions of members of this epsilon. So what we have concluded for epsilon, epsilon does not cover X, so this will also not cover X. So that will complete the proof. If this union of all elements in H is contained in the union of all elements in this epsilon and epsilon does not cover X, so H is also not a cover of X. If H is not a cover of X, remember H was having some maximal property, in particular F is contained inside H. So if H does not cover, F does not cover, okay. So that will complete the proof. So we have to prove two steps here. In the first claim, we have to prove the John's lemma, whatever hypothesis needed, that is the first part and then you have to prove this claim, okay. Once we prove this claim, claim two is over once that the proof will be completed. So it remains to prove claim one and claim two. So let us do it one by one. Look at claim one, let me just recall, okay. This theta has a maximal element is what we have to show. For that, what is the ingredient that you have to put inside the John's lemma? Take any chain inside theta, we must show that that chain has an upper bound. This is what I will show. This part is very easy. As usual, quite often let phi be a chain in theta. Recall what the meaning of chain? Chain is a totally ordered subset of theta under the usual inclusion here, okay. Consider the family G, which in the union of all members of this chain, obviously under the inclusion that will be an upper bound for this one. But don't hurry, that must be an element of theta. Then only it will be justified. The larger set of all subsets, it is an upper bound, fine, okay. So clearly G contains F because each member of this member of this chain, they are members of theta. All the members of theta contain F, okay. So it is a subfamily of B also because each member of theta is also subfamily of B. So if you take unions of all these members, you take each, all the members are inside B. So that is not a problem. Moreover, suppose finitely many members of G say G1, G2, Gn cover X. This is the last part which I have to show that it is inside theta. That no finite family covers X, that is what you have to show, right. If not, suppose there are G1, G2, Gn belonging to G, which cover X. Remember what was G? G is just union of members of this chain. So G1 will be inside say some lambda 1, G2 will be lambda 2, G3 lambda 3 and so on. But these are all one contained in the other. You take the maximum of these. When you have finitely many of them, you have the maximum. That will contain all the G1, G2, Gn, okay. But that is a member of this theta. So it would cover, so it is a contradiction, okay. So I repeat all these Gis are in one single element G prime of B. But this G prime is a family which belongs to theta by definition. It has no finite sub cover for X. So this is observed because we assumed that G admits a finite sub cover. So G does not admit finite sub cover. That qualifies it to be a member of theta. Therefore, every chain in theta has an upper bound in theta. Once you have satisfied this property, John Slema tells you that theta has a maximal element, okay. So first claim is done. Second claim, take an element which is a maximal element for theta. Any maximal element, fix that, that is call it as H. Clearly, F contains H. So our aim is to show that F does not cover X. So you showed H does not cover X. In fact, what we will show is the following. Namely, the claim which says that claim is even stronger than what we need. Namely, union of all members of H is actually contained inside union of all members of is epsilon, okay. Remember this epsilon comes from the sub basis, the sub family of sub basis, okay. And we know that this does not cover. So this is what we have to prove, okay, which is stronger than just showing that F does not cover, okay. Let us prove this one now. Take a member here, U inside H. By the very definition, U is a member of the base. A member of the base means it is intersection of finitely many elements S1, S2, SN from the sub basis. We claim that one of the SIs is actually inside H, okay. See start with some U inside B, okay. We want to show that that U is concerned in the union of these things, okay. Now what we end up is saying that one of these SIs is actually inside H, okay. If this is not the case, let us say it is not inside H. So this is another sub claim. You may say claim 3, but claim 2 is not yet done. It is part of claim 2. So suppose one of these S1, S2, SN is not inside H. If this is not the case, then consider family HI, which is H union singletonite. Add this one more member. You get another family. Add H1, add S1. You get one family, H1. Add S2, you get H2. Add S3, you get S3 and so on. So you get H1, H2, H3, which are all larger than H. If I have put one extra element and that I am assuming is not inside H. Each of them is a sub-family of B, okay. So every element inside this curly S is inside B also and contains H, which of course contains F. But by maximality of H, these HI are not members of theta. You see, they are larger than because each H is a maximal element. This can happen only if each of them, each of them admits a finite sub-cover. There are members in this family. Just that union of them is a finite cover for H, okay. So what are those members? If you pick up only members for a match, that is not going to cover. So each time you have to put SI also. But just SI may not cover. Along with some members here, finitely many, this SI will cover. That is the meaning of that, okay. So I get, for each I, see UI1, UI2, UI, UI, PI, 1 less than K less than 2, up to PI, UI1, UI2, etc. These are the members of HI, union, actually they are members of H, union 1 SI. All these UIKs are members of H. And what are they? They are, this union is whole of X, okay. SI has to be there. SI is not there. That will be give you H is a cover, finite cover. That is the assumption that H is the, H doesn't have a finite cover. Now look at all these UIKs from H1, H2, H3 and so on. Along with, instead of S1, S2, SN, okay, you just take the intersection, okay. And now the Kruskal's matter is that this will be a finite sub cover for X. Clearly it is finite. It is a finite sub cover for X. But these are all members of H only now. These are the members of H. You, you starting with the, I remember H. Why this is the cover for X? Take any point, if they are inside UIKs, any fix one PI here, fix one I, if they are UIKs, that is fine. Otherwise, they will be in the corresponding SI. Each time they are here, that's okay. But if it's not here, none of, none of these UIKs contain a point X, then this point must be inside SI for every I, which is the same thing as saying that point is inside U. Therefore, this cover, okay. So this violates the fact that H is belongs to theta because it is a finite sub cover. So what we have proved is a substatement here that one of the SI's, let us call it as S1 is already in H, okay. So that just means that this S1 is an element of this epsilon, right. See, S1 is already inside H. S1, where we started with S1, S1, Sn, okay. They are intersections of, just intersections of, sorry, yeah, this each SI's are members of curly S, okay. This is the base, this is the basic elements here, all right. But what is epsilon? Epsilon is H intersection S, right. If it is in H also and instead of sub basic, instead of basic, it's a basic open set that will be inside epsilon by definition. Therefore, U is inside S1, it just means that U is contained in the union of all these, this is the second part that we wanted to show. The RHS here, RHS here, one of the members contains U because that element is inside epsilon, okay. So U is one of them. So started with this one. So we have shown that this union is contained inside that one. So that completes the proof of the claim to and therefore Alexander's sub-base is proved. Therefore, we have completed proof of Tyknoz theorem also, okay. One question that immediately opposed to our mind is, what happens to the analysis of Alexander's sub-base theorem for Lindelof property? Remember, we have already seen that Lindelof is not even finite productive. Namely, we have seen that the L, R comma L, where L is the semi-interval topology has the property that the product is not Lindelof whereas RL is essentially Lindelof, okay. But Alexander's sub-base theorem may still be true. Why, why that is not true? What is happening? Is it not true at all? So this is what we want to question. Does the imitation of the proof in the case of compactness work if we simply try to replace the phrase finite sub-cover by countable sub-cover wherever finite, you know, the doesn't admit finite cover is sort of countable, countable, why, where do we go wrong? You can figure out where it is, but here is a concrete example which says that even the Alexander's sub-base theorem, you know, the statement will not be true if you recall, if you replace the, you know, compactness by Lindelof property, okay. So that makes the Alexander's sub-base theorem more important in some way. See, it's just a narrow thing and still he was able to prove here. That is the whole idea, okay. So for this one, I will just quote this xy 3.91.2, okay. But let me give you a little bit of this one. What is it? Look at the family S of closer intervals a comma b, where a is strictly less than b, okay. Then S forms a sub-base for a topology on R, any family of subsets of a, subsets of a given set x will form a topology as a sub-base, okay. But this topology is nothing but discrete topology, okay. This topology is a discrete topology. Why? Because given any x, I can take something x minus 1 to x and other one x to x plus 1, both close intervals. Intersection will be just singleton x. So every singleton x is open mean it's a discrete space. On the other hand, so once it is a discrete space by the way, an uncountable discrete space is not leaned alone, okay. On the other hand, the exercise there asserts that every cover by a sub-family of S, this is S, admits a countable sub-cover. This is for the usual topology of R, okay. With the usual topology of R, you show that take a and b are close, use the topology of R, that is a finite sub-cover. We just show. The finite sub-cover part is just set theory. So it covers R. Even this may be, sorry, this is not finite sub-cover, countable sub-cover. This may be uncountable sub-cover normally, but you can have a countable sub-cover. So this was the exercise. The point is, if you take open intervals a, b, then of course, you know that it is because R is lentil of. You have to use that also. But now we have to show that even if you cover it by closed intervals, it will have a countable sub-cover, okay. So granting that exercise, what we have is the following. Namely, Alexander sub-based theorem is not true for Lindelow's property. There is another remark which I would like to do, not with Alexander sub-based theorem or Cauchy's theorem and so on, but another property for whether something is productive, namely quotient maps. Take two quotient maps and take the product of these maps. So x1 to y1, x2 to y2, then you have a product map from x1 cross x2 to y1 cross y2, okay. More generally, you can ask for families xi to yi or qi is a quotient map. Then you can take the product map here from product of xi to product of yi. It is a quotient map and then you can study the properties, okay, quotient, straight forward quotient may fail. But suppose you take open maps, suppose you take open quotient maps or closed maps or closed quotient maps and so on. So there are a number of such problems. So I will just sum it up. We will not go to deeper study of these things. They are not too difficult or they are easy or too difficult, right. So some of them I will take it in the part 2 of this course. But right now you can observe that openness is finite productive. If you take an open map, two open maps, product will be an open map that is very easy to see, okay. And if you have open quotient, means open and surjective map, that is an open quotient, then it will be productive, no restrictions even you can take arbitrary products, okay. Three and four are much harder. Just if you take arbitrary quotients, just quotient map, even two of them will not be quotient, will not be a quotient map, need not be quotient map, okay. Unless you assume some more hypothesis, okay. So that is an interesting case here, which is needed in many other places also but that will be done in part 2. So we will stop here with productive properties and so on, whatever so far properties which we have studied. So next time we shall start studying some more topological properties. They will be, in general they will be called as what, what is the name, largeness properties. Thank you.