 Welcome to module 45, Ticknoff's theorem. So, Ticknoff's theorem is a land mass result in the development of point set topology. The key result that we are going to use as a step for Ticknoff's theorem is that this is called Alexander's Sub-Based Theorem, an expiatorological space and fix a sub-base S for its topology. Then X is compact if and only if every cover of X by sub-family of S admits a finite sub-cover for X. We have seen that if X has a sub-base such that open covers from this sub-base, from this base they admit a finite sub-cover, then X is compact that much we have seen already. But Alexander's Sub-Based Theorem goes one step ahead. It says instead of base you can just use a sub-base. The proof of Alexander's Sub-Based Theorem itself is not very straightforward. So, we shall postpone the proof of this one, but use this one to prove Ticknoff's theorem. Okay, so once again I repeat that Alexander's Sub-Based Theorem essentially cuts down the verification of compactness of a space from arbitrary open covers to open covers coming from a single sub-base. So, that is the whole idea. Ticknoff's theorem can be stated as follows. Start with a family of topological spaces, each of them non-empty. Then the product space X i is compact if and only if each X little i is compact. Okay, what we have seen is that if the product space is compact, each factor space is compact. Of course, for this you have to use that that X i's are non-empty. Therefore, the projection maps are surjective. A surjective continuous function takes compact space to compact. That is very easy theorem that you have proved. Using that, you will follow that if product space is compact, then each X i is compact. Now, we have to prove the converse. Assume that each X i is compact. Let S be the standard sub-base for product topology which we have been using, namely consisting of all p i inverse of u i for all open sets u i inside X i and for all i inside i. By Alexander's sub-base theorem, if we show that an open cover from S, you know members of S, that admits a finite sub-cover. That is enough. Okay, any arbitrary open cover but members are from S, if that admits a finite sub-cover, then X will be compact. So, that is what we are shown now that starting with an arbitrary open cover with members of S, we will show that that is a finite sub-cover. So, for each i in i, put u i equal to dose u i inside X i such that this p i inverse of u i is in S prime. This S prime is the sub-cover that we have chosen, it is a cover that we have chosen. Members of S prime cover X. From this, we want to extract the finite sub-cover. So, first I define u i, this is a sub-family members of open subsets in X i. Those u i's in X i such that p i inverse of u i is contained inside this S prime. Then each member of u i by the very definition is open in X i. Okay, we claim that for at least one i, u i is a cover for X i. Okay, we have not claimed that all the u i's will cover corresponding X i here. At least for one of the indices, this must happen is what belongs there. If not, what happens? There exists some X i little X i belonging to X i minus the union of all the u i's in u i because this is not a cover. For every i, this will happen. Okay, so pick up one point X i in the complement of this. So, this gives you one element X belonging to X i to be such that p i of X equal to X i. Now, X is in some member of F say because this is my this S prime here. Okay, say X belongs to pi j inverse of u j for some pi j inverse u j belonging to some S prime that is what for some j. This means that if X belongs to this one means its jth projection is X little j, which we have chosen must be inside u j and with u j belonging to some curly u j. That will be our prediction because we have chosen this one to be such that they are in the complement of all this. Okay, so therefore one of the u i's will cover the whole of X i. Say u 1 just for definiteness sake, u 1 is a cover for X 1. Since X 1 is compact, this gives you a finite sub cover which you can call it a u 1 1 u 1 n. Okay, it follows that pi p 1 inverse of u 1 j's, you take p 1 inverse of these things will form a finite sub cover for X sub i from S prime. There are members of S prime that is how we have chosen. But this is now cover because X 1 is the union of all these things. So, p 1 inverse of all these things will be the whole inverse of cover for X i and these are members of this one. So, that completes the proof. Okay, that completes the proof of Tyknoff's theorem. There are several proofs of this important theorem. Indeed, it is a fashion with every aspiring topologist to give his own proof of Tyknoff's theorem. Never mind that only a few of them may succeed. Nevertheless, there are quite a few, quite a few proofs of this theorem. Okay, now let me give you an example here. Namely, our favorite example are with semi-interval topology, lower limit topology or sorry, general fry topology and so on. There are various. This we have seen earlier several times. This space is a Lindelow space. So, let G be a family of semi-open intervals which cover R. So, I am trying to prove how it is a Lindelow space. It is not very easy to see. It is enough to show that this admits a countable sub cover. So, I am using that this S is actually a base here, open subsets of the form a comma b, a closed and a b open. That is the definition of this topology L. Okay, take a cover by these open sets. There is no need to take unions of these things and so on. This is a base so we can take this one and then show that it is as a countable sub cover. Okay. So, what do we do? We will use the property of R in the usual topology and then compare it with this one. So, put u equal to union of open intervals a, b where this a comma b, a closed, these are members of this G. Okay. So, we started with G, a family of semi-open intervals which covers R. Now, you drop out the first point here in each interval. Okay. And take only open interval a, b. Then this union will be clear in open subset of the usual topology in R. But now the usual topology is second countable. Therefore, it follows that there exist a countables of family a and b and contain inside G such that this u is union of countable many open sets. So, what I am using is every open subset of a second countable space is second countable. Therefore, it is a Lindelof. Okay. If I just use directly R u is Lindelof, I do not know how to, this one is Lindelof because it is an open subset not a closed subset. Okay. But if you say second countable, then every subspace is second countable and second countable spaces are Lindelof. So, you get a countable sub cover for this subspace u which is an open subset. Therefore, why I have put y or u whatever it is, why u is open. Okay. Now, I put back all the a ands but only taking from this countable sub cover, put y equal to union of bracket a and n, put back these points. This may not be the whole of R. If that were R, you are fine. See, we dropped out all these starting points of the interval. The initial point of the intervals you dropped out. Now, you put back. But that may not be the whole of R. However, it is not difficult to see that whatever is left out, namely put f equal to R minus y, what are they? They are all the starting points of intervals a comma b coming from g. Some of them, some of them are already here in this countable family. Some of them are left out. That space f equal to R minus y is definitely a closed subspace. It is a discrete set. No open interval will be contained inside that one. The open parts have been taken care by this a and b n, open a and b n. So, that is easy to see that and hence it follows that f is countable. Once f is countable, you pick up open subsets which cover f for each one of them, one open subset from g. Along with that, you put all these a and b ns also, which you have got already. Together you get a countable family that will cover the whole of R. So, that shows that the semi open interval topology is Lindelof. But now I am going to show that the product is not Lindelof. See that was my idea. To show that we showed products, even infinite products and so on of compacts which is compact. And we are all the time telling that Lindelof property keeps tagging on. So, this is one phrase where it does not. Even product of two of them need not be Lindelof. So, take x equal to R s cross R s R l. I should be taken l because I have used this notation l here. So, look at that is not Lindelof. For if it were, then the anti diagonal x comma minus x, x belong to R. We have used this one earlier. The anti diagonal being a closed subspace will be also Lindelof because closed subspace of Lindelof space are Lindelof. On the other hand, given any point x comma minus x in delta hat, delta at Videl, consider the open subset you know first closed x comma x plus 1 or x plus R whatever cross minus x comma minus x plus 1, some positive number minus x plus not necessarily 1. You take these open subsets. These are open subsets in this x namely R l cross R l ok in the product pathology. Alright. What is the intersection of this one with the anti diagonal? It will be just the first point x comma minus x ok. So, intersection with the subspace delta at Videl is just the single term. That means all the single terms are open in delta at Videl in the subspace topology. It just means that delta at Videl is a discrete space. However, it is its cardinality is the cardinality of R. It is uncountable ok for each x inside R that is x comma minus x. So, an uncountable discrete space cannot be lindeloff ok. So, product is not lindeloff even product of two of them. So, this is the picture I have taken. This is the anti diagonal x comma minus x in R cross R. Open subsets of this product space are of this nature x comma minus x plus s ok. It is x comma minus x, x plus R comma minus x and then these the dot dot dot these parts are not there in the open path. But these two lines are there intersection of that one with the anti diagonal is just this point ok. So, that is a picture for showing that the anti diagonal is what is a discrete subspace. So, we have given an example of lindeloff. So, we have shown that product is compact if each coin is compact, but the lindeloff property is not even finite productive ok. Coming back to this Alexander sub base theorem let us get a little bit familiar with one of the central results in set theory which is very useful especially in topology and algebra. Of course, we are going to employ this immediately in the proof of Alexander sub base theorem ok. So, a little more point set topology here today and then we will wind up today. Tomorrow we will again proceed with Alexander sub base theorem. So, this is about partial orders and so on. So, start with a set which is partially order preferably non-empty ok do not take empty set and by a maximal element in x we mean x belong into x such that x is less than or equal to y for some y will imply x is equal to y. So, there is nothing sitting over x. So, that is the meaning of maximal element alright. Note that there may not be any maximal element inside x. Like if you take r with the usual order it has no maximal element that means it is not bounded right. So, there is no maximal element. Also there can be more than one maximal element ok. So, you can think about you know if you take some subsets of any set then put a inclusion map it is you can have bigger ones one bigger one bigger here those two are not comparable and so on there are lots of such examples alright. So, maximal elements may exist and also there may be plenty of them also either of them can. Now there is another simple definition by a chain in a partial order set we mean a subset of the form whenever x and y are inside this subset either x must be less than or equal to y or y must be less than or equal to x ok. Of course, both of them happen that is also allowed but then x will be equal to y that is by definition. In other words why is a chain if and only if under the restricted order it becomes a totally order set that is another name for it totally order set always has this property ok. A totally order subset of a partially order set will be called a chain. So, we have just introduced another word here which is popular in the in the point set you know point set theory that y be a subset of some partially order set. An element z inside x is called an upper bound for y if y belongs to y implies y is less than or equal to z. So, this is an upper bound. So, all these things are very you know very straightforward definitions. But these definitions are now made in arbitrary partially order set that is what you have to be careful not inside r or q or integers and so on. They will of course apply to r, q all those things also. So, this is a general partially order set alright. Now, here is what is called as John's lemma which is almost like one of the axioms of point set theory you know set theory as such equivalent to axioms of choice. John's lemma says that start with any non-empty partially ordered set. Suppose every chain in this x has an upper bound then x has at least one maximal element. It does not say anything about uniqueness of course there may be plenty of them. It just assures you there is a maximal element okay this is John's lemma alright. So, next time we shall use this one very effectively to prove Alexander's sub base theorem. Thank you.