 Hello, welcome to NPTEL NOC, an introductory course on points at topology part O. So, today we will study Wallmann Compactification, module 34. As motivated last time, we shall make a blanket assumption that the space x tau is a T 1 space. As before, let W x denote the collection of all ultra closed filters on x. We also have introduced this map function phi from x to W x. Namely, taking any point x, take the atomic filter f x, which we know that is ultra closed filter because x is a T 1 space. We propose to topologize W x so that the pair phi comma W x becomes a compactification of x. Recall, the compactification is actually an equivalence class, but we keep on saying that you pick up any representative that is also compactification. And the compactification satisfies that the map phi from x to whatever space you are taking is an embedding such that the image of that embedding is dense in the whole space. The whole space W x is compact. So, these are the condition I am just recalling that is all. So, going towards topologizing W x, let us introduce some notation here. For each subset A of x, define A plus to be all ultra closed filters in x which contain A. A must be an element of that. A is an element of this ultra closed filter. So, this collection of all such ultra closed filters will be denoted by A plus. The following lemma on this notation is very simple, but very useful also for us. A is in F if and only if f is in A plus. So, this is a definition of A plus. Empty set plus is empty because no filter contains empty set. X plus is the whole space because all ultra closed filters in particular all filters contain X. So, these are straightforward. A intersection B plus is A intersection A plus intersection B plus. So, this needs little bit seen how take a ultra closed filter which contains the intersection. Then just by being a filter it will contain both A as well as B. Therefore, that closed ultra closed filter belongs to A plus as well as B plus. Conversely, if a closed ultra closed filter contains A as well as B, then the intersection again being a filter, the intersection will be also inside that filter. Therefore, that filter belongs to A intersection B plus. So, this also there is no ultra closed filter involved here, but here when you come to open sets we have that one. Namely, if u and v are opens of sets of x, then u union v plus is u plus union v plus. So, once again if f contains this element u union v, then we know that one of the u must be inside f which is same thing as saying f belongs to either u plus or v plus. The converse part is v v once f contains u, it will contain u union v which is a super set. Similarly, if f contains v it is super set u plus v v there. So, this way is easier. This way needs that the criterion that we have for ultra closed filters. So, this is what I have written down here. So, let us go through one and two are obvious just set here dicting definition. The third one is what I have seen I will repeat it f belongs to A intersection B plus. By one this means that A intersection B is in f that is definition anyway. Since f is a filter this implies A belong A and B are inside f because they are super sets and f is inside A plus intersection B plus. So, these steps are completely reversible because if A and B are in f intersection is there that is the word. So, the fourth statement as I told you it is a consequence of that proposition which character which gives you that u and v are open subset and u union v is there then either u plus is there or I am not either one of them u or v must be there. So, this was proposition 7.64 which was proved separately for ultra closed filters. So, property this especially the 3 property 3 it just means that the family of all this A plus is where A is an open subset of the topology at topology on x. So, this property as you see it is closed under finite intersection. So, that is the important property for us. So, look at B hat such that all u plus where u is at tau now I am not taking arbitrary subsets I am taking only open subsets look at this collection. This collection has properties closed under finite intersection. Therefore, it forms a base for a unique topology tau hat on w x namely all that I have to do is take arbitrary union of members of B hat that will be topology for x topology for tau hat namely topology tau hat namely on w x. So, there are so many ways of given topology we have to just justify that this topology is the right one right. So, that is the task now the set w x with this topology tau hat generated by the base B hat is a compact T 1 space. We are not so much worried about T 1 space, but compactness is the first thing we need. Moreover the function phi from x to w x which is saturated injection already defines an embedding of v x inside w x and the image v x being dense with phi x contained inside w x being dense. So, that will complete the statement that phi comma w x is a compactification of x actually this term says the T 1 compactification. So, the first thing is to see that w x is compact to see that at w x is compact for any space compact it is enough to prove that an open cover from a base you fix one base for the topology take any open cover from that base members of that base. If that admits a finite sub cover then that should be compact this is a standard result that we are going to use now. So, we have fixed this base we will take that base itself B hat take members of B hat which cover w x and let us see that it admits a finite sub cover u i I will try to be a family of open subsets to such that w x is union of u i plus in case I am making a sub case here in case there is a finite subset j of i such that x itself is contained inside the corresponding u j's. Then it follows that by property 4 this is a finite collection so w x is equal to union of u j plus because x plus will be equal to union of u j plus we have proved for 2 but that can be immediately generalized to n any finite number that is what you take x plus x plus is w x so that will be equal to union of u j plus now this is a finite cover. So, we are done the second case is suppose this is not the case what is the meaning of that that there is no finite family out of the u j's that u i's that we have chosen which will cover x remember the u i plus is are covering w x. Now I am talking about u i's being covering x why should happen there is no need we are not assuming x is compact after all right so suppose there is no such finite sub cover so that case remains so that is the case we are going to address in a standard way you know this kind of argument is used several times in the college especially when dealing with compact space so we may assume that no finite sub family of u i cover x by De Morgan law this just means that the complement of this family u i's right members of this family is a family of closed sets with finite intersection property no finite intersection of this complement can be empty because the corresponding members would have covered the x right. So, it has finite intersection property therefore there exist an ultra closed filter g on it which contains the closed filter generated by this family as a base any family of member which have finite intersection property will generate a filter that is automatically closed filter because these are closed subsets every closed filter is contained in an ultra closed filter and that is what I am denoting by g. Let us say that this g after all is a member of w x so it must be inside u i plus for some i inside i because u i plus is a cover for w x what does this mean by the very definition of u i plus this means this u i is inside g remember g is a super you know it is a larger filter which contains the filter generated by these elements therefore all the u i plus is inside g in particular whatever u i plus comes from here that is also there but u i and this u i plus and its complement cannot be both be inside g. So, g belongs to u i plus means that u i is inside g but g contains u i complement so that is not possible because no filter will contain both set and its complement. So, compactness of w x is proved let us now prove that w x is a T 1 space that comes very easily right now if it is not a T 1 space what does that mean there exists a pair of points inside the space now they are now ultra closed filter distinct points means f 1 is not equal to f 2 such that every open set u in x because now I am looking at basic open subsets around f 1 ok they will be like open subsets in u and then u plus f 1 will be inside u plus but f 2 is also inside u plus there will be such a f 1 f 2 whenever you take a neighborhood of f 1 it will be a neighborhood of f 2 also so there will be such a pair it may happen starting with f 1 f 2 it may not happen just because it is a T 1 not a T 1 space somewhere it is happening that is the whole idea somewhere it is happening ok. So, f 1 is in u plus implies f 2 is in u plus ok now these are b plus elements of b plus so b hat right I have not b hat so do not get confused now you take b as a base for f 2 consisting of closed subsets of x because by definition every ultra closed filter must have a base consisting of closed subsets ok. From earlier theorem we get if we take a member of this base see then the complement which is an open subset that cannot be in f 2 because the member is in f 2 f 2 is not in u plus is the same thing as saying that u is not in f 2 right if f 2 is not in u plus by this choice you know f 1 cannot be in u plus right f 1 is not in u plus it just means that u is not in f 1 if u is not in f 1 this is where I have to use this criteria given any open set u or u complement must be inside the closed ultra closed filter. So, u belongs to f u is not in f 1 implies u complement which is c inside f 1. So, what we have proved here we have proved that this base which is for f 2 is completely contained inside f 1 every member is here therefore the filter generated by b will have to be contained inside f 1 because f 1 is a filter ok we started with two ultra closed filters right. So, one cannot be contained inside the other ok or they are equal. So, whichever way you are contradicting because you have started with two different ultra closed distinct ultra closed filters ok. So, if you are just watch out this one ok if you are just doing these things with just ultra filter alright could you have done this one can you get this you keep examining these kind of things. So, where is the really the role of ultra closed filter is coming up ok to show the continuity of f is the next task and then there will be one more namely to show that f is a closed map or an open map that will complete the embedding ok finally we have to show that it is dense also. So, to show that f is continuous it is enough to show that inner image of basic open sets is open ok. So, if you take a basic open set will look like u plus where u is an open set of phi ok and u is a open subset sorry u is a open subset of x right what is phi inverse of x phi what is phi inverse of u plus remember what is phi phi is x going to fx where fx is the atomic filter. So, start with an atomic filter it will come from x because there is no mapping right. So, phi inverse of u plus will be just a point there and that point must be inside u right because u belongs to fx if and only if x belongs to u. So, these are just you know phi inverse of u plus is just the original set u alright. So, in the u is 1 so continuity of u not only that you observe this just means that u plus intersection phi x is precisely phi of u because inside phi x we only have the atomic filter. The atomic filter containing the whole of u just means that x is inside u. So, it is phi u phi u is equal to u plus intersection phi x is stronger than just saying this one. But because this is just an embedding these are all ok. Now this I can use this one again it remains to show that phi x is dense. But once again it is enough to show that phi x means every member of b hat see phi x is dense means it should meet every non empty open set. But inside a non empty open set there will be some member of b hat right base. So, if it means basic open set that is enough right this follows from this equation take any basic open set phi of x intersection u plus is phi of u so it is non empty. So, this is so this just equation proves density continuity as well as openness. Take an open set any open set of phi image of phi is u plus which in open set intersection phi x. So, it is inside phi x if it is open phi u is open by definition phi x the image of phi x takes the topology or the induced topology everything is open if it is intersection of an open subset inside the larger space. So, this proves the openness it proves the continuity as well as density. So, those things are very easy to do. So, let us complete now the definition of Walman compactification as I told you earlier a compactification is a certain equivalence class right. But here we are we are usually taking one particular representative all the time ok anything which is equal to this one will be also called Walman compactification in general. So, here we have first of all what we have starting with any T 1 space we constructed this W x which is just the collection of all ultra closed filters on x with a very specific topology with the base consisting of u plus where u is open subset of x ok. So, that compact space along with the embedding of x inside that this is called Walman compactification of the T 1 space you can remember that W x is automatically T 1. So, here are a few remarks the T 1 s of x is only used in getting the function phi from x to W x ok why because I start with x and then I construct f x. But why should f x be an ultra closed filter it is an ultra filter why it is closed that is where I have used that x is T 1. So, singleton x is closed. So, singleton x is the base for the whole of f x any atomic filter it is a base. So, that is one way immediately to get the function itself we have used automatically it is injective ok there is no problem whereas the topology on W x is always T 1 the T 1 s of x is not used in this one ok. One does not know when this topology on W x is Hausdorff why I am making this comment usually one would like to know because we are in some sense obsessed with Hausdorff's. So, Hausdorff under some suitable nice condition if it is freely T 1 by putting little extra condition may be you will get this Hausdorff right. But here I am saying that one does not know when this topology on W x is Hausdorff. So, next time we shall continue to study the universal property of Wallmann compactification remember the universal property of stone check compactification and in that is a universal property of even some kind of universal property of one point compactification also this will be somewhat similar to that of course, in each case it will be slightly different ok. Thank you.