 Welcome to NPTEL NOC, an introductory course on points at topology part 2, continuing with the study of filters, today module 33, ultra closed filter, filters and ultra filters on a given set, have nothing to do with a particular topology on x, though they control the behavior of all the topologies on x to a large extent, namely via the notion of convergence, right. Now we shall introduce a subclass of filters which depend on the given topology tau on x, so that is the difference between these filters and ultra filters and ultra closed filter, they have the name says that closedness comes because we are referring to a particular topology tau on x, a filter f on x is called a closed filter, if it has a base consisting of closed subsets of x, so I started the topological space, so closed subsets make sense, it is called an ultra closed filter, if it is maximal in the correction of closed filters, so that is the definition of closed filter and ultra closed filter. Now if an ultra filter is closed filter, then it is ultra closed filter because it is already maximal in the collection of all filters, on the other hand you start with an ultra closed filter, this may not be an ultra filter because it is maximal only in a subclass, there may be larger ultra filter which are not closed, so this difference you have to keep in mind that is all. Now let me have some examples, singleton capital X is always a closed filter because singleton X is always closed in whatever topology you take with respect to any topology, on the other hand every filter is a closed filter on a discrete space because every subset is closed also, so you can take the filter itself as a base if you like, no problem. So these are some easy examples of closed filters, the co-finite filter on an infinite set is not a closed filter, for if it were then it would contain a non-empty finite subset because the only non-empty finite, non-empty closed subset are the whole space or the finite subset, so if it contains X as a base then it has to be, if X is a base then X is the only element, if it is a co-finite filter X cannot be a base, so it must have finite closed subset as a base but the moment it is non-empty finite subset that is contradiction because it is complement will be also there, as a member all the open subsets are here, non-empty open subsets are here that is not possible, so conclusion is that this co-finite filter which consists of all open subsets other than the empty set in the co-finite topology is not a closed filter, so I have given you both examples, easy examples of closed filters as well as not closed filters, okay, let X be a P1 space, then I can give you more examples, for each X binomial to X consider the ultra filter FX, singleton X being a base for it and singleton X is closed, this FX is a ultra, I know it is a closed filter, already it is ultra filter therefore it must be ultra closed filter, also note that FX converges to X and X alone, why because FX contains an X, alright, so these are the important ones and easily available ultra closed filters, of course you may expect that there are and there may be many other ultra closed filters on an infinite set, okay, every closed filter is contained in an ultra closed filter, so this is similar to every filter is contained in ultra filter, of course again the proof is John's lemma you have to verify something, the verification is very easy here also, the proof is as usual appealing to John's lemma, take A to B the collection of all closed filter containing a given filter automatically it is non-empty, if lambda is a chain inside this inside this family A, each element of this chain has a closed base B I, B I is a closed base for F I, F I is a chain, okay, then take the union of all the members of all the elements in B I, so union of all B I that family will be a closed base for union of F I, this is what you have to verify, okay, because it is a chain this is impossible, okay, again union of F I is a filter is easy just like in the earlier case, so it is a closed filter, therefore it is a member of lambda, if member of this A and it is an upper bound for all members of this one, so that means condition for John's lemma is satisfied, therefore conclusion of John's lemma says that there are maximal elements, so every filter is contained in an ultra closed filter, okay, similar to the theorem 5.7.5 to 0, which is nothing but characterization of filter, ultra filters, the 3 characterization we had given, okay, we have the following characterization, it is not so strong a little weaker because we have only 2 of them here, okay, A and B, exactly similar, F is an ultra closed filter, the second one is not arbitrary subset but only for open subset, every open subset U in X either U or U complement must be inside F, if this condition is satisfied it must be ultra closed filter, start with a closed filter of course, you do not prove that it is closed filter, only ultraness is proved, okay, the proof is more or less same but you have to see why the openness is coming here, this condition is obviously weaker condition, it is not for every open subset, right, after all it is maximal is also inside the family of closed filter, so that is the trick, so let us see, let us go through the proof correctly, A implies B is not very easy, we are not very difficult, start with a closed filter, ultra closed filter, let B be a base for that, B consists of only closed subset and it is a base for F, that is the meaning, given an open subset U of X, consider the family B union complement of U, that is a closed subset, so it is a closed family, family of closed subset, right, if F union U complement has finite intersection property, in fact you do not need the whole of F, if B union F, this is finite property is enough, then B union U c also has finite intersection property, so these are actually, these are different only, and hence it would generate a closed filter F prime containing F, okay, the moment you assume this as a finite intersection property, there will be a filter F prime, that will contain this F, because it contains B, okay, now F is ultra closed, therefore F must be F prime, which just means that U complement is enough, okay, if it is not the case, means what this thing A has, it does not have finite intersection property, then what happens, that is otherwise, there exist an A inside F, such that A intersection U complement is empty, okay, that is the finite intersection property is violated, this just means that A is contained inside U, therefore A is inside F, so U is also inside F, so either U is there or U complement is there is what we have proved, okay, so B is proved now, now let us prove B implies A, okay, suppose there is a closed filter F prime, such that F is contained F prime, we want to show that F is equal to F prime, let B prime be a family of closed sets, forming a base for F prime, okay, there is one, because F prime is a closed filter, for any B inside B prime, suppose this B is not in F, okay, I am just supposing this is not in F, then the complement of B must be in F by condition B, but F is contained F prime, so B complement is in F prime, B and B complement both are in F prime, that is a contradiction, because intersection will be in F prime, therefore this B must be inside F, since this is true for all B, this B prime, the curly B prime must be inside F, okay, but B prime generates F prime, so F prime is inside F, so therefore equality holds, okay, so the proof is more or less the same, but you know the flavor is different because you have to use the openness and closeness and so on here, that is all, but condition C is missing here, okay, see now you understand why last time we put this condition at the last, okay, and then proved A implies C, C implies B and B implies A, this condition is nothing but given A union B inside F, A is there or B is there, in the case of ultra filters, now similar condition but slightly different can be expected, but no, this only one way, one way implication is there, okay, the corresponding condition C is not equivalent to being ultra close filter, so let me state it separately, do not get confused with the earlier theorem, this proposition says suppose F is an ultra close filter, then given U and V, two open subsets of F, this condition holds, what is this condition, what is this statement, this statement is if and only statement, U Union V is inside F, if and only if U is inside F or V is inside F, actually one way is obvious because F is a filter, okay, if U is there or V is there, U Union V being a superset is there, so what we have to show is that this larger set is there, why the smaller, one of the smaller one is there, okay, under the hypothesis that F is a close filter, ultra close filter, okay, so one way is obvious, this implication is F is an ultra close filter, U and V are open subset such that the Union is inside F, we have to show one of them is inside F, this is not true means what, let us examine that, suppose this is not true, then it follows that the complements U C and V C must be inside F, this is not true means what neither U is there nor V is there, but then the complements C must be there because U is an ultra close filter and we can use the previous theorem, okay, therefore the Union, U Union V complement which is nothing but U complement, intersection V complement that must be there, but now you have a problem, we already assumed U Union V is inside F, but you are assuming that intersection, U intersection V intersection that is also inside F, this is empty, okay, it is complement and this one both of them cannot be there, that is all, by a simple induction we get the following important result, if X is written as a Union of finitely many open subsets V i and F is an ultra close filter on X, then one of the V i must be in F, okay, so this is very easy because what we have proved is that if U and V are open subsets and U Union V is inside F, then one of the U and V must be inside F, right, to begin with every filter contains the entire set X, therefore you can just write X as I raise to 1 to n minus 1 V i Union V n, apply this criteria, either V n is there, then we are done or I raise to 1 to n minus 1 V i must be there inside F, now you apply induction, okay, so one of the V i must be inside F, alright, so from this one let us reduce another important result here which is characterization of compact spaces in terms of ultra close filters, remember in earlier theorem we had characterized the compact spaces in terms of ultra filters, every ultra filter is convergent, that is the condition which ensures that X is compact and conversely, so exactly same result is known, space X is compact if and only if every ultra close filter in it is convergent, okay, so what is the idea, let X be a compact space and F be an ultra close filter in it, suppose F is not convergent to any point, then you will get a contradiction, very easily, namely it does not convert to any point means none of this nx, the neighborhood is contained inside F, that is the meaning of that X does not convert to F, this entire neighborhood contains at F is the same thing as X converges to F, F converges to F, what does this mean, for each point X inside X we have a neighborhood Vx of X such that this Vx is not in F, one neighborhood for each point, but when you vary this point you get a covering for X, but now X is compact, so you get a finite covering, okay, as soon as you have this one, this corollary says that one of the VI must be inside F, but that is a contradiction just now none of the VI Vx is inside F, okay, so compactness implies that every ultra close filter converges to some point, okay, it may convert to more than one point also, nobody ensures the uniqueness, now let us put the converse, assume that every ultra close filter on X is convergent, let C be a family of closed subsets of X with finite intersection property, okay, we have to show that the entire intersection of members of C is non-empty, so that is enough to show that X is compact, right, so let F be the closed filter generated by C, because any family which has finite intersection property generates a filter, but that will close filter because this will be members of all this here they are all closed subsets, okay, so we get a closed filter contained in a ultra filter, ultra close filter, so let F prime be an ultra close filter contained by John Slema, now X be a limit point of this F prime, because we are assuming that every ultra close filter is convergent, so let X be a limit point of limit of F prime, this means that an X is contained inside F prime, if X is not in C for some C inside C, that would have meant that complement of C is a neighborhood, that complement is a neighborhood, the neighborhood is contained in F prime, that means complement is in F prime, okay, but all the members of C are inside F and hence inside F prime, therefore X must be inside C, for every C inside C, that just means that the intersection has the common point X, so that is non-empty, okay, so characterization of a compact space in terms of ultra close filter comes out, see we have to prove this F fresh both ways because if every ultra close filter is convergent, this does not mean immediately that every close filter, every filter, every ultra filter converges, right, because there are many more of them, on the other hand if compact space shows that every ultra filter is convergent, but none of the ultra close filters may be or may be a ultra filter or some of them may be some of them are not, right, ultra close does not mean that it is ultra filter, so either way it is not true, but so we have to prove it fresh, but on the other hand if you look at the proof, it is not all that different, of course this is little harder, we have to use the comp, we have to use the closeness and generate close substrates and so on, okay, a non-convergent ultra filter is supposed to indicate the presence of a hole in the space which is making the space non-compact, okay, because if it were compact then every ultra close filter would have been convergent, okay, so this ultra close filter there is no space for it to be convergent because that point is missing, that is the idea of that there is a missing point in the space, so and that must be making it non-compact, so this idea is similar to the presence of non-convergent Cauchy sequences in an incomplete metric space, by the very fact we know that in incomplete means a metric space incomplete means there is a Cauchy sequence which is not convergent, so not convergent Cauchy sequence indicate there are holes inside a metric space, following a similar track as in the case of you know completion of a metric space, we may try to fill up these gaps by including all ultra filters on X, okay, so that could be one way of looking at it, but then you are warned already by the above theorem that you do not need all ultra filters, if all ultra close filters are convergent then the space will be compact, therefore it should be possible to get a compactification by just taking care of all the ultra close filters, so they should be enough to fill up all the holes, so these are all just surmising, I mean we are just you know loud thinking may say trying to do oh this may be true then finally you have to prove all this, let us concentrate our attention on the set WX of all ultra close filters on X, okay let us not give up this idea that is all we have to work of course, so WX is nothing but you know set of all ultra close filters on X, okay, next going back to the analogy again of completion of a metric space recall that eventually recall that eventually constant sequences correspond to the points in the space itself, right, if you have some a18 an and then all an plus 1 etc are equal to X, X, X, X, X that sequence automatically converges to X and then what we do in the completion the element X was identified with this Pochise sequence or the other way around whichever we will do, so that was the map from the space X to the collection of all sequences and then of course we introduce there some equivalence relation and so on, right, so that was the idea, so we can try to do that namely instead of sequences first of all you have to go to next eventually constant nets is the key word now here, on the other hand we do not want your nets but we will use the bridge from nets to the filter which you have defined earlier, right, right, so what do you get if S is an eventually constant net the corresponding FS what is that that will be the singleton atomic filter FX, okay, so you see we are coming to this atomic filter FX it is supposed to represent you know a prototype of eventually constant sequence that is the whole idea, oh that is how it is convergent also all the time, okay, in our definition it is also an ultra close filter provided X is T1 till then we do not need that, okay, all FX are closed filter if it only if X is a T1 space that brings us to make a blanket assumption that our space X is a T1 space because we just do not want to give up this WX, right, ultra close filter, so we better assume X is a T1 space before we go further, okay, in which case all happens if this subset of all atomic filter will just correspond to the points of X, okay, so this way WX can be thought of as a set which contains X by this embedding by this injective mapping by this identification X going to FX, so we have enlarged our space what remains we must put some topology on this one so that it becomes compact that is the first thing then we must examine that the map X going to FX is injective that is okay continuous and an embedding so that the image is dense this all we have to, okay, right, it is also clear that each FX converges exactly to one single point that is very important for us, right, that is an extra bonus though we are we were not bothered about that, okay, but that is because of some strange reason there is no T1-ness or T2-ness involved we do not want to bring the T2-ness here usually a filter will convert to a unique point if you have a host of space, okay, that is not the reason these are very special filters FX they will convert to only one single point we cannot expect such uniqueness behavior by other ultra filters unless we are ready to make further restriction that X is host of, right, so therefore it seems that we may need to introduce some kind of equivalence relation on the set WS, okay, so that the equivalence classes may be better qualified to become a compactification just like in the case of completion of metric space, okay, luckily it turns out that we do not have to worry on this point this process of compactification is much simpler than the construction of a completion of a metric space that is going to be the topic of discussion for us now which we shall carry out in the next module, okay, so this is the motivation for considering the so-called Wallmann Compatification as we will do next time, thank you.