 Hello, welcome to NPTEL NOC, an introductory course on point set topology, today we will continue with our study of filters, especially ultra filters today, module 32, as a easy consequence today we are going to derive another proof of Tyknoff's theorem. Start with any set by an ultra filter on X, we mean a filter which is not contained in any other filter. In other words, in the family of all filters partially ordered by the standard inclusion, ultra filters are maximal elements. All atomic filters effects, experiments are ultra filters, however there can be other ultra filters also, you never know. Remember this effects means all subsets of X which contain the little x. If you have one more element there, that means that member does not contain little x, then singleton x and that member will be, intersection will be empty, that is not allowed. Therefore, FX is maximal, that is the idea. So, these atomic filters do play an important role, every filter is contained in an ultra filter. So, this kind of result we are now very familiar using John's Lemma. For what we do, start with a filter, look at the family A of all filters on X which contain the given filter F. This family is partially ordered as usual by inclusion. If lambda is a chain in A, then the standard arguments will give you that union of all members of this chain is again a filter which will contain all of these things. Therefore, that will mean upper bound for this family. Now, you can apply John's Lemma, there must be maximal elements. As everywhere else, the existence of ultra filters is assured by John's Lemma, but John's Lemma is very strange thing. Even though nobody may be able to explicitly display one such, we are guaranteed that they exist. For example, this is not a joke. For example, consider an infinite set and the co-finite filter that you have considered, namely collection of all subsets such that their complement is finite except the empty set. Of course, empty set will not be there anyway. So, that is a filter. Clearly, it is not an ultra filter. It is not maximum, you can put extra elements there. Of course, you have to be careful, but apparently nobody knows an explicit example of an ultra filter containing it. John's Lemma says there is one. In fact, there may be many, but if you want to know, you want to construct one. Let X be any set, F be a filter on it. Then the following statement A, B, C are all equivalent. So, we have two different criteria other than the definitions which are equivalent to F. So, in all, we will have three definitions of ultra filter. So, F is an ultra filter. The statement B is for every A inside X, either A or A complement belongs to F. Both of them cannot be there. So, that we know. So, this either A or A complement belongs to F. So, this is quite a strong condition. The third one is a weird one, but this is quite helpful also. A and B are in X. The union is in F implies A or B is inside F. Remember, by the very definition of filter, if A is there or B is there, union will be there already, because any superset will be there. But now some superset is there which has written as union of two subsets. Then one of them must be inside F. So, this is the condition for an ultra filter. Let us prove this one for some, you know, pedagogical reason which we will come to know later. I have put this A, B, C in this order. Maybe I should have put A first and C second and B, B third. But in my mind, this is easier to understand after the definition of ultra filter. Ultra filter itself is easy to understand. Namely, it is a maximal element. It is one of the maximal elements. So, that is easy to remember. And then this is simpler and easy to remember also. But while proving, I will prove it in a different way. That is economical. A implies C, C implies B and then B implies A. That is what we will do. So, first assume F is an ultra filter and union of two subsets is inside F. First we claim that F union A or F union B has finite intersection property. If this is not the case, then we get two subsets B1 and B2 inside F, subsets of X, but they are members of F, such that A1 intersection A is empty. Similarly, B1 intersection B is empty because each of these family has violated the finite intersection property. If you remove B, it has finite intersection property. If you remove A, it has finite intersection property. So, only because you have put A here and B here, it might have violated the finite intersection property, which just means there are members A1 and B1, which when intersected with A and B respectively are empty. But then look at A union B, which is already member of F. That is the hypothesis here. Intersect with A1 intersection B1. A1 and B1 are both elements of F. Therefore, intersection is also an element of F. So, look at this intersection. This is one element. These two are two other elements. Take the intersection. That is also another element. So, this finite intersection, this is nothing but A union B intersection A1. A intersection A1 is empty. So, this is just B intersection A1. Then intersection is B1 again. B intersection B1 is empty. So, intersection of this finite intersection, intersection of this is empty. That is a contradiction because A union B and intersection B are already in F. So, it follows that these two have finite intersection property. Any family with finite intersection property will generate a filter. So, both of them will generate a filter. A intersection B or one of them or the one generated by F intersection B containing the other filter. Both of them will contain F. Since F is maximal, this proves that either A must be inside F or B must be inside F. Remember, we said this or this one has finite intersection property and then assume that none of them have. That is what we have got this time. So, since that contradiction, we get one of them has finite intersection property. Generate a filter which is larger than F, but there is no filter which is larger than F because F is ultra filter. So, either A or B, at least one of them is inside. Here it is not either A. Both of them may be also there. I do not care. So, statement only says that the A or B is inside F. There is no either A or F. C implies B is one line proof because what I can do? I can take this A complement as B. Then A union A complement is the whole space X. It is always there. That is free. Therefore, I can apply C to conclude that either A or A or A C belongs to F. Of course, I will know that both of them cannot be. So, we can put either A or B. So, C implies B is very cheap. That is why I have proved A implies C first. Now, let us prove B implies A. Suppose F is not maximal. Suppose, according to this definition, F must be maximal filter. That means what? There is a filter F prime containing it properly. So, take an A inside F, F prime minus F and member which is in F prime but not in F. Now, C implies A C, the complement must be inside F because A is not inside F. Therefore, A C is inside F prime also because F prime is larger than F. Once A C is inside F, A C A C is A complement is inside F prime but A and both A prime are inside F prime. That is the contradiction. So, F prime must be equal to F. Therefore, we have proved A implies C implies B implies A. As a corollary, we prove something very important, very useful also. Take any ultra filter and take a function from F from X to Y, any function. Then F check is an ultra filter on Y. So, ultra filters behave very nicely. No condition on F. F check, remember F check of F is a unique filter which contains F check of A for all A inside F. How is that obtained? It is generated by those things. In fact, all that you have to do is take all supersets of F, F of A where A is inside this curly F. That is the definition of F. So, let us see how this comes. All that I have to do is I will use the criterion C or in fact, I can use criterion A here. What each one is I have put here? Actually, I will use criterion B here. See, I always think of this as C but now I would like to think of the second one A or A complement. So, let B contained inside Y, we have to show that either B or Y minus B is inside F check of F. Take A equal to F inverse of B that is subset of X. Since F is an ultra filter, A or X minus A is inside F. If A is inside F, F of A is inside FF therefore and also we know that F of A is contained inside F because A is F inverse of B. Therefore, B is inside F check of F. The other case is if X minus A is inside F, then F of X minus A is inside F of F by definition. It is X minus A is a member here. F of X minus A is in its family. Also F of X minus A is contained inside Y minus B. Therefore, B is a superset. Y minus B will be inside F check of F. So, we have shown that B or complement of B is inside F check of F for arbitrary subset B of Y which means F check of F is an ultra filter. Now, X be a topological space F be an ultra filter on X. Then every cluster point of F is a limit point of F. Remember, for a filter every cluster point was a limit point of a sub filter. What is the meaning of sub filter? Sub filter is a filter which contains that but F is ultra filter so there is no larger filter than that one. Therefore, it must be F itself. This F prime which is which has X as a limit must be F itself. I repeat if X is a cluster point of F, earlier you have seen that there is a sub filter F prime. Sub filter means what containing F. For which X is a limit but F is ultra filter so there is equality here over. A topological space X is compared if filled only if every ultra filter in X is convergent. So, we have arrived at a very good theorem combining various earlier remarks. So, let me go through all these earlier remarks. 4.40, 7.40 there is a remark this remark has actually three parts there. Theorem 7.50 and above theorem and theorem 7.46 which characterizes the compact spaces in terms of filters. If you do not remember maybe I should start just recalling this one. So, first let us look at this remark. It says that if F is containing F prime then F converges to X implies F prime also converges to X. Conversely if F prime converges to X then extra cluster point of F. So, this is what just we have used we will keep using this one again. So, next we have this one every a filter is contained in an ultra filter. So, that also we will be using. Next we have this criterion for compact spaces that every filter has a cluster point or every filter has a convergent sub filter. And finally we have just now we have proved this theorem namely every ultra filter then a cluster point is a limit point. So, if you combine these things what you have is start with a compact space and F prime ultra filter on it. As a filter F has a cluster point. But just now we have proved that F prime ultra filter it has this cluster point must be a limit. So, one way is over. Converse it is likely more complicated. Suppose every ultra filter on X is convergent starting with any filter we can put it inside an ultra filter G right just now I quoted that theorem. Now a limit of G because ultra filters have limits limit of G is a cluster point of F by remark 7.40 which you have just seen. Therefore again you apply this theorem 7.46 every filter has a cluster point. So, that is that condition is equivalent to X is convergent. So, what we have a need theorem now a compact swag if and only if every ultra filter is connor in giant. Later on we have to improve upon this one. One more improvement is there in a special case I will come to that one I am just you know preparing your mind for such an improvement. Now we will prove Tickenhoff's theorem in a very easy way a very canonical way. Recall what is Tickenhoff's theorem if you have a family of compacts compact spaces then the product is compact and conversely. Converse part is usually it just put for completeness of statement result that is not part of Tickenhoff's theorem actually because that can be proved much easily everybody knew that namely image of any continuous function under a continuous function image of a compact set compact space is compact. The projection maps are continuous so each X j will be compact if the product is compact that is not what we have to prove what our aim is if X j's are compact X must be compact. So, what we shall do is every filter on this product is convergent. For each j look at pi j check of f just now we prove that pi j check of f is an ultra filter on X j. But X j is compact so it converges to some point X little little X j inside X j. Now it is a matter of just easily verify using the product topology that this point X whose jth coordinate is X j is a limit point of f. See what you have to show a limit point the neighborhood of this one is contained inside f that is the meaning of a filter converges to X right converges to that point. So, maybe I will leave this as an exercise to you ok I mean there is no much I am not much exercise but you have to write down the detail that is all unlike the case of topologies it is easy to determine all filters and ultra filters on a given finite set. You might have known that the study of finite topologies is indeed not a part of you know point set topology at all as such it is maybe number theory, combinatories all sort of things are there it is quite a different flavor different kind of things and as far as topologies are considered to some extent it provides some concrete example that is all beyond that topologies do not use it much. But the problems are there and it is easy to find problems but difficult to find solutions here of course. So, here similar questions if you ask for filter and then you can combine them also so I am going to tell you one single thing namely counting the number of filters and ultra filters on a finite set that is very very easy suppose f is a filter on x x is a finite set then you can take intersection of all its members that must be a member of f right of course this must be non-empty because it is a finite intersection and it follows that that member b is contained inside every other member it just means that f is fb the atomic filter once b is there all supersets are there everything is there already so f must be fb atomic filter ok. So, every filter is like that means what there is a one one correspondence between subsets of f subsets of x and filters on x except b equal to empty. So, throw away the empty set you exactly have 2 power n minus 1 element the number of filters on it. Similarly let us look at when is this filter an ultra filter that is also easy the moment rather more than one point it will not be ultra filter because I can take a smaller one namely singleton b here for example that f of fx will be contained inside fx will contain this one. So, there are larger filters right therefore the only ultra filters are fb where b is singleton. So, how many singletons are there precisely n elements over ok. So, that is just for your flavor, but I am I just want to I am myself not an expert in this kind of thing I want to ensure that there you can find many interesting questions here, but answer should not be easy. On the other hand if x is infinite then the answer is much more difficult anyway we do not know all the filters nor or ultra filters on a given set yes take infinite in particular take a countable infinite set natural number even there we do not know. As observed earlier we do not know an explicit example of an ultra filter which contains the co-finite filter on the set of natural number or any countable set. So, one more remark here suppose a filter contains a subset b then obviously the atomic filter is contained inside f because f contains all supersets of b suppose further that no proper subset of b is contained inside f see f may have even smaller subsets right. Then what happens given any a inside f b is there a is there a intersection b is there but a intersection b is not a proper subset of b just means that b is already contained inside therefore f must be f b alright as a special case assume that f contains a finite subset then clearly we can choose b to b such that cardinality of b is the smallest among the all finite subsets contained inside f it follows that once you choose that smallest cardinality just cardinality smallest it follows that f must be f b so you cannot have several b such that all of them have same cardinality that is also clear anyway. So, in particular we have proved that every ultra filter which contains a finite subset must be an atomic filter ok. So, this is something I would like to say some trying to understand what are ultra filters and so on on an arbitrary space on arbitrary set ok. So, the darkness is still there once you do not have this hypothesis namely a ultra filter which may not contain any finite subset then you will not know how to characterize them ok. So, let us stop here next time we will study ultra close filters.