 Welcome to module 31 of NPTEL point set topology part 2 course. So, having studied the some general properties of filters, today we shall study convergence properties of filters. For that, we start with a topological space not just a set. Remember that filter can be talked on any set whereas, now we have a topology x comma tau be a topological space and f be a filter on x. A point of x is called a limit of f or you can say it is a limit point no problem limit of f that is with respect to the topology tau. So, limit will not be spoken unless you have it above that is important. If the neighborhood system the entire neighborhood system n x the neighborhood set of all neighborhoods at the point x that itself is contained in the filter. In this case we also say that f converges to x. Similarly, we define x to be a cluster point of f if every member of n x intersects every member of f. So, here you can see that this is a weaker condition than that one because if n x is contained inside f then this properties also prove because f itself has that property finite intersection of finite limit numbers is nonemptive. So, that is why so that is a weaker condition this every member intersects every member of f. Indeed all these are copied from what happens for the case of when you take a sequence take the corresponding limit of a sequence or a cluster point of sequence and look at what is happening in the associated filter. So, that is the motivation for these definitions. Some observations which are immediate from the definition are the following. If you have one filter f contained in another filter f prime then f converges to x implies f prime also converges to x. The larger filter so called larger in this sense also converges to x. So, all the points can limit limits of f are limits of f prime also. Conversely, if f prime converges to x then you cannot say that f converges to x, but f will be x will be a cluster point of f. So, that is a weaker corner system. For this reason the bigger filter f prime is called the sub filter. So, this is some terminology which you have to a strange terminology which you have to live with because this is borrowed from what happens to sequence and subsequence. If you have a sequence which converges then every subsequent will also converge to the same point. If a subsequence converges then the point of convergence of subsequent sequence will be only a cluster point of the original sequence. So, that is what is happening here. Secondly, every limit point is a cluster point that I have already told you because cluster point is a weaker condition. But the converse is not true just like in the case of sequences. So, we shall not bother to discuss it any further. Limit point is a cluster point, cluster point need not be limit point. The third observation is let x be a limit of a filter. If a is any member of f then x will be in the closure of that member x belongs to a bar. So, this is very easy to see, but I have added explanation here. If not what happens? If x is not in a bar will mean x is inside a bar complement, a bar is closed so a bar complement is open that means that a bar complement is in nx but nx is inside f. So, a bar complement is in f but a and a bar complement, a complement bar, a bar complement either way they are disjoint. So, therefore, that will be a contradiction to the property finiteness property of f. So, that cannot happen. Therefore, if something is a limit then it is easy to check that for each member it must be in the closure of that member for each member of the filter. There is one more important remark. In fact, it has three parts here and we will use it. So, pay attention to this one. Now, I am looking at a net which is a generalization of a sequence. Start with a s from d to x that be a net in a topological space s, x is topological space s and let us take fs to be the associated filter. Then the first thing you can check is that x belongs to x is a limit point of f. The task take any point that is the limit point of this s in the sense of when the net is a for a net. If we do not leave x is a limit point of the corresponding filter. The second thing is a cluster point of f similarly, if we do not leave that point is a cluster point of fs. So, this is precisely what I have already told you that both the definition of limit point and cluster point have been copied from what happens to the associated filter. The third one is very important here. The association s to fs has the naturality property. For any function f from x to y, we have f check of fs is equal to f of f you know curly f of f composite s. You take f composite s that is another filter another net. Take the associated filter that associated filter is nothing but f check of fs. So, 1 and 2 are easy to verify. The third one is actually a general property has nothing to do with the net. So, I should have done it earlier. In any case, let us check that this is also true. So, I will take some time to check that f check of fs is equal to f of f composite s. So, here is the proof. Start with any subset b of y function f is from x to y. So, things are happening inside y. So, take a subset of y that subset b is in f check of fs if and only if this is by definition of what is f check of fs. If and only if there exists a belonging to fs such that f of a is contained inside a b. So, that is the definition how to generate f check of fs. It is contains all the supersets of all the images of fs under the map f. That is the definition. So, b is inside f check of fs if and only if there exist a such that which is inside fs such that f of a is inside b. Now, this is same thing as there exist d belonging to d such that sd the what you call the right ray you can say sd is contained inside a. So, this part is same thing as a belonging to fs. Second part is same thing.fa is contained inside b. So, but this is same thing as there exist a d inside this such that f of sd is contained inside b f of sd is f of a right contained f of a f of a is contained inside b so directly back is in the f of sd is containing a b. So, that is same thing as saying there exist a this such that f composite s of d which is nothing but f of i which it is containing that bit. But now this is a condition for f composite s. So, this just means that V is inside the SOC active filter of F composite S. So, that is all. So, it is purely point set topological point set thing here that is no topology in this one, but this will be useful. So, I so this is important and this this the map S to FS is going to be very important. It is a one way bridge from Nets to filter. The filters have much more generality you cannot always come back to the Nets here. So, now let us come back to study of limits and cluster points. A point in X is a cluster point of a filter F. If it only if there is a sub filter F prime containing F, I will read F prime but I will read it as sub filter F prime containing F of which it is a limit. It is a limit for every sub filter this may not happen, but if it is a limit this will be sorry it happens that for every sub filter also it is a cluster point. So, now we are putting a I am actually giving you a converse of that one also. So, if part we have already seen in the earlier remark to see the only if part suppose X is a cluster point of F. This just means that NX the neighborhood of system union F this has finite intersection property. See within NX finite intersection property between neighbor systems within F it has finite intersection any member here will intersect any member here the condition for being a cluster point. So, NX union F will also have finite intersection property. Hence we can take F prime to be the filter generated by this family. Any family with finite intersection property will generate a filter namely you now take all the sets which are supersets of some members of that that will be our F prime. I have to produce one F prime there exists some filter there may be many more no problem. The next theorem is about function take two topological spaces and a map a from X to Y a function actually sorry just a function from X to Y. Let X belong to X be any point then the following conditions are equivalent. So, I told you it is a continuity of functions now giving characterization in terms of filters. So, start with any function take a point in the domain F is continuous at that point in the first condition. Second condition is for every filter F on X which converges to X F check of F converges to FX. So, this is a statement. The third statement is for every net S from D to X S converges to X implies F composite S converges to FX. So, these are three statements this statement I would have proved when doing NETS and so on. But I deliberately postponed it because these things can be done much easier once you go along with the filter source. So, the three conditions are equivalent is the statement of the theorem. So, we will as usual we will 1 implies 2 implies 3 implies 1 that is what our plan. Assume that F is continuous at X now take a filter F which converges to X this is the same thing as say N X is contained inside F. Let now we belong to N of FX I have to show that we belongs to F check of F that is what I have to show. Start with a neighborhood of FX by continuity of F it follows that there is an open substitute a neighborhood of X such that F of U is contained inside V that is a continuity. But now U is in N X but N X is inside F. Therefore, F of U will be inside F of first of all all members of F therefore it is in F check of F. So, what we have shown is that F U itself is in F check everything above everything above bigger than that will be also in F check because F check is a filter. So, V is in F check. So, we have shown that the neighborhood system at F X is inside F check of F. So, that just means that F X F check of F converges to F F X is a limit. So, 1 implies 2 is over. Let us do 2 implies 3. For this we use remark 7.41 both 1 and 3. So, let me just show you 7.41 what it was just now we have done that but let me show you that one namely 1 and 3. The association here is a limit point of if and only if X is a limit point of F S. So, this is the statement for from a net to a filter. A limit point of S if and only if it is a limit point of F S that is the first one. The third one is association F check of S equal to this one. So, both of them will be used what I am going to say what I just told you that is all. So, let me go back to the proof of this thing 2 implies 3. Start with a net S from X D to X. Let S converge to this point X. Then we know that the filter F S converges to X by 2 this implies F check by 3 actually F check of F S converges to F X Y 2 that is fine. F check of F S converges to F S. So, by this remark that we have seen F check of F S is nothing but F of no the filter associated to F of F S composite S. Therefore, again by this remark we get F composite S converges to F. F X is a limit point limit of this filter the same thing as that F composite S the function converges to F. So, that is the 2 implies 3. Now, let us go to 3 implies 1. So, here we prove it by contradiction namely if F is not continuous X then we will construct a net now which does not converge to F X, but it converges to X. So, that is the way we are going to prove 3 implies 1. Start with assumption that F is not continuous at X. This just means that there is one neighborhood V of F X such that no neighborhood of U no neighborhood of X F of U will be contained inside V which is same thing as saying that F U intersection Y minus V is not at it. So, it is not contained inside V means Y minus V intersects F V. So, this happens for every neighborhood U of N X every neighborhood U of X. Now, I choose my directed set D to be N X itself with its usual inclusion. Reversed inclusion you have to remember that. And S from D to X I am going to define the net now is such that each SU is chosen from this non-empty subset U intersection F inverse of Y minus V. So, F U itself is non-empty just means that there is some U some point U inside U intersection F inverse of Y minus V then only this will not end. So, check that point. So, by the example of choice there is such a function. So, that function is going to be our net. So, what is the property of S for each U inside D T is nothing but N X SU is inside U also but it is actually inside F inverse of Y minus V all where V is fixed at U is ranging over all neighborhoods. To show that S converges to X we have to show that S is eventually in every neighborhood of X. So, if W is any neighborhood choose W itself after that means what W less than or W prime which is same thing as W prime is contained inside W. S of W prime is inside W prime by the choice but W prime is inside W. So, this means eventually this entire net is inside W. Thus we have shown that S is eventually in every neighborhood of X this means that S is converging to X. On the other hand it is very clear that F composite S as a net what is its property S composite S of U is inside Y minus V for all U V is fixed. So, V is a neighborhood of F X some after something you should have been inside eventually inside this one no but it is not possible because all the time for all U its SU F of SU F S of U is inside Y minus V is outside. So, this means that F of F composite S is not converged to F X. So, these were some fundamental properties. Now, we shall prove some useful corollaries to these things about products and convergence properties for products. Take a family X, Xj of topological spaces and take X to be the product. Xj is a topological space is F is a filter on the product. Take a point X inside the product the filter converges to this point if it only if look at all the projection maps look at the corresponding the filter the image filter pi j check of F. You take pi j of F and then take the filter generated by that that is pi j check of F. So, pi j of X is a point of Xj, pi j check of F is a filter on Xj what we want is for every j, pi j check is a limit of pi j check of F. So, this is if and only if one way we have already seen namely because pi j's are continuous functions general statement we have proved if F converges to X then F composite F check of F converges to F F that we have already seen. So, we have to see only the converse part. So, only if part we have seen now we have to see only the if part. Let us now suppose that pi j check of F converges to pi j of X for all j what does this mean the neighborhood system at pi j of X this is inside the component Xj the topology space Xj that is contained in the pi j check of F this filter this should happen for every j. But what we want to show is the neighborhood system Nx in the product space of X that is contained inside F. For this it is enough to show that the sub base for Nx you have to remember what is the topology? Topology is product topology the product topology neighborhoods at a point are generated by a sub base pi j inverse of Uj where Uj range is over all the neighborhoods of Xj Xj is simply the pi j of F. So, Uj's are in N of pi j of X you take pi j inverses take finitely many of them as j ranges over different indices here take the intersection. So, that will be those will be the basic neighborhood. So, this is a sub base if this sub base is inside F finitely many intersections will be there and anything bigger than that will be also there. So, whole Nx will be there. So, that is why this is enough to show that for each j if you can fix j here this is all j's are taken if that is contained inside the if it only for each fixed j it is contained. So, I have to show that pi j inverse of U belongs to F if U belongs to N of pi j of X. So, that is very easy for each Uj inside the neighborhood the 26 whatever we have we have put this one what is this one pi N of this one is contained inside this one. So, what is the definition of this pi j check of F? So, that is what you have to show when is a member here is here that is what I have to interpret. So, that implies that this Uj is inside pi j check of F if and only if there are finitely many members of F such that their intersection pi j of F i intersection F 1, F 2, F k then to take pi j's of that take the intersection that is contained inside Uj because these things are first of all there in the generating thing sub base to base and then every super steps must be there this is the condition for members of pi j check of F. So, if such a thing is there then intersection of F i's pi j of that is contained in this one. So, intersection of F i's contained in the pi j inverse of I am applying pi j inverse on both sides here I can just take intersection which is pi j of this is contained in the intersection of pi j's that is not equality need not be equality this is contained inside this one pi j of that one pi j of that one this whole thing is contained in the pi j inverse of therefore, pi j inverse of Uj is inside. So, next theorem is a topological space is hostile if filled only no filter F converges to two distinct points of F. Remember when we defined a limit of a filter we did not say that the limit or something there may be many limit points in general, but you take a hostile space it can have at most one limit point that is not the point the point is if filled only if. So, we get a we get a characterization of hostile spaces maybe this is the fifth or sixth characterization remember that we had several characterizations of hostileness hostileness being very important. So, this will keep coming again and again. So, a topological space X is hostile if filled only no filter F converges to two distinct points inside X. So, let us prove this one suppose X is not equal to Y and they are points of X and F converges to both X and Y then what happen n X is inside F and Y is inside also F. On the other hand X is also as implies that there exist a U inside X and a V inside Y and Y such that U intersection V is empty. So, this is all darkness both U and V are inside F. So, that is cannot happen because then empty set will be inside F or you can just say U does not intersect V that cannot happen that far. So, converse suppose X is not hostile then I will produce a produce a filter which will have two distinct points which are limits. So, not hostile just means that there are points X and Y not equal to each other such that every member of n X intersects every member of n Y. If there is one member here one member which does not intersect will be hostile not hostile means precisely this one. This just means that n X union n Y has finite intersection property as soon as something has finite intersection property we know that it is contained in a filter namely the one is generated by this filter generated by this family. So, that filter will contain we will have both X and Y as limits because both n X and n Y are there that is all. Finally, we have this big theorem here like X be any topological space then the following conditions are equivalent like host darkness has many characterization we want to characterize compactness also X is compact is the first condition every filter on X has a cluster point every filter has a convergent sub filter that is for filters again I have promised you to do similar things for nets also. So, two different characterizations here similar one every net in X has a cluster point every net in X has a convergent subnet. So, A, B, C, D, E are equivalent namely these are so four more characterizations of compactness in X what we have already proved is B implies and implied by C. Look at here every filter has a convergent sub filter. Convergent sub filter means it is a cluster point and conversely this part we have seen similarly net also this part also we have seen. So, B implies and implied by C D implies and implied by D these much we have seen. So, we shall now prove A implies B implies D implies A automatically C and E will be taken care. So, let us prove A implies B. Suppose F is a filter on X with no cluster points then I have to show that X is non-compact. So, that is the contra positive of A implies B. This means that for every X inside X there is an open subset U X such that X is inside U X and U X does not meet some member V X of F. If U X meets every member meets every member then you are done. So, there is one member here and one member here we do not meet that is the negation of that F has no cluster points for every X it must happen. For some neighborhood this happens then that will be cluster point for every one member U X here and one member here U X is a neighborhood of X and V X is a member of F they are they do not intersect they are not meet. By compactness we get a finite collection X 1 X 2 such that X is inside union of U X size because for each X I have got U X is a neighborhood. So, that is an open covering when you range X over X. So, take a finite sub cover here that is what I would take but then you look at intersection of all V X size corresponding to each X I here your V X size here finitely many of them intersection of V X size what is it I see that is empty because union of V X size is the whole of X. If there is a no if there is a member here which will be inside each V X size it will be in one of the V V correspond U X size here each V X I U X are disjoint. So, there is there cannot be intersection of V X size empty which is absurd because these are finitely many members of a filter. So, that is not okay. It is not very hard to prove it directly also you can try to prove it instead of you know reduction and absurd instead of assuming F is there is such a filter then you prove that X is not compact. So, I have used a compactness and if as we say has filter which is no crucial point then I have arrived at a contradiction. So, B implies D now. So, I have to go to net now starting with a net S consider the corresponding filter F S by B this filter has a cluster point. But this is same thing as S as a cluster point as union 7.41 which we have just done now. So, B implies D is all day almost done for us. Finally, I have to show that D implies A. So, this we have to work a little harder. So, let us so please pay attention. So, this way I am saving time for instead of proving B implies A and then D implies A, D implies A, B implies D is very easily done. So, D implies A I have to start with a family of closed subsets of X with finite intersection property. We have to show that the entire intersection of members of C all of them taken together that is nonempty that is equivalent statement for compactness. So, this is what we have to show. As in the previous example, now consider D the family of all intersections of finitely many members of C which is directed by the reverse inclusion. Where is this one? Get this one the family of all intersections of finitely many members of C. In particular, it will contain this C also C 1, C 2 are members of C, C 1 intersection C 2. C 1 intersection C 2. In other words, you saturate this C put by putting all finite intersections. So, take all finite intersections of these members. They are all closed subsets by the way and you take the reverse inclusion. So, that is the partial order there now. Let me case that we have done several times like in 7.5 and so on. So, that is the directed set curly D for us. So, what is the relation? D is less than or equal to E if and only if D contains inside it D and E are subsets of set of this topological space X and there are members of this now members of this family. Since C is contained inside curly D and I want to show the intersection of members of C is nonempty, I am going to do something more intersection of all the members of D is nonempty because this is a larger intersection. So, intersection set will be smaller. Intersection is taken on a larger family. So, intersection will be actually smaller. If this is nonempty, if this itself is nonempty, then that will be also nonempty. So, intersection of D such that D is inside curly D, this is nonempty is what I am going to show. Now, look at each D inside D, they are nonempty by definition because I have taken finite intersections of members of C which has finite intersections of it. So, they are nonempty. So, we can take the net to be from X D to X to be such that S D is inside D. So, once again here we are using axiom of choice. So, such a function is there. So, let that be our S now. Suppose X is a limit, a cluster point of S, suppose it is a cluster point of S. We claim that X is inside D for all D inside capital, inside this curly D. This we have actually seen, I will repeat it because you might have forgotten how it came. Suppose it is not true, suppose X is not inside some member, D belongs to D, then X minus D is a neighborhood of X because each member in the curly D is a closed subset, that is the point. It is a neighborhood of X. Hence, by the definition of a cluster point, there exists an E belonging to this D, curly D, such that for all members which are following A, following this D, that means E is contained inside. Remember, this is E contained inside D. All the S E's will be inside X minus D. X minus D is a neighborhood and the X is a cluster point. So, it should have. But then this implies D is inside S and would imply D contains E and S E is inside E all the time. So, if S E is inside E, S E is inside D, X minus D is contained as X minus D. Therefore, S of E is inside E as well as inside S of E, X minus E also, that is observed. How can it be in X minus A as well as S E? X minus D was a neighborhood who contained X minus E, right? Whenever D is contained inside. So, these two are contradictory. I have started S E inside E but S E is now X minus E, that is it. So, for this net which is cleverly chosen, this will fail. So, what we have proved is that X belongs to D for all D inside D. So, intersection of this one is not, all right? So, that completes today's plan, Model 31. Thank you. We will meet again next time.