 Welcome to the last module of point set topology course part 1. So, we shall continue the study of Tuplio-Eichel vector space. Several important basic results you have derived already. So, in this last module, I will prove three important results about Tuplio-Eichel vector spaces. The first one is every finite dimensional vector subspace y of v is linearly homomorphic to k power n. Further, if v is Hausdorff, which is same thing as just assuming t naughtness, then y is a close subspace of v also, okay. Every finite dimensional vector subspace is linearly homomorphic, if linear isomorphism to k n without without assumption of Hausdorff-ness, it may not be a close subspace because you can always take indiscreet topology on any Tuplio-Eichel vector space. Then the only close subspace will be empty set of the whole space, okay. So, that is very easy to see that the condition is necessary here, okay. So, here is the proof. Start with the basis for y, say u and u to n. Then the map f of e i going to u i extends linearly to an isomorphism from k n to y, right. So, linear part is done. Already we have seen that in the last corollary of the previous module that any linear map is continuous. So, f is continuous. So, the only thing that is missing is f inverse should be also continuous from y to k n, okay. So, this is what we have to show. Now, you take s to be the unit sphere in k n, okay. I am taking k n which is Euclidean topology I can take. So, talk about unit sphere there, there is no problem. Then we know that s is compact. Under continuous in a compact subset put compact subset. So, f s is compact. Now, f is injective. So, f 0 goes to 0. So, nonzero vectors won't go to 0. That means 0 is not in f of s, okay. Now, b be a balanced neighborhood of 0 which does not intersect f s. See, f s is a compact subset, okay. So, I can choose 0 is not there. So, you can choose some neighborhood here which is disjoint from that. It is a regularity or whatever. But I want to choose it as a balanced neighborhood, alright. Any two and compact subset and a closed subset can be separated. That is what we have already seen. So, for that you don't need outsource necessary. So, see slowly I am using all the results we have proved in the past two lectures for topological vector space. So, choose a b, a balanced neighborhood of 0 which does not intersect f s, alright. Put a equal to f inverse of b, okay. Which is same thing as f inverse of b intersection y because the image of f is the over of y. It is contained inside y. That is why it is a f inverse of b intersection y, alright. a is a neighborhood of 0 because f is continuous. Since f inverse is linear also, linearity is no problem. What we are trying to prove is continuity of f inverse. It follows that this a is a balanced set of, okay. Some scalar multiple times something is contained inside, etc. You have to check. So, from the balanced neighborhood of b, it follows that a is also balanced. So, scalar multiplication comes here because f is a bijection, okay. Alpha times some vector is alpha times f of some vector. So, that is all I have to use to show that b is balanced implies a is balanced. In particular, if z is a point inside a, then the entire line segment 0 that will be inside a. This is the property of balance that we have seen all the time, okay. So, this implies that a is star shaped at 0. All the line segments are there. Therefore, in particular, a is connected, okay. Now, a intersection s is empty because I have started with b intersection f of s is empty, okay. So, we know that the unit ball inside kn, okay, whether it is k equal to r, r, k equal to r or z, unit ball always separate the whole k power n, okay. Namely, those things which are strictly inside and those things strictly outside. So, there are two disjoint open subsets, right. So, since a intersection s is empty and a is connected, it follows that a must be contained in the open unit disk in skn with central 0, right. Now, given any epsilon pole tail, it follows that if you take f inverse of epsilon b, that will be epsilon times f inverse of b by linearity of f inverse, okay. But this epsilon b, so epsilon times f inverse of b is epsilon b is a, that is epsilon a. So, this epsilon a will be contained inside epsilon d because a is contained inside this unit ball and d, okay. I can write this as d power n but that is not correct because I do not know whether k is r or z. So, it depends upon it is maybe d2n or dn. So, I have just written as d unit disk here. This implies that now we see f inverse of epsilon d is contained epsilon d, f inverse is contained at 0, okay. So, 0 is an element of y. So, y to k power n, we have shown that this linear map is continuous. But y is a linear space, y is also a topological vector space. Continuous 0 is enough to show that f inverse is continuous everywhere, okay. See how connectivity finally has been used to show that the inverse is continuous, alright. So, this is one important theorem. Well, I have to show as the other part. Namely, suppose v is outside, then I want to show that y equal to y bar. Namely, y is closed also. Why it is closed? That is what you have to show. Given p belong to y bar by continuity of scalar multiplication it shows that there exist t positive. So, said p is inside t times b, okay, alright. See the whole of y actually there are various ways of seeing this. y bar is contained inside the whole of v. The whole of v, take any point at this one. The larger, the bigger and bigger multiples of b will cover the whole of it. So, some t times b will be, will contain p or every p. Therefore, p will be inside y bar intersection tb. p is already inside y bar and p equal to tb. So, it is inside this one. It can take a closure, okay. So, that closure is contained inside f of t a bar, right, because a is nothing but f inverse of d, okay. So, if we apply f, f of, f itself, y is for what? y itself is f of the whole thing, right. So, the whole thing is in the image f of this. So, this f of t a bar. So, y is nothing but f of kn. After all, everything is there. So, there is no need to write y at all. If it is a v, there was a problem. So, the whole thing is inside the image. So, this contains f of t a bar now, okay. So, but this f of t a bar is contained inside, you can write t a bar here and then take the closure, okay, f of t a bar and then take the closure. But t a bar is compact, okay, complex subset of kn, right. The a is a bounded subset already, inside, contained inside the unit is, its closure is compact, okay. So, it is complex subset of kn. f of t a bar, once a bar is compact, t times a bar is also compact because t is anyway some scalar. So, f of t a bar will be a closed subset of v because it is compact and v is house dot. So, this is where we are using first time that v is house dot, that closed subset, compact subsets are closed, that is what we have here, okay. So, therefore, p which is inside f of t a bar, there is no need to take the closure here. So, p is inside f of t a bar, but f of t a bar is after all inside y because f of whole kn is y, okay. So, p is inside y, that is what we have here. So, y bar is contained inside y, therefore, y bar is equal to y, all right. So, closure property follows by again Inaval theorem that closed subsets of kns are compact. So, all very important things have been used here, all right. So, let us go to the next result for which I will make a temporary definition of local compactness. Later on in part 2, we will study local compactness on its own for arbitrary topological spaces. Here it is a tentative definition for a topological vector space v is locally compact if there exists a neighborhood o of 0 such that the closure of o is compact. So, there is a compact neighborhood for the, the zero element states on, okay. So, that is the meaning of local compactness, all right. With this one, so we make this final result about topological vector spaces and finite dimensionality. See, we have already proved that a normal linear space is finite dimensional if and only if the unit sphere in it is compact. In the topological vector space situation, there is no unit sphere, there is no way, there is no concept of unit sphere because there is no metric. So, the local compactness is, is an indirect way of bringing that, okay. Once unit sphere is compact, the disk was compact. So, having to telling that a normal linear space unit sphere is compact, it is equivalent to saying that it has a compact neighborhood, the zero has a compact neighborhood and that compact neighborhood makes sense. So, that is the definition of local compactness here. So, we have brought back, you know, avoiding the metric, the compactness property can be always generalized. So, local compactness has come here, okay. So, that is what it is and the expected theorem here is that now every locally compact vector space is finite dimensional and converse, converse we have already proved. If and only if converse we have already proved, why? Because any finite dimensional vector space we have just proved that it is isomorphic to Kn, linearly isomorphic to Kn. So, in particular homeomorphism therefore, we know that it is locally compact, okay. So, we have to prove that if it is locally compact, then it is finite dimensional. Now, the proof of this one, maybe it is little so slightly longer, I am not very sure, but it is much more elegant than the proof for an ordinary spaces, okay. It is more or less canonical proof here. So, just observe that, okay. The if part is an easy consequence of or where theorem that we have proved that every finite dimensional vector space, ethypological vector space is isomorphic to Kn. So, let us prove the converse. Now, let V be a hostile and locally compact, okay. Did I prove that hostileness forward? Hostile and hostile, locally compact and hostile, okay. Hostileness is a must here, okay. So, locally compact and finite dimensional vector space is automatically hostile, right. So, I do not have to worry about that. So, locally compact and hostile. So, assume that and choose a neighborhood B such that B bar is compact, that is local compact. So, we get finite number of vectors V over V to Vn belonging to V such that B bar is compact, it is contained inside finitely many, you know, translates of half bar. Half the bar that is an open subset of the neighborhood, xi plus this one we neighborhood of xi, xi's are taken all over V, they will cover B bar. But then you can choose finitely many V1, V2, Vn such that I relate to 1 to n, Vi plus half B will cover the whole thing, okay. So, let V prime be a linear span of V1, V2, Vn. Then V prime is a finite dimension because it is span by only n elements. Therefore, by 5.56 theorem, just we have proved V prime is a closed subset of, here we have used hostileness of V. So, V prime become finite dimension is closed subset, this is the previous theorem, part of previous theorem, alright. Now, look at B which is contained inside B bar obvious and that will be contained inside now V prime plus half B, because what is V prime? It is contained inside all this iron into 1 to n, this V1, V2, Vn plus half B. But this is, each of them is contained inside V. So, union is contained, sorry V prime, union is contained to V prime. So, this whole thing I can replace it by one single element V prime which is very huge actually compared to what I have written there. It is contained V prime plus half B, alright. Now, we keep using the property that V prime is a vector subspace, which is a strong thing, okay. So, it follows that if you take half of B which will be contained is half of V prime plus half of half of B, right. But half of B prime is V prime itself, it is a vector space, right. So, it is V prime plus half of B. So, half of B is contained inside V prime plus one-fourth of B. Therefore, start with B which contains V prime plus half B, but half B is contained inside this one. So, I can write V prime plus V prime plus one-fourth B. But V prime plus V prime is V prime. So, V prime plus one-fourth B. So, what has happened? B is contained inside V prime plus half B, but we have improved it to V prime plus one-fourth B. You repeat this process, next time what we will get? It is contained to V prime plus one-eight B. And next time it will be one-fourth, two-fourth four times B and so on, sixteen and so on. So, you just repeat this process, that is all, right. So, what is this? This is that because B is contained in V prime plus one by two-fourth N times B for every N. That is the meaning of B is contained in the intersection of all these. This is a decreasing sequence of things, right. So, it is intersection of all these things because it is contained in everything. But you remember that this forms a neighborhood system for zero. This is what we have proved last time. Therefore, this intersection is nothing but the closure of V prime. Closure of V prime, what is that? Closure of V prime is V prime itself, right. This V prime is closed and we conclude that B is contained inside V prime. The moment an open subset, non-empty open subset is contained inside a vector subspace, that vector subspace must be the wall of the space. So, this follows from other lemma that we have already used. Namely, B is contained inside powers of two times B union of all those things. Any increasing sequence of numbers will do this job, ok. Increasing sequence, strictly increasing sequence or increasing to converging to infinity, I should say, ok. So, that is same thing as now. It is contained inside union of all two power N times V prime because B is contained in V prime. But two power N times V prime is V prime itself. So, its union of V prime is just V prime. The entire vector space is contained inside the subspace, that means they are equal. So, what we have done? We started with V, we produced some finitely many elements with certain properties and then we took the linear span of that and then we showed that V is equal to V prime. Anyway, so what we have done is that V is finite dimensional, ok, under the assumption that it is Hausdorff and locally compact, alright. So, two very important theorems we have done, very nice and elegant theorems. The third one is again curiosity, ok. See, you have proved that a topological vector, topological group is a regular space without any assumption, with any further assumption. So, what is the best thing you can say about a topological vector space? So, here is one, namely every topological vector space is completely regular. By the way, it is very important computer regularity. We cannot, in this course, you cannot see any of these important. One is about inviolability theorems because continuous functions are coming there, right. Another one is what is called the uniform, uniform topologies. So, this we will not be able to discuss here. So, there is something important in saying that something is a completely regular and that is what is it namely every topological vector space is completely regular. If we assume T 1 as it will be T 3 and half space, ok. So, this is just false shot of normality if we have proved the normality then you would have been even great, but that is not proved generally perhaps, ok. So, I am not very keen on that one. Now, let us prove this far, ok. It takes a little more time, but let us prove that that every topological vector space, ok, given any point and a closed subset, I must find a continuous function. But you do not have to do any other way point because we are working at a topological vector space. A point can always be specialized to the 0, ok. Given any neighborhood view of 0, we must find a continuous function f from entire vector space to 0 1 such that f of 0 is 0, ok. And f of the complement of u is singleton 1. The entire complement of the open set is goes to one point, the 0 goes to another point. Those two points can be chosen anything arbitrarily, but we will conveniently choose them 0 to 1 and the interval also is the domain to be also 0 1, ok. You could have definite a comma b and so, there is no problem. So, this is what we have to prove, ok. So, somewhat like in the theorem of Eurydon, we have to work, but it is is easier than that perhaps here. Start with u naught equal to u. Inductively, we are going to find a sequence of open subsets, ok. Now, we are using topological vector space condition here. So, it is easier than doing it in arbitrary normal space. Inductively, choose symmetric open neighborhoods u n of 0 such that u n plus u n is contained inside u n minus 1. Remember, symmetric means u n inverse is u n, but here inverse is minus u n. So, I could have put here minus also then that also you choose the same thing, ok. u n plus u n or u n minus u n is same thing because u n is symmetric, symmetric here with respect to the addition, ok. So, u n plus u n is contained inside u n minus 1 for every n. So, what I do? Start with u naught, take a u n which has this property u 1 plus u n is contained inside u naught and keep doing that, ok. Now, let d denote the set of all diadic rationals including 0 to 1 over. What are diadic rationals? Each r inside d has a unique representation r range from 0 to infinity C i r 2 power r, ok. 1 by 2, 1 by 2 plus 1 by 1 by 4 or just 1 by 4 and so on, ok. So, you can take any anything like that. So, it is a unique representation here, ok. Where all these C i r the denominators are either 0 or 1, ok and only finitely many C i of r are non-zero. So, this is my definition of this diadic rationals. I have given you complete description of this. What are these? Put a r equal to v if r is greater than or equal to 1, ok. So, this r I do not know already it is not rational r if they are not equal to 1 I put a r. And if r is inside d 0 to 1 open, ok that is inside d then put a r equal to these look at these numbers 0 and 1 they are the scale r, right. So, I can put C i of r times u i this makes sense because we are working in a topological vector space where u i's are subsets of v capital v. So, 0 times u i means just 0 that is a 0 vector C i r times C i is in C i r is 1 it will just u i, ok. Then you are taking the sum of all this, ok. So, what I should mention is here you can take these things as only just finite sum only finitely many C i r's are non-zero. I have put infinity here as if convergent sequence and so on, but diadic rationals are only finitely many C i's are 0 non-zero, ok. I do not want to stop here and so on then n could be infinity. So, this is just 0 to n and then n going to infinity all of them are there and so on. So, C i's are non-zero. So, most of them are 0 here. So, this is a finite sum, ok. I do not want to something u 1 maybe u 3 maybe u 5 and so on. So, these are these are finite sum, but this makes sense u i's are open and each of them will be open subsets, all right. We first make a few observations on the subsets a r. Each a r is a open neighborhood of 0 and a r is contained inside u 0 contained inside equal to u that is by definition. The u 1 plus u 1 is contained inside, right, u 1 plus u 0 you also contain inside u 1 because u 0 is already contained and so on. So, a r is contained inside u 1 this is important. If r is less than s, ok, inside b then a r will be contained inside a s, ok. So, these are similar to what you have done when proving your reasons lemma somewhat similar, but easier steps here, ok. So, why this is true? Choose the smallest k such that c k of r is 0. Once you r there are 1 c 1 c 2 c k's, right. Choose the smallest c k's c k of r is 0 and c k of s is 1. Among the r and s you keep comparing, ok. So, some 0's you may get. So, first time this means s and these things are 0, that is what I have to say. Then a r will be contained inside a 1 by 2 power k. Once 1 by 2 a or 1 by 2 power k a s will be even even bigger than that, the more elements will be there. That means this has u k already and then some plus something so on. So, that will be contained inside a s, ok. a s will have u k plus something more. So, that is why it is contained inside a 1, ok. Recall that if a and b are symmetric never roots of 0, then a bar union b bar is contained in a union b a a plus b, right. This is what we have seen for topologically groups itself, ok. Given r less than s, choose n such that 1 by 2 power n is less than s minus r which is some positive number. r is less than s, s minus r will be some positive number. So, choose n sufficiently large. Then a r bar is contained inside a r plus u of 1 by 2 power n, ok. The u of this notation, what are u naught, u n and so on, I could have just put a suffix here, but I have put a bracket here, does not matter. u power 1 by 2 power n contained inside a s because of this property. u 1 by 2 power n is less than s minus r. What you are adding here is less than s, alright. So, a b c we have proved. Now, let us define f x equal to infimum of r for all x belonging to a r. You see, if you have datacrational then you have to choose it to be smaller than that, you see infimum of those things. But if there are no datacrational then remember this will be the whole of a and this will be r equal to 1, ok. So, there were points of v are there like that, alright. So, f x will be 1. By a, it follows that f of v minus u is 1 because the moment it is outside u, you have defined it to be r to be 1. So, the infimum of this will be 1. So, specify definition. Also, since u 0 is inside u n and u n is a of 1 by 2 power n, ok. If you just take a r, what is a r? Let us just recall all these properties I have told. But what is a r? a r is the sum of all these. But when r equals to 1 by 2 power n, there is only one element there, right. So, that will be u u n or u whatever, notation. That is the notation u n comes. So, u n is a 1 by 2 power n. For all n, this is the case. Therefore, if 1 by 2 power n is there for all n, infimum will be 0. So, f 0 is 0. It remains to prove that why f is continuous. We have proved that f 0 is 0. f of p minus u is 1. That is fine. Continuity of f is what we have to do. So, there is full analogy between Eurydon's Lemma and this one. If you had not understood the Eurydon's Lemma, you would not have been able to prove this one, okay. I do not claim that if you have understood the Eurydon's Lemma, you would be able to prove this one. But now, you have adopted those ideas in the case of Doppler-Eichler vector basis. So, why f is continuous, okay. Ready? Start with any point x in a and put f x equal to t and epsilon positive, okay. Choose r and s belonging to d such that t minus epsilon is less than r, less than s, less than t. The effects equal to t, what is the definition? It is infimum of one side, okay. Therefore, you will have r and s inside the random ration such that t minus epsilon, t, t minus epsilon is less than f, right. It is less than r, less than s, less than t. So, between r and t, t is larger, t s is smaller, okay. So, you can choose one some r here and then smaller than that one, t minus, between t minus epsilon, t plus epsilon, you can choose two numbers because this is all then subset of 0 comma 1. That is all I am using here. That is nothing more than that, okay. So, then a r y will be continuous at a s and x is not inside a s because f x is t, it is smaller than, is bigger than t. So, this cannot be inside a s. If it is inside a s, the value of this one will be smaller than that. So, this means that if you take w equal to v minus a r bar, okay, a r bar is closed. So, v minus a r bar is an open neighborhood of x, okay. So, this implies that f of w is contained instead of r to 1. So, the least thing is a r. It cannot be anything lower than that is not there. So, it has to be bigger than r. So, it is r to 1. It cannot be bigger than 1 anyway. All of them are less than equal to 1 only, okay. So, f w is going to be r to 1, okay. This already completes the proof of continuity. In the case t equal to 1, we have formed a neighborhood. If t were, I do not say t is equal to 1. Start with f x, start with x, f x put f x equal to t. That is all, okay. It may happen that this t is 1, then this argument says that already that the continuity of the function f at points x such that f x equal to 1 is proved, okay. That is a special case, all right. Now, you assume t is less than 1. So, we are inside now rational, okay. Choose p belong to d such that t less than p less than p plus epsilon. Given any epsilon, I will produce a neighborhood, okay. That is what I have to do. Then it follows that x is inside ap because t is even smaller f of x is smaller. So, it must be inside ap, right. Now, f of ap points inside ap, okay. They must be between 0 and p. Once it is already p, it may be even smaller. So, but it may be 0 and p. That is what it says. So, it follows that the same w, okay, the same thing here intersection ap is a neighborhood of f such that f of w intersection ap will be contained inside now t minus epsilon to t plus epsilon. Once t is smaller than epsilon 1, you may assume that this epsilon is smaller than 1 minus t. So, that even after adding t, it does not go out of that one. For such epsilon, it is proved fine. Otherwise, anyway, it can go out, but it is inside t minus of t plus epsilon is what you have to show for every t such that, you know, there is some epsilon is what this what you have to show. There is some neighborhood f of this w intersection ap will give you a neighborhood. Remember, all these ap's are open subsets, okay. So, that completes the plan as to whatever I have done to this part one course. I just repeat what were the reference books here. It was completely influenced as a student and even today by the book of Siemens, George Siemens. And in spirit, may not be in content, I have followed this book. For example, the definition of t0 space, t1 space, normal space and so on, normal and t3, t4 and so on. Those things differ from the Siemens book. So, be cautious about that. I have already told you earlier, be cautious about that. The spin and see back, this book I refer to counter examples in topology. This book is now available online and whatever I had seen, it has expanded much. So, since 40 years back, it has expanded a much. So, this is good for not learning topology, but as a reference book, it is nice. That is what I have shown. So, my second book in topology, we may say is Kelly's book, which is a very fantastic book but difficult to read as compared to Siemens. Siemens was a pleasant man, but I enjoyed Kelly's book also. Later when I joined IIT Bombay, my colleague had written a nice book and K. D. Yoshi, that is introduction to topology. So, I have borrowed material from this book also. So, I have high regard for this book, but it is okay. So, in fact, the existence of this book stopped me writing a book on my own, on points of topologies. All right. Before that, even in TIFR, this was my third book in topology, Hurwitz and Wallmann, dimension theory. This is a very specialized book, but it teaches you topology like anything. So, this is a wonderful book. Okay. This is specialized book in dimension theory. And then there are nice books, which I have just browsed to. I have not studied them and so on. Armstrong, this book by C. Weinepati. Okay. So, here you can see Rudin's function analysis. These are used for these topological vector spaces and so on. Okay. So, that is what it is. You can also look into. So, there are some papers here from which I have borrowed material. All these books, they are good book, Dugunji and what are the other ones here? Well, that is it. Satish Hiral is a nice book on matrix spaces. Okay. So, that is what it is. So, however, I want to tell you, I remind you that I am going to give notes and you have already the notes with you. So, just to get through this whole course, you do not need anything else. You just go through the notes properly work out and assignments, you are done. Okay. So, yeah, that is all I wanted to tell you about this. So, finally, I would like to end up this one with a lot of thanks to the NPTEL team, my own team of tutor whose help has been extremely useful, extremely great moral support and so on in bringing out this course to you. And also big thanks to all of you. If you have stayed with me so far, I hope I will see you in the second part also. Okay. Thank you.