 Welcome to module 60 of point set topology part 1. We shall continue the study of topological vector spaces today. As earlier throughout this section we shall fix V to be a topological vector space. I will not use this notation for anything else. We have results parallel to the fundamental lemmas we had for a topological group obtained merely by changing the multiplicative notation to additive 1. Because if you have topological vector space V plus and the 0 element there will give you a topological group which is abelian. So, we shall denote them by additive notation plus instead of the multiplication dot. The multiplication will be there only not from V cross V to V, but only from k cross V to V now. So, all the statements which were multiplicative for the topological group that we have seen will now be obtained by just changing them to additive notation. In addition because of the scalar multiplication and the continuity of that we will have many other important results. So, all those things the new time I will give you a hint or a full proof of those things, but those which we have already got for a topological group as such those I will only restate it at all. So, with this the first theorem here is again a list of statements ab, c, d, e, f which are all easy consequences of the statement that we say topological vector space. You start with any abc subsets of V, then all these statements are true. So, let u be a neighborhood system of 0, then for any subset a, a bar will be equal to intersection of all a plus u, u inside u. Remember there was exactly similar result even stronger intersection of a u, multiplicative notation in the case of topological groups a u or you can take u a and so on ok. And you do not have to take all the neighborhoods just a neighborhood system will do ok. So, that is also an easy consequence anyway ok. So, the next one is a bar plus b bar is contained inside a plus b bar. Once again the corresponding thing was for topological groups we had a bar into b bar contained inside ab bar. So, same thing is true for this one I do not have to elaborate this one. Next thing comes this is new to topological vector spaces only because now I am using convexity. This convexity notion was not there in a topological group. Let c be a convex act then both c bar and interior of c are convex. b is balanced implies b bar is balanced further if b is 0 is in a b interior which is same thing as this b is a neighborhood of 0 then interior of b is also balanced. See unless you have 0 here you cannot claim that one ok. The second thing is this is Eth 1, 5th one if b is bounded then b bar is bounded. So, I have taken convexity, balance in this and boundedness ok. If a is a vector subspace then this closure is also vector subspace ok. So, on c, d, e, f are new to topological vector spaces a and b are old one only right. Let me let me just go through them again. In the context of topological groups you can check lemma 5.4 ok. We have proved that a bar is equal to intersection of a plus u such that u is a neighborhood of 0. If you take all the neighborhood then this is just in the multiplicative notion you have done. So, it is true for this one also. But then you can just restrict it to a neighborhood system. So, that will be a smaller family of same members of this family itself that is smaller family. So, intersection has to be larger ok. But here the smaller family has the property that each member is contained in some member here therefore the equality occurs. So, the b is directly from lemma 5.14 for products namely a b bar a a bar into b bar if contained inside a b bar which will become now a bar plus b bar contained inside a plus b bar ok. Now, come to statement c. If c is convex I must show that c bar is convex. For any t belong with i ok we have t times c bar from part that you do not have to worry about that also. This is just multiplication by t right which is a translation if t is 0 both sides are 0 if t is not 0 then it is a homeomorphism. So, t c bar is t of c bar. Therefore, in particular what happens is t c bar plus 1 minus t c bar 1 minus t will be also inside 0 1 right. So, I can take these two which is nothing but t c bar plus 1 minus t c bar right from the first observation. But now this is a bar plus b bar. So, it is contained inside a plus b whole bar right. So, there is t c plus 1 minus t c. But for each point here t c plus 1 minus c is a point of c. So, this whole thing is just c bar it is contained inside c bar no problem ok. So, this proves that if you take some u here and some v here t times that u plus 1 minus c times v is contained inside c bar which means c bar is convex. The proving this statement here is in here convex it ok. Now, prove the convexity of the interior observe that t c interior plus 1 minus t c interior is already contained inside c because c is convex some point here t times that 1 minus t times some other point here is already inside c by convexity of c. But this one t c is now t of c interior that is an open set 1 minus t times this one is also an open set maybe one of them may collapse to single point that is the caution. If it is a single point when t is 0 or 1 minus t is 0 it is not an open set. But it is already inside the inside the interior of c bar. Therefore, this part is always an open subset of v and hence it is contained inside the maximum open subset that is interior ok. First is contained inside c, but this is open there were it is contained inside c interior. So, c interior is also convex. Now, let b be balanced then for any alpha in scalar with mod alpha 1 we have alpha b is contained inside b. So, it is contained inside b bar also ok. So, b bar the closure it is a closed set right. Therefore, if you take alpha b bar which is same thing as alpha b bar ok. Alpha b bar is the smallest closed subset contains alpha b must be contained inside b bar ok. So, that already assumed that always proves that b bar is balanced alpha b bar is less than contained b bar means b bar is balanced. Further now we assume 0 is in the interior of b. Then alpha b interior is contained inside alpha b and then contained inside b. If alpha is not 0 this will be an open subset. Then alpha b interior is open subset and hence it is contained inside interior of b. On the other hand if alpha is 0 then this will reduce to single point ok. 0 is there otherwise there will be problem. So, alpha b is 0. So, 0 is already there in the interior. So, you are done. So, this proves that b interior is also balanced ok. Alpha b interior in both the cases is contained inside inside b, but then it is contained inside b interior. So, b interior also balanced. Now, come to a what was a statement e let us go back. Statement e here is if b is bounded then b bar is bounded ok. What is boundedness? Given a neighborhood v of identity ok. I must produce I must produce as m such that s bigger than m imply b is contained as s of b. That is what I have to do. That is the meaning of that for b bar, b bar is contained as s 1 that is what I have to do ok. b bar is bounded means that ok. So, first I apply regularity of the vector space, topological vector space which we have told ok. Given a neighborhood v of a by regularity we can choose a neighborhood, close the neighborhood u such that e is contained inside v contained u contained inside u bar contained inside v that is about it. So, I am taking u bar as u now ok. u I am calling I am not calling it as open subset, but it is a neighborhood which is closed. Given any neighborhood there is a closer neighborhood contained inside that to take this one to a close neighborhood ok. Now, apply the boundedness of b to get an m such that for all s bigger than m, b will be contained as s times u for every s bigger than m ok. Therefore, b bar closure is contained is a s u bar means nothing the s times u bar, but u bar is u because I have my choice is that u is a closed neighborhood. It is contained as s u ok. Now, s u is contained inside v for every s greater than or equal to m ok. So, sorry s v here, s u is contained as s v for every s greater than or equal to m. So, that is what my aim was. So, the proof of that b bar is parents it is done ok. Now, the last one it is a statement that if something whatever notation I have taken same notation. So, a is a vector subspace then a bar is a vector subspace the similar proof we had for the pliable groups also ok. But here, here this you can look at this one this is only for the scalars between 0 and 1. But entire discussion you know can be applied to any t any any square area ok. Because now I am assuming that instead of c I am assuming a is a vector space already. So, lambda times a will be already inside a. Therefore, the same proof will work here for to give you f. If a is closed a is a closed under addition and scalar multiplication that is a vector space sub space then a bar is also closed under scalar multiplication and addition that is what we mean ok. If u and v are there lambda times u plus nu times v is there that is what you have to show right instead of t and 1 minus t. So, proof is the same thing there with the extra assumption that instead of c arbitrary convex set. So, c a is now vector subspace let us go ahead let o be a neighborhood of 0 in v. Now, the first claim is there exists a balanced open neighborhood b of 0 such that b is contained inside o. So, a arbitrary neighborhood we can improve it to become balanced open neighborhood. Openness is obvious anyway balanced neighborhood that is the whole idea. Further if this o is convex then inside a in the statement a we can choose b to v convex as well that means it is balanced and convex ok. That is every topological vector space v as a local base consisting of balanced under the neighborhood that is the statement a. If v is locally convex that means it has convex neighborhood like this then it has a local base consisting of balanced convex neighborhood. Here the important thing is you just assume one convex neighborhood then there is a whole system of convex neighborhoods ok that is the whole idea here. Sir this balanced open neighborhood this concept is it something similar to symmetric neighborhoods in topological groups? It is just a symmetric it is just one single thing right inverse, inverse corresponds to minus 1 here right. So, that is that is my that is much weaker there even in topological groups you could not do anything the scalar is only minus 1 just comes there it is all right inverse becomes minus right. So, here is a stronger symmetry right you may say only for unit scalars also that is also symmetry but this is much more stronger ok does imply that but it is much much stronger ok. So, I told you that this is almost like bringing the balls inside a metric space ok. So, these things will be played the whole of open balls in a metric space. So, we are bringing them in a back door entry it is like that ok. So, they will play the role of open balls that is much more stronger than just symmetry ok. So, by continuity of scalar multiplication at 0 moment you have some neighborhood there exist an open neighborhood V of k such that if delta is positive such that alpha w is contained inside O for all alpha with mod alpha less than equal to delta ok. O is a neighborhood 0 goes to 0 under scalar multiplication ok. So, you must you can control the scalar that is the whole idea there is a positive delta say mod alpha as a delta would imply the whole image alpha w is contained inside O. So, this is by continuity of scalar multiplication all right. So, all that we are doing is the use of scalar multiplication here. Now, let B equal to union of alpha w where mod alpha is less than equal to delta. The first part says that this family is non empty that is at least 1 w. Now, you take all such w which satisfies this mod alpha is less than equal to delta. There may not be any right. So, you have taken all such n where is 1 such that alpha w is contained inside O ok. Now, you take union of all of them then B is an open neighborhood of V ok. So, B is contained inside O and B is balanced. You must put that mod alpha is less than equal to delta I would like to have this one contained inside O, but that is the condition here on alpha ok. So, so I would like all these alpha w's also contained inside O. So, in any case union of all open subsets alpha w therefore, B is open B itself will be now balanced why because if multiple say beta mod beta is less than equal to alpha beta times B will be contained inside union of alpha into beta times B, but if multiply alpha by beta where beta is less than mod alpha will be automatically mod alpha beta will be less than equal to delta. So, those things will be contained inside this one again that is the water. So, so beta is balanced and it is an open subset of contained inside O and containing 0. Now, in the second part I want it to be what second part what is it I want it to be convex as well right. So, suppose one of them is convex that O is convex you have already got. Now, you take A to be intersection of all alpha O where mod alpha is except equal to 1 ok. So, in the case of if the vector space is real vector space this just means O and minus O that is the way you have done to get a symmetric neighbor would remember that. So, here you have to take mod alpha equal to 1 only to saturate it intersection of all of them ok. Now, check that A is convex because O is convex to begin with. So, this you have to check that this will be also convex ok. Hence interior of A is convex that is what we have seen right. If C is convex interior of C is convex was one of the part of the previous theorem. Now, let B contain itself will be a balanced neighborhood of E. Inside this one by part A we can take a balanced neighborhood of E. Then if mod alpha is equal to 1 then you can write B as alpha inverse of alpha of B alpha and alpha cancelled out. But alpha B being the balance is contained inside B. So, it is contained alpha inverse of B, but alpha inverse is also mod 1. So, it is contained inside B. So, all these things must be equal right right. Therefore, B itself is contained inside this intersection because for all alpha is contained inside B. So, intersection B itself will be contained inside all of them ok for all each alpha ok. Therefore, interior is a neighborhood of E ok because B is a neighborhood and 0 belongs to E. So, if you take the whole of interior that will be a neighborhood of E because that is one such. It remains to show that interior is balanced alright. Interior is always convex. Now, interior balanced is what I would show. For this it is for suffice to show that A itself is balanced. Again we have shown that if A is balanced A interior is balanced ok. So, to show that A is balanced take mod beta less than equal to 1 and beta is equal to t times alpha where mod alpha is equal to 1 ok. Any beta can be written as e power 2 pi i t right times t. So, that is why you can write t where mod alpha is equal to 1 and t will be between 0 and 1 positive, non-negative ok. Once you write like that beta A is t times alpha A, but alpha A is equal to A by this property its intersection that is why t alpha A is equal to A that is equal to now t alpha A equal to t times A. Now, what is A? A is intersection of all I can write you do not write alpha it may confusing. Now, write another variable gamma O where mod gamma is 1 ok. This is just A just t times that no, but now you can push the t inside to intersection of all gamma mod gamma equal to 1 gamma of t O ok. But O is a convex set to begin with ok assumption is that one containing 0 and then t times O is contained inside O. t times O consists of all the line segments from 0 to points of points of O t times that one. So, all the t times O will be also inside. So, therefore, what if its whole thing is inside O? So, beta A is contained inside instead of this t I can just write O right. Once I write O, this whole thing will be nothing but A right beta A is contained inside. Thus what we have shown is that in general balanced neighborhood form is fundamental system. If it is locally convex, then convex and balanced neighborhood form is fundamental system ok. Now, we will slowly these are fundamental results. Now, we will slowly you know arrive at some concrete results. Every compact subset of K is bounded. Every compact subset K of V is bounded ok. So, you see I told you that these balanced neighborhoods are playing the role of matrix I mean balls in a metric space. So, now we can talk about compact subset being bounded just like in a metric space ok. Of course, our notion of boundedness is also different here. This boundedness is stronger. That is what we have already remarked in the last time. What is the proof? Proof is very easy. Given a neighborhood O of 0 inside V, we have to we have to find some m positive such that S is bigger than m implies K is contained instead S O. For any neighborhood you find an m which is this property that will mean that K is bounded. This is the definition of boundedness ok. Now, the previous theorem says that part A says that I may assume O is balanced in neighborhood. Suppose I do it for a smaller neighborhood balanced in neighborhood smaller than O. The same statement it will be true for O also ok. So, I can assume that O itself is balanced, but I am replacing it by B. B is a neighborhood of 0 which is balanced and contained inside O. So, we shall now prove that find an m such that S bigger than m then K is contained either S times B. If we prove that then this B is contained inside O it will be contained S O also ok. But since this is a balanced neighborhood now B will be contained inside twice B contained inside 3 times B and so on. So, this will be an increasing frequency it is because expanding, expanding ball is just like balls you see. See in R n what you have done? You would have taken a ball with radius 1, you will be content is concentric ball right. So, contained inside twice of that contains 3 times that and so on. This is precisely the property that I was hinting at. And from lemma 5.241 whatever we get K which is you see increasing union of this one ok. So, remember if we have an increasing union of any neighborhood where these numbers go to infinity then this whole thing is covering the entire vector space just like in the case of R n and so on anyway. So, this is what is happening in a topological vector space also that is 5.24. K is a subset of V anyway, but V is contained in the union of N this. So, that is 5.24. But K is compared and these are open subset. So, we will get one of them covering the whole thing because it is increasing union ok. So, we get N such that K is contained in N B. Now, you take M equal to this N if S is bigger than that again using balance these balance of neighborhood K will be already insert N B, K will be also insert S times that one ok because S is bigger than M here right. So, this M can be chosen as N. So, one way increasing union you have seen. Now there is another parallel decreasing union, decreasing intersections balls of smaller and smaller radius right. So, similar thing here if B is a bounded neighborhood of 0 then 1 by 2 power N B are N range over all the natural numbers forms a neighborhood system for 0. It is just like the ball B epsilon where epsilon times 0 is a neighborhood system. Instead of that you can write 1 by B 1 by N also or then you can B 1 by 2 power N also right. So, similarity this is what we are going to prove 1 by N etcetera epsilon times 0 etcetera are more difficult just this is nice easy thing and it is enough because this sequence converges to 0 ok. So, let U be any neighborhood recall by definition of boundedness what implies boundedness of B implies that there is another way of looking at it instead of there exist M such as bigger than M etcetera you can invert the whole thing there exist epsilon positive such that delta less than epsilon implies delta B is contained inside U ok. If B is contained inside S U where S is bigger than M is one definition equivalent to delta less than epsilon implies delta B is contained inside U this epsilon can be chosen like this. Once you have that all that you have to do is 2 power minus N less than epsilon ok. Then 1 by 2 power N B for some N large they namely this much that will be contained inside U for every neighborhood some member is here is inside U means that this is the neighborhood system over ok. So, we shall use all these things in a neat way to you know derive some nice results. Another easy and fundamental result here is take any 2 vectors ok take any 2 vectors let X be any topological space and alpha beta from X to K be any 2 continuous functions ok. So, alpha and beta continuous functions into the scalar the case is scalar field on which V is a vector space ok. Then you take the linear combination of alpha X times U 1 that is a vector inside V plus beta X times U 2 it is a linear combination right, but it is a function now as X varies on X capital X into V this function is continuous. These are elementary things which we have observed inside R n and so on right. So, same thing we are observing in any any topological vector space what is the proof proof is precisely this one namely scalar multiplication and additions are continuous that is all. Now, we have an easy corollary here every linear map K n to V is continuous where n is any positive integer is a special case now ok. X was an arbitrary space ok we produce alpha and beta continuous functions we assume then the linear combination is continuous. Now, I am going to say every linear map from K n to V is continuous. So, for to talk about linear maps you have to have the domain also as a topological vector space, but you cannot repress this one by arbitrary topological vector space. An arbitrary topological vector space to arbitrary topological vector space linear maps may not be continuous ok that is why this theorem this corollary is important. It says that K power n which is nothing but a finite dimensional vector space over n then every linear map is continuous. Understand the importance of this one. If you replace this one by arbitrary where topological vector space then it will not be it makes sense, but it will not be true. Proof is easy similar to what we do for R n and so on ok. Once we have that lemma ok, let F be any linear map alright. Any linear map into vector space from vector space to vector space is determined by its values on a basis ok. So, let E 1 e to E n be standard basis you could you could have chosen any base no problem. Then I will look at the image of E 1 e to E n under F. So, put U k equal to F of E k ok. Now, every element F z belonging to k n can be written as z 1 E 1 etcetera z 10 E n where the dyes are scalars inside k right. Then we know that by linearity F of z is nothing but z 1 times F e 1 plus plus plus z n times F e n right. Not only that the coordinate function inside k n to k z going to z k this is a k th coordinate of z right. They are also continuous function. Therefore, all that I have to do is iterate you know the previous lemma for two of them ok. z going to z 1 is a continuous function times E 1 plus z going to z 2 is continuous times E 2. So, add them. So, first two addition is continuous, next third one is continuous and so on. So, what you get is z going to z 1 E 1 is what? F of E 1 etcetera z n into E n is what? F of E 1. So, that is continuous, but that is precisely the value of F that right. So, F is continuous, F is expressed as sum of n functions each of them is continuous alright. So, easily proved, but a non trivial result, an important result that finite dimensional vector space is linear maps are continuous. You know as a student for me because we were all the time studying finite dimensional vector space is linear maps are continuous linear maps are continuous. But suddenly when it goes in infinite dimension and linear maps may not be continuous that was a realization of ok. So, there is no way. So, it took some time I really thought that I can prove that every linear map is continuous, but that is not the case ok. So, I won't emphasize that fact here. So, let us stop here today. So, next time we will reap a good harvest and prove three important theorems and that will be the end of the course. So, today we will stop here. Thank you.