 Welcome to module 9. Today, we will concentrate on topology of Rn. Of course, we cannot cover the entire topic as such. A few glimpses of what is happening inside Rn including R, R2, R3, etcetera, all together one single thing. I am not making any separate cases here. So, let us recall that n dimensional Cartesian coordinate space consists of n tuples of real numbers x1, x2, xn, xi's are inside R. Euclidean metric you have denoted by d2 of xy is nothing but the norm of x minus y then the two norm which is the square root of the sum of the squares of the differences xi minus yy. Recall that a closed disk was defined as set of all points which are less than or equal distance less than or equal to r from a given point dr of x. So, all points in y such that now I am using this distance I can drop out this one because for now onwards unless I mention this it is always the Euclidean norm. So, I could have done just d of xy, but here I am elaborative writing d2 of xy less than or ok. Similarly, the open disk is defined d2 of xy less than less than or only ok. And when that you just take the equality that will be the boundary sphere it is called a sphere of radius R inside everything is at Rn ok. So, it should be n minus 1 dimensional sphere all y belong to Rn d2 of xy equal to r. So, we do not take when r equal to 0 we do not take this as a sphere. So, r is positive is important here ok. The special case when the radius is equal to 1 we will have a separate notation because we have we have opportunity to use it again and again. So, I will just denote it by dn instead of d1 0 is a decorated dn Euler font ok. This is a closed disk of radius 1 centered at 0. Just to indicate that I am working inside Rn I am putting this n here at the top. Similarly, bn is the open disk and Sn minus 1 is the sphere of radius 1 ok. So, this is a short notation. So, the first thing is that the d2 metric here the Euler the Euclidean metric which is coming from the norm has this obvious property namely if you add a vector to both of them namely x as well as y d2 of x plus z comma y plus z is the same thing as d2 of xy. So, that is called the invariance of d2 under translation. The adding z constant vector is called translation. This I denote by Tz. Tz is a map from R into Rn given by x1 to x plus z. Its inverse will be x1 to x minus z and obvious is continuous map which is obviously T and T, Tz and Tz inverse which is T minus z are inverses of each other. So, it is a homomorphism. Likewise for each non-zero scalar you can talk about this Mt multiplication by T the scalar multiplication x going to T times x. This time you have to have T non-zero if you want it to be a homomorphism. If T is 0 this will take an entire thing to 0 is constant function 0 right it will not be homomorphism that is all. So, what is the inverse? Inverse is T going to x going to T inverse. So, if you look at T of z minus x maps dr of x dr of x center x subtract x and add z that will go to into dr z. The radius does not change that is precisely the point here namely d2 of x plus z y plus z d2 of x y. So, this ball this disk closed disk of radius R same equal to this radius R but centers have changed. So, all closed disks with radius R they are homomorphic under Tz minus x. Similarly, you can fix 1.0 center and multiply the use of multiplication by arbitrary positive numbers m s by R here then it will be dividing by R and multiply by s. So, the radius R will become s. So, dr0 is mapped to ds0 and vice versa you multiply by m R by s that would be inverse. So, dr of x s ds of y now you can take they are also homomorphic to each other because you can compose it first multiply and then translate and so on. So, the conclusion is that any 2 disks any 2 open disks are homomorphic any 2 closed disks are homomorphic any 2 spheres are homomorphic. So, first you understand what is going on here. If you put equality everywhere you will get spheres you put strictly less than or strictly less than s you get everywhere open disks less than required to be get closed disks. Argument is the same same maps Tz minus x and m s by R will work for all of them. So, they are all homomorphic to each other. By the way all these are defined on the entire of R n these functions these homomorphisms are defined on the entire of R n. Just to see that this is homomorphism we may not need it there may be many other homomorphisms. But this is the easy way of seeing that they are homomorphic to each other. Now I come to another important one namely from the whole of R n to the open ball of radius 1 centered at 0 I have this homomorphism. What is it? It is x divided by 1 plus norm x square this is again 2 norm I have taken 1 plus that I have taken then I have taken the square root here raised to half. You can just directly check that this map here y going to y divided by 1 minus norm y square raised to half this is the inverse of that. This map is defined only on the open ball strictly open ball because as soon as norm y square is 1 this is not defined you can 1 minus y norm y square will be 0. So, that indicates that as y approach the boundary this map this value goes to infinity. So, they are not there in the domain. So, this is the inverse of that that is directly you can check. So, that will give you that R n is homomorphic to b 1 you combine it with whatever we have seen here all open balls are homomorphic to the whole of R n. It is very important here that you take R the open ball not the closed ball ok b 1 0 not d 1 0 ok. So, b 1 0 had the notation here but we have no we did not have notation for that ok. Now, I come to different sets instead of round spheres and discs and so on now I will take squares rectangles ok all these things I call in general in R n as boxes. So, in this terminology an open interval is a box open interval cross open interval is another box open interval cross open interval cross open intervals is usual box in R 3 but I am using the same word for all n ok. So, I rank to 1 to n a i comma b i open these are all open balls open boxes ok they happen to be open subsets of iron also very easy to see similar to what we have seen you know take any ball take any point inside that at that point you can give an open ball completely contained inside the original ball centered this time at the point you have chosen. Similarly, for these open boxes also you can do therefore, these are all open subsets first of all ok. So, what I want to do is that any box like this any two of them they are homomorphic to each other. So, first of all you should check that we have already seen that n equal to 1 any two open intervals are homomorphic remember that ok. So, you can use that put several of these a i b i to say c i d i here ok then take product of these f i's which are homomorphic from a i b i to say i d i that will give you homomorphic in this way ok yeah. So, instead of that I am going to write one more namely this is just the way we have done earlier also namely from minus 1 to plus 1 to a i b i you write down this one f i ok then take f is product of f i. So, if all of them are homomorphic to this j n this another notation I am going to use namely interval minus 1 plus 1 taken n times product taken n times ok. So, if that is homomorphic to any open interval any open box like this any two open boxes themselves will be homomorphic to each other ok. So, finally, I want to say that open boxes and open balls closer boxes and closer balls they will all be homomorphic to each other ok. So, here is a thing here I should have taken open ball here I do not have notation for this one we have here b b 1 nice yeah b 1 0 would have been the right thing here ok yeah. So, I am going to give you a homomorphism from j n to b n 0 to put d n it will be only into the interior because j n I am taking this as open here ok. Of course, I am taking usual topology on both sides the Euclidean topology on both sides ok I want to define a homomorphism here, but this should be the b 1 0. So, we shall do this geometrically illustrated by fine you can write down a formula also there is no problem by using convexity of both the spaces. So, this method will be used later on for different things also ok. What is the meaning of convexity inside any vector space a subset is called convex if given any two points inside that subset t times this vector plus 1 minus t times that vector which is line segment the line segment joining any two vectors the entire line segment must be inside a subset that is the meaning of convexity ok. I am just recalling it what we are going to do is use the fact that they are both star shaped with the apex point at the origin the star shaped means little less than convexity here one of the point is fixed set that is called the star from there all the lines which are emerging from that point to any of the point of the space of the given set just entire line segments must be there like if you take the union of minus 1 plus 1 open interval along with minus 1 plus 1 on the y axis x axis and y axis the whole of x axis y axis union that is not a convex set, but it is star shaped at the origin ok so that is star shaped definition of star shaped with a apex point at the origin that means I am writing it down here every nonzero element of dn lies on a unique line segment 0 u where this u is a unit vector ok. So, this is the property I am going to use this property can be stated a star shaped if you do not understand this property will tell you what I am going to use so that is enough given any point p in dn p not equal to 0 p equal to 0 is fine p not equal to 0 there is a unique line segment the line segment 0 p will do that ok that is contained inside dn because for all these points the norm will be less than 1 the distance between 0 and that point will be less than 1 so that is the criteria here moreover ok if every point in x minus actually every point the x minus t minus 0 that is not equal to 0 lies on a line segment like this ok for a unique p belonging to s minus 1 s n minus 1 namely take p equal to x by norm x ok if you take x by norm x what is the modulus of this this again you say 2 norm here ok this modulus will be 1 so that is a point of s minus 1 the norm will be 1 so it is point of s minus 1 ok 0 to p will contain this point x because to come back to x what you have to do you have to multiply by its norm of x fact so here is a picture ok this I am going to use later on so I will come back to this picture a little later on ok every so what I want to do is now the the same statement which I did for the sphere I can do it for the rectangle a closed rectangle ok take a point on the boundary of the rectangle join it to the center every point in this inside this rectangle other than 0 will actually lie on a unique line like that so that is a claim ok exactly same way you cannot do that for the the j n bar this j n bar is minus 1 to plus 1 raised to n ok that is the closure this I am taking close the close intervals here if you replace s n minus 1 by the so called boundary of j n what is the boundary of j n you see in r 2 the boundary consists of this line this line and that line right the 4 sides so how do you define that here the x 2 coordinate is 0 right here x 2 coordinate is 1 here the x 1 coordinate is 1 here x the coordinate is 0 so that is what you have to do one of the coordinates is either sorry this whole thing I would take so this is minus 1 to plus 1 not 0 ok this is minus 1 and that is plus 1 so that is what you are going to do so xi is plus 1 or xi minus 1 xi plus minus 1 is 0 take the product that is 0 the same thing as one of them at least must be 0 which is the same thing as xi is plus 1 or xi is minus 1 ok that is the boundary of this j n the boundary of j n bar also ok even if you do not know the word boundary in general does not matter this is the definition here for the boundary of j n notice that j n can be defined in a different way all points x belong to r n such that the maximum norm is less than 1 ok this rectangle here in r 2 can be defined as see minus 1 to plus 1 cross minus 1 to plus 1 or you can just defined as all x comma y such that the maximum of modulus of x modulus of y is less than 1 so that is precisely the maximum norm here ok and j n bar is all those which is less than or equal to 1 the maximum norm is less than equal to 1 the boundary is all those in which maximum norm is exactly equal to 1 ok at least one coordinate must be exactly equal to 1 I mean modulus of that coordinate so it could be plus 1 or minus 1 that is precisely what I told you here how to how to get this j n which I have wanted here ok so these are some elementary things descriptions of some interesting subspaces of r n ok so one thing which you can observe is the l infinity norm taking the maximum of modulus of x 1 x 2 x n or taking the square root of the sum of the squares etc both of these are continuous functions in r n what is the meaning of r n here with the usual topology and these are coming into r plus 0 to infinity so there also I am taking standard topology with that these are continuous functions because each modulus of xi is continuous taking the maximum continuous that is what here the difference is continuous taking the square is continuous taking the sum is continuous again taking the square root is also continuous because these are non-negative ok wherever you are taking the square root the inside thing is a non-negative quantity so this part is all continuous ok now comes look at the fraction here 1 x divided by 1 divided by norm x infinity ok these things are continuous wherever the denominator is not 0 you take a function f which is continuous 1 by f will be continuous wherever f is not 0 that is the thing I am using here from elementary real analysis ok these are real value functions take x going to 1 by norm x infinity or x going to 1 by norm x square they are continuous on r n minus 0 because at 0 they are 0 everywhere they are positive so it follows that the following two formula eta 1 x which is x divided by norm x square now I am multiplying this by x x is continuous so multiplication of these two that will be continuous eta 2 similarly is x by norm x square that is also continuous so they define continuous functions now where are they I start from the unit sphere then I take eta 1 unit sphere means norm of x is 1 the two norm of x is 1 but it is l 2 l it is l infinity norm may be something different I have divided about that means what if I take the l infinity norm of this one now it will be equal to 1 therefore it will be in the boundary of J f similarly here start with anything on the boundary ok its l 2 norm could be different so divide by that now we should take the l 2 norm of this whole thing it will be equal to 1 so you have landed inside s n minus 1 here so these two maps are continuous ok what I want to say is it is elementary geometrically to see that they are inverses of each other in fact take eta 1 of this ok now what is eta 2 of this one I have to take whatever here and divide it by its l 2 norm ok this is some real number so l 2 norm of this is l 2 norm of the numerator divided by this number ok but what is the l 2 norm of this one see I want to say that it take eta 1 of whatever from starting s n minus 1 then apply eta 2 on that one you come back to s n minus 1 ok so we want to say x goes to x ok when you divide anything by l 2 norm it will become back to x because norm of x were already equal to norm of l 2 norm of x was already equal to the 1 there here ok so here is the picture that is what I wanted to say picture demonstration of this one take a point on the sphere unit sphere ok extend this line segment so that it hits the boundary of j n how do you get that on this line you have to take only one element there is only one element in the same line segment in same ray the positive ray which has its l infinity norm equal to 1 and that is this point ok because there is a unique point which will intersect this one so x goes to eta 1 x now start with this one if you divide it by its l 2 norm you could find this one because that is the only point which l 2 norm equal to 1 this is the only point so that is the inverse of eta 2 of this point would have been this one so this is this is why eta 2 of this one ok so geometrically it is very easy to see but you can do just you know computation loss ok so from the boundary to the boundary I have a homeomorphism here because they are inverse of each other ok now what I do I look at phi 1 from the interior the entire thing d n to j n bar so I define this by a formula phi of t x remember every point here looks like t times x where x is a point on the boundary right except when the whole thing is 0 this x will be indeterminate you could have taken any point otherwise it is a unique point everything can be written like this so x is an s n minus 1 and I do not want the 0 element here so 0 less than equal to t less than strictly less than t less than 1 ok 0 strictly less than t less than equal to 1 so this will give you all points other than 0 those points you take t acquired by the infinite norm infinite norm of x at point 0 I have to define something so for that you define it as 0 so it is nothing this which nothing but remember it is eta t times eta 2 of x so I have defined defined by extending this this eta eta 1 eta 1 I am extending I am getting phi 1 extending how by putting 0 going to 0 ok the problem is that whenever you define it by two different formula you have to verify the continuity of this now continuity is now what why because this is only one point I am worried about by taking the limit of this as t tends to 0 if the limit is the same is the same value here then the function is continuous ok so that is what I am using here elementary calculus so we need to check the continuity of phi 1 only at the origin ok elsewhere it is given by this formula which is which is continuous already we have we have seen so at the origin so what you want to do consider sequence t m x m converging to 0 see x m's are inside s n minus 1 they are not fixed they may be different points ok but t m's are real number when does this sequence convert to 0 if and only if t m converges to 0 ok because norm of x m is bounded by just 1 it is a quite equal to 1 there ok it is bounded sequence this will be 0 only if t m goes to 0 ok these are inside s n minus 1 they do not converge to anything or they may not converge to anything they may be converging if at all inside s n minus 1 if and only if t m is 0 therefore since x m is s n minus 1 this implies norm of x m infinity you see this infinite norm of x m is bigger than 1 by square root of n ok there are n points ok n coordinates each of them is less than 1 by square root of n the sum total square of the sum total will be less than less than 1 right so so at least one of the coordinate must be bigger than equal to 1 by square root of n once one of the coordinate is bigger the infinite norm will be bigger so that is all I am using so the infinite norms are all bounded away from 0 this n is fixed here by the way the m is the one which is changing here the sequence is with respect to the variable m ok and hence fever now take t m x m or t k x k ok does not matter that l 2 norm of that by this definition it is l 2 norm t k t k x k divided by x k norm x k which is which this part is bounded by above by 1 by square root of m so divided by that will be bounded by less than equal to square root of n so it is bounded therefore when t tends to 0 this goes to 0 ok t k over x k by norm infinity the denominator is is what is bigger than 1 by square root of n therefore 1 divided by that is less than equal to square root of n exactly similarly we can define phi 2 as an extension of eta tau here ok and verify that it is continuous the same proof will work there and it will be inverse of phi 1 all all over ok enjoy writing down the full details here all right so this will give you a homeomorphism from square to the circle ok the entire square including this one into the days and then generalize this one this is the picture for r equal to 2 but generalize this one I have never used in this r equal to 2 only illustration in a diagram this is there ok so all these things are varied for all r n all right we shall later on show that the unit disc in r n with respect to any norm remember we have several norms here all these LP norms and so on right is homeomorphic to the unit disc in the in the d n means what now the l 2 norm ok the Euclidean norm here they are all homomorphic to each other but that will take time we will do it next time maybe maybe a little later not next time thank you there are some exercise here which you can go through by yourself they will be rewritten to you as they will appear as assignments and also they will be in the notes so you do not have to worry about that to try your hand at these all these exercises ok so that is it thank you