 Welcome to module 49 of point set topology course part 1. Last time we introduced the notion of half-dough spaces and proved some very important theorems. One of them was the characterization of homeomorphisms. Characterization means in a particular cases that's all namely if you have a bijection from a half-compact space to a half-dough space then it is homeomorphism. So there are other cases also that we have proved. So let us try to give some illustrations of usefulness of these concepts and theorem. So we will go back to the study of quotient spaces now. Let X be a topological space and J be any interval of the form A, B closed or A closed and B open and B could be infinity also, okay. So I will instead of writing A, B, A, B and so on I will just write it as J it will represent any one of it. By a cone over J we mean the quotient space of X cross J by the identification X, A is identified with X prime, A for every X and X prime inside X. And no other identification that is the meaning of when you define a relation by declaring some rule, the rule is only this one. After that by reflexivity, transitivity, symmetry, etc. the relation is completed, okay as an equivalence relation. So in this case you know by reflexivity every point is connected that you have to take it as a definition, it is not there, symmetry is already there here and transitivity is also obvious. So only points X cross X cross singleton A, they are identified as single point, a single class here, okay. So that is the meaning of this cone over X, the quotient space will be denoted by CX, okay. So the image of X cross A under the quotient map is called apex of the cone, okay what single point it is then entire X cross 0, so that will be called apex of the cone. The image of X cross T for any T not equal to A, see there is a point other than A, they are all as it is, as they are there is no identification. So X cross T to its image is a homeomorphic copy of X, okay and it will be contained inside the cone under the quotient map it is again homeomorphism and this may be selected to be called the base of CX, why I am saying selected to be because there is no definiteness, you could have taken any T other than A, all of them you can say that that is the base of the cone. In practice especially when J is a closed interval we can take T equal to B, okay we select X cross B as the base, so there there is a definiteness, okay in the case of open cones namely when B open or B infinity and then also it is open anyway there is no definiteness, so there is also these terminologies open cone and closed cone depending upon whether you have taken a closed interval here or an open interval, not an open interval half open interval JAB, corresponds to this A, B open will correspond to open cone and open cone does not have a unique base you could have taken any T they are all equal and they are same purpose whereas in the closed cone you can take the last point here B and X cross B will be the base, the prototype of this one is when X itself is a circle then you must have seen all the pictures of right circle cone and so on in 12th standard you study the cones also, okay, conic sections and so on. So the definition is generalised from instead of X being circle you could take any topological space and do this, okay the cone construction is one of the very very important thing in especially in algebraic topology also in topology, okay it has some canonical properties which we shall describe what is the meaning of this canonical property, suppose you have a function from X to Y, okay any function then it will define CF from CX to CY, so you have defined the cone over a space now I am defined the cone over function here cone of CX, so what is CF? CF is a map from CX to CY, first you define it from X cross J to Y cross J to be f of XT equal to fx, t and then they take the class, okay here also you have take the class CF of XT going to fx, t remember this class in the same point XT unless t is the first point A and when it is first point A all these XT represent the same point but here also they will represent same point, so there is no problem here to in the definition of this function CF, okay the point is if f is continuous then f cross identity continuous then the induced map CF this will be also continuous not only that there are many other properties I want to list them, the CF piece has problems all the following properties, if f is continuous then so is CF, if you start the identity map from X to A, C of the identity is identity of CX, if you have X to Y and G from Y to Z then the cone over G composite F is nothing but cone over G composite cone over F, if F is a homeomorphism then so is CF, okay. So 1, 2, 3 are all easy, the 4 follows from 3 by taking F from X to Y and G equal to F in moths, if you put F in moths from Y to X then C of G composite F is identity, so that is identity on this side also it means CG is the inverse of CF, therefore CF will be homeomorphism, okay. So these things are easy to verify there is no problem about them, okay these are called canonical property, now I will come to specializes specific examples, suppose you start with S naught that is the zero dimensional sphere which is just unit vectors unit numbers inside R is minus 1 plus 1, the topology is discrete topology on that, what will be the cone over that, to be very specific now I take J equal to the closed interval 0 to 1, okay. What will be the, what will be the cone of CGC S naught, it is easy to see that first you take S naught cross J, so that will consist of two copies of the interval, closed interval 0, 1, 0, 1 minus 1 cross 0, 1 and plus 1 cross 0, 1 they will be disjoint copies, right. But when you identify which one what is the identification, I have to take A to be minus 1 here, right, A to be 0 here, right, so minus 1 all the points, minus 1 plus 1 cross 0 they will be identified, nothing else, that means what, the bottom points of both these are 0, 1, 0, 1 is the R, you hold them vertically, okay, so minus 1 copy is there, plus 1 copy is there, those minus 1 and plus 1 cross 0, 0 they will come together, so you will get a V shape, right, but V shape is homomorphic to minus 1 plus 1 closed interval from minus 1 plus 1, okay. So, this is the very simplest case, as such I have already told you that if you take S 1, S 1 as a circle, then the cone over that one will look like actually an ice cream cone or a funnel and so on, okay. If you flatten it out, what you will get is a disk, the two dimension disk, so that is that point I will explain it will be more clearly and in general generality for all n, so more generally put X equal to Sn minus 1, so n equal to 2 this should be just the circle, but I am now considering general case together, all of them together, n equal to 1 is over and J equal to 0 to infinity instead of the closed interval I take 0 to infinity. Consider the map theta from X cross J to Rn given by theta Xt goes to t times x, so this is a unit vector I am multiplying by a scalar, the scalar varies from 0 to infinity, clearly this multiplication map is a continuous surjection, every point inside Rn, okay, take a vector divided by its norm that would be unit vector that will be inside Sn minus 1. What is the corresponding t? It is just the norm, you do multiply it by the norm you get back the vector v, okay, so this is surjective map, okay, it is a bijection if you take nonzero vector and and don't take t equal to 0, that means on the open interval 0 to infinity cross X it is a bijection, okay. And where does it go in images? Rn minus 0, okay, so this is precisely what you call as polar coordinates in the case of n equal to 2. At 0, you know 0, any vector v that will represent 0 into v which is just 0, so that point is over represented, okay, there are too many points which represent that point, but everywhere else there are unique representations, so eta of X0 is 0 for all X in X and hence eta induces a continuous bijection when you pass on to the cone. See whatever eta was taking several points to sink same point, all those several points, various points namely X, 0, they are all identified in CX, two single points, therefore the eta hat from CX to Rn, this is injective map also, it is already surjective, this eta is surjective, so eta hat is also surjective, so this is a bijection, okay. It is a continuous bijection by the very definition of how to check continuity on quotient spaces, because eta is continuous, so if you write Q as X cross J to CX as your quotient map, so eta hat composite Q is eta, so it is continuous means eta hat is continuous, okay, so eta hat composite Q is eta, Q is the quotient map, that is what I am denoting this one, okay. So why this is a homeomorphism, right? If you knew that this is compact, this you know it is hostile, then you are done, right. In particular, instead of taking infinity here, suppose I take a closed interval 0 to 1, 0 to a whatever a positive, okay. Then I would get a homeomorphism, but I will not get on to Rn, what I will get? Suppose I take this one, it is 1, then I get all vectors of length less than or equal to 1, which is nothing but dn. If I take R here, I get all vectors of length less than or equal to R, which is the closed interval Br0, right. So our theorem already says that all these balls are cones over, what are the basis there, what is the last thing, when this second coordinate is equal to R, that will be the sphere of different radius, not Sn minus 1, Sn minus 1 you get only when R equal to 1, okay. I want to say that even this infinite cone Cx is homeomorphic to the entire Rn, okay. So there are different ways of seeing this one. The point is continuity of the inverse, this part, okay. If you throw away the 0 from the 0 from here, 0 from here, that is obvious because there it is eta xt equal to tx unique. So how do you get xt? You see it is just the first coordinate is x whatever we have got a norm v, the second coordinate is norm v. So both of them are continuous there. I can divide by norm v because v is a non-zero vector, but if it is non-zero vector, you cannot do that. So there is a doubt when you extend it to Cx, why this is continuous, okay. Sorry, this continuity is okay, why the inverse is continuous? There is no justification there, okay. So let us prove that one rigorously once for all so that you want a very doubt left in your mind. So we claim that eta has a homeomorphism. For this we need to check eta hat is an open mapping or a closed mapping, okay. Let u be an open subset of Cx, okay. Put v equal to q inverse of u. See I start with a subset here, but then I go to q, q inverse of that. I am coming to x cross 0 infinity, okay. Close and down 0 infinity. Here I am coming, okay. That is my v, v equal to q inverse of u. We make two cases. Suppose the apex point x cross 0 is not in u. That just means that v is completely contained inside x cross open interval of 0 infinity and q from v to u is a homeomorphism. This case is very easy. This we have already analyzed. Since x cross 0 infinity to Rn minus 0, eta is a homeomorphism, eta v is open, okay. And in this case eta v is same thing as eta hat of u, okay. So this part is okay. Second case is the important case. Namely suppose x cross 0 the apex point is inside u. The singleton x cross 0 is a closed subset of Cx, okay. Why? Because the rest of them is open. That is easy to see. Or inverse image of x cross 0 is just x cross 0 the entire sector. Something is open, something is closed if you notice inverse image is closed, right. So x cross 0 singleton is closed. q inverse of that is x cross 0. So that is closed. So this is closed. Therefore it suffices to show that eta u is a neighborhood of 0 in Rn. This point is coming to that point, right. So rest of them is no problem. So its only problem is why it is a neighborhood of this x cross 0, okay. It is a neighborhood of 0 in Rn. We now use the compactness of x equal to Sn minus 1 and Valais theorem to get an epsilon positive such that this w is equal to x cross 0 epsilon, okay, is contained inside v. See you have a axis Sn minus 1 is compared. Cross 0 to infinity you have, okay. So this is your y and then use Valais theorem. If you have an open subset containing x cross some little y, then you have a neighborhood v, okay. If you have neighborhood v, then you have a neighborhood of what it is? A ball, epsilon ball here such that this x cross 0 epsilon is contained inside v, okay. X cross 0 is contained inside v. It gives you a small epsilon. So this is the Valais theorem applied to x equal to Sn minus 1, okay. Under eta the image of w is nothing but the open ball v epsilon 0. This is a unit vector. I am multiplying it by some number between 0 and epsilon. So it will give you a vector of length less than or equal to epsilon less than epsilon. All vector less than epsilon are inside this one. So this is an open ball. Image of this one is open ball. Which is clearly contained inside eta v which is eta hat of v, okay. So once this is inside v, eta of that one will contain eta v which is eta hat of v. So starting with an open subset u inside cx, we have shown that eta hat of that is open, okay. So open bijective continuous map is an open map. So I am making this remark which I have already told you. Yeah. Would you please explain it again how Valais theorem is applied there? Valais theorem, we have to remember what is Valais theorem. If you have a topological space x which is compact and any other space y, okay, then you look at x cross singleton y inside x cross y. Suppose it is contained in some open subset v, x cross little y contained inside v, okay. Then there is a neighborhood of this y, okay, one single neighborhood of this y such that x cross u is contained inside v. Yeah. So here that neighborhood is 0 to epsilon. Because it is 0 to infinity, this space y here I am implying it is 0 to infinity, 0 closed open infinity open. So how do you get a neighborhood whatever neighborhood of 0, they will contain 0 epsilon, okay. Yes. So I have already told you that when n equal to 2, this is the popular you know representation of r2 or complex numbers in polar coordinates. Unit vector in the complex numbers you can write it as e power 2 by i t. But if you r3, r4 and so on you cannot write that way, just write a unit vector. So then this is also polar coordinates inside rn. Every non-zero vector is uniquely termed as what v equal to t comma u where u is a unit vector. What is this u? This u is nothing but v by normal. What is t? It is normal. As soon as the vector v is 0, you can we have to take t equal to 0, but u could be anything. So that much of ambiguity is there in polar coordinates. But if you think of this as a cone, there is no ambiguity. Now I want to do one more serious thing here. The above discussion is valid verbatim in any finite dimensional normed linear space m comma norm. So we are doing it in rn, but the same thing holds for every non-linear space which is finite dimension. Where we replace x by the unit sphere in m with respect to the norm. If the norm is l2 norm, you will get the u standard sphere. If it is l1, what you get? You get diamond shape. If it is l infinity, you get a square and so on. So all these things we are saying. Even if it is arbitrary, any norm, none of these LPs, this statement is true is what I am claiming. What is the missing part? The only missing part is that perhaps is y axis compact. X is what? Unit sphere. Why x is compact? Because m is finite dimensional, we want to say that x is compact. So I will, we might have already seen this one, but I will complete this one, this argument. First of all recall that by elementary reading algebra, it follows that any finite dimensional linear space is isomorphic to some rn, where n is the dimension. Also we have proved that any two norms on rn are similar. Similarity preserves boundedness and closeness. Therefore, this norm, whatever norm I do not know is the unit sphere with respect to this norm is closed and bounded subset with the l2 norm also. In the l2 norm, closed and bounded subsets by Heineborel theorem is compact. So there is nothing missing. There was apparently missing information, but everything is available. Therefore what we have is that whatever analysis we did namely c of x is homeomorphic to the whole of rn that is valid inside any norm linear space, finite them in norm linear space. Now we come to the converse of this one, namely if the sphere is compact in a norm linear space, then what? Then m itself is finite dimensional. That is the next theorem that we are going to prove. This is a standard result in function analysis. So let us prove this one as an application of whatever we have done so far. A norm linear space m comma norm is finite dimension if and only if the unit sphere in it is compact. So let us denote s, denote by s the unit sphere. Suppose it is compact, then you can take half, balls of radius half around each point that will open cover. So that should admit a finite cover. So there will be finitely many points x1, x2, xn inside s such that s is contained inside union of all the balls of radius half centered at xi i raised to 1 to k. I have got some points inside a vector space m. So I can take the linear span of them. Let L be the linear span of x1, x2, xn. So this is a linear subspace of m. We want to claim that L is equal to m. If we prove this one then it follows that m is finite dimensional. Dimension may be equal to n or less because these may not be independent. They span L. So dimension will be less than okay. So why L is equal to m? So the proof here is very important the step. I would not have much time to spend on that one but in functional analysis they do much more elaborately about first order quadratic approximations and so on. So all those things will be easy for you once you learn what is going on here. Namely suppose you have a proper subspace. This part itself is an independent thing now. Suppose L is not equal to m. That is L is a proper subspace okay. Say x is in the complement of L. I want to find the nearest point to x inside L. So that is the approximation nearest point okay. So I do not want to elaborate that one. So what we want to do is L being a finite dimensional okay. This is a complete and a close subspace of m. This part we have seen okay. Every finite dimensional non-linear space is complete. Once it is complete it will be automatically a closer space okay. So consider the function, the distance function given by the same norm nothing else. From L to R given by distance of y and d of y is distance of y from x. x is fixed x. So this is a continuous function. It is just equal to norm of x minus y okay. Take d to be the infimum of all these d x y's where y range is over L okay. So infimum y, infimum is make sense. First of all see the distance to some point is already finite. So that is fine right. So all they are all bounded fine. So these non-empty and so on. So and distance function is always bounded below okay. So infimum makes sense. Why infimum is positive? If the distance is 0 okay then we know that x will be inside L because L is closed. So distance must be positive okay. So this d which is infimum of d x y is actually the distance of x from L okay. I have just recalled this on here. So that is positive because L is closed and x is outside L. So then there exists a sequence y n inside L okay such that distance between x y n converges to d. What is d? It is infimum. By the definition of infimum you must have points here converging to that point right. So those points are nothing but d of x y n. They are real numbers d of x y n converges to d okay. It follows that y n is a Cauchy sequence. See d x y n converges. These are real numbers but this implies y n is a Cauchy sequence okay. But y n's are inside L which is complete. So y n converges to some y naught belong to L. This y naught is the one which realize is this distance. It follows that d equal to d of x y naught. The limit will be d, t equal to d x y naught okay. When you take the limit y n comes to y n. So infimum is actually minimum and it is attained all right. So in fact now you see that this y naught in as such may not be unique in general but in this case if I only say linear space it happens to be unique also but we are not interested in that part. So we have got a point such that d equal to d x y naught with the infimum. So here the picture. So what I have taken? This is L. I am assuming that it is smaller than the whole space m. It is not the whole space. This is the origin okay. This is symbol of maybe positive radius whatever all right. This is my x and on L I have located y naught. So these are y 1 y 2 y n converging to y naught okay. It turns out to be if you know what is the meaning of perpendicular and so on. It happens to be like that but this happens only in an inner product space. If you have just a normal near space you do not have the notion of angle. So there is nothing like perpendicular and so on. So you do not have to draw these pictures perpendicular and so on okay to be very precise. But what you can do is you look at x minus x x minus y naught that is the positive that is the non-zero vector right. You divide by its norm that will be a point here on the unit sphere okay. So this z is nothing but x minus y naught divided by its norm okay. So that will be a point of the unit sphere. So this is what I am doing here okay. So next slide put z equal to x minus y naught divided by norm. Then norm of z is 1. Therefore, z is one of the b 1 by 2 xi's right. It must be one of the b 1 by 2 xi's because the entire of the sphere s is covered by these balls for some i equal to 1 to k. Also for any y in L we have norm of z minus y we can rewrite it as z is x minus y naught divided by norm of x minus y naught right. So I am pulling x minus y naught is here but I am pulling out the denominator outside okay. Then I have to multiply by that number x minus y naught times y. But now y minus this one is a linear combination of y naught and y. y naught as well as y are inside L. So this whole thing is inside L right x minus y naught plus this thing. So that is an element of capital L. Therefore, this norm minus this one must be bigger than or equal to d and then divide by this this numerator denominator is there. d divided by norm of x minus y naught okay. So that is equal to 1 because d is nothing but the norm of d is equal to norm of x minus y naught distance between x and y naught okay. So what I have shown is that for every point y the distance between z and y y inside L is bigger than or equal to 1. So that is the absurdity because all these x i's are inside y right. We have chosen those things inside y. So that is the problem. So that is why this picture is funny because it is absurd picture. So this ball should contain one of these balls should contain z but they are all of half radius but this is at the distance away from that okay. So this is the contradiction. So that contradiction because it is not the whole space. So once again this method of minimizing this norm etc is important in elsewhere but we have used it to prove that compactness of the unit sphere implies that the norm in space is finite dimension okay. So next time we will continue with applications of this nature okay. More and more examples study. Thank you.