 Hello, welcome to NPTEL NOC on introductory course on point set topology part 2. So, we continue with the study of compactification. Today module 19, Alexandrov's compactification continue. Last time we introduce this special one point compactification of a locally compact house door space, which will be automatically compact house door space. This is what we have seen. We have seen that any closed subspace of a compact space is compact. This is not the case with an open subspace. However, it is not hard to prove that every open set in a compact house door space is locally compact. This you can do directly. Our last time whatever we proved theorem tells you that every locally compact house door space is an open subset of a compact house door space. It is a corollary to the Alexandrov's compactification that we produce here the other day. Here we shall prove the converse now. Take it of like a space x, x is locally compact and house door. It is not only if it is an open subspace of a compact house door space. So, this is what we want to prove. The only part all that you have to do is why in the larger space you take one point compactification namely Alexandrov's compactification of x as in the previous theorem. Now, let us come to the converse part if part. Suppose x component y is open and y is compact and house door. Clearly x is house door. Every subspace of a house door space is house door. So, we are only left with proving local compactness. So, given x belonging to u, u open in x, we need to find an open subset v. So, that x is inside v, v contained inside v bar, v bar also contained inside u and v bar compact. Since u is open in y, y minus u is compact subset of y because it is a closed subset. Therefore, we can find disjoint open subset v and w in x as that x is in v and y minus u is contained inside w. Now, I am using that y is compact house door space. Therefore, it is regular normal and so on. This means that this v is contained in y minus w is contained inside u. v and w are disjoint open subsets. Since y minus w is a closed subset of y because w is open. So, y minus w is compact. Therefore, v bar will be compact. If it follows that v is as required. v bar is contained in here is closed set which is compact. So, if v is contained here v bar is also contained in here. So, it is compact that is all. So, it is easy part, but we wanted to record this one what happens to an open subsets of a compact house door space and vice versa. As a corollary to this theorem, this is theorem that we have earlier. Let k be a compact subset of an open set u in a locally compact house door space. Then there exists a continuous function f from x to 0 1 such that f of k equal to 1 and f of u c equal to 0. This theorem we have proved earlier. Now, that time also I had indicated that we will have a different proof. Now, this becomes an easy consequence because starting with the k compact subset of an open subset u of a locally compact house door space k contains a locally compact house door space is a situation. But you put one point compactification you are transferring the whole situation into in a compact house door space. Then this becomes an easy corollary that is what it is. So, I repeat we need to consider the case when x itself is not compact. If it is x is compact then there is nothing to prove. So, consider one point compact in x star of x which is which we have seen this one point compact is the Alexander compact is intake. Then k is compact in x star also because k is already compact subset of x. So, it is compact subset of x star also and it is closed because x star is off dog. Anyway u is open in x star since x is open in x star since a compact house door space is normal that is all all the time I am using by Arizona's lemma you get this. So, instead of local compactness you are able to convert the whole thing into compactness by compactification. So, this was one of the themes I had told you why people study compact spaces, compactifications. So, things which cannot be solved inside an arbitrary space you can solve it go into the compact space and then solve it and come back. So, this is just a small illustration of that. In any case this is this theorem itself we had proved directly in 2.14. So, let us examine a few simple examples first. We have already seen that the simplest example of one point compactification is any half open interval contain the closed interval. Note that a, b is a compactification of the full open interval also, but that is a two points are in there two point compactification it is not one point. What is then the one point compactification of this open interval in order to answer this we will actually answer a much more general question. Remember a open interval a b is homomorphic to r itself. So, in general what I am asking is what are the compactifications of r n. One point compactification in particular since r n's are locally compactosed or we are actually talking about Alexandrov's compactification here. So, let us have a clear geometric understanding of these compactifications. Consider r n which is both locally compact and also by the above theorem we know that the one point compactification I am talking about the one point remember is compact or in the geometric description of this space is possible in this situation. We claim that for n greater than or equal to 2 that is an embedding of r n minus 1 into s n minus 1 such that. So, we are only interested in you know s r 2 s 1 single point there is no point. So, r n minus 1 n if you do not do an n equal to 2 which is r r r square r cube etcetera we are interested in such that the image of this embedding is this is just one point namely I am I have chosen the specific point here namely the north pole 0 0 0 the last coordinate 1 ok it is usually called north pole and which minus 1 it will be south pole. That will show that eta comma s n minus 1 is a one point compactification of r n minus 1. All that you have to do is put an embedding of this one in s n minus 1 minus this one clearly this is an open subset which is dense ok singleton n is a closure point of the whole thing all right. So, that will show that it is already an embedding and now we know that s n minus 1 is compact all right later on we will actually show that it is alexander's compactification alexander's compactification ok. So, recall some notations here s n minus 1 is a unit sphere summation x i square equal to 1 x 1 x 2 x n x i square equal to 1 right and n is north pole and u is the complement of the singleton n ok. So, what I am going to do here this is geometrically I will explain it what I am doing it here look at this north pole take any point other than the north pole that is a point of view it just means that this line segment extended whatever to whatever extent you want will hit the the r n minus 1 cross 0 plane here the nth coordinate being 0 why because this line will never be parallel to the r n minus 1 it will not parallel to this one because the point x its n minus 1 coordinate is smaller than this one here only 0 0 0 1 is the only point on the sphere with last coordinate being equal to 1 all our things are less. So, when you join these it will go down somewhere it will hit the this one namely the nth coordinate becomes 0 exactly at one point this line segment the end point is a point on the exact for instance suppose I take a point here when you join this one it will be somewhere here the intersection point of this line segment and the plane ok. So, this is got by merely this picture right. So, what is this function x goes to phi x this is called the stereographic projection the stereographic projection usually, but you can specify this from the north pole ok. So, let us now work out a formula for phi x ok. So, how to write down the formula you should start writing down equation or formula or parameterization whatever is convenient to you of this line an entire line this line is a line joining two points which are which are distinct point let us solve. So, t times this into 1 minus t times that will give you the entire line as t range from 0 to minus infinity plus infinity ok. So, here what I have done t times x plus 1 minus t times capital n as t inside r what is the point that we want we want this point to have its nth coordinate 0 that is the point of intersection of this line l x and the plane r n minus 1 cross 0. So, what is the nth coordinate of this? So, remember this is vector and n vector right t times x is t times x n that is the nth coordinate what is the nth coordinate of this one nth coordinate of capital n is 1 here right it is 1 minus t. So, t times x n plus 1 minus t is equal to 0 that will give you a unique value t which is equal to 1 divided by 1 minus x n. Remember this x n is not equal to 1 that is very important here right. So, this is a valid solution ok you know go back here put t equal to 1 divided by x n minus 1 you precisely get the value of p x ok. Therefore, p x is nothing but the first coordinate divided by 1 by x minus second coordinate divided by 1 by x minus x n minus 1. So, x n and so on last coordinate x n minus 1 to 1 minus x n. So, all coordinate divided by 1 minus the nth coordinate is 0. So, I do not have to write this is a point of number r n minus 1. If you want to write r n minus 1 cross 0 then you have to put last coordinate 0. In order to compute the inverse map we can reverse this geometric argument. So, go back here start with a point here start with a point on this plane. So, for this suppose this is a point now how to get back x very easy namely join v x and n ok line segment intersect the sphere exactly in one point other than n n is already a point of that one ok how to do that in the parameter equation parameter form of this line segment you have to put the condition that the point x whatever we want to solve for lies on the sphere. So, that will give you two solution x and n we already know one solution. So, the other solution is very easy to determine ok. So, that is the geometric way of saying as well as how to solve the how to get the formula also. So, again I am writing y is an element of r n minus 1 cross 0 as an thought of as an n vector with the last coordinate being 0 ok t times y plus 1 minus t times n t belonging to r is the line ok. When you take coordinates take their square take their sum that is equal to 1 is the condition I want this point to be on the sphere that gives you t square plus summation y i square plus 1 minus t square equal to 1 ok all the other coordinates of n are 0 here right. So, only the last coordinate will give you this one. So, this is the same as if you factor out 1 minus t square you can write 1 and 1 cancels out. So, you get t into t into i rank 1 to n minus 1 y square plus 1 minus 2 equal to 0 ok. So, the solution t equal to 0 corresponds to n. So, I do not want that the other solution I want namely cut down this t what you get is this t equal to 2 divided by 1 plus summation y i square which is norm y square. For this t if you plug in here what you get is the point x where y is phi x. So, this is phi inverse y. So, while I am writing a eta this is notation now equal to phi inverse namely eta y is nothing but 2 y 1 divided by 1 plus norm y square etcetera 2 y n minus 1 divided by 1 plus norm y square up to n minus 1 coordinate this will give you that one. The nth coordinate has to be solved now by what by the equation summation y i square equal to 1. So, that will give you norm y square minus 1 divided by 1 plus norm y square ok. Since we have actually obtained this one by 1 1 correspond on this geometrically there is no need to verify that these two are inverses of each other ok that is how I have put this one. If you are not satisfied you can plug this formula for x inside here and see that phi composite eta as well as eta composite phi will be identity maps ok. So, look at the formula they are not only continuous they are differentiable as many times as you want. Therefore, what we have is that both eta and p are diffeomorphisms in their respective domain namely eta from r n minus 1 to u ok e is a diffeomorphism on to u. So, it is embedding inside r n r n minus 1 ok. Thus what we have proved so far is eta comma s n minus 1 is a one point compactification of r n minus 1 ok. So, let us also verify that this one point compactification is actually a alexander of compactification. There are several ways of doing it one simple way is to get a homeomorphism from s n minus 1 to the this is alexander of compactification this homeomorphism which will be eta when you restrict it to r n minus 1 sitting inside here ok. So, all that I want is phi x which is defined on this open subset I extended by just putting tau x equal to star the extra point or whatever you want if you want write it as infinity you can write it as infinity ok. Whenever x is equal to n whenever x is not equal to n it is just p x. So, phi x has been extended like this. So, this function is well defined no problem because for x not equal to n I have phi x phi x is well defined is of continuous I am taking the north pole to the infinity here or star here that is all the only thing we need to verify is continuity of this one when x equal to n ok. So, that is very easy to check that I will leave it to yourself ok. Once you check that continuity it follows that this tau is a homeomorphism because it is a bijection phi is already a 1 1 mapping and though extra point goes to one single point here that is all. So, this is a bijection right bijection from where from a compact space to a half dot space therefore it is a homeomorphism. Moreover tau composite eta is what eta of any y right what is it eta is phi you know now phi of that y phi of eta y is y. So, that is the identity map eta tau composite theta is identity similarly phi composite eta is also once it is phi x and then you composite eta that will be also identity map. So, this proves that eta Sn minus 1 is equivalent to the alexander of compactification of Rn minus 1 ok. So, both ways so here Rn minus 1 is sitting as an inclusion map here Rn minus 1 is sitting via eta which is the inverse of phi all right. So, here we have made a homeomorphism which is an equivalence of of what of compactifications that we have defined. So, the two compactifications are treated as one single because what we have is a compactification is an equivalence class ok in that sense it is the same equivalence same compactification that is all ok. More generally instead of taking the north pole you could have taken any point p on Sn minus 1. The geometric argument will be exactly going d tau the only thing you should not take all the time Rn minus 1 cross 0, but you should take the corresponding whatever vector you take say p take the subspace which is perpendicular to that that is all and then do the Schrodinger projection again as we have done earlier ok. Formulas will be more complicated that is all geometries as as good as in this special case there is no no change there ok. So, all of them would have given you different embeddings of Rn minus 1 inside Sn minus. The case n equal to 3 is of special interest because we can then express eta in terms of complex numbers identify R2 with complex numbers. Then eta can be written in a different way in terms of z suppose z is y1 comma y2 or y1 plus iy2 then our formula here this formula eta will become much simpler actually all these norm etc can be written nicely it is z plus z bar by mod z square plus 1 z minus z bar by mod z square plus 1 mod z square minus 1 mod z square plus 1 ok. So, we are working now with n equal to 3 all right. So, this picture will actually give you the the so called extended complex plane C union infinity will be identified by this map with the S2 that the unit sphere inside R3 ok. As a subspace of R3 eta C gets the Euclidean metric from R3 which when expressed in terms of the parameter z can be thought of as a metric on C itself. So, you are matrizing the complex numbers in a different way ok. So, what is the formula distance between so I am writing as a DC instead of the usual distance DC is the tell you what DC is z comma z prime which is nothing, but the standard R3 norm here the Euclidean norm eta z minus eta z prime norm of that if you compute that eta z using that formula it will be twice z minus z prime divided by 1 plus z square into 1 plus mod z prime square the whole thing square root. So, this is called the called metric and that is why I have put a C here ok. So, when you take two points of C sitting inside S2 all that you have to do is join that chord ok they are inside S2. So, the point join that chord look at the length that length is precisely the Euclidean distance. So, that is why it is called chord metric ok. In particular this metric on C is a bounded metric topologically it will give you same topology and that is so whole idea why it is different metric ok. The same thing you can do for any Rn, but you will not get this nice formula because here we have used specific algebra of complex numbers that is all ok. We shall meet another interesting compactification of Rn minus 1 in chapter 10 namely example 10.15 ok. There are I have told you many examples many compactifications of a given non-compact space ok. So, it is not possible to discuss all of them. So, here is an exercise the hint is what I want to draw your attention to namely we proved a big theorem about locally compact half door spaces ok in 2.40 about the connected components of such a space and so on. So, if you use that one ok. So, you can solve this problem and when n equal to 3 this has a special significance in the case of complex numbers ok. So, it will give you a characterization of simply connected domains in terms of purely in terms of you know without going to the without going to the compactification between inside this one namely c minus say g is a connected domain c minus g has no bounded components. So, this is the characterization which is purely in terms of topology point set topology. So, solve this exercise and enjoy it. Thank you next time we will study proper maps.