 So, let us begin this lecture. So, in the last few lectures we discussed factification of a non-compact and locally compact topological space. So, let us say let me make a remark on what happens when x is compact. So, if x is compact. So, in the results we prove we are assumed that x is not compact, but x is locally compact. So, but what happens when fx, x is compact. So, if x is compact then x is also locally compact. This is because of silly reasons as for each x in x we can take the open set u to be x. So, remember the definition of a locally compact topological space. We need that every point small x in our space should have a neighborhood u says that the closure of u is compact. So, if we take u to be x in this case since u closure is then going to be x and x is compact. So, this will do the job. So, compact topological space is obviously locally compact and if you consider the one point compactification of x. So, then so, we construct the one point compactification in the same way. So, our x hat is going to be x disjoint union one more point p naught and the topology is given in the same way that we did that we did in the non compact case. And we can easily check that it is easily checked that this point p naught will be an open and close subset. So, if our x is for instance this sphere which we know is compact because it is closed and bounded inside r 3. So, then x hat will simply be the sphere disjoint union a point away from this. So, the conclusion is that we have this inclusion of x into x hat then I of x is not dense in x hat. So, in any case we wanted to compact if I think. So, that we get a compact space and so, if our space is already compact. So, then it is not very interesting. So, this is just a remark. So, in the previous lecture we proved the following. So, we have proved the following two results a x is everything ok. The first result we proved is let x be non compact and locally compact topological space. There exists a host of and compact topological space x hat and an inclusion from x to x hat such that the following four things happened x hat is simply I of x disjoint union one more point p naught I is continuous I of x is open in x hat and 4 I of x is dense x hat. So, if x is already compact. So, then everything else will happen only the fourth condition will fail and another result we have proved is the uniqueness of the one point compactification. So, let t. So, let me just correct this. So, in the. So, I had in the previous lecture where we proved this result b I had said that x has to be compact, but that is not necessary. So, let t be a host of space. So, I just want to correct this. So, in the previous lecture had stated this to be compact. So, compactness is not required we only need t to be a host of space ok. So, assume that there is a continuous map j from x to t such that j of x is an open subset and this map x j from x to j of x it is bijective and is a homomorphism. So, then we considered the following map f from t to the one point compactification of x. So, x is a non compact and locally compact topological space. So, x we have j over here and we have x hat this is our I and we are going to define f as follows. So, the picture we had made was the following. So, we took this open disk and let us say we put it into this closed disk this is j. Let us say I take an equator over here. So, this is the one point compactification these are additional point p naught this i. So, this equator is going to look like something like this all the points at infinity all these boundary points like everything towards the boundary is going to get collapsed to p naught right. So, as we move further and further away as we move towards the boundary of these are x out x. So, all those points are going to get collapsed to p naught ok. So, we define the map f as follows if t is in j of x right. So, then define f of t to be equal to i of j inverse t. This makes sense because j is a bijection from x to j of x. So, if t is in j of x then there is a unique j inverse t and I can just apply we can just apply i to it and if t does not belong to j of x then define f of t to be equal to p naught ok. And the assertion of our proposition was then f is continuous. So, let us see some applications of this proposition. So, this proposition we have proved in the previous lecture. So, as an application. So, let us prove that the one point compactification is unique right. So, what do we mean by that? Suppose there are two compactifications. So, we have we have x over here and here we have this inclusion i from x hat i from x to x hat. Now, suppose there is a j from x to t such that we have the following properties t is compact j of x contained in t is open 3 t minus j of x is just one point. So, let us call that point p infinity over here or let us call the q naught one point q naught and 4 j from x to x j of x is a homomorphism. So, in this case t is another compact space which is obtained by adding 1.2 which is another compact topological space and t minus j of x is just this one point and j and x is embedded inside t as an open subset. So, then by our previous proposition there is a map f. So, we may write t as j of x disjoint union this point q naught and we may write x hat as i of x disjoint union this p naught right. So, by our construction definition of the map f we have this map f right. So, here it is i compose j inverse it is a bijection and here it takes this is one point outside j of x and that is sent to p naught right by definition right. So, this implies that clearly f is a bijective map and by the previous result by the previous proposition f is continuous right. Moreover using the result as t is compact and x hat is host of we get that f is a homomorphism. So, here we have used the result use the following result. So, f from x to y is a bijective continuous map where x is compact and y is host of. So, then f is a homomorphism. So, let me make a remark it is important that y should be host of important that y is host of right which is an assumption throughout in the rest of this course unless it is stated explicitly that a space may not be host of because otherwise we can take the identity map from x to x right and give this x the trivial topology right. For instance we can take x to be Sn. So, here we can give x the usual topology and here we can give x the trivial topology right. The identity map is obviously going to be continuous because here the target has the trivial topology right and but obviously this map is not host of even obviously this map is not a homomorphism even though x is compact right. So, before we proceed let us remark that there is an important class of spaces which is not locally compact. Many of the spaces we have encountered in this course they are all subsets of Rn and Rn is locally compact and many of the spaces we will examples we have seen in this course are locally compact yeah, but there is a big collection of spaces which is not locally compact. So, namely so let us see why this is as an application of what we have learned let us see why this is not a locally compact. So, if x is locally compact. So, if you do not know what Hilbert space R you can just forget about this example. So, then there is an R positive such that the open ball the closure of this ball around of radius R around the origin this those x in x such that the norm of x is less than equal to R is compact, but this is not possible. Now, this is a compact metric if this is compact then this would be a compact metric space and therefore every sequence will have a convergent subsequence, but this is not possible as the sequence. So, we can take these vectors Eis the Eis former or the normal basis for this Hilbert space REI does not have a convergence of sequence. So, this contradicts the result we proved that every compact the result we proved every in a compact metric space every sequence has a convergence of sequence. So, this brings us to an end of our discussion on locally compact logical spaces. So, next we want to describe what the quotient of all g's. So, let us begin with a very an example from group theory which you would be familiar with. So, let g be a group or rather a concept from group theory and let n contain g be a normal subgroup. So, then we have the first we have the set g mod n which is set of equivalence classes g mod equivalence where we define x x comma y x is defined equivalent to y x and y are in g if in relief y inverse x is in this normal subgroup. So, instead of writing g mod equivalence we often write g mod n. So, moreover we can give g mod n or g mod equivalence whichever you want to write it a group structure such that the natural map. So, let us call this pi this is just a map of sets a priori to g mod equivalence which takes an element x to its equivalence class. So, in this case the equivalence class is the coset of n. So, this natural map. So, we can give g mod n a group structure such that this natural map becomes a group homomorphism that is one thing further. So, that is one nice thing which happens when n is a normal subgroup, but there is another nice thing which happens this map pi has the following property. If we have a group homomorphism which is constant on equivalence classes. So, then we get an induced map of sets g. So, we have g to h we have f and f is constant on equivalence classes. So, which means when we look at g mod equivalence we get a map of sets let us call it f bar and the important thing is that f bar from g mod n to h is a group homomorphism. Note that since f is a group homomorphism f is constant on equivalence classes if and only if the normal subgroup n is contained in the kernel of f. So, in the next lecture we will see an analog of this concept for topological spaces. So, before we end this lecture let us see one very standard application of this corollary that we saw of the uniqueness of 1 point compactification. So, let x be equal to r n right. So, then x is not compact x is locally compact. So, now we know that. So, recall the stereographic projection. So, what did we do? We had r n and we took the sphere and we deleted the north pole for instance p and for any point x on the sphere we joined p with x and we get a unique point y. So, the stereographic projection phi is from s n minus the north pole p to r n right. We had seen that phi is a homomorphism. In particular if we let t to be s n and j to be equal to the inverse of this homomorphism. So, j is phi inverse from r n to s n right. So, then the hypothesis of the previous corollary of this corollary of the above corollary are satisfied right. Because we are taking t to be s n s n minus the image of the stereographic projection is just the one point the north pole and this will imply that this will force the map from the induced map f that we get from s n to x hat is a homomorphism. Thus the one point compactification is s n. So, we will end this lecture here.