 So, in the previous lecture we saw that given a topological space given a locally compact topological space space X if it is not compact then we can compactify it. So, what did it mean by saying that we can compactify it meant that. So, we can construct a compact topological space X X hat such that X is sitting inside X hat as dense. So, recall that when we say compact topological space X in our definition of compactness we need when we say compact topological space X hat in our definition of compactness we need X hat to be host of and we had proved that X hat is host of as a dense subset as a dense open subset and in fact X hat was obtained is just X disjoint union a point P naught. So, there is a picture we can keep in mind while we. So, let us take the real line the real line is locally compact but it is not compact and we can take this point P naught over here. So, when we compactify the real line let us see what we get. So, instead of making the picture like this we will make the real line like this. So, recall that there were two kinds of open subsets one kind was type A open subsets these were open subsets of X right and the type B open subsets were as follows we take any compact subset of R. So, for instance we can take this union of two compact subsets of the real line. So, that is going to be compact and we just take the complement subspaces. Compliments in X hat compact subspaces or in other words what we have specified is what are the open subsets the open subsets which contain P naught are exactly complements in X hat of compact subspace of X. So, in the same way we can make see another example of R 2. So, instead of making R 2 like a plane we can make it as a sphere. So, so from this in the picture it is clear that one point compactification it seems that it should be S 1. And similarly if we take R 2 make a dotted line. So, this is a R 2 and we can imagine the point P 0 over here. So, R 2 is also a locally compact and therefore, we add this point P 0. So, the one point. So, what are the neighbourhoods open subsets which contain P 0 these are the open subsets of type B. So, these will be we can take any compact subsets subset of R 2. Let us say this is D. So, then D will be somewhere over here is D and we just take the complement of D. So, this X hat minus D. So, it is useful to keep this picture in mind. So, let us begin today's lecture. So, let us write down what we proved last time. Let X be locally compact topological space which is not compact. There is a compact topological space X hat along with the inclusion from X to X hat such that X hat is just I f X union a point which we call P 0 I is continuous I f X is an open subset. So, let us make the following easy remark I from X to I f X is a homomorphism. So, we have already seen that I is continuous and the image of I is lands inside I f X and it is a bijection. So, we have already seen we know that I is a bijective. So, thus to show that I is a homomorphism it is enough to show that if U containing X is open then I f U is open in I f X. But now note that I f U is the same set U and as U is a subset of type A is an open subset of X hat of type A this implies that I f U is open in X hat and as I f U is contained in I f X and so also. So, thus I is a homomorphism. So, this theorem we had proved last time and in the previous lecture I had mentioned that this particular one point there could be many one there could be many compactifications of a locally compact topological space, but this particular compactification has a very nice and special property. So, which is what we will see in this lecture. So, in the topological space X hat has the following property proposition that T be a compact topological space which contains X as an open subset. So, what is the meaning of this sentence right. So, this means that there is an inclusion J from X to T it is an inclusion of sets such that J of X containing T is an open subset J from X to J of X is a homomorphism. So, there is a inclusion says that let me also write J is continuous ok. So, then there is a unique map. So, what we have is we have X over here and we have this inclusion J, J is a continuous map from X to T and the image. So, let us say the image is J of X and this is a subset of T there is an open subset and we have this map like this and we also have the map I also satisfies these properties right there is X hat there is also open an X hat right. So, then there is a unique map F from T to X hat is a unique map F from T to X hat such that the following two things happen F compose J is equal to I and F of T minus J of X is equal to this P naught. So, X hat has this pointed infinity which is P naught. So, let us try to see what this means by means of an example. So, let us say X is this open disk let us say I embed the open disk to this closed disk right. On the other hand the 1 point compactification is J this is I the 1 point compactification looks like. So, I have to add a point at infinity. So, it will look like something like this. So, this point is P 0 right. So, what is F doing? F is collapsing all of the boundary F. So, F collapses the set T minus J of X which in this case is this black boundary to the point P 0 right and F is completely determined at other points F compose J is equal to I. So, we want to say. So, this in this example F does this, but ok. So, let us say that let us try to prove this proposition which is that there is a unique such map F ok. So, of course F is continuous there is a unique continuous map that is the assertion. So, the map F has been defined set theoretically on all of T right. So, this this map has been defined set theoretically on all of T. So, how do we do it? So, on J of X define F to be. So, J of X J is a homeomorphism from X to J of X. So, we apply J inverse is X and then we apply I. So, this goes to I of X this makes sense as J from X to J of X is a homeomorphism and ok. So, we do not need homeomorphism right now. So, we just need bijective is a bijection right. We just need bijection right now because as of now we are really trying to say that F is defined set theoretically right and outside on T minus J of X we have defined F to be the constant map P naught. So, thus the map of sets F is completely determined right and so, it is unique. So, we only have to show that F is continuous. So, for this so, let V contained in X hat be an open subset. So, let us first assume that assume that P 0 P 0 does not belong to be right. So, in that case the open subset is contained somewhere over here right. So, so then it is easily checked this is an easy trivial check that F inverse of V is completely contained inside J of X. So, as a result so, this shows that F inverse of V is actually equal to F restricted to J of X inverse of V. Now, notice that so, note that F restricted to J of X is from J of X to X J inverse and this compose with I to I of X right and as J from X to J of X is a homomorphism this implies that J inverse is continuous this implies that I compose J inverse is continuous. So, this implies that F restricted to J of X is continuous. So, this implies that combining these we get that F inverse V is equal to F restricted to J of X inverse V which is open. So, thus if P 0 does not belong to V then F inverse of V is open right. So, now let us consider the situation if P 0 belongs to V then X hat minus V is compact and contained once again contained in I of X. So, once again F inverse of this since it is contained in I of X is equal to F restricted to J of X inverse of X hat minus V, but this is a compact subset I of X right and this set is in fact equal to F restricted to J of X is equal to I compose J inverse and both these are homomorphisms. So, this is from J of X to X J inverse and this is I of X right and both these are homomorphisms. So, this implies that F restricted to J of X is a homomorphism. So, which implies that F restricted to J of X inverse is a homomorphism is continuous which is continuity. So, this implies that F restricted to J of X inverse of X hat minus V is a compact subspace J of X. So, also so this shows that F inverse of X hat minus V is a compact subspace and so also a close subspace. So, this implies that, but F of X hat minus V F inverse of X hat as F inverse of X hat minus V is equal to T minus F inverse of V this implies that F inverse of V is an open subspace. So, this proves that. So, this completes proof. So, we will end this lecture here.