 So, in the previous lecture we had used the following two exercises. So, one I will show that the real line is homomorphic to the interval 0 1. So, let us see how to do this I just want to sketch a quick proof. So, we will give a homomorphism from the real line to minus 1 1 and you can easily check that using a linear isomorphism that the interval minus 1 1 is homomorphic to 0 1 to the interval 0 1 ok. So, let us give a map from r 2 minus 1 1. So, we simply send x 2 x divided by mod x plus 1 ok and easily check this is bijective and the inverse of this map is given by y maps 2 and this is also continuous. So, let me so check that these bijective and continuous both these maps that is left as an exercise. This is one exercise the other exercise I had mentioned was. So, we take let x and y be topological spaces fix a point small x and x and consider the inclusion from y to x cross y given by y maps to x comma y ok. So, clearly this factor is like this. So, I and let us call this f naught let us call this map f yeah. So, then f naught is a homomorphism. Obviously, x cross y has the subspace of already. So, let us quickly check this. So, f is f is continuous as both projections. So, recall we had seen that to give a map to a product it is enough to give a continuous map to a product it is enough to give continuous maps to each of the factors and both the factors the first factor is just the constant map y to x that is obviously continuous and the second factor is the identity. So, both these are continuous and since the image lands in x naught in x cross y it follows that f naught is continuous right. Here we are using the following exercise. So, let us say not exercise result which we had proved when we talked about subspace topology. Suppose a to b we have continuous map f and let us say the image lands inside a subset c of b and then this f factors as f naught. So, if we give c the subspace topology then this map f naught from a to c is continuous. So, f naught is bijective is clear. So, therefore, to show that f naught is a homomorphism is a homomorphism it is enough to show that f naught of v is open for an open v in y right, but f naught of v is exactly equal to x cross v and this is equal to x cross v this open set intersected with x cross y this subspace right. Thus f naught of v is open which implies f naught is a homomorphism. So, this completes both the exercises which were mentioned in the previous lecture. So, now let us proceed with our discussion on compactness. So, last time we proved that close subspace for compact space. So, this time let us begin today's lecture with the following proposition let x be a topological space y contained in x be a subspace which is compact. So, here y is a subspace of x and therefore, it inherits the subspace topology and this topological space is compact that is our assumption with the subspace topology. So, then our claim is then y is close to x. So, let us prove this. So, we shall show that the set x minus y is open. So, once again let us make a picture let us say this is our x and say this is our y. So, let us pick up any point x over here. So, let x be in x minus y it is a point outside y. So, we need to show that there is a small neighborhood of there is a small open set around x which does not meet y. So, we choose any y in y. So, let us say this is y then as x is host of. So, as I had mentioned it is from now on we will always be assuming that our topological space is a host of if not mentioned as x is host of there exists open sets u y containing x and v y containing y. So, this is our u y and this is our v y such that u y intersection v y is empty. So, we can do this. So, we can do this for all y in y right. So, then the sets v sub y form an open cover for y. So, that is we can write y as. So, y is contained. So, y is contained in this union y in y v sub y. So, this implies that y is equal to union y in y y intersection v sub y and since this is an open cover this has a finite sub cover. So, this is equal to union i equal to 1 to n y intersection v sub y. So, this implies that y is actually contained in the union of finitely many of these. So, now let u be equal to the intersection i equal to 1 to n u sub y i. Now, each of these u sub y is contained x. So, therefore, this implies that and this is a finite intersection. So, this implies that u is an open subset of x x is in u. So, we claim that. So, let us look at u intersected with the union of y i's i equal to 1 to n v y i's. This is contained in the union i equal to 1 to n u intersected v sub y i's which is contained in the union i equal to 1 to n u is contained in u sub y i u sub y i intersected v sub y i, but each of these is empty. So, this implies that u intersected this union is empty. This implies that u intersected y which is contained in u intersected this union v sub y i as we saw that y is contained in this union is equal to empty. So, thus given this point x in x minus y we have found u contained in x open in x such that x belongs to u and u is contained in x minus y. Thus x minus y is open. So, this completes proof of the proposition. So, as a result of this proposition using this proposition we will use we will do this very useful theorem. Let y contain an R n be a subspace then y is compact if and only if y is closed and bounded. So, I have not defined what bounded means, but it means the obvious. So, let y contain an R n be a subspace. We say that y is bounded if there exists some m positive such that y is contained in the ball of radius m around the origin. Let us say in the Euclidean metric. So, let us prove this theorem. So, let us first assume that y is compact. So, then we can write y. So, then y. So, first note that R n we can write as the union n greater than equal to 1 balls of open balls of radius n. So, this is a open cover for R n because every point is contained in one of these balls open balls right. These keep getting larger and larger. So, this implies that y is equal to I can we simply intersect both sides with y, y intersected b 0 1 b 0 1. This is an open cover for y right and since y is compact this has a finite sub cover. These balls are contained. So, b 0 1 is containing b 0 2 is containing b 0 3 and so on. So, this implies that there exists some n such that y is contained in b 0 1 thus y is bounded. Because of this previous proposition this previous proposition says that R n is a host of topological space and we have y which is compact. So, y is also closed right. So, the previous proposition implies that y is closed. So, thus y is closed and bounded ok. So, this proves one direction. So, conversely assume that y is closed and bounded. So, since y is bounded we can find a m large very large such that y is contained in this product of these intervals right. So, y is bounded means y is contained in some large open ball and this open ball we can put inside some inside a square like this inside a closed square. So, or y is this thing. So, more over so note that. So, we prove that 0 1 is compact as 0 1 is homomorphic to we can just in fact even write a linear homomorphism. This implies minus m comma m is compact this implies minus m comma m to the n is compact. So, as y is closed in R n this implies y is also closed in subset and by what we proved in the previous class previous lecture we prove that closed subspace of a compact space is compact. So, using that as closed subspace of a compact space is compact this implies y is compact. So, this completes a proof of theorem. So, as an application of this theorem let us prove that S o n is compact. So, recall. So, as an application. So, recall S o n or the same proof for o n will work S o n is the set of those matrices A with real entries. So, is that A transpose A is equal to identity n cross n identity. So, claim S o n this has the subspace topology from M n r right and M n r we can identify with R n square. So, is compact. So, our claim is that S o n is compact. So, it suffices to show that S o n is closed and bounded. So, let us look at this map. So, let us first show that S o n is closed. So, let us consider the map from M n r cross r. So, this map is given by A its map to A transpose A comma determinant of A sorry I forgot to put the determinant condition. So, we claim that this map is continuous right. So, to show that it is continuous it is enough to show that first we look at the map from M n r to M n r A goes to A transpose A this map is continuous and the map from M n r to R A goes to determinant of A. So, this map is continuous is clear because the determinant of A is a polynomial in the entries of A and we know that polynomial the entries of A are simply the projection maps and projection maps are continuous from this product and polynomials therefore polynomials in these continuous maps are continuous. Let us look at this map over here. So, once again M n r is identified with R n square with the product apology right. So, to say that this map is continuous it is enough to say that each of the coordinate functions is continuous. But once again when we look at A transpose A the coordinate functions are polynomials in the entries of A. So, thus A goes to A transpose A is continuous right. So, this shows that phi is continuous. So, now the point identity n cross n comma 1 in M n r cross r is a close subspace is a close subset. In fact, in any r m the a point. So, we can just take any point A 1 A 2 up to A m is this just the single set is a close subspace. So, this implies that phi inverse of this identity n cross n comma 1 is a close subspace, but this is precisely is precisely. So, therefore S 1 is closed and next let us see that S 1 is bounded. So, this is easy because when we look at the condition A transpose A is equal to identity. So, if A is the matrix A 1 1 A 1 2 and so on. So, then A transpose A is equal to identity. So, this will imply that. So, when we look at the i i th entry of A transpose A this is equal to summation j equal to 1 to A j i square ok. And since this A transpose A is identity this sum has to be equal to and when we take sum over all i's this implies summation i comma j A i j square is equal to n right for every A in S 1. So, thus S o n is bounded right. I mean the topology on R n square is given by this Euclidean metric and what we have just proved is that every element A is contained in the ball around 0 of radius square root n ok. I can add a square root n plus 1 ok. So, we will end this lecture here.