 Welcome to model 57 of points at topology part 1 course. We will continue our study of topological groups today which we have started last time, right? So here are some notation which I am going to fix up at least for today, maybe tomorrow for next time also, so these notations will be different for this section. Take any topological group or just a group to begin with, for any group you can have just notation A and B are subsets, okay? A followed by B, A, B you can write, A is equal to all little a little b where a is little a is in capital A and little b is in capital B, okay? Similarly, A inverse A will be all A inverse where little a is inside A, all right? I want to draw your attention, this is not a subgroup generated by something and so on. It is inside the group but these are just sets and this is just A, B, okay? But you have to combine them with the group log, here you have to take the inverse and take the corresponding set. So the following easily proved fundamental results are at the heart of various special topological properties of topological group that we are going to obtain, okay? So I will not keep saying that G is a group, G, mu, E is a group and so on. So this will be G will always denote A now, topological group for some time. The first thing is this lemma says for E, G and G, look at the left multiplication map given by G, namely L G of H is GH, so it is a map from G to G. Similarly, I can take the right multiplication R G of H is equal to H times G, okay? So one is multiplying on the left, multiplying on the right, okay? So both are maps from G to G, they are self-homomorphisms of G, okay? It is very clear that they are invertible, the inverse of L G being equal to L G inverse. So multiply G inverse, G inverse G is identity, so it will be H. So they are bijections, you know, L G and R G. Why they are homomorphisms? Because they are also continuous, okay? Multiplication, you are just here, you are freezing the element G, okay? Only H is variable. So if like a two-variable function, you are taking one variable, one the other one means fixed set, so that is also continuous, okay? Since G, L G and L G inverse both are continuous, both of them are homomorphism, similarly R G and R G inverse. So an easy consequence of this is that you take any open subset, okay? Any open subset, okay? And X be any arbitrary subset of G. Then U X, what is U X? U consists of little U comma little X, okay? Multiply, multiply together and collect it together. So that will be an open set. Similarly, X U, now multiplying on the other side. Or you can take X U, X inverse, or you can take many other combinations, go on multiplying, just like writing alphabets, okay? One of them is open, that is necessary. Then all these, all these subsets will be open. Can you see why this is true? Look at U, that is open set. When you multiply on the right by X, what is that? It is just taking R G of U, R X of U. R X is a homomorphism. So R X of U, one single X here, okay? Namely U little X, that is open. Because image of U, now U X is nothing but union of all these things where X varies over X. Therefore, this U X is open. Similarly, X U. So once this is open, you can apply to, you know, X U times X inverse. X inverse is some other set, right? You can go on applying finite many things all the time. So there will be all opens of sets, okay? I am repeatedly used that R G and X G, L G are homomorphisms, okay? So this is the fantastic thing happening here inside a topological group. The second thing is the inversion I G, which I have denoted by I G, that I G equal to G inverse. This is a self-homomorphism of G and if you take again, I composite I, namely I square, that is the identity because inverse of G, okay? Inverse of G is G itself, okay? So why this is again, this is continuous what we have seen already. This is a, because I square is identity, this is homomorphism. So I composite, I is identity. So this is bijective. So both I and I, its inverse is I itself. So it is a homomorphism, okay? It is a very special homomorphism. It is of order of two. Next, here is a easy thing which you will keep using without even mentioning it. So here I have mentioned it for the first time. Maybe, you know, it will be used several times, namely take any three subsets, okay? Suppose A B intersection C is empty, then and then only A intersection C B inverse will be also empty, okay? It is very easy to verify. If A B intersection C is non-empty, it would imply it is non-empty. That is the way you can prove it. Take a point here, which means you should look like A into B but it is an element of C. So C equal to A B, right? But then A will be equal to C B inverse. So there is an element A here and the same element is here, C B inverse. So it is also non-empty and convergent, okay? A group homomorphism G, G to G prime F, okay? What is a group homomorphism? F of some xy is equal to fx into fy for all x and y inside G, right? A group homomorphism of a topology group is continuous if it only is continuous at a single point, namely e belong to G. Any other point will also do but let us prove for e belong to G, okay? So you must have learnt it elsewhere but let me sum it up here because of the importance of this little result here, okay? We need to prove only if part because only if part, only if not only if part because what you know is if it is continuous, it is continuous at G also. Just continuous here at a single point makes it continuous everywhere that is the part which you have shown, okay? So let F be continuous at e and G be any other element, some other element. Let us look at the image of that namely G prime equal to FG. Since F is a homomorphism for every edge inside G, we have FH you multiply G inverse on this side, F of that F of G inverse will come out, right? F of G inverse again by the same multiplication it is FG, F sorry FG inverse. So FG, FG inverse cancel out. So only FH remains same from here. So it is FH is equal to this one. What I have done is I multiplied G inverse on the left inside a bracket. So outside a bracket if it will be FG inverse because of the F is a homomorphism. So I have to compensate it. So I am multiplying again by FG so that it cancels out, alright? But now what is this inside thing? It is left multiplication by G inverse, right? And FG is a point of G prime. So it is a left multiplication by LG prime inside G prime. So this whole thing will look like LG prime composite F, composite LG inverse operating upon H. Therefore F itself is equal to LG composite, LG prime composite F composite LG inverse, okay? So I have written F itself like this complicated way but this will help us now. Why? Because I know LG inverse is continuous. LG prime is continuous. To show that F is continuous at a point I have to only show that F is continuous at some other point. What is that? That is precisely what is happening here. See I want to show this map F is continuous at G, okay? So I apply LG inverse to this. This G comes to E. Now I apply F but F is continuous at E. So I can go all the way up till here. When I come to F of that if it is E prime and LG prime will take it to G prime. That is nothing but F orange, okay? So middle thing is continuous at E. These two are continuous functions. The composite is continuous and starting point here is G. So the composite is continuous at G, okay? There are many different ways of writing it down. So I find it elegant way of writing it. So if you learn this kind of writing down it will make other concepts very clear at elsewhere also. Now I introduce terminology here because this inversion keeps coming again and again, right? You take a subset A contained inside G and call it symmetric if A equal to A inverse. That is you should take the iota which is the inversion. Under inversion it is invariant. i of A is A, okay? So such a thing is called symmetric now, okay? This is a temporary notation. This has been used by others. So I am also using this word terminology. Now here is a lemma for symmetric things. Let O be a neighborhood of the identity element in G, okay? Then the following is true. There exists symmetric open some sets U of G such that they contain the pointy and such that they are inside O. So starting with any neighborhood, you can improve it to become a symmetric neighborhood, okay? There are many ways of doing it. So there are many of them. In fact, if this happens to everything, you can go on taking for every, you know, there is a symmetric neighborhood systems. So this means just that instead of all neighborhoods, you can just take symmetric neighborhoods to form a fundamental system at the origin. So that is the profound thing here, okay? Similarly, even more profound is there exists symmetric open sets V of G containing the pointy and such that V is a subset of VV that is obvious anyway, okay? But VV itself is contained inside. Now since V is symmetric, I can replace V by V inverse at my will. So this will imply a lot many things. That is what I told you. These are elementary observations to build up the topological theory. So you have to come back here. If you have made mistakes or you have not understood these things correctly, then you will have problems. You will see that, okay? So starting with any neighborhood of identity, I can get such, you know, beautiful neighborhoods, neighborhood is symmetric, neighborhoods which are contained inside their own product and the product is contained inside over and so on, okay? So how to prove that? Proof is also very elegant in sense that, okay? Here I am actually proving even stronger thing here. Namely, instead of single point, you can do it with any neighborhood. That A contained inside O be an open subset of E, okay? See, I started arbitrary neighborhood only. Now I am taking A to be open subset. A neighborhood means after all there is an open subset, right? So E belongs to A. A is open. A contained inside O. Let us start that. Since the inversion is a homeomorphism, it follows that A inverse is also open, okay? Because it is eta of A. Take U equal to A intersection A inverse. A is open. A inverse is open. Intersection is open. Inverse image of E is E itself. The inverse of E is E itself, sorry. So E is in both of them. So that was U is a neighborhood of E, open neighborhood of E, okay? What is U inverse? It will be A inverse intersection with A. It is same thing as U. So U is symmetric open set containing A and U is contained inside O, okay? Because U is A intersection A inverse and A is also contained inside O, okay? So this is one way of obtaining a symmetric neighborhood. Use continuity of the function for the next one, namely xy equal to xy inverse. So at the point E comma E belongs to G cross G. Where does E comma E go to? It will go to E. So take a neighborhood O. O I have taken. By continuity, there will be open subset here and open subset here where the product is going inside this open set. So that is what I am writing here, namely E comma E belongs to B1 cross B2. B1 and B2 are open subset, okay? This nu of B1 B2 will be equal to B1 B2 inverse. That will be contained inside O. So this is my continuity of this map which I had denoted by nu earlier. Apply A this part to get a symmetric neighborhood V of E such that this V is contained inside the intersection of B1 and B2. B1 and B2 are neighborhoods of E. So intersection is also neighborhood of E. So now I can improve to become symmetric also. So V equal to V inverse is the extra hypothesis I can put, okay? So once you have that V is always contained inside V into V because identity is there, right? So V equal to E V that is contained to E V. So but I can replace this V by V inverse. So this should be V into V inverse also because V equal to V inverse. But now this V is contained inside B1 and it is also contained inside B2. So first one is B1, second one is B2 inverse hybrid, that is all. And that is contained inside O, okay? So proof is over. So we will have opportunity to use this one, no problem, let us see. Now a little about closures and so on. Let A and B be any subsets of G again. Okay? Let this curly new be the set of all neighborhoods of A in G. Let curly as be the set of all symmetric neighborhoods of E. So this is a smaller, this is a smaller family, okay? This is all neighborhood. This is only symmetric neighborhood. Then the claim is A bar is an arbitrary subset. I am making a statement about A bar. A bar is intersection of all A V where V range is over symmetric neighborhoods. It is also equal to intersection of all A V where V range is over all the neighborhoods. Then the other way around instead of A V, I have V A here and V A here. The other two things are similar here. So either I can write it on the left side or write on the right side. If I take intersection of all of them, what I get is A bar, okay? The second thing is much easier. A bar B bar is contained inside A B bar. So it needs statement. Okay, let us see how these proofs work. Let us first prove the equality, the first one. What is the first one? A bar equal to intersection of V belong to S A V. So your S is only symmetric, okay? What is the meaning of A bar? Take a point in A bar. Its closure means every neighborhood of that point intersects A. That is the definition of closure, right? We shall use that directly. The definition of closure, no other properties, directly use the definition. Okay, suppose X is not in A bar, then I won't show that it is not in one of the sets here. Therefore it is cannot be an intersection. So not in A bar means not on the right hand side. And again not here means not here. That is what I want to approach, okay? So suppose X is not in A bar, then there is a U inside V. What is V? Neighborhoods, neighborhoods of identity element, okay? Such that X times U intersection A is empty. So what is X times U? X times U will be neighborhood of X. X is not here. Every neighborhood of X will look like X times U, where U is a neighborhood of identity. This is what I am using here. You see neighborhoods of any other point, you don't have to look anywhere. You just translate them. Left translation or right translation, it doesn't matter. Starting with identity element and a neighborhood, multiply by X on the right or X on the right, that will contain X. And it will be neighborhood because these are homomorphisms, multiplication on the right or left, right? That is what I am using here. X U in the DC, any neighborhood of X will look like this, okay? It is most general one. So there is no specialty here. And one of them intersection A is empty because X is not in A bar. That is all I am doing. Now by part A of the previous lemma, we can assume that U is symmetric. You can go back to a symmetric neighborhood. It will be smaller thing contained inside U, so that that will be still empty. So I can assume U would still be symmetric, okay? Now it follows that X is not in A U. See I have A, B intersection, C is empty, implies A intersection, B is C inverse empty. That is what I am using here, okay? So if this is empty, X cannot be A U. Now this is symmetric. Therefore it is one of the elements here, one of the members here. So that means that X cannot be in the intersection, okay? So one part I have done. Now start with the point which is not here. Then you can show that it is not in the closure. This is what I have, right? Conversely, suppose X does not belong to RHS. That means that is a symmetric neighborhood now directly because I have put S here. Say let us call it as V, okay? Such that X is not in A, B. See V is a symmetric neighborhood. I have A times that I am taking, right? So X is not in A, B. So that is the meaning of this. It is not in the intersection. That just means that again going back here, X V intersection A is empty. X V is a neighborhood effect. This means X is not in the closure. So part A is done. The next one, what we have to do? A, B, A bar, B bar, containing inside A, B, bar, okay? So let us see why. So first thing is we know in a product topology if you have a subset A and a subset B of the two X and Y, then A cross B closure is the same thing as A closure cross B closure, right? That is what I am using here. A cross B bar is equal to A bar cross B bar, okay? Now all that I have to use is the continuity of the multiplication. That is all. The mu from G cross G to G, XY going to X into Y. So how? Let us see. A cross B is obviously contained inside mu inverse of A, B because mu of A cross B is nothing but A, B, okay? If you take A inverse, A cross B will be contained in mu inverse of A, B, all right? But A, B is a smaller subset than A, B bar. So this is contained in a mu inverse of A, B, okay? Sorry, mu inverse of A, B bar. But that will imply if you take the closure, this is a closed subset. Why is this a closed subset? It is a bar of something and then mu inverse, mu is continuous. So mu inverse of this one is a closed subset containing a set. So its closure will be contained inside that closure being the smallest closed subset containing the set, right? So I am using something which we have done in a long, long way perhaps. But A cross B bar is A bar cross B bar. So A bar cross B bar is A cross B bar that is contained inside mu inverse of A, B bar because this is a closed subset containing A cross B, okay? So mu of this will be contained inside A, B bar. Now apply mu on both sides, this will be A bar, B bar and that is contained inside A, B bar. So do not make the mistake of you know that A, B bar is contained inside A bar, B bar, okay? First thing you should observe is that A bar is closed, B bar is closed but A bar, B bar may not be closed, okay? So these are the questions. Now we are going to do something quite deep suddenly, okay? We have everything, every machinery is ready for it. So start with a compact subset and C be a closed subset of it, okay? No other assumptions, K is compact and C is closed, okay? Suppose they are disjoint, then they can be separated by open sets that is just but we can do better namely there exists one single open subset V of V of open neighborhood of A such that Kv that would be neighborhood of K that we know, right? Kv intersection Cv is empty, Kv is a neighborhood of K, Cv will be a neighborhood of C because V is a neighborhood of identity but these are quite large open subsets for that matter but they contain K and C, the intersection is empty. So such a neighborhood of identity I can find. So this is the claim. In particular it will follow that K and C can be separated by open sets. It is similar to normality but far away from normality because I assume K is compact, not a closed set, okay? Not just a closed set. If it is a closed set then this would have to give you normality. So quite near normality it comes, okay? So that is why I am doing this. We are suddenly proving this such strong results. So now do you understand that topological groups, that topology and a topological group has to be special, alright? Though it ranges from discrete to indiscreet, okay? Anything in between but it has to be special, okay? Let us prove this one. If K is empty there is nothing to prove because I can always take, you know K times V when K is empty is just empty set. Empty set intersection anything is empty, that is no problem. So let us assume that K is non-empty. Similarly we can assume that C is non-empty, alright? So I start with X belonging to K that is why I am justifying that let K be non-empty and so on. Take a point X inside K put O equal to X inverse of G minus C. The C is closed so G minus C is open translated by X inverse that is also open. See look at this one, X is a point of point of K, okay? So I am taking K X inverse of this one. So that is an open subset is all that I know, okay? Now from Lemma 5.30 whatever we get a symmetric neighborhood V X of E such that V X X V X V X is contained inside O. See why this is true because O is a neighborhood of identity. Why? Because X is inside K therefore X is inside G minus C because X is not in C. K and C are disjoint that is all, okay? If X is here X inverse of that will contain identity element. So this O is open and it is a neighborhood of identity. You can improve it to get a symmetric neighborhood V X such that V X V X is inside O. So this was part A and B both combined here of this Lemma which we have proved just now. This implies that X times V X V X intersection C is empty. It is contained in the old X inverse of G, right? So this X I bring it on the left here. So what I have? What is G minus C on that side? Okay? So it is contained inside G minus C but intersection with C is empty there, okay? So for each X we have found out X V X V X intersection C is empty. So this already tells you that X A is G is regular that topology is regular. Why? Because suppose you instead of K compact and so on, K if this singleton X was, this K was singleton X, okay? Then if I prove whatever the statement that would be regular and that is all, that is all I have proved. X V X V X is a neighborhood, okay? Contained inside the complement of C. So here itself the proof is over of regularity, okay? So we have already proved that every topological group is regular on the way to proving this lemma, okay? Let us continue. Now applying the lemma 5.13 again we get another symmetric neighborhood which I have denote by U X of E such that this U X V X is contained inside V X, okay? Because V X is a neighborhood of identity, all right? If you combine this one X V X V X intersection C is empty. Now I can replace V X by U X U X which is smaller neighborhood of and then V X. So U X U X, U X U X intersection C is empty, okay? So do not go on doing this. Now we have already arrived this one. Now U X is a neighborhood of E, right? So U X is contained inside U X U X and since U X is symmetric, okay? This immediately implies that X U X U X is contained is X U X U X intersection C X, U C of U X is empty. Now one U X on this side I am translating it to the other side, okay? When you translate you have to write U X inverse but you do not have to write U X inverse because U X is symmetric. Out of the three you can translate to the one other side and out of three you can combine two of them and that just take one of them. U X is contained as U X. So X U X U X is intersection with C of U X is empty, okay? We have not yet completed. So one single thing we have done this much. From regularity we have improved it to this much, okay? The same neighborhood here repeated twice comes on the C part also, okay? To prove regularity as I had just the X V X intersection C is empty that was enough, okay? Now we go, now we use the fact that K is compact. So if K is compact there will be finitely many X 1 X 2 X K belong to K such that K is contained inside X I U X I, I did not want to okay? Why? Because these X I U X I X U X or X V in so over K that will cover K and that is an open cover. So out of that you extract a finite cover. Now we put V equal to intersection of U X I because it is finite intersection it is open because E U X I is symmetric. Symmetric means what? Eta of U X I is eta, okay? So intersection will also have that property. It follows that V is symmetric neighborhood of E open neighborhood. Moreover we have K times V is contained inside union of X I U X I I am replacing K by this union times V then I push this V inside the bracket. So it is a union of X I times U X I V, okay? But then each U X I V, V is what? V is intersection of all these, right? So it is contained inside each U X I I can replace V by U X I depending upon what this X I is. So I write one to a X I U X I U X I, okay? So K V is contained inside this one finite union. Therefore if I take the intersection now K V intersection C V I want to show that this is empty, right? That was my finite thing. Look at K V intersection C V. This K V is contained inside this one large thing, okay? So you take this thing intersect with C V push it inside the bracket that is all, okay? So it is union of all these intersections, finitely many of them. But that is contained inside union of X I U X I U X I as it is but C of V what is V? V is intersection of these things which contains the C of U X I. These are larger open sets, okay? But now what are these things? Each of them is empty by the choice. Remember that each of them is empty. I have taken only finite many connection X I's are coming from capital K, okay? So each of them is there are finite many of them anyway. Each of them is empty. So union is empty. So we have proved that a topological group is strongly regular in that sense compact and disjoint close sets. Contract set and the close set can be separated by open sets in a neat way. One single neighborhood of V and its transit K V intersection C V is empty, okay? So I am just subbing it up for posterity. Every topological group is regular that much is fine, all right? So here are some elementary X I U's again. You can work out them. But these are serious examples, serious exercises. If you work them in that order, you will be able to do all of them. So they are built up in that fashion, okay? And then there are some other very nice things happening. I have no time to discuss them fully but we will discuss them provided you show equal interest and come up with some solution, maybe wrong, maybe right, whatever it is. So these are left to use exercises and maybe specifically mentioned assignments, okay? So let us stop here. Next time we will continue the study of the colloquial groups. Thank you.