 Welcome to module 54, we shall continue the study of separation actions this time concentrating on checking whether they are productive. So the very first theorem here is that regularity is productive, okay. So this is stated as product Xj of is regular if and only if each factor X little j is regular. Just as in the case of freshness because regularity is hereditary, if Xj is regular, okay if now if Xj is regular then each Xj is regular, why? Because you can always think of each coordinate space as a subspace, each factor space as subspace of the coordinate space, right. X cross Y is contained in X cross singleton Y goes inside X cross capital Y, so that is subspace, you know X can be identified with that, that is what we have been using already. So if the product space is regular, the subspace is regular that means each factor is regular, that is easy. We shall now prove the converse, pick up X belonging to U where U is open in Xj, okay. So we must produce another subset, open subset such that its closure is contained inside U and this open subset contains X, so between X and U I must squeeze another open set, okay. So in the product apology something is open means, it is a neighborhood of this means there is a basic open set and basic open set look like what we have a finite subset I of J and open subsets V i contained inside X i such that the ith coordinate is inside V i for all i inside i, this i is a finite set and look at P i inverse of all this i ring from 1 to n like that finite product i is not a capital I V i's, that obviously contains X okay and that will be contained inside U, this is a basic open set, all basic open set look like this. So actually regularity can be checked by just using basic open sets, if I show that inside this I can get another say W, such that W bar is contained inside this one, that is same thing as doing it for you, okay. So I have come to assuming all the you can you could have assumed that U itself is basic open set, that is all. Now for each X i being regular for each X little i contained inside V i, V i's are open, you can find an open subset W i in X i such that X i is inside W i inside W i a bar entire V i between X little i and V little i, we can screen another open set W i. So this I can do for all i inside i and then I take product of these finite W i's that I am calling W, then one thing which we have already checked all the time is the closure of this product is the product of the closures, each closure here is contained corresponding V i, so this product will be contains a product of V i's, okay. Now if you take P inverse, P i inverse of this one, X will be inside P i inverse of W because each the ith coordinate X capital I is inside W by the very choice, X i's are inside W, X is inside P i inverse of W, automatically this will be contained is closure, what I need to show it is contained inside U, right. Now this is a closed subset containing this one and this is also a closed subset, P i is an open map, it is actually continuous map, so P i inverse of closed, closed subset is closed and it contains P i inverse of W, therefore this closure is contained inside this one, actually these are equal here but I am just taking the easy way, this is contained inside this, that is no doubt and this has been chosen such as this inside U because what is this P i inverse of W bar, see W bar is inside this one, so P i inverse of W bar is P i inverse of these and these things are contained inside U, that is a choice of P i's, so that completes the proof, so not much hard work here, regularity is productive, so in that sense it is nearer to house darkness, right, indeed there is two schools of thought in topology, one people use house darkness whenever they have difficulty, other school uses regularity whenever they have difficulty and they achieve similar results, in fact identical results, okay, so in that sense regularity is quite nearer but in a slightly different way they work, nearer to house darkness is what I wanted to say, okay, but somewhat surprisingly normality which we tend to think like nearer to regularity, it has weird properties, normality is not even finite productive, the reason may be because it is not even hereditary, okay, even one way is, even if X cross Y is normal, it does not imply that X is normal, Y is normal, that is a funny thing here, so we shall see now second example, the example that we are going to see is one of the beautiful examples which you are already familiar with and we have studied quite a few properties of that, that is the semi open interval topology, X is the real line with the semi interval topology, namely the base consists of half closed intervals, A closed, B open, okay, very specific because that is why it is called L semi interval, if you could have write it as R also, that would be A open and B closed, so that is somewhat due to that, these two are homomorphic to each other, so if you study one of them, the other one is same, by just writing X cross to minus X, you will get the other topology, so that will be X equal to minus X will be homomorphic, okay, so concentrate on the semi interval topology, each open interval can be written as a union now semi open intervals of this form, right, you suppose you want to write open A comma B, then all that I have to do is close A plus 1 by N, okay, comma B union over all N will be equal to open AB, right, so all open intervals are also can be written as this one follows that this thing is finer than the usual topology because all the open subsets in the usual topology will be also open in this topology, however this closed interval and half closed intervals are not open in the usual topology, so this is finer, strictly finer than the usual topology, anything finer than a hostile space will be also hostile, therefore this space is half star, okay, now observe that every semi open interval is also closed in, closed in L, okay, so when you have a base consisting of closed resets, you know it must be base of topology, therefore they are open also, open and closed, okay, that itself is a very strong property and people have studied such things, so whatever you study though, this example will feel completely different to that, okay, so all those properties will be there for this RL, okay, closed intervals are closed inside a usual topology, but here even half closed intervals, okay, they are both open as well as closed in this topology, why this is closed? Because what is complement B to infinity, so that is by definition is an open set minus infinity to A open, that is also an open subset because you have just seen that every open subset in R is also open here, okay, so you can take X belonging to such a basic open set AB, that is an open set, then you can choose AB itself as a subspace also, its closure will be itself, so regularity is satisfied, we actuously here, you don't have to choose another W at all, you have already seen that this space is Lindelof and the product is not Lindelof, you remember that this was as a remark, let me just show you that we have already done this one by taking, by taking the product and showing that the diagonal is discrete, remember that this was the example, right, so here we have shown that the product is not Lindelof, therefore you know you can deduce a lot of things, namely of course this itself is Lindelof we have seen, so this was not all that trivial, okay, you have to use the second accountability of usual topology and so on, okay, so but I don't have to repeat that, okay, we have seen this one that the space is Lindelof but the product is not Lindelof, all right, now I want to tell you that this semi-interval topology is not second countable, you see second countable would be imply Lindelof but this is an example of Nanga, Lindelof space which is not second countable, why, if it were second countable the product will be also second countable because second countability is finite product invariant, okay, finite product, that also you have seen but once x cross x is second countable it will be Lindelof also, so that will contradict the previous observation that we have done, all right, so this is a space which is Lindelof but not second countable, okay, now here is Rimal, we have used finite product invariance of second countability in a peculiar way to see that it is not second countable, okay, a space is not second countable, so you see you can use certain theorems in a negative way also, so they also help in this way, all right, now let us continue with this space, finally I want to show that the product is not normal, okay, non-normality of the product space, just now we showed that this is a regular space, right and the product is a product of two regular space regular, so you will get this is a regular space but not normal, okay, so how do you show that this seminal typology product with itself is not normal, once again we go back and see that the diagonal is a close of space, all right, okay, in the product it is actually discrete, so that is what we had seen before, okay but now we want to use it very crucially, all right, so we have seen above that the induced ecology on delta, delta twiddle, usual delta will denote the diagonal x, x, this is x square minus x, this is discrete, okay, why because you could take a half open interval across half open interval just touching the point on the diagonal, the stuff it is contained inside the one side of the diagonal, one side of the whole diagonal, so if that is a open set intersects with the diagonal will be a single point, so that single point is open, that is how we are now this will help us in showing that the product is not normal, we take a to be the set of all points x comma minus x with x being rational, okay, any subset is closed now, okay, similarly you take b to be the complement of a that will be also closed, if it is closed because it is a discrete set, it is closed inside the delta twiddle, delta twiddle is closed in the whole space, so these two are now disjoint closed subsets of x, okay, r comma minus r where r is rational, other one is r comma minus r where is your rational, now we claim that there exists no open sets u and v containing a and b respectively such that their intersection is empty, in other words we start with u and v open, a contains u, v contains v, then we show that u intersection v is not empty, assuming on the contrary that means what, suppose you have two opens of sets a and b containing a and b respectively u and v, once they are open it follows that for each x inside r, okay, there exists a positive real number epsilon x such that x comma x plus epsilon x open cross another one minus x comma minus x cross epsilon x this open, see these are points of our RL and this is the product apology, right, so I am taking a product neighborhood which is the basic neighborhood in the product policy, this is contained inside u or contained inside v according as x is rational or irrational, if x is rational it will be inside u, x is irrational it will be inside u, okay, of course this length of this interval x to x plus epsilon x, this epsilon x will depend upon x, alright, so now fix a rational number s not belonging to q, it follows that for all irrational numbers t belonging to s not minus epsilon s not, what is this epsilon, we have fixed these epsilon x, remember that I am taking the same thing here, s not minus epsilon s not comma s not the x, the y coordinate of s not is this one but the x coordinate is slightly smaller, okay, actually I should take this one to be minus s not because I am interested in the points on that antidiagonal, okay, irrational numbers t belonging to this open set, this part is okay, this is just s not here, we have epsilon t, the corresponding epsilon t must be less than epsilon s not by 2, okay, it follows that for irrational numbers we have epsilon t less than or to s not minus t because otherwise you will collide with the, it will go inside v, so this is rational number, so this epsilon t is less than s not minus t, okay, because I have taken t to be inside this one, namely small, slightly smaller than s not but inside this interval, okay, the corresponding epsilon t cannot be, cannot exceed that one, okay, u and v are disjoint is used here, okay, so that is why this must be less than epsilon, now inside this one now choose an irrational number t not such that s not minus t not is less than epsilon s not by 2, see, so I am further taking, cut it down, you can show another irrational number here, okay, irrational number will also have same property, okay, in particular you have epsilon of t not will be less than epsilon of s not by 2, that is, that is all I am using, now you interchange the role of the irrational number and rational number, you started s not, you got a t not with all this property, you fix this one, now apply the same argument to t not to get a rational number with the same property but this time you write it as s one, repeat the above argument, okay, with t not in place of s not to obtain a rational number s one, in the interval t not comma t not plus epsilon t not such that epsilon of s one is less than epsilon of t not by 2 which is epsilon of s not divided by 2 square, okay, so this is the consequence of epsilon t not itself is less than epsilon s not by 2, so one stage of construction is over starting with a rational number, you get an irrational number with some property closer closer and then irrational number you get again a rational number, one cycle is over, now repeat this cycle, repeat it, repeat it, so what do you get, you get a sequence, okay, so I will show you first, I will show you what in the first stage what I am starting with a rational number here, it has some epsilon s not here, epsilon s not here, okay, and t not is just little small, little you know on this side, corresponding length of this one should not come over here, you see but otherwise these two not be disjoint, so it has to be at the most this much, right, so this epsilon t not will have to be smaller than the difference between s not and t not, so that is all I have put, okay, yeah, whether I choose it here or here it is the same thing, all right, but I have meticulously chosen it behind here, that is all, okay, now I can choose the t not the next rational on this side, so s not, t not, s 1, t 1, s 2, t 2 just like in the proof of Leibniz series, okay, alternatively they will be between t not and s not, okay, so that is how I am going to choose those numbers here, that is for one cycle is clearly stated and you repeat it, okay, so what do you get, repeating this process we get two sequences s n and t n such that each s n is rational number, each t n is an irrational number, t n's are inside s n minus epsilon s n by 2 comma s n and s n plus suns are inside t n comma t n plus epsilon t n by 2, so this is this minus, which is on the left side, this is on the side, on this side for all n, okay and epsilon t n will be less than equal to epsilon s n by 2 and that is less than equal to epsilon t n minus 1 by 2 square and so on, all so on, epsilon s not divided by 2 to the power n, 2 to the 2 n, by the nth stage you would have got 2 square, 2 to the power 4 and so on, 2 to the power 2 n for all n, for all irrational numbers t between s n and s n by 2 comma s n we have epsilon t is less than equal to epsilon s n by 2, once you have chosen that one for everything in between also the length of those intervals which you have chosen has to be short otherwise they will collide with the other one which you have chosen for the rational numbers, okay, the basic assumption is that you and we are disjoint, therefore you see that, all right, so that is all now you see that there is a contradiction here, for all rational numbers for the same reason s belonging to t n to t n plus epsilon t n by 2 we should have epsilon s less than its number, so it follows that both the sequences tend to a common limit, okay, say r is clearly s n is decreasing and t n is increasing, I told you this is just similar to the proof of you know alternating series why it is convergent is similar to that s n is decreasing and t n is increasing monotonically and hence the limit will be between t n and s n, t n is less than equal to r less than equal to s n for all n, but then from 6 and 7 combined together whether r is irrational or rational it follows that the corresponding epsilon r has to be less than epsilon s naught by 2 power 2 n, so everything in between see it has to be, it has to satisfy this far because if it is rational number it is this silver irrational number the other one will work, okay, so these are two statements here, therefore what we have is this is true for all n this r is independent of n, so when you take the limit of this it will show that epsilon r is 0, so contradiction is to the fact that we can choose you know open rectangles around each point on the diagonal such that one set of rectangles are inside you other are inside we they are disjoint, so one small observation is that this semi internal topology is not matrizable why because if it is matrizable it is a product will be also matrizable matrizable means what there is a metric the product topology is given by the product metric, right, if it is product metric then it will be normal also that is an absurd thing because just now we proved that it is not normal, okay, so that is the contradiction just the way we have just defined that when it was not at all clear that something happens, okay, here I have put a few exercises to you try them they will only illustrate and more and more about this familiarity you will get familiarity with the concepts here a complete regular is hereditary, okay, so hereditaryness goes t0 sorry host darkness then regularity complete hereditary normality you will say that is what topological space as she says a base for a complete regularity, okay, you have to verify for all point x and an open subset you do not have to you have to just verify it for only a subbase you have to prove that each point x belong to x v inside only the subspace sub subbase x belong to v there is a continuous function such that fx is 1 and f of the complement of v is 0 enough to verify all these statements for sub basic basic open sets is easy sub basic open sets you have you have to just be careful and be done with it and you can verify this here sure that will be helpful especially while dealing with products so I have put that one as a illustrative example a exercise which will help you to solve the next exercise show that the product of completely regular spaces is completely regular one way is obvious because of the hereditaryness okay after that show that the lower limit topology RL is completely regular and hence conclude that RL cross RL is completely regular okay in particular this gives you an example of a completely regular space which is not normal we just prove that it is not normal okay so let us stop here and take up this study next time thank you