 Welcome to module 43 of point set topology part 1. As promised last time, today we shall study productive properties or whatever topological properties we have made, we will verify which one is productive, which one is not. That is the whole idea of today's talk, okay? So when we are studying products, I would like to recall a few conventions and notation that we have said earlier and also introduce some more convenient notation. My fundamental idea here is that when you have finite products, it is okay to use the order, the order tuple notation, Cartesian coordinate space notation. But when you go to infinite product, it is totally inconvenient and whatever you use, it will not be exactly rigorous. You will have to keep on using identifications, okay? So we will try to reduce that kind of notation as far as possible and use a logically consistent notation and that is possible only if you think of the product space as a function space, okay? So having said that, let me start telling you what are the notation. J will denote a non-empty indexing sector. So this is all for the entire, entire of this course you can say, but definitely in this chapter. I denote a subset of J and Ic will denote the complement of i inside J, okay? For each J in J, xj denotes a non-empty topological space. So I have used non-emptiness twice here. Pay attention to that. Whenever i is also non-empty, xi denotes the product space over these indexing set i, xj. This i here, the suffix i is the indexing set, okay? Remember I use x power i, superscript i only when all the xj's are equal to x. So that is a different notation. I am not going to use that one here at all, okay? The underlying set of xi, I want to recall namely it consists of functions now, x. So I will call it as element of the product, but it is a function, okay? From where? From the indexing set to the disjoint union of all these set size, okay? So any function like this will be an element of this product. So that is the underlying set. Then we have these projection maps p little j from xj to x little j. What are these? pj of any x, x is a function, remember? You evaluate x at the jth element here, j element, element j, so x of j. And this is another notation, xj. This is our usual coordinate notation. So I do not want to give up that one, but I have given up the ordered n tuple, ordered supers and so on. That I have given up, ordered pairs and so on, okay? So let pi, now i is capital. This also denotes xj, the entire product space to xi, which is a partial product because i is a subset of j. This denotes the projection map defined by the same principle, namely pi of x is a point in xi. So it is a map from i to the disjoint union. On the point i of i belong to i, it is xi. Other coordinates are forgotten, that is the meaning of this projection map. We also use the notation x suffix capital i to be image of pi of a consistent with this one. Whether it is a singleton or a whole set of bunch of indices, you can use that notation. Moreover, suppose xi is a point of x capital i, everything, little x i is a point here and y, now the suffix is, prefix is i complement inside xi complement. So i and ic are disjoint, okay? So we are complementary thing. We shall use the notation, this ordered pair xi comma yic to represent the unique element in xj with the property that pi of xi yic is xi and pi c of this point is yic. If I give pi and pi c, the point is uniquely determined and that point has a property that pi of this point is xi and pi, pi c of this point is yic. So that is the meaning of this one, okay? So note that for any proper subset i of j and for any non-empty subset ui of xi, we have the jth, little jth projection of pi inverse of ui is the whole of xj for all j in the complement of, complement of i, okay? Compliment of i will pick up all the elements xj, that is pi inverse of ui. Observe this one, okay? So these are basic things which you have to be familiar with, then only you can make sense out of the product space. The central fact which will be used again and again is that the collection S, pj inverse of uj where uj's are inside xj, they are open and j is inside capital J. So these are all variables here. Look at all pj inverse of uj to care that collection S, this forms a sub-base for the product topology on xj, that is the definition of the product topology on xj, okay? So I am just recalling that. Now something more I am going to introduce here, this notation, let F denote the set of all finite subsets of j, all finite subsets. I do not need empty set, so you throw empty set, it may cause unnecessarily problem, that is all. Then the collection B which is pi inverse of ui where ui is open in xi and i is inside F. So i is a finite set, ui is open in xi which just means you know you can take it as to begin with u1 cross u2 cross un where ui, little ui's are open in xi, but there are more open sets, okay, than just product open sets. You take all of them, take pi inverse of those, so this will automatically contain all the finite intersections for members of here, so it will be a base, it will have more open set, no problem, so it will be a base. So this base is more convenient for me, so I will use this one, but this is sub-base, that may be also used, okay, these two things I will keep this. It should also be noted that each pi is an open map for every non-empty subset i of j, the empty subset this pi does not make sense, so that is what you have to say on non-empty, okay, all the projection maps are open maps. Moreover, take any member of this s, what are they, they look like pj inverse of vj, okay, for v belonging to s, pj of v is equal to xj for all but finite number of gj, okay, this is true actually for members of v also, here it is only one j, if i is not equal to j it will be whole thing, but here it will be like this, so this is true for members of v also, okay, so with these conventions you will see that many proofs can be written very, very closely, idea becomes extremely clear. The first thing I want to prove is connectivity and path connectivity are product invariant which is at the bottom of our list 6 and 7, okay, but these concepts were the first thing which we are considered in our chapter, all right. Now each factor xj is a quotient of xj, right, open quotient actually, any subjective map, continuous subjective map, connectivity, path connectivity of xj implies that of xj, this we have seen several times, okay, so this part is all right, it is the proof of the converse that needs to be worked out, namely if xj is connected, path connected then the product xj is connected, path connected, once again path connectivity is straight forward, suppose each xj path connected starting with any two points xy inside x capital j you can choose path omega j 0, 1, 2, xj continuous function joining these points xj to yj, they are inside x little j, right, so there you can join them, so you have got all these families of paths then you define one single omega from 0, 1 to xj such that its jth component, see w omega t must be a function, right, from capital j into the disjoint union of all these xj's, so on the jth thing you take this path omega j t, okay, so automatically this will be continuous function, we have verified a function is continuous if and only if all its coordinate functions are continuous, so these are continuous or these continuous, when you put j, when you put t equal to 0 this will give you omega j 0 which is xj, t equal to 1 it will give you yj, right, for each j that is the term that means omega t of 0 is x and omega t of 1 is y, so you say path joining x and y, that proves the product is path connected, so let us now concentrate on connectivity, recall that you have already proved that finite products of connected space is connected, so this was Corolli 3.35 or whatever, okay, I do not have to show it to you, we know this one very clearly, not very long we proved this one, all right, indeed we proved a stronger result namely if y is connected and the fibers are connected then x is connected under any quotient map, right and using that we can prove two product of two of them is connected, then three of them is and inductively any finite product is connected, all right, so I am going to use that one, now I go to arbitrary family, so xi, im, j be a family of connected topological spaces, take any point z, what you should show is the connected component of that point is the whole of z, whole of, okay, so this is my xj, I am taking it as z here, that matter, okay in short notation product be any point, okay, put a z equal to, so I am taking a point in the product space, I want to show that after all I do not show the product space is connected but what is my idea, idea is to show that the connected component of any point in the whole space, okay, so this is a first step here, take a z to be all pi compliment inverse of this z i c, remember what is z i c, z i c is a point inside x i c which is the projection of z, okay, take p i c inverse of that which is nothing but x i cross singleton z i c that is a notation for this, if you want to use the Cartesian coordinate notation, I do not want to use it, so I have written it carefully like this, p i inverse of z i c, the coordinates inside i i, okay, they are dropped out, there is no mention, they could be arbitrary but when j is inside i c, that means j is not in i, the coordinates of that point must be the coordinates of this point, they must coincide, so that is the meaning of this p i inverse of this one, okay, take union of all these where i is a finite set, i is beyond f, finite subsets of j, okay, each p i inverse of p i c inverse of z i c, this will be homomorphic to x i, just now I told you, it looks like x i cross one single point in the complement namely in i c, okay, they are homomorphic to i c, they are connected why because i is finite and the finite product is connected we have used, so these are each of them on the right hand side is connected, moreover z is a common point to all of them, therefore it follows that az is connected, so this argument we have used before also, okay, it is very easy to see that az is connected, take a separation if possible, suppose az is b separation c, where will be z, it will be in one of b and c, the moment z is inside b, all these things will be inside c, okay, that means b is empty, so if z is inside c, b is empty, if z is inside b, c is empty, so there is no partition, there is no separation, so this is a connected set, this is not the whole of xj, but it is very close to that, namely what is this closeness here, if you take the closure of this in the whole space that is the claim, okay, we shall show that az closure is xj, okay, so half the step we have proved that this is a connected set, the second half is that its closure is the whole space, okay, so that will complete the proof, why? Because closure of a connected set is connected, all right, so what is the meaning of closure is the whole space, every non-empty open subset in xj will intersect az, so that is what I have to prove, that az intersects every non-empty open subset u of xj, what is the meaning that u is an open subset non-empty, inside xj it contains an open subset of a form p i inverse of ui for some open subset ui in xi, these are the basic open sets, that is the first thing I am going to use here, where i is itself a finite set, okay, there must be some such thing and ui must be non-empty open inside xi, i must be finite, take a point wi inside this ui, this ui is non-empty, right, so it follows that look at this point wi, z i c, what is the meaning of this, all the i coordinates i inside i are like w, w little i, if i is inside i c then it is z little i, that is the meaning of this point, we have defined this one, this point is clearly inside p i c inverse of z i c because it is i c coordinate at this point, i c coordinate at this point, so it is inside this one, also it is inside p i inverse of ui because this wi is inside ui, so it is p i inverse of ui, okay, so it is in the intersection but the first set is contained inside a z, the second set is contained inside, second set I have written just same thing but it is contained inside u, so the whole thing, whole this point is contained a z intersection u, that is the end of the proof, why, because starting with u I have shown that a z intersection u is non-empty, okay, here is a picture, if you want picture to explain whatever I have told here, so this is an arbitrary point z, then I have bunched up this x i, i is a finite set of indexing sets, suppose it is just x 1, then this looks like the x axis and the rest of them are y axis, the rest of them are i c I have put, so you have divided the entire set into i c, i and i c, so draw this, this is one plane, this is a finite product of finite things, so this is connected, now if you take an arbitrary point w inside this u1 inverse image, this dot dot dot u1, okay, it is inverse p i inverse p i c inverse of this ui, so if you take any point like this here, p i inverse of ui, you can project back here, actually this y I have started with a point here w and this is w i c over that length, this y is nothing but w i here in the notation, okay, its other coordinates are just the coordinates of this point, you can think of this as y coordinate here, y coordinate of certain this point y are the same, okay, so i c coordinate of this were same, so that is precisely the point here which will be in the intersection, this was an arbitrary u, okay, so every u contains some such set is what we have to use namely what are the base for the product apology, that is all, okay, so you see the proof that product is path connect, product is connected has just two ideas, one is you look at a point and then take all these coordinate planes, finite coordinate planes passing through the given point, so i am using the terminologies for r, r, r square, r cube and so on, okay, that is the only way you can picturize, you can imagine, coordinate planes and so on, okay, there are no planes here, all the all propolisative spaces are arbitrary to apply in spaces, okay, so that set A z is what it is dense, so that is all you have to remember, proofs are not at all difficult, okay, so let us do the other things next time, this itself is a something both we have done, other properties we shall do next time, thank you.