 Welcome to module 44 on point set up as you course. So, today we will continue our study of properties which are preserved under products, productive properties. So, let us begin with this proposition. You start with a arbitrary product of topological space is Xj, pick up a point in it. So, Xj will be first countable at that point, if and only if each factor Xj is first countable at X small j for all j inside j. So, all coordinate points are having a first countable base at that point, that is the first condition. Second condition is that the subset Sx of the indexing set all j belong to j such that this Xj is not a Sierpinski's point in Xj. Look at all that point, all that the indices that set Sx must be countable then the other part he will find only if so then the reverse is also true which is what the proposition says. These are somewhat not very straightforward. So, let us go through the proof carefully. So, first suppose Xj is first countable the product space at the point X belonging to Xj that means that we have a countable base, local base for Xj at the point Xj, that is the meaning of this first countable. Then A follows from the fact that all the coordinate projections are open and surjective. So, if we take P i of B where B range is over the countable base that will give you countable base for our Xi at the point Xi. So, this we have seen that first countability is weakly hereditary in the sense that under open surjective maps it is preserved. So, part A is proved. The second part is something some peculiar thing. Suppose that for some uncountable subset i of j Xi is not you know what is this I have taken Sierpinski's point that is what I want here. This is this is next theorem sorry what I am doing here yeah to prove P suppose B is a countable neighborhood of system of Xj at this point and Sx is not countable. You want to prove that Xx is not countable. That means what there are uncountably many points X sub j and they are not Sierpinski's point means that they have proper open subsets as neighborhoods. Sierpinski's point is defined to be such a point wherein the only open set containing that point is the whole space. So, not a Sierpinski's point means Xj belongs to Uj Uj open and Uj is a proper subset of Xi Xj for every J inside Sx with Sx is uncount Sx is not countable set of such points which are Sx means what they are Sierpinski's points. So, now I am assuming that this is not countable. Now what happens this is countable that is not countable or something happens. Then for each Pj inverse of Uj Pj inverse of Uj is a sub basic open set. There must be a V belonging to V that is V is the countable base at the point capital X at the point this X belonging to Xj. So, this V must be inside Pj inverse of Uj for J in Sx. By Pj null principle for each V there is such a thing I mean for each but number of points in V is only countable but these Js coming from Sx are uncountable. So, it follows that one of the V inside V for which this happens for an uncountable subset I of Sx. If all of them are countable countable even a countable subset Sx will be countable. So, that means that Pj of V when you project J coordinate that is contained inside Ui because V is contained in Pj inverse of Ui and this is happening and Uis are not the whole space. So, Pj of V are proper open subsets contain proper open subsets for every J that is a contradiction to this basic fact that we have observed or we have. This must happen at most for finitely meaning. All other things it must be equal to whole of Xj. But now we have got a countable thing. So, there is a contradiction. So, that proves that proves what one way. So, I have to prove convo. Suppose A and B are true. Then I have to show that Xj has a countable base at the point X. So, that is easier actually choose a countable local base Bi at Xi for every Xi where I is inside Sx. Remember what is Sx? Sx is set of points wherein the Xj, these Xj's are not shipping this point. For those things I am choosing a countable base. For others what should I choose? It does not matter because the only open set containing Xj will be the whole of Xj. It does not matter because Xj is the countable base itself. So, you can ignore them. That is the meaning. But now I have chosen for each J I have chosen a countable base at Xj and Sx is countable. So, this Sx, currently Sx is set of all Pi inverse of B where B is inside this countable family and I raise over all Sx. This is countable. So, these are countable many on countable many sets. So, they are countable many subsets here. So, this is a countable family. Therefore, if you take finite intersections of members of this that is Bx that will be also countable. But once you take Bx finite intersections that becomes a local base at Xj for the product space. The Sierpinski points where all those indices they do not trouble you at all because the entire space as soon as you take some open subset around corresponding Xj there around Xjj it will be entire space. So, for those things you do not have to take finite intersections only on this family you have to take. So, that will become a countable base at X. Now, why I have proved this one so carefully is this is point wise. Now, the same proof will go through if you want to do globally, okay, proof will be the same thing, but statement will be slightly different. What is that? Because now you are taking for all points this is happening. Okay, you want to do that. That is for first countability at all the points. Why is this first countability at a single point, right? First countability at all the points. So, that is our next theorem Xj is first countable if and only if each Xj is first countable that is the first part this is point wise which is also point wise is first part. So, the second part says that Xj is in discrete space for all but a countable number of j inside j. If all the points are Sierpinski's points in a space, right, that is an indiscreet space that in the difference between condition B here and condition B in the previous proposition and that should happen but for a countable setup for countable subset anything can happen other than that it must be all of them must be in discrete spaces. Indiscreet product with indiscreet space does not disturb the rest of the things that is what the theme is here. So, let us go through this one but this is more or less like the previous theorem now, previous proposition that Xj be first countable then first countability of each Xj for j follows from the above proposition, same thing like open surjective maps preserve the first countability. So, this process next suppose for an uncountable set i of j Xis are not indiscreet. This means that we can select X i belong to X i which is not Sierpinski's point for each i inside i and this is an uncountable set okay. So, having chosen X i you take a point at which has these as the coordinates, ith coordinate is equal to X i. If choose an X belong to Xj any point is set P i of X is equal to X i on this uncountable set other things can be anything. Then from the proposition above it follows that Xj is not first countable at that point X because condition B of the proposition is not satisfied. So, it is by contradiction you can prove that the property P B follows okay. The converse is also easy here suppose i contained inside j be such that i contained inside is countable and Xj is indiscreet for j in the complement. This is the condition we want away from a countable set it should be you know indiscreet all of them should be indiscreet. And for each X in Xj we select a countable local base Bj at Xj for each j inside i. See again you have to do it point wise now construction okay. The indiscreetness gives you for all point the Sierpinski condition B is satisfied therefore the same conclusion that we have done there will work for this okay Pj inverse of Uj where Uj is inside Pj and j is inside i this i is countable. So check that this forms a local base okay. This is repetition of the previous part but only thing is having an indiscreet space away from a countable set will give you the Sierpinski points okay just naturally out of those things the Sx the condition Sx is countable will be satisfied. Now we just make a statement and it is the proof is straight forward let Xj okay is countable product of an arbitrary product of topological spaces. This will be second countable if and only if each factor Xj is second countable and Xj's are indiscreet except for a countable number of j and j. So exactly same thing has first countable rating. So this time you do not have to worry about point wise to take a base go back to go back come back and so on the same proof will work. Let us come to now separability in our list. Separability is a tricky business okay one has to be bit careful here but whenever things happen it happens easily also. It is easy to check that if each Xj is separable and j is countable then Xj is separable okay all that you have to do is product of ai's the closure is equal to the closure of the each closure of product of ai okay. First take the product then take the closure is the same thing as first take all the closures and then take the product. So if ai's are dense in Xi ai bar will be the whole of Xi okay. Then you take the countable product then take the closure that will be dense. The only thing is this is always true but when the product of countable set is countable it has to be used. If you take uncountable product it will not be countable alright. So countability is has to be preserved so you have to take j to be countable that is all. But there is a curious phenomenon here namely even if the cardinality of j is the first uncountable which is noted by the C is cardinality of R for example okay first continuum okay. Then Xj turns out to be separable though the argument I have given doesn't work and it is not an easy argument here. So we have no time to do that one and you know it is not used by us anywhere. So I have given you a reference you can look into that namely Willard's book. However if cardinality of j is bigger than C then even this will fail. Each of Xj may be separable but the product may not be may not be separable okay. So let us stop here for today just summing up what we have done so far. So after hereditary property and co-hereditary properties checking for various things namely connectivity, path connectivity, compactness, window loathness, first countability, second countability and separability. Then yesterday and today we checked about first countability and second countability and separability okay. So this is the gist of the various properties of topological property that we have studied so far alright. So we have still more thing to do with product properties namely compactness and window loathness we have yet to worry about that. So that is another topic so that will be taken next time okay. So thank you.