 Welcome to chapter 4 of points at topology course part 1. So, this I have named as largest properties. Essentially, we are going to study properties of topological space such that whenever a topology has it, something which is larger than that will also have it. So, that is the rough definition, formal definition we had already given. But there are some more here which we feel that just like the case of first countability and second countability, they are the opposite here. Something called regularity and normarity are not exactly fit into this definition. So, to begin with, we will take the celebrated fresh air spaces. So, that is module 47. So, we begin with a Burbaki style presentation here. Starting with a topological space X, several statements are equivalent to each other. That is the statement of this theorem. What are these statements? Any one of them can be taken as the definition of fresh air space finally. So, that is the whole idea. So, X, Y are different points. There exists an open set U in X such that X belongs to you and Y does not belong to you. So, read it carefully. You take two distinct points. Then the statement says there is an open subset containing X but not containing Y. So, this statement also implies that there is another open set V which contains Y and not contains X. So, that is the logical conclusion of this statement. You have to understand that because X and Y are two different points. Just because I have written first X here, second Y, it is not an ordered pair. So, main statement X belongs to you and Y does not belong to you. There is such an U. You can interchange that in also. So, that is the first statement. The second statement is any subset of X is the intersection of all open sets containing A. Take intersection of all open sets containing A. That will be equal to A, any subset. If you take empty set what happens? You can take empty set also as an open set. So, intersection will be empty. Given any X belonging to X, the intersection of all open sets containing this singleton X is equal to singleton X. So, this is a third statement. From arbitrary open subset, arbitrary set you have come to singleton set. So, it is obviously 2 implies 3 is clear here, right. I have some more statements. It is only three of them. There are more statements here. The theorem continues the statement. The fourth statement says for all X belong to A, singleton X is closed, relaxed. So, this one statement is very easy to remember. All the singleton sets are closed. Okay? The fifth one says every finite subset of X is closed. So, that is an easy consequence from 4 because finite union of closed sets is closed. The sixth statement is every subset of X is a union of closed subsets. So, that is also easy from 4 directly. In any case, it is also 5 because every subset is a union of finite subsets. The seventh statement is every non-empty subset of X contains a non-empty closed subset of X. So, this doesn't seem to be immediate, but we will see all these things are very easily implying one, one implies the other and so on. They are all equivalent to each other. That is the statement. Okay? So, let us go through the proof once again because I have not indicated all the proofs. One implies two. Given two distinct points, okay, there is an open set containing one and not containing the other is the condition one. Okay? Take any subset A of X and a point Y which is not in A. Then we must find an open set U which contains A, but not Y. Okay? So, let us look at statement 2 here. Any subset A, the intersection of all open sets containing A, okay, is equal to A. So, if you want to show that it is equal to A, you must produce some open set containing A, but not containing the point Y. Where Y is a, not a point of A. So, that is what we have to prove, right? So, that is what we are proving here. One implies two. Okay? By one, for each X inside A, we have an open subset U X such that U X I have given because it depends upon X. X belongs to U X and Y is not even U X. So, you vary X is but keep Y as it is. Take U equal to union of all U X's are X values over A. So, this will contain A since none of them contain Y. So, union is also not contained Y. So, when you take intersections of all open sets containing A, this Y will be left out. So, since every point away from Y will be left out, the intersection is exactly equal to A. So, that 2 implies 3. Okay? Now, 2 implies 3 by merely taking A equal to singleton X, this I have already included, I have already indicated. So, what is the statement 3? Take any, take all the collection of all open subsets containing X, the intersection is just singleton X. Okay? That is the statement 3. 3 implies 4. We want to show that singleton X are closed. That means take singleton X, take the closure. I should show that it is singleton X itself. Okay? This is the same thing as saying that if Y is not in X, then Y does not belong to exposure. Right? Exclosures singleton X is what you have to show. But by the hypothesis applied to the point Y now, you see for every point intersection of all open sets is just that point. Okay? It follows that there exists an open set V such that Y is inside V, X is not in a way because X is not equal to Y. Okay? So, I am not using number 1 here, I am using number 3 here to conclude this one. Okay? So, if this open set does not intersect the singleton X, that Y cannot be in the closure. That is all. So, this is the definition of the closure. So, 3 implies 4 is over. All the singletons are closed. Alright? Now, 4 implies 5. Since every finite union of closed sets is closed, every finite subset will be also closed. Okay? 5 implies 6 is again obvious because every set is the union of finite sets. Every set is the union of singletons is one thing. Next one is every set is the union of finite sets. Now, 6 implies 7. This shows this follows since every non-wide set has to contain a singleton. See, let us go back and see what is 7. Every non-empty set of a contains a non-empty closed set. Singleton is closed set we have proved. Okay? Every finite set is closed that we have proved already. So, non-empty implies what? There is a point. That point singleton point is closed. So, the, so this implies 7. Okay? You see singleton set from 6 every subset of X is union of closed sets. Singleton X, how can it be union of anything? It is union of singleton X itself. Right? So, that must be closed. So, it is already implies that this implies 6 implies already 4. Okay? So, anyway, you can jump from 4 to 7. Now, 7 to 1 I will complete. Okay? Finally, that will complete the whole chain. 7 implies 1 is whatever. So, apply 7 to singleton Y to conclude that it is closed. Singleton Y should contain a closed subset. Right? The singleton Y is closed. That means what? Complimental singleton Y is open and that will contain X. So, you, you can take U to be X minus Y. Okay? So, all these statements, implications is one implies other is very easy. So, we have proved it in a single chain. Okay? Now, the definition is the following. A topological space which satisfies anyone and hence all of, all the above conditions namely in the above theorem such a thing is called a fresh space. Incidentally, you know, he is one of the founders of this points of topology. He had his own definition of topological spaces in which he, he put this extra condition for topology. Just like Hossdorf also puts his own extra condition. So, present day definition later on whatever he has been adopted, this is what is called as general topology, general then fresh as well as Hossdorf and so on. Which doesn't have these extra conditions? Sir. So, all the metric spaces will be fresh spaces, right? Obviously, they are fresh spaces. You know, they will satisfy many more properties in this chapter now. You watch out for them. Fresh space is still our motivation after all. But we will see is that lots of fresh spaces are there which are not metric spaces. Very first thing you notice is that freshness is a topological property. If X is homeomorphic to Y, X is fresh A implies Y is fresh A. Okay? You can take any one of them. For example, I will take singleton sets are closed. Under the homeomorphism, two sets will go to closed set. So, singleton sets the other will also closed. That's all. Even though it is good to remember all these requirement conditions so that we can use whichever one is most convenient to us. Okay? You may find it somewhat confusing to remember all that. Whereas, condition 4 that just now I used namely singleton sets are closed. It's the easiest thing to remember for me at least. Okay? Therefore, I remember fresh A space by this condition. Other things I can enter whenever I want. Okay? This is a question of, you know, which one to remember. So, fresh A space is for me that every singleton is closed. That's it. One of the important properties of fresh A space is that now we have a good characterization of an accumulation point of a subset inside a fresh A space equivalent to what we do in metric spaces. A point x pilon to x is an accumulation point of a subset if and only if every open set containing x contains infinitely many points of A. For a limit point or a closure point, for a closure point, all that you needed is intersection is non-empty. One of one point is enough. But if it's a accumulation point in particular if inside a fresh A space just like in a metric space, okay? Every point which is a accumulation point should have the property that every open set containing x intersection with A must be infinite. Okay? Not very difficult to see. You have to carbon copy metric, but don't use the metric. Use the property that given x and y different, there is a point open subset here, blah, blah, blah. Singleton sets are open, complements are singleton sets are closed, complements are open. That's all. Okay? That I have linked in assignment. Another important property is also an easy consequence of the definition is the following. If f from y to x is continuous, remember y to x not x to y, huh? x is fresh A here. Inverse image f inverse of x, that is those are fibers, they are all closed. Why? Because singleton sets are closed. Okay? This is very useful thing. So that's why I have, so for example, we have been using this inverse image, continuous functions, real value, continuous real value functions, inverse image of a point that of all f, that of all x such that fx equal to 0. It's a closed set. So this is because r is fresh A. Okay? So that is why it was working. So now you can, you can, you can see that inverse image of singleton set is closed, singleton is closed, so inverse image is also closed. Okay? A typical example of fresh A space is any set with co-finite topology. See, I don't take matrix space as a typical example. Matrix space is too good. Take any set with co-finite topology, especially take an infinite set. If you take finite set and co-finite topology, then it is just like a discrete space. Okay? Discrete spaces are obviously fresh A spaces. But co-finite topology on an infinite set is a typical example of fresh A space. It is not matrizable. Okay? It doesn't come from a matrix. Okay? So this is what we have to, that is why this is nice. Observe that on a given set, co-finite topology is the smallest topology among all fresh A topologies. So that's why I take this as a typical example. So to get an example of a space which is not fresh A, you have to take some topology which is smaller than co-finite topology. Indeed, fresh A is a largest property. Once some topology is fresh A, if top prime is a larger topology than top, that will be also fresh A, which is very easy because singletons sets are closed here. So they will be here also. Right? Here means what? If something is closed here, something is open here, something is open here. That is the meaning of that. Of course, every matrix space is a fresh A space. Fresh A space is clearly hereditary. See singletons sets are closed in intersection. Take A is a subspace of S. Take A is singleton A. It will be closed in X itself. Right? So intersection with A is A, singleton A itself. So all singletons inside A are also closed. So every subspace of a fresh A space is fresh A. Very easy. That's hereditary property. It is not co-hereditary. So here is an example. You can give lots of examples later on. Let us begin with some example. Take X to be the quotient space of R where all the points of an open interval, any open interval, let us say minus 1 to plus 1 is identified to a single point, collapse the open interval to a single point. Minus 1 will remain separately. 1 will remain separately. All the points outside minus 1 plus 1 will remain separately. No identifications there. Only whenever X and Y are both between minus 1 and plus 1 strictly, they will all be equivalent to each other. So that is one single class. So let us look at that point, that class S star. This is just a notation. In the quotient topology, the star and the image of one, every member has an image. Everything between minus 1 plus 1 is denoted by star now. One single element. So in the quotient topology, the star and image of one will be distinct points. And every neighborhood of the star will contain the image of one. In the other way around also, every neighborhood of one will contain this point. So you cannot separate them in that way. Every neighborhood of one will contain the image of star. That means every neighborhood of one intersects minus 1 plus 1, the interval. So there will be some point here. So that point is, when you go down, it is just a star. So this is not a fresh space. So here is another property. Fresh space is productive. Take a family of topological spaces X, J. The product is fresh space. You should know if each X, J is a fresh space. The first thing is, suppose the product is a fresh space. Suppose X, that will take two of them only, X cross Y. Then I can take X as a subspace of X cross Y by choosing some point Y, then look at X cross little Y sitting inside X cross capital Y. That is a subspace. Since I am assuming X cross Y is fresh, this subspace will be also fresh. But X cross little Y is homeomorphic to X. It is a copy of X under the map X going to X comma Y. So X is fresh. Similarly, Y is fresh. Argument is similar. This is the case in the infinite case also, but we have to write down it carefully. That is all. Every factor X, J can be thought of as a coordinate subspace just like X going to X comma Y. So that is what I have done here. Pick up any point Y. Then forget about the Jth coordinate of that point. So that is Y Ic. Singleton I, capital I is Singleton J. So Y Ic is complement. So all the other things are there. That is in a point of X Ic. Product taken over this complement. Only J is missing here. Then if you go to X comma X, X going to X comma Y Ic, this will give you an embedding of XJ. XJ inside product of XJs. So each XJ is subspace of X capital J in that sense. By hereditaryness, each XJ is fresh. We have to do the converse. Suppose each XJ is a fresh, then Y this one is a fresh. That is much easier. Conversely, suppose each XJ is fresh. Take any two points, distinct points inside XJ. What is the meaning of the two points are distinct? At least one of the coordinate is different. If all the coordinates are the same, then the points are the same. So there exists a J in J such that XJ is not equal to YJ. Now list space, this space XJ is a fresh space. Therefore, you will get an open set UJ inside XJ such that XJ is inside that and YJ is not there. Now you get PJ inverse of UJ that will contain X and Y will not be there. Y is there precisely if the YJ coordinate of Y is inside UJ, but YJ is not inside UJ over. So one of the coordinate space you have used, but to conclude that X is fresh, you have to say all the coordinates because starting with X not equal to Y, you do not know which coordinate is different. So that is the whole idea. I think we will stop here. We have done good work with fresh spaces. Next time we will study house door spaces. Somehow the house doorness has taken the limelight. It is more important than T1 spaces. So next time.