 Welcome to module 48 of point set topology course part 1. So, today we will continue the study of largeness properties, one property study namely fridgespaces. Now, today it will be Hausdorff spaces. Once again similar to what we did for fridgespaces, but this time only four equivalent conditions. So, I will make a statement on a topological space X, the following conditions are equivalent. Given two distinct points X and Y, there exist open sets Ux comma Ui in X such that X is in Ux, Y is in Ui and Ux intersection Ui is empty. So, this is also stated like every pair of distinct points can be separated by open sets. The second condition is for every topological space Z and for every pair of continuous functions alpha beta from Z to X, the set Z belonging to Z such that alpha Z equal to beta Z is a close subset of Z. Third condition for every topological space Y and for every continuous map from Y to X, the graph of F which we denote by gamma F is closed in Y cross X. Remember graph of which is nothing but set of points Y comma Fy. The fourth condition is a very simple condition the diagonal X comma X as X belongs to X inside X cross X with product topology is closed. That is a close subset. So, let us go through the equivalence of these four statements first. So, starting with the two points, distinct points can be separated by open sets. I want to show that set of all points that such that alpha Z equal to beta Z is closed whenever alpha and beta are continuous functions from Z to X. Same thing as saying that set of points wherein alpha Z is not equal to beta Z that is an open set. Alpha Z and beta Z are points of X. Alpha Z is not equal to beta Z implies you can apply the previous condition one. You will find a U containing alpha Z and V containing or U alpha Z and U beta Z opens of sets. Sets of the intersection is empty. If the intersection is empty, there are inverse images will be also opens of sets which are empty. Intersection will be empty. There will be also disjoint. That just means that for any point inside alpha inverse of U Z U alpha Z and any other point inside beta inverse alpha will never be equal to beta. So, that is how you get that the complement is open. So, that is I am repeating it here. Put A equal to all Z such that alpha Z is not equal to beta Z. So, we want to show that it is open. Take a point Z inside A. By one we get opens of sets U and V such that alpha Z is U, beta Z is V and there it is joined. Alpha and beta are continuous. If it opens of set alpha inverse U, beta inverse V inside both containing the point Z and the point Z because alpha Z belongs to U and beta Z belongs to V. Therefore, when you take the intersection, Z belongs to intersection, okay. Then take a point, take alpha of that, it will be inside U, beta of that will be inside V. So, they are not equal. So, that is a subset of A, okay. So, points G is meant to set. Then this is what happens because I am not taking inverse image and the same map do different maps. But that is a common point. That is a neighborhood now. That neighborhood is contained inside A. That is the whole idea. Second statement implies third statement. For every topological space Y and for every continuous assumption F from Y to F, the graph gamma F contained inside Y cross X is closed, okay. So, this is what we want to prove. So, recall that the graph is just nothing but points Y comma F Y, Y belongs to Y. This is a subset of Y cross X. So, let us take Z as Y cross X itself, okay. And consider two functions alpha of Y X is equal to F Y. This is second projection here and beta of Y X is equal to X, okay. Y X is F Y from here gamma and beta of Y X is X, okay. Second projection followed by alpha and for alpha and beta. So, these are my alpha and beta now and Z is Y cross X. Both are continuous and we have put Y comma X such that alpha Y X is equal to beta Y X. Then what is it? First coordinate is Y. The second coordinate the Y comma X is X, this beta, this is second coordinate beta here. That is equal to F Y. So, this coordinate is F Y means it is gamma MIF, okay. Points were in alpha Y equal to beta Y. So, there is a close subset of Y cross X, okay. See, you can directly prove 1 implies 3. I want to prove 2 implies 3. That is why I have to do this. Take a special case, 2 is true for all Z and take a special Z equal to Y cross X and alpha equal to this way. Now, 3 implies 4. I have to show delta X is the diagonal. The diagonal is nothing but graph of the identity map X comma X. So, apply 3 to the case where in Y equal to X and F is identity. So, diagonal is closed, okay. Now, 4 implies 1. So, that is also easy. If X is not equal to Y, we see that X Y is not on the diagonal, okay. The diagonal is closed means the complement is open. So, X comma Y is in the open subset. It is open in Y in the product quality, right. So, there must be an open, basic open sets U comma V, right. Such that X Y belongs to U cross V contained inside delta X cross X minus delta X. That will mean that U cross V intersection delta X is empty. This is the same thing as saving that U intersection V is empty, okay. What is the meaning of U cross V intersection delta X is empty? Take a point here. The same point here you have to take to be inside delta X, okay. Then only you will get non-empty set, something non-empty. If you cannot do that, that means any point here cannot be taken as point here. They are disjoint. And it is if and only if actually this can be used in many other places also. This is purely set theory, but it is quite useful here, okay. So, that completes the definition of, they complete the equivalence of these four conditions and we make a definition. A space is called Hausdorff if it satisfies any one of the conditions and hence all of the conditions, okay. So, let us make some immediate remarks here. The very first thing is you know that every metric space is Hausdorff. The topology coming from a metric is always Hausdorff. Because given two distinct points, you can take the distance between them and take the half of that and take to open balls, they will be disjoint. The second example is the co-finite topology on an infinite set is not Hausdorff. Indeed, the co-finite topology has a fantastic property that any two non-empty open sets intersect. Of course, I have to take infinite set. The finite set, co-finite topology is not very interesting. That will be Hausdorff of course, okay. As soon as infinite set, an open set, non-empty open set means its complement is finite. Therefore, two non-empty open sets cannot be disjoint. An important property of Hausdorff space is that every sequence in it has at most one limit point. A sequence may not convert that you know. But if it converges, the limit is unique. So, this was one of the properties which you have been all the time using and you are familiar with from the real analysis and from metric space also we have. So, that is one of the motivations to keep this Hausdorff's property of metric spaces and make it axiom for some topological spaces. If not the metric, let us at least keep this property. That is the motivation. Now, we start mixing up some of these properties. The first thing is every compact subset of a Hausdorff space is closed, okay. This is a very good thing. But later on, we will keep on mixing the compactness and Hausdorffness. So, this is only a starting point. So, how does it prove that? You take a compact subset of a Hausdorff space X, look at X minus A. I want to show that that is open. So, take a point X in X minus A to each point inside A, okay. We can get an open subset UA comma another open subset VA, UA is a neighborhood of A and VA is a neighborhood of X, UA intersection VA is empty. So, this I do for all A then I get an open cover for A. This open cover will have finite sub cover because A is compact. So, let us assume that A is contained inside UA1, UA2, etc. UAN union. Correspondingly, you take VA1, VA2, VAN and intersect them. V is intersection of VA1, VA2, VAN. You should watch this game. Here, I get a finite cover. There, I get intersection corresponding to intersection. So, this technique will be used again and again, okay. So, what is this V good for now? V is open. If X belongs to V, V is open, okay. Why X belongs to V? Because X is inside each of VAI, VAO, X is here all the time VAI, okay. The point is now V intersection A is empty. Why? Because take a point inside A, it will be one of the UAIs. If one of the UAIs, then UAI intersection correspond to VAI is empty. But this is even smaller. We see V is contained in the VAI. So, UAI intersection V is empty, okay. So, that is why V intersection A is empty. That means this V is contained inside X-A. So, this we have done for each point of X-A. Therefore, X-A is open, all right. So, every compact subset in a half star space is closed. This we knew in metric spaces. Right? The proof is exactly same except that we should not use open balls. There we are all the time using metric and open ball and so on. If you there also, you can use the same proof because this proof works in general, okay. Now, we are in a position to derive one of the most important results in detecting homeomorphisms. It is quite application oriented. Result. Yeah. A continuous bijection from a compact host star, compact space to a host star space is a homeomorphism. The domain must be compact. The core domain must be host star. A continuous bijection is a homeomorphism. What is missing? Either you should prove that this map is open map or this map is closed map, okay. So, what we shall prove is that this map is a closed map. Since there is a more general result but not so popular as this theorem 4.8, but theorem 4.8 is a near consequence of that one. I will prove, I will state and prove that one, okay. Then you will see that this while proving that you will get a proof of this one also. So, this is 4.9 any map f from x to y from a compact space to a host star space is a closed map. You see, I am not assuming injection, projection thing, okay. So, this is a more general result. What is it says? You take any continuous function from compact space to a host star space. It is also a closed map. Further, if it is surjective, then it is a quotient map. This part we have already seen. Every surjective quotient, surjective closed map is a quotient map. Surjective open map is a quotient map. All that we have seen, okay. The second part we have seen. So, this one we want to show that why this f is closed map. Start with a closed subset in x, say f, okay. Be a closed subset of a compact space. That was remarked 3.58 whatever. It is compact, right. We have proved that one while studying compact spaces. And hence, by just the proved theorem here, this theorem compact subset of a host star space is closed, okay. So, what I am going to do? I am going to do one more, one more theorem namely image of a compact set is compact. So, I have come to know why. So, f of f is, is what? A compact subset of a host star space. So, that is the end now by that theorem which we have proved just now. It shows that f of f is closed, okay. So, I repeat by a result about compact subset, closed subset of a compact set. ff is compact, f is compact. By a theorem on continuous functions that preserve compactness, ff is contains a y is compact, okay. But from theorem which we have proved just now is y is host star will play now. It was ff itself is closed. So, closed set goes to closed set, f is a closed map. Okay. So, I repeat now, if it is surjective, a closed surjective map is a quotient map. If it is bijective, closed, we have proved, injective we have proved, okay, surjective we have assume, sorry, injective and surjective we have assume. Therefore, it is a closed map. Therefore, inverse image shows on that sort, inverse image, inverse function is also continuous and so on. So, so the proof of both of them are over now, okay. Here is a remark about mixing compactness and host starness. Suppose you have some set and three different topologies, one contained in the other but not equal, tau 1 contained inside tau 2, tau contained inside tau 2, okay. So, tau is trapped between tau n and tau 2 and equality does not hold, that is what we have assumed, okay, they are distinct. Equality there is an equality, that is all, there is no more to say, okay. Suppose these are topologies on x and the middle one is compact as well as host star, okay. You have mixed it up to somewhat opposite nature properties, compactness is a smallness property and host starness is a largeness property. So, you have mixed that them, something wonderful happens which justifies the largeness and smallness property. Then tau 1 will not be, sorry, tau 1 will be compact but not host star. Why it is compact which is smaller than tau, which is compact, but it will not be host star, that is the claim. Similarly, tau 2 will be host star because it is larger than tau, but will not be compact, okay. So, how to see this one, all that you have to do is start with x tau and take the identity map, okay, or inclusion map, whatever you say, identity map, identity because x to x is same set here, okay. But topologies are different. If x, if tau is larger than tau 1, the identity map will be continuous. This is compact, okay. If this is host star, then what happened? This is, sorry, this is all host star is already given, okay. Tau 1 is, yeah, tau 1 is, we do not know, x tau is host star compact. So, use the fact that this is compact. If this were host star, then this will be a homeomorphism, which means tau is equal to tau 1, but tau is not equal to tau 1. Similarly, x tau 2 to x, x tau 2 to x tau, you take again the identity map, okay. This I know is host star. So, if tau 2 is compact, then this will be a homeomorphism. But tau 2 is not equal to tau. So, this is not a homeomorphism, okay. So, tau 2 fails to be compact. You go above tau, which is compact host star. Everything above fails to be compact. Everything below fails to be host star. So, this is like optimizing both of them. Whether such things exist always or not, that is a completely different question, okay. On a given set, such things may not exist. One does not know. Just like freshness, house dorsiness is a largeness property. So, that is what we have just seen and a nice illustration here. It is hereditary, but not co-hereditary, okay. Very easy to produce examples. All through we will have such examples. A product x j is host star if and only if each coordinate space is x j. So, in this sense, it is a productive property, okay. All these proofs are exactly similar to what we have done for freshness and they are straightforward. There is no hitch in them. Only thing is you may not be able to see immediately why it is not co-hereditary. Exactly similar example. The same example we will do of collapsing an open interval, say 01 inside R, okay. The real line is host star. The quotient will not be host star. So, don't make the mistake that quotients are host star, okay. Quotients are very fussy. So, there is many things, there are many things to do about the largeness, the host dorseness as such. In particular, we are now going to mix up these properties with other things like compactness and so on. And also, we may go back to linear, some metric space theory again and so on. So, at this point, let us break. So, that is enough for today. Thank you.