 Hello, welcome to module 47 of NPTEL NOC, an introductory course on point set topology part 2. So, today we will introduce the order topology, so far you are only studying the partial order sets, totally ordered set and so on only satirical aspect. So, today we are bringing the topology, in a sense we had already done this in part 1. So, some part of it will be just a recalling that example it was an example there of a topology. So, let us recall that, start with a linearly ordered set or whatever you call totally ordered set. For any point x in x, let us have this notation Lx is the left ray. So, here I am taking open ray all y less than equal to x, Rx is the right open ray all y bigger than x. These are receptively called left and right ray. They are also called respectively initial segment and terminal segment. So, this terminology also we have used. Now, let us have this L to be the Lx where x is inside and this is a family now, family of all left rays, R is family of all right rays, S is S L union R. So, family of various subsets of our x. Take this as a sub base any subset can be taken as a sub base for a topology right. All that you have to do is take all finite intersections and then take all possible unions that is the topology. So, take S as a sub base for topology which you will denote by tau less than less than equal to that denoting and this one denoting that this topology corresponds to the partial order here. For x less than y let us also have this notation this interval x comma y is not an ordered pair here set of all z in x such that x is less than z less than y. So, all points strictly between x and y it may be empty also we do not know. Borrowing the terminology and practice from real analysis Lx, Rx, xy etc are all called open intervals. Likewise we define closed intervals and half closed intervals also all that you have to do is you have to put the less than or equal to here or only one of them places and so on. I do not want to go into those details those things are common practice. So, I will also use them there is no need to again spend too many should for example if I take here if I take S than or equal to then that will be the called closed array similar closed arrays right closed left array and closed right array and so on. So, we shall list a few important properties of stetoplocal space some of the proofs being either trivial or left to you as an exercise because some of these things we have seen before also. If you have a order preserving bijection from x less than or equal to y less than or equal to prime. So, these are two different you know linear orders suppose you have an order preserving bijection then this thing as a map from topological space corresponding topology I have put now it is a homeomorphism of the respective order topological spaces indeed it will preserve the base sub basic open sets themselves. So, it is a very strong homeomorphism right. So, if x has no least element nor greatest element then the family B of all open sets such as b x y x less than or equal to y x and y belonging to x this form a base it is not the same if there are least element and greatest element. If you take intersections of left rays and right rays you will get these things but you know to get a base you will have to include those elements also. So, that you have to be be careful about that one alright. For example, 0 to 1 both closed if we only take open rays there will be problem right. So, you will have to see what you have to take 0 closed 1 say half open and so on that is why all that you have to do is exactly same kind of thing you have to if 0 prime I would like to put a prime here not to confuse with real number it is a least element of this x suppose there is one I am not assuming there is one if there is one all that you have to do is put this kind of half open intervals also in the base. Similarly, if there is a greatest element infinity prime here put half open intervals closed on the right also as inside the base if open then that will be a base whereas defining the sub base there is no such problem now first thing we observe is this topology is housed off. So, given x less than y of course given x not equal to y x is less than y or x bigger than y. So, I can estimate less than y if there is a z such that x less than z less than y then we can take u equal to l z and v equal to r z as disjoint open subsets left ray and right ray open left ray then x will be inside you and y will be inside you. So, you are done I am giving you an argument to show that this topology is housed off on the other hand if there is no z at all between x and y this can happen for example when you take natural numbers included in the largest space and so on right. So, such that there is no element between them then what happens then all that I can take is take the open r x open left ray r x and open left ray r u l y ok l y will contain x and r x will contain z they will not they will not have any problem. So, there is nothing in between so there is intersection will be empty right. So, on the other hand if there is no z such that x less than z less than y then you take u as l y and v equal to r x. So, there will be disjoint open subsets contain x and y. Now, let b be any nonempty subset of x if supremum of a respectively infimum of b infimum of a you can take a at b or whatever exists then supremum of b is inside v bar ok. So, I should say supremum of b exists infimum of b exists it is in the closure see disclosure is with respect to the topology now respectively infimum of b is inside b bar supremum infimum the proof is exactly similar actually much simpler if you think do not use any algebra no need for algebra real number ok definition of the infimum that is what you have to choose and what is the meaning of open set here and what is the meaning of closure and so on. You may assume that b is not a single term ok because supremum of a supremum of b will be singleton that singleton itself it will be already inside that one that is nothing to prove. So, we have to show that every neighborhood u of x which x is I am denoting by supremum of b intersects b that is the meaning of that x will be inside b bar ok. This u contains in the open interval s to x x closure ok u contains u is an open subset some s to x will be there inside the open subset containing x right. So, for some x that it will be there may be on the other side also I do not know that one ok I do not I do not care it is enough to prove that this sx intersection b is non-empty I want to show that u intersection b is non-empty. But if this is the case if this is empty ok what happened this s will be smaller you know x will not be supremum s will be smaller than that one that will be upper bound for b right. So, this is contradiction so it has to intersect that is all I am using the moment you come this is supremum moment you come a little lower there must be elements of b right that is what supremum means all right. So, similarly infimum of b is also inside b if b is closed otherwise inside b bar ok. So, in part 1 we have proved that if x is connected then it is order complete all I am just recalling that one if x satisfies the property that between any two distinct elements there is a third element then the converse is also proved. So, these two things we have proved there right now recall that we use connectivity prove that every closed interval in R is compact that is what we have done there. But we can use you know we then used it to prove Heineborel theorem first of all prove the closed interval is closed intervals in R are compact and then pass down to R and and so on yeah yeah closed intervals are closed intervals yeah minus infinity plus infinity do not call closed interval forget it open intervals can be bounded or unbounded in the same way we can prove something you can say which is Heineborel theorem now ok. So, start with a totally order set take a to be a compact set then a is closed in x and a is complete and bounded in the bounded means what in the induced order the ball is on it ok completeness comes there. Convergly if x is order complete then every closed and bounded subset of x is compact ok see it is completeness order completeness comes by taking the sub this this this the same relation restricted to a that is why I have to in the induced order ok. So, this is if and only if part compactness implies closed and bounded closed and bounded implies compact let us go through it carefully start with the compact set a suppose a is not bounded above if a subset is not bounded above original space is also not bounded above right. Therefore, the family L x x belong to x becomes an open cover for x itself. So, in particular it will cover a also, but since a is compact we will get a finite cover a is contained inside i 2 1 2 and L x i's take x to be the maximum x 1 x 2 x n they are only finitely many of them ok in a totally order set. So, you can take maximum it follows that a is bounded by x right because L x i's were all contained inside L x now which is contradiction we just said a is not bounded similarly we can show that a is bounded below this is this what we have to bounded above similarly we can show problem to be exactly same thing instead of L x it take i x ok. So, it is bounded now x is off-dorff space we have seen just now it follows that being a compact subset it is closed off we start with the assumption that a is compact. Finally, the order completeness of a follows easily from what we have just observed namely infimum and supremum of this set which are well defined because a is a bounded set they will be inside a they will be inside a bar, but a is a bar is equal to a because a is closed. So, they are inside a over. So, that is precisely meaning of the order completeness right for the converse part let x less than b order complete and a be a closed and bounded subset it follows that with respect to the order restricted to a a itself is order complete and bounded right. Therefore, without loss of generality we may assume that x itself is bounded and prove that x is compact order completeness and bounded and it is compact instead of sub proving that for a subset. So, that is how I stated if x is order complete and bounded then it is compact. Here we invoke Alexander sub base theorem ok. So, what we will do take you to be an open cover for x by members of a sub family members of a standard sub base a sub family of a sub base you have to fix a sub base then take sub family of that which will cover x that cover should have finite sub cover. So, that is enough for by Alexander sub base theorem ok. So, here what we do we take the family of all left rays and right rays that s equal to l union r remember that. So, take a sub family from there which covers x and get a sub sub cover then Alexander sub base theorem says the space x must be compact. So, put alpha equal to minimum of x, x is bounded ok the immediately we are using that then alpha cannot be in any right ray because I am taking only open rays here and hence it must belong to one of the left rays because left rays and right rays are going to cover I mean I have chosen some family u of left rays and right rays right that covers it is not all all the left rays and right rays some left rays and right rays it is cover. So, alpha must be in one of them. So, it must be in a left ray ok put l prime equal to all those x for which l x is inside you. So, this is a non-empty this set l l prime is non-empty why because alpha must be in one of them ok one of the l axis. Then what we have seen just now I am not sure that l prime is non-empty. So, once your l prime is non-empty you put beta equal to supremum of l prime l prime x is supremum of l prime because l prime is subset of x after all. So, it is bounded bounded above all. So, supremum axis order completeness issues then this beta prime cannot be in l x because it is bigger than all of them right. So, if it is here then then beta will be less than equal to x actually less than x. So, beta will not be in l x for any x inside l prime. Therefore, beta must be all these are left rays are all taken here. So, what is left out right. So, beta we must be around to r y for some r y inside you this means something belongs to r y means that beta must be bigger than y ok strictly bigger than y since beta is the supremum of l prime you see I have chosen y is less than beta there will be some elements here it follows that there is x belong to l prime between y and beta y less than x less than you could beta it may beta prime ok there must be some element like this ok in l prime beta itself is not in l prime ok perhaps beta is not in l x for any x prime. So, y is I bet I have used to supremum here. So, this is the property of supremum y less than x less than over beta. But then we have these two members l x and r y ok belonging to you l x union r y must be equal to x ok everything must be between the supremum and supremum after all minimum and maximum of this. Next we have seen that connectivity implies order completeness this also we have seen just I recall that in the part 1 that is what one of the things we have closed if x is well ordered see this all totally order subset and then a topology connectivity implies order completeness that is one way. But if it is well ordered it is highly far from connectivity but that also implies it is order complete much easier way much simpler way right because it is truly satisfied the condition that every subset which is bounded below has greatest lower bound as a least element actually least element belongs to that set and so on that is the well ordering ok. Once bounded below sets have least I mean greatest lower bound then the other one also follows these two are equivalent. So, order completeness follows very easily. On the other end well ordering implies that topology is totally disconnected that is what I said is far away from being connectivity ok. So, how do we see we will see this one namely x belong to x be not a maximal not the maximal element ok pick up some element is not maximal that is Rx not equal to empty set there are elements bigger than x this means Rx is non-empty we take the infimum of Rx and denote it by x plus 1 and call it the immediate successor of x immediate successor by the way this immediate successor is the key to the entire I would say all these pianos axioms and Zernelow, Frankl, Zernelow and so on this was actually goes back to Grasman in modern set theory perhaps his name does not appear but this immediate successor you know he pointed out this is the one which you have to know this is the idea which you have to use. So, finally it was piano whose axioms became the best there were many many trials in between several people had tried it you know one of the other guy whose name is quite quoted is Dede kind ok. So, pianos axiom is based on the entire algebra of natural numbers is constructed out of this is the first time it appears this plus sign this is algebra right x plus 1 we are not going to do any algebra we are stopping here x plus 1 so all that I want ok. So, if Rx is non-empty for any x then you take the infimum because it is a total order well order the infimum exist and it is unique that infimum is called x plus 1 it follows that the half closed ray Rx1 bar I am I am putting now this bar denotes the closure also so there is no contradiction here if you take the topology this will be the closure you can just define this set to be all x plus all y which is bigger than equal to x plus 1 ok x plus 1 less than equal to y strictly less than y would have been open Rx plus 1 right open ray close ray this one so it is equal to the open ray and hence is a closed subset so x plus 1 is a closed ray but if you take the open ray Rx what happens there is no element between x and x plus 1 right so open ray Rx is also equal to that therefore yeah therefore this is both open and closed this is a closed set right therefore if x is less than y if you take any y then this Rx plus 1 bar ok this Rx plus 1 bar would be a closed set containing y and not containing x so there is a separation you can actually write down separation Lx plus 1 you know bar Rx separation of this one so this is strong totally disconnected namely it actually satisfies our S1 that that what we have studied earlier ok so is stronger than out of this is stronger than totally disconnectedness also in any case every point other than the least and the greatest element are cut points this cut points is also used this is a this a contribution of today kind cut points means what you throw away one point the space become disconnected ok such points are called cut points so I think this much topology is good enough for one day so indeed we will have many other topological aspects of this one because finally what our aim is to produce lots of examples out of one single example right so we will meet again more topology next time we shall construct the ordinals the example that we are interested in so far thank you