 Welcome to NPTEL NOC, an introductory course on points at the party part 2. Today we shall start a new topic, module 45, partially ordered sets. Let us recall some basic theory of partial order. Some of them must be already familiar to you, so we shall be somewhat quick here. By a partial ordering, which we usually written as less than or equal to, on a given set X, we mean a binary relation which is reflexive, transitive and anti-symmetric. Unlike in the case of equivalence relation which is reflexive, transitive and symmetric, so that is the big difference here, anti-symmetric. Anti-symmetric just means that if X is less than or equal to Y, and Y is less than or equal to X, then X must be equal to Y. A set with a fixed partial order is called a percent. Partial order has been shortened. Strictly speaking, it is the ordered pair X, less than or equal to. But as usual, topology, matrix pair, we will let X be a matrix pair. Let X be a topology, that is what we have seen. Similarly, we will say that X be a partial order, X be a partial order set, let X be a percent. But then we immediately will mention what is the partial order there whenever we are dealing with it, that is it. We also use the symbol X less than Y, do not put that equality part here. If X is less than or equal to Y in this sense, but X is not equal to Y. Less than or equal to Y will include X equal to Y case also. Because if X is less than or equal to Y, Y is less than or equal to X, then X is equal to Y, that is what we have seen. And reflexive just means that X is always less than or equal to X, so that is already there. Given a subset A of a partial order set, this element X is called an upper bound for a If A is less than or equal to X for all A inside A, so you may say that X is the biggest among all elements of A. But X itself may not be an element of A, so that is called upper bound. There are many upper bounds, there may not be any upper bound either. So it is just a definition, no existence or uniqueness. It is called least upper bound, if X is less than or equal to Y for all upper bounds of A. For all upper bounds Y, X must be less than or equal to Y. So then it is called least upper bound. Note that least upper bound if it exists is unique by anti-symmetry. Because if X and Y are both least upper bound, X is less than or equal to Y, Y will be less than or equal to X. So that is the anti-symmetry used here. Then least upper bound will be unique. So then we will call it supremum of A. We shall denote it by supremum of A. Similarly the terms lower bound, greatest lower bound, infimum etc are defined. So all these things we are using and we are used to the study of real numbers. Same definition, same thing. Only thing is in real numbers there will be addition, subtraction, multiplication. Nothing will be used here. They are just arbitrary sets. How far you can do, what analysis you can do, what topology you can do. We just order them. That is the topic here today. A poset is called a linear order or a total order. Both the terms are used. If given any two elements X and Y inside A, we must have X is equal to Y or Y is less than or equal to X. Or of course X equal to Y is also allowed. So you can say this is law of trichotomy. Of course by anti-symmetry, if both X is equal to Y and Y less than or equal to Y, then we have X equal to Y. An element X inside X is called maximal. If X is equal to Y, Y inside X implies X equal to Y. Take any partial order set. Maybe you take a partial order set and take a subset of that with the induced partial order, restricted partial order. Then you can talk about maximal element inside that. So this is just like similar to supremums and so on. Maximum element here by the way in arbitrary partial order need not be unique. A linear order or a total order if that is there, then there is a different thing. So we have just maximal elements also we have defined. A maximal element means what? If anything is bigger than that, it must be equal to that. I mean different things I am defining. This total order may not have anything to do with each other. A posset X is called well-ordered. If every non-empty subset A in it is bounded below, suppose you take any non-empty subset which is bounded below, that is not the case. Every non-empty subset must be bounded below. Infimum of A exists and that infimum must be inside A. So this is well-ordered. This is a very strong condition. Simple example of well-order is a natural number with the standard order. Given any subset of natural number, there is one which is the smallest. So that property has been excimatized here. So natural numbers are also totally ordered. But well-ordering, what is well-order? Automatically it will be totally ordered. Why? Take any two elements X and Y. That is a subset. That subset must have an infimum. That infimum must be inside that. That means infimum means one of the elements X or Y. If it is X, then X will be less than or equal to Y. If it is Y, then Y will be less than or equal to X. So one of the orders is there. So automatically a well-order is a total order. A total order may not be well-ordered. Just you can take the example of all the integers. That is that a subset may not have an infimum, which we know. A subset A of a totally ordered set X is called an initial segment. So just pay attention to this. An initial segment inside X. If you have an X inside X such that the set A is all elements which are less than X are inside A. Or all elements which are less than or equal to X are inside. So see this may not include X. But it is possible that this A includes X also. So there are two different cases. So both of them are called initial segments. Quite often this A is called an open initial segment. And this part will be called a closed initial segment. Just like the closed ray minus infinity to 0, 0 closed. Or here the closed ray minus infinity to 0, 0 open. So these are the standard examples. Fallen result is one of the most useful result from set theory. And it is equivalent to Axiom's choice. We take it for granted. What is the John's lemma? So John's lemma states the following. Take any partially ordered set. Non-empty. Suppose every linearly ordered subset X has an upper bound. Then X has a maximal element. Every linearly ordered subset. Every totally ordered subset of X. Means what? Under this same order. You have to take the subset order. You have to take the restricted order. Then take a subset, take the restricted order. That must be linearly ordered. If such linear order have an upper bound. Then X has an maximal element. There may be many maximal elements. This is just a subset. If it is total order, there will be only one maximal element. X is only a subset. Note that the lemma does not assert any uniqueness about such maximal elements. It is very important. Using John's lemma, we can easily prove Sermelo's another very important theorem. It is also an axiom. Every set X can be well-ordered. We have just seen that the set of integers is not well-ordered. With the usual order, it is not. So what this Sermelo says is that there is another order in which it will be well-ordered. Some existence of some order, some partial order which will be total order, which will be well-ordered. That is the meaning of this one. To prove this one, you can do it independently. But you will have to use an axiom of choice. That has to be there. There is no other independent proof as such. Because this statement, John's lemma, both of them are equivalent to axiom of choice. So we are assuming this one. It just means that we are assuming axiom of choice in the background. But now what we will do? We will use this one to prove this one. So that way, our task will be simpler. How do you do that? In order to employ John's lemma, you have to have some family of order set and so on. Then you say something, say maximal. And that maximal satisfies whatever you wanted. So I start with a family lambda of all ordered pairs a, some partial order where a is a subset of x. Look at the family px, the set of subsets of x. Of course, you do not need an empty set. Empty set you are throwing away. Okay. And put a well-ordered on that. Take that one. If a well-ordered is different, that will be a different element. Okay. So it is an ordered pair of such things. A is a subset and this less than equal to is a well-ordered on that. Why this lambda is non-empty? Why is lambda is non-empty? Because singleton sets can be given only one partial order. That partial order is automatically well-ordered. And they are elements of lambda. Therefore lambda is non-empty. You start with x non-empty, it can be non-empty. Okay. So here I am assuming x non-empty. I do not worry about empty sets here. All right. Now on lambda, we will put a partial order. What is that partial order? A1 less than equal to 1. A2 less than equal to 2. If you are related by this, I will read it as prick or you can read it as precedes. A1 precedes A2 if and only if. One condition A1 may be equal to A2. Along with what? The partial orders are also same. That is the one. Or A1 is an initial segment of A2. The way we have defined initial segment. Okay. So that is the condition. So I am expressing that in the second part. Once A1 is equal to A2, then what you should have? The order A1 and order A2 are the same when taken on A1. You take the A2 order, restrict it to A1. It must be A1 order. Which is same thing as saying x, y belong to A1. x, y less than equal to 1. If and only if, x is less than equal to y with the relation 2 here. Okay. Then lambda, this one is a percent. What are the things you have to verify? What are the members? Members are look like this one. They are less than equal to themselves because once they are equal, this, that is okay. So reflexivity is there. Okay. Transitivity is also obvious because if A1 equal to A2, A3 is fine. But if this is an initial segment here and A2 is an initial segment in A3, then the same point will give you A1 is also initial segment in A3. Reflectivity is here. Details are left to you as an exercise. Do that one. Otherwise, all these definitions may become a bit, you know, go in the air. So you have to spend some time. So do this exercise by yourself. Next, I have to show that, see, it is a partial order line. So what I will do with that? Take a linearly ordered subset of this lambda. So gamma be linear. I will show that this has an upper bound. Okay. All elements here are dominated by a member of lambda. That will satisfy, that will mean that conditions for John Flemmer are satisfied. Okay. After that, we can conclude that there is a maximal element here and that maximal element is going to give you whatever you want at namely a well-ordered set on X. So let us see how. Start with a linear order subset. Put b equal to union of all these members inside this gamma. Okay. Remember a sets. What are they? All these are ai contained inside aj or aj contained inside ai. This total order is there, linear order. Okay. You take the union. That is a subset of X. Now if you take any two elements in b, there will be one a1 belonging to gamma such that both of them are will be in a1. Why? Because X may be in say a and Y may be inside some a prime. But a is less than or equal to less than or equal to a prime. One is contained in the other. Therefore you can take the bigger one. So there is a a1 for which both XY's are inside a1. Now you define the new relation on b which is less than or equal to prime. This new relation I am going to define. How I am going to define? X is less than or equal to prime Y. If we don't leave, X is less than or equal to Y inside a1 because they are members of a1. Now you have to see that this is well defined. Does not depend upon what a1 I have chosen. If I have chosen some a2, then a1 is a segment of a2 or a2 is segment of a1 which means this relation will be the same for as far as X and Y are concerned. That is why this is very defined. Automatically it will be a order, partial order. So what we have to say is that it is a totally order. B lambda prime is a totally order. Actually we want to show that it is a well order. Total order is obvious. Just now we have just shown this one. So this is a well order. Then this will be a member of this lambda. And then we have to show that this is an upper bound for all the elements inside this gamma. It is an upper bound for gamma. So quite a bit of work to do. So we claim that B less than or equal to prime is well order. What does that mean? Take any non-empty subset C of B so that it has an infimum. The moment this non-empty intersection with one of the members of gamma must be non-empty because what is B? B is union of all members of gamma as a subset. So this is non-empty. As a subset of A1 less than or equal to which is well ordered, a C intersection A1 must have an infimum. Let us call it as T. This T is an infimum of C intersection A1 inside A1. We claim that T is the infimum for C inside B prime which is very few T. So that is what we have to show. Once we show that well orderness of B, less than or equal to prime is proof. So here is the proof. Take any element X inside C. Say X must be inside some A2 belonging to gamma. Because all elements of C are all members of some B. Since gamma is totally ordered, this A1 is preceding A2 or A2 is preceding A1. In precedence there is one way namely A1 may be equal to A2. Then obviously this T which is minimum of C intersection A1 will be less than or equal to X in the larger B in the B prime order. Because these two are equal and X and T are both elements of this A1 equal to A2. So we shall assume that A1 is not equal to A2. Now there are two different cases. Either A1 precedes A2 or A2 precedes A1. Suppose A1 precedes A2. Now there are two cases again because A1 is not equal to A2. A1 may be an open ray, open initial segment like this. Where this Y comes from A2 for some Y inside A2. Now if X is less than Y under this A2 then by the definition X is inside A1. T is the least among all these. Therefore T is less than or equal to X. To begin with you have taken X to be an element of C remember that. The moment is A1 it will be in C intersection A1. So therefore T will be less than or equal to X. But this is the relation in the larger B prime also B also. So T is less than prime X. Now second case is Y is less than or equal to X. But then T will be less than or equal to Y less than or equal to X. So it follows that T is less than or equal to X which in terms of T is less than or equal to prime X. The second case is instead of open ray you have an closed ray. That means I start with A2 all A1 is all A2 such that A2 is less than or equal to Y here. The proof is exactly same you have to just put less than or equal to here. And here you have just put less than that is all. So identical proof ok. Now the second case is the other way around A2 precedes A1. Then it follows that C intersection A2 is a smaller subset than C intersection A1. It may be equal also I don't care. And hence T will be less than or equal to this X also. Because for all these elements T is smallest. So T is smaller on this one also. But again T is less than or equal to prime X. So in all these cases you have shown that T is less than or equal to prime X for all X inside C ok. No restriction C intersection A1 now. So this means that B is well order. Now we shall prove that this is an upper bound. So far it is well order it is a member of B that member of lambda ok. Now it is a upper bound for all and members of gamma ok. Start with a member of gamma. Then for X and Y inside A1 we have X less than or equal to Y if and only if X is less than or equal to Y. Prime by definition of this prime less than or equal to prime by this is the rule we have already. Therefore one thing is clear namely the order is fine. We have to show that this A1 less than or equal to 1 precedes B less than that is what we have to show right. So this condition is satisfied. Now suppose A1 is the whole of B that is one case as a set. Then clearly these two are equal because we have already shown this one. So as posits also they are equal. Now otherwise otherwise means what A1 is a proper subset of B. That means there must be A2's inside this gamma such that A1 is not equal to A2 but contained inside A2. If everything is contained inside A1 union will be A1 which is B. So that is not the case now ok. So A1 is contained inside A2 but not equal. Therefore A1 less than or equal to 1, A2 less than or equal to 2 must be related somehow because they are in the total order this total order set gamma. Since this cannot be a subset of this one. So it is this one A1 less than or equal to precedes A2. Therefore A1 is an initial segment either open or closed does not matter is initial segment of A2. Therefore it is initial segment in B prime also. B also same. Y will give you the same property alright. So we have proved that every A is an initial segment. It precedes B ok. In either case we have shown that this is an upper bound for lambda this gamma with respect to this relation. So far what we have proved is that zoncelam condition for zoncelama satisfied. Therefore there is a maximal element y, this one in x. What is the meaning of this? This means y is a subset of x ok. And this less than or equal to is a well order. And if there is another y prime less than or equal to well order subset of x ok. Then that cannot be bigger than this one. That y cannot precede that one. This is a maximal element. Then they must be equal. So that is the meaning of maximal element. So now we claim that this the underlying set y is the whole of x. That will complete the proof. We have found a partial order is actually well order ok. So y, y, y should be equal to x. That is what we have studied. If not we can pick up some element x inside a complement. Put z equal to y disjoint in an x. That means one extra element ok. Extend the well order on y to z by declaring the only thing that we have to extend is how x is related with other element. So declare y less than x for all y inside y ok. This way z becomes a partial order. It is automatically well order because if a subset is already inside y it is well order. If it is not it will contain x but x is largest. So there is still the minimal element coming from y part. y part will be again give you minimal element. Finally if you have just singleton x then that is singleton x itself is a minimal element. Therefore this is a well order. Now what we have what? Z is a well order set. But y precedes that one. This is a bigger thing. That is a contradiction. A contradiction because we assumed y is smaller than x. It is not the full x. If y is x there is no problem. So that completes the proof of Zermelo's axiom. Here we have made it a theorem. Namely every set can be well ordered. As I told you the original proof of Zermelo which was proved 1904 more than 100 years 120 years is directly from axiom of choice and is certainly more complicated. The above proof is simple and short because we have directly used John's lemma and not so obvious fact that axiom of choice is equal to John's lemma. Try to prove John's lemma using axiom of choice that it will be again a horrendous task. So we have avoided that one. Next time we will do another important landmark result in set theory principle of trans-finance induction. Thank you.