 Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2. Today we shall begin our study with principle of transpionite induction. As a sample, we will also employ this one and prove Ticknoff's theorem. Of course we have proved Ticknoff's theorem elsewhere. So that is a good way of learning a new tool. So test it on something which you already know where it works. So the principle of transpionite induction says the following. Start with a well ordered set with a least element, we will denote it as 0. This is for convenience. Suppose p alpha is a statement about alpha inside x and we use the symbol p suffix a to denote that p alpha holds for all points alpha inside a, a is a subset. Now suppose p0 is true and for any alpha inside x, p0 alpha, this is the segment consisting of all beta which is less than alpha inside x. So p of 0 alpha implies p alpha holds. Remember this means that p of beta is true for all beta less than alpha that is the meaning of p0 alpha. Then p alpha holds. If this is the case then the principle of transpionite induction concludes that px is true that is for all elements of x the statement will be true. Suppose on the contrary px is not true. What does that mean? This just means that if you take elements inside x for which the statement is not true that is a non-empty subset. This b is a non-empty subset. A non-empty subset in a well ordered set has a minimum. An infimum alpha and alpha is inside this one that is why we can call it as minimum. Since p0 is true in the beginning assume p0 is true and this implies this one. So p0 is always true that means this alpha which is the minimum is not equal to 0 and moment is smaller than that it must be true otherwise alpha will not be infimum. So p of 0 alpha this is true. Every beta less than alpha p of beta is true that is the meaning of this. Therefore p alpha is true. That means what? Close if you can take the closed segment 0 to alpha. But then what? Then alpha cannot be inside b that is the contradiction. So that is the proof of principle of transpionate induction which is as you see is an easy consequence of well ordering. So that is the whole idea. So which in a well order thing this always holds. The point is you are already very very familiar with the principle of mathematical induction. That is the special case of this when x is the natural numbers along with the usual order. Of course that is a well order. So this principle of transpionate induction is a far far generalization of mathematical induction. This is true for any x as soon as you have a well order there you can use it like this. So our next aim is to illustrate the use of principle of transpionate induction with one single example but that example is going to be something very very important namely product of arbitrary non-empty families of compact spaces is compact. That is a Tickloss theorem. We have proved we have given several proofs of that. Now we will give another proof of that one here. So this is again in here I am going to use one of the theorems that we proved in part one. I would not have time to do that again here but I will recall it for ready reference. This is the theorem that we proved in part one. At x be any topological space then the following two conditions are equivalent. The first condition says that for every topological space z the projection map z cross x to z is a closed map. The second condition is x is compact. So this theorem gives you a characterization of compact spaces. So what we are going to do is employ this one both ways in proving that Cartesian product of compact spaces is compact. Of course you have to take non-empty product and non-empty. If you take empty product it is assumed to be a singleton or some system there is no problem but each space must be non-empty that is important. Every non-empty product of compact spaces is compact. So the proof seems to be just a small trick and using the principle of transpionate induction correctly which amounts to more or less you know selecting appropriate notation here otherwise it is extremely easy proof. So I will tell you the idea and then tell you how it works in simple cases. Rest of them is just pure notation sender so on. So given an indexed family of compact spaces you want to prove that x prime is a product of each xi i inside i prime is compact. The plan is to show that for any topological space z the projection map z cross x prime to z away from x prime to the first factor this is a closed map and use the previous theorem. To prove this itself we will use the principle of transpionate induction. So here is the first step which will tell you what kind of tricks we have to do with employing principle of transpionate induction. Put a well order on the set that is important set namely the indexing set i prime. Let us denote the least element there by one. One belonging to i prime is the least element. Take any order no problem it must be well order. Extend this to a well order of i which is disjoint union of i prime with two extra elements are denoting by 0 and infinity by declaring that every element in i prime is between 0 and infinity. In other words now 0 is the least element and infinity is the greatest element. There may be a least element here that is always there and that element I have returned to one but there may not be any greatest element. So I have put in a greatest element also. So this is the first trick you can say. Clearly 0 becomes the least element and infinity becomes the greatest element inside this i which is two extra elements from i prime. In the first line we are using the well ordering principle for i prime and now on i you have defined an order. It may not remain the well order set. It is a well order set. Take a subset if it contains 0 0 is fine. If it does not contain it will be inside it may be containing infinity. Infinity is the largest it cannot be smallest. If you throw infinity also there will be a smallest element there. Because i prime was well ordered. Yeah well order we have even in the proof of you know using John's lemma proof of that one we have extended this one and so on you see. So the extensions of ordering for one element two element three elements infinitely many elements is obvious. Wherever there is infinitely many then you need we have to go back to John's lemma. You have to go back to principle itself ordering. In other words what your objection may be that suppose i prime is well order and i prime is contained inside a larger set j prime will there be a well order on j prime which extends the well order on i prime even that is true. So but we have not proved such a thing. We can whatever proof we have given for existence of well order. You can modify it to prove such a thing also. So that is not of any use for us. I am just I have only two elements. I prefer I want it this way not an arbitrary order. 0 is the least element and i is the infinity is the largest element. In fact we want to work inside i prime only but these two things help us in putting the correct notation you know instead of getting instead of buggy notations that is all. Now for each j inside i let us denote the segment 0j which means all elements strictly between 0 and j including 0 and j. So this I will denote by capital J. Note that 0 infinity is nothing but the whole space i. Now another small trick I want to prove that for arbitrary space z z cross x prime to x prime where x prime is a product that is z cross x prime to z sorry is the projection map is a closed map right. So I want to change the notation here put z equal to x naught now and x infinity I choose another space namely a singleton u which is a harmless thing okay it is anyway compact. This is not compact and it is not supposed to be compact anyway okay. Now to take x to be product of all these x sides remember if I only take i prime then this x prime that is what we wanted to prove is compact. Proving one more factor cross with a singleton is compact the same thing as proving the original thing is compact. What we are going to prove is now x prime cross x infinity with a single singleton that is compact okay some more notation for any subset a of i that x suffix a denotes the product of xi where i range over a itself. For each pair of subsets a contained inside b contained inside i in inclusion equality allowed okay let pi b a denote the projection map from x b is b is larger thing to x a which just means that drop out all the indices which are not in a that is the projection map right. One more simpler notation pi of i to a will be just denoted by pi a when the the top thing is the full thing you just write as pi a so that is from the whole space x to x a okay these are our projection maps note that with this notation xi what is xi it is x okay and pi i is the identity map nothing is dropped out alright one more notation pi not is what it is just projection to the x not coordinate alright we want to show that this is a closed map since x not is arbitrary this will prove that x prime cross x infinity which is singleton is compact that same thing as proving the x prime is compact okay so this is what we want to prove so that will prove that x prime cross infinity is compact and x prime compact okay. Now the next step is so start with a closed subset of x okay you have to prove that the image is a closed set so take a point in the closure and show that it is inside the original set pi not of f you want to show that this is closed so this is equal to the closure so I have I will take a point in the closure and show that it is inside that's all alright now suppose you have just two two spaces say x is just z cross sum x1 okay then how do you have done this one x x projection map is compact okay so the second factor is compact projection map is closer is what you had proved right so this would have been done similarly for finitely many cases we have already done the product is compact and so on but if you don't use it how you do that you just reflect what we have to do okay successively first you get a point in pi 1 first first projection say suppose there are three x x is z cross x1 cross x2 cross x3 let us say so you have started with a point in x x not which is z right so which is inside the closure take a point x not comma x1 okay inside pi 1 of f closure the projection of that one to pi not will be same three as this one next take a point x not x1 x2 in pi to pi 1 2 1 and 2 both of them pi 1 2 3 etc when you take all those indexes it is same thing as identity map which is pi of which is just pi i of f is just the same thing as f f bar is f because start with a closed set okay so you would have it this one namely you would have proved this one of course this is different prove I am giving you we already know that product finitely many of them are are compact so this would have been another proof in the case of when product we have taken over a finite family right so once you have this idea okay now everything is just set of notations and finally applying the trans finite induction okay I think that makes the idea clear in the proof what is going on alright so start with a point x suffix top x not because I would like to use the lower suffix for the coordinates okay this is some point in pi not of f bar bar denoting the closure okay so for each j inside i I will make a statement here sj sj denotes the statement that there exists some x suffix j belonging to xj such that the xj is actually inside pi j f bar okay you have to start with f which is in the product space x project it to this xj and takes closure so this element must be in this closure then it must also satisfy that all its projection j prime projection of this one agrees with xj prime so this is an inductive or inductively xj prime denotes the element here with this property okay so it must be true for xj prime also here xj prime belongs to pi j prime of that one that is the inductive statement one single statement meaning all these okay so the jth projection of this one must be the same thing as the j prime projection element here which just means that if you take the j say ith projection of xj where i is less than little j then it must be same thing as ith projection of this element where i is actually smaller than even j prime that is the meaning of this projection map projection of these points which you are choosing next one should be the same thing as the points which you have chosen already okay so this is what I did in the case of you know in the illustration case first you choose x0 and then x0, x1 and then x0, x1, x2 so that kind of notation is not possible that is what I have tried like this that is okay so this will be true for every j prime inside 0 to j this statement makes sense even when j j prime is 0 there is no problem that x0 has been already chosen so we know this statement is true otherwise it is just a statement now why I am making this statement examine the statement okay for s infinity what is the meaning of the statement as s infinity that means there is a point x suffix infinity inside x this this infinity will be now meaning 0 to infinity which is just the whole space I have told you so x infinity will be inside x such that x infinity is inside pi infinity of f which is again f bar but f bar is f so x infinity is inside f and this one means that in particular pi0 of this x infinity is x0 okay so what does that mean that means this point it is in the closure but it is actually inside f so we have shown that this x0 is inside pi0 of f right so that will complete the proof so that is the idea of making this statement this statement for j equal to infinity is the one which we want finally what is the starting point the starting point will be s1 for example s0 is too obvious s1 we can prove then we have to prove that sj for all j by what obvious tricks namely suppose this is true then x1 is true that step where prove that is the hardest step here okay all right so let us prove s1 okay this is where the induction begins what is x1 x1 is a compact space x0 is z x0 cross x1 to x0 it is a closed map because x1 is compact therefore if you take pi01 this I could have written as the you know square bracket also here no problem both of them are same there is nothing else between 0 and 1 by the way so pi01f take the closure p of that p is next projection projection okay that will be a closed map because this is a closed set I am taking so p of that one will be a closed subset of z equal to x0 and contains this subset right p of pi of f but that is pi0 of f therefore it contains pi0 of f bar because it is a closed subset but it contains x0 okay this means that you have an x1 inside x1 such that this x0 x1 belongs to pi0 bar this proves s1 so this is why I have I have explained it already before now I am using this new notation I have explained it so this is the way from x1 to x2 x3 and so on we can go keep going very easily alright but now we want to use induction directly so s1 is true now next thing to prove that suppose for some k inside i okay not equal to 0 this should be 0 it is not 0 sj is true for all j less than k alright we want to prove that sk is true why once we prove this that sj for all j less than k implies sk is true the condition for trans finite induction is over that will that will prove that s infinity is true so all that we have to do with sk is true okay define y belong to 0 to k x0 to k product up to taken up to 0 k open to be the element such that this yjth coordinate of y is the jth coordinate of this xj for every j less than k then it follows that this yj which is nothing but the projection pi j of 0k see j j is something smaller than k right for each j you have for each j less than k capital j pi j 0k to this one of y is xj all that you have to verify is take the ith coordinate here and ith coordinate here for i less than equal to j then these are these are the same because I have just projection maps there okay so that is what I have done here for to begin with if you take j little jth coordinate of yj that is yj that is same thing as the jth coordinate of x power j provided j prime is less than j okay I should j prime here then from 34 this property okay what what do we get pi j j prime of xj is xj prime so this implies the jth prime coordinate of xj is equal to jth prime coordinate of this one also okay for all of them so that means that jth prime coordinate of y itself is equal to the jth prime coordinate of x power j okay so I have defined what y should be this y will be in pi of 0k operating upon f the closure of that so all this is happening inside x 0k okay so we have to prove this one we have got a y which projects correctly but why it is in the closure so this is where something has to be used from the product of all what is that the definition of product apology that is what we are going to use let w be a neighborhood of y okay in the product apology what does that mean there exist a finite set i contained inside a contained z i and in opens of set v contained inside x a see a is a finite set okay such that y is inside v cross this x taken all the indices in the complement of a okay so that is the basic neighborhood sub basic neighborhood so there will be some such neighborhood contained inside v contained inside w okay choose j to be the maximum of a a is a finite set okay that is why maximum will exist and put u equal to v cross x b contained inside x j okay now all the elements in a where b is 0 to j i have taken full okay which are not inside a of course okay so we want to be inside this finite set but in between there may be something so you take take j to be the maximum of a and put u equal to v cross x b see the v is inside purely inside a so i am taking 0 to j which is a larger set then i take this opens of set inside x j then we automatically get y belonging to u cross what is this one j complement is only 0 k minus 0 j 0 j part will be coming from u okay other things are here which are full spaces x j okay so this anyway subspace of that one so this will be also contained inside w all right so we have got a neighborhood like this for each point y for each neighborhood of y we have got a such a neighborhood these are basic neighborhoods therefore what happens what is the meaning of this one if you take the jth projection of w for w is inside x 0 k right that contains this sector this subset contains inside w so the projection will contain that one the jth projection will contain u cross x j what is the jth projection of this one all these things will go away so it is just u and u contains the jth projection of y okay that is how we have chosen that first we choose only finitely many terms afterwards we allow the rest of them to be free so that the y will come inside that okay so this y is now inside u that is the whole idea there but this is nothing but x j okay from 35 we have to use this right because once j is there y is chosen such that its jth projection is x j here we have chosen jth projection is x j for here okay so from 34 we know that this x j is inside pi j f bar and u is open in x j it follows that u intersection pi j of f is non-empty this implies u intersection x j complement that is non-empty intersection with pi j inverse of the projection map pi 0 j to f is non-empty but that is contained is a w intersection the projection of f on 0 to k so for each point you see y neighborhood of y it intersects this one therefore this must be in the closure of this okay so this is these notations may be new to you but this kind of thing you have proved while proving product is connected etc so similar argument this is not new to you the next step is similar to S1 what is that we have yet to prove that you know this implies next one so that is what now consider the projection pi 0 k okay namely full one more element this is one more element is there right so capital K is 0 to k all of them so omit that one element take the projection okay we want to show that this is closed but you have to use that this is closed but that is that becomes because this x k is compact right so we have to use each x k is compact after all so because x k is compact this is a closed subset here the image pi k 0 k of pi k f bar okay that is closed subset I am taking here is a closed subset of 0 k but it contains pi k of f right its projection of pi k of f is there which is nothing but pi 0 k of f and then it contains a closure therefore there exists x k belong to x k from here I can extend it to this one that is a whole one more element I can take that is a whole idea x k belong to x such that y comma x k what is the meaning of this notation now it just means that the earlier projections of this element is the old y the kth projection is x k okay this is element of pi k f bar so here I am using that x k is compact okay one at a time that step is again used here so clearly for all j less than k if you take the projection of this y comma x k that will be x j okay it is being same thing here projection of y so it is a x j okay so this proves the statement x k assuming that it is true that namely for every j less than k s k is true that closes the transfinite induction over the proof is over so this proof is worth repeating quite often you read it carefully next thing is we want to go ahead with this order topology and so on so how far we can imitate the model namely the real line with its order so we go back to that and introduce a few more terminologies all based on what we know already what we are familiar with with real line so let us recall the notation of least upper bound and greatest over bound in the context of a totally ordered set we are recalling it so that everything is devoid of all algebraic properties of real line only the order properties are being used so that is why we are recalling emphasis is on that so start with any totally ordered set a subset a is called is set to be bounded above with respect to bounded below if there exist alpha belong to such that for all a inside a we have a is less than equal to alpha or the other way around is alpha is less than equal to a okay bounded above and bounded below this alpha will be a lower bound here it is alpha is upper bound here in that case alpha is called an upper bound in the other one it is lower bound okay same definition that a be bounded above and alpha be an upper bound we say alpha is the least upper bound for all for a or another name is supremum of a if for every upper bound beta of a we have alpha is less than equal to beta that means first of all the set of upper bounds must be non-empty then you take alpha to be the least element in the set of upper bounds for all upper bounds are bigger than alpha so then it is called a least upper bound it is very descriptive name supremum is another terminology exactly similarly you can define greatest lower bound and infimum also okay the notation will be infa and sup a note that we have not claimed that supremum and infimum exist okay however it is easy to check that they are unique if they exist and that is where you have to use the anti-symmetry property that's all in a totally order set the following two conditions are equivalent what is that every non-empty subset a of x which is bounded above has a least upper bound in x okay that upper bound may not be an element of a in x every non-empty subset b of x which is bounded below has a greatest lower bound this is just the dual of the statement but these two are equivalent so let us prove one implies two the proof of two implies one is just the other way around you know just you have to reverse the inequalities and so on right so they are similar so one implies two if you prove when you are done okay let us prove one implies two start with a non-empty subset of x which is bounded below look at all the elements x a inside x such that a is less than it could be for every b inside b it just means that all the lower bounds a is non-empty is the assumption okay and because b is non-empty to begin with a is non-empty and it is bounded above every element of b is a bounded is a upper bound okay so it is bounded above also the set a itself is bounded above so let us be supremum of a which exists because x satisfies one every bounded above set has a least upper bound so that supremum is there so this is the hypothesis one now we have to conclude b right so this s is clearly less than equal to b for all b inside because because of what we have taken the least element here okay hence this s itself is a limit here because s is less than equal to b if beta is any lower bound for b then by definition beta is inside a therefore beta is less than equal to s therefore s is the the greatest lower bound it is infimum all right so by upper bound exist and then the supremum exist if the lower bounded below lower bound exist is what we have low infimum exist what we have two implies one I have told is similar okay few more terminologies is a totally ordered set is said to be order complete if it satisfies condition one and hence condition two because they are equivalent of the above limit I repeat what is the mean of order complete every bounded above set must have least upper bound which is same thing as saying every bounded below set as a greatest lower bound so if that exists inside this x tau x partial order okay this must be first of all a totally ordered set such a thing you will call a order complete so you may be already knowing that this this property we have been extensively using for the real numbers right so we have made it a definition here if x is a linearly ordered finite set then it is easy to see that it is order complete right therefore order completeness axiom is relevant only when x is infinite next time we shall use all these okay and then study order topology so far we are not mentioning a topology at all okay all the time we were talking about the combinatorics of this partial ordering total ordering well ordering and so on now the topology we write down okay thank you