 Welcome to NPTEL NOC and introductory course on point 6th apology part 2. We continue our study of separation properties S0, S1, S2, S3 and so on. So today we will just study one example known after Knaster and Kuratovsky. Kuratovsky you must be familiar, he has done a lot of mathematics which you might have learnt some of them and so on. I am not very sure whether you know anything about Knaster. However, the example that you are going to give is named after Cantor also, it is called Cantor's leaky tent, so which is quite descriptive, tells you how it looks like, later on I will give you the picture also. So the point of this example is that it is a compact connected subset of R2, so it is everything that you want to have, you know, yet what is happening is if you delete just one point k minus a, it is totally disconnected. So that is why it is very starting when it was produced, it was a sensation, okay. So let us start the construction, but it will take some time. Right now we are going to do something in between, you start with Cantor, Cantor is there, you start with the deleted middle one-third Cantor set C contained inside 0, 1. The only property of this Cantor set that we are going to use is that it is totally disconnected, it has a countable subset which is dense, okay. So if you do not remember anything else, you should get remember this much, it is a countable, it is a complete, it is a close subset of 0, 1, it is complete matrix set because it is a close subset of 0, 1, okay. It is totally disconnected and there is a countable subset dense, okay. So we are going to use only this much actually. So now let P be the set of all end points of deleted open intervals in the construction of the Cantor set from 0, 1, okay. That P is a countable set and it is dense, okay. Also C is a perfect set that means every point of C is a limit point of C. Indeed every point is a limit point of P itself that is P is dense. Also note that P is countable, okay. So you can forget about what P is, I have produced such a P namely the end points of the deleted middle one-thirds. Remember start with 0, 1, you are deducting the open interval one-third to two-third that is the first step. The end points are one-third and two-third, they will be there. Of course 0, 1 to begin with they are there, 0 and 1. Then one-third, two-third will be there. Then 1 by 9, 2 by 9 all those things will be there. That is the meaning of this capital P here. Indeed what I have said is this P itself is dense, okay. Now take Q to be the complement of P. For each P inside P and Q inside Q, I am going to define a certain subsets of I cross I. Close interval 0, 1 cross 0, 1, okay. So what are they? The first coordinate is P, the second coordinate is between 0 and 1 of course is P cross I but S must be rational, okay. So you can easily remember this countable set can be thought of as representing rationality. They are rational, okay. So the second coordinate is also rational. The complement for LQ where Q is inside Q, same way you can define Q, T but the second coordinate is irrational except the two end points Q, 0 and Q, 1. So they are allowed, okay. So it is a union of Q, 0, Q, 1 and Q, T where T is irrational. So for each point in the counter fact, I have defined vertical line segment, subsets of the vertical line segment C cross I in two different ways, okay. When P is inside Q, I am calling it LP, P inside Q, I am calling it a LQ, that is all. Now take S to be the union of all these LCs. So it is actually subset of I cross I, it is actually subset of C cross I, okay. Now I make a small modification in S, namely S hat that is the union of S along with the line segment parallel to the X axis, 0, 1 cross 1, okay at the top, this is the top phase of I cross I. So the construction of S and S hat is over, no definition of topology, topology comes from I cross I, okay. We are not going to change the topology. So we claim there are three things we have to do now, S satisfies S0, totally disconnected that is the union of S1. S does not satisfy S1, okay, though it is half-door, you should see that S hat is connected. Why is it connected? Just because we put this connected interval at the top, so that will take care of connectivity. We have to prove it, it is not that easy, alright. So these three things we are going to prove now, okay. Definition is clear, for P inside this countable set, the second coordinate is rational, in the complement the second coordinate is irrational except the endpoints Q cross 0 and Q comma 0, Q comma 1, they are allowed, that is all, okay. And S hat has more points, namely the entire line segment 0, 1 cross 1, alright. So let us proceed. The first part is not so difficult to see that S0 is totally disconnected, each point is a component, alright. Suppose A is a connected subset of S, I want to show that A is a single term, right. Look at the two projection maps, pi 1 and pi 2, restricted to S into 0, 1, right. But where are they contained inside, that is what we want to show. If you take the first projection, the image will be inside C, right, because the line segments are taken only based on the points of C, okay. So pi 1 of A being connected subset of C must be a single term, why? Now we have used the fact that C is totally disconnected, alright. So pi 1 of this A is a single term, what does it mean? This A is contained in the line C cross I, alright. Now look at pi 2, okay, pi 2 of that, okay, this means A is inside L C, that is inside one, one vertical line segment. What is pi 2 of C? If C is inside P, then pi 2 of L C is Q intersection 0, 1, okay, see all the points are, second coordinate are rational here, right, so it is Q intersection 0, 1. And if C is inside Q, pi 2 of L C will be 0, 1 minus Q, all the irrational number, but end points will be there, 0 and 1, does not matter. In either case, both pi 2 of L C and, in either case, this L C and this one, pi 2 of L C, in both cases, they are what, totally disconnected, Q is totally disconnected, so complement irrational number disconnected, their subspace is adding two extra points, does not change the total and disconnected here, in this particular case, okay, the end points have been added, okay, inside that, if you add one point, it is not going to change the total, this is very easy to see that one, alright, whereas the same thing we are going to do for this asset, not asset for K which is coming, adding one extra point is going to become connected, so that is the startling point here, however here it is easy to see that, it is adding fewer and one, it is still disconnected, totally disconnected, so the complement, the projection maps, pi 2 projection maps are also singletons, if both the projections are singletons, the set must be singletons, so that was easy, now for the next B and C, what I am going to do is, first I will have a convention that, any subset Y of S, Y bar will denote the closure of Y in the ambient space I cross I, rather than the subspace S or S, okay, so be careful that, the closure notation is taken in the larger space I cross I, that is all, this is just a convenient notation, otherwise each time I have to say closure in Y, closure in S and closure in S set or closure in I cross I and so on, right, the closures of a subset in different larger spaces could be different, so that is why you have to be careful here, our plan is to get the proof of both B and C in one go, however we have to break up the proof into a number of easy steps, alright, let us see how easy they are, step one is, let A be a non-empty clopon subset of S, for each C inside C, such that C, 1 is inside A, this is hypothesis, a subset is clopon and a subset of S and C 1 belongs to A, only then I am going to define this subset G, C, A, depends upon the C as well as A, such that S belonging to 0, 1, such that singleton C cross the entire line segment S to S intersected with S, all points which are inside S, A must be contained inside A, after all A is a subset of S, so I am going to take intersection with S1, okay, so this is definition of GCA, by the very construction this is a non-empty set, why, because C, 1 is inside A, that is the starting assumption for this A, C, 1 is inside A, therefore C, 1 will be inside GCA, so this is non-empty, once you have non-empty set, take tau C equal to, I have an elaborate notation here which I may not use, tau C A depends upon A after all, it is infimum of this set, GCA, okay, look at this set, this is a bounded subset, it is a subset of 0, 1, it is non-empty, so infimum makes sense, the infimum will be between 0 and 1, okay, now use the fact that A is open, this is started with a clopon set, it follows that tau C must be less than 1, why, because once it is open, C, 1 is there, a small neighborhood of that point is there, which means some line segment here epsilon to 1, intersection S will be inside A, then the infimum will be less than that epsilon, okay, which is less than 1, so tau C is though as such it is between 0 and 1, it is actually strictly less than 1, okay, because A is open, next can it be 0, yes it can be 0, but what happened, if tau C is 0, then L C is contained inside A and conversely the entire line segment intersected with S is inside A, then obviously what happened, 0 will be inside A, inside GC, therefore the infimum will be 0, conversely if this infimum is 0 means the open interval 0 to 1 intersected with S, that is contained inside A, therefore this entire line segment L C minus the point 0, minus the point 0 C, that is inside A, just because the infimum is 0, but then 0 C will be a limit point, okay, and A is closed, see now I am using the A is closed part here, so 0 C is also, C comma 0 is also inside A, that just means that the entire L C is inside A, so remember this criteria now, tau C is 0, if you will down with L C is contained inside A, hypothesis on A is, A is closed one and to begin with C comma 1 is inside A, so this is our first step, the second step is instead of just one clopon set, we are taking a non-trivial separation of S, there may be many separations, actually there are many, namely you take a separation of the canter set, then raise the whole thing above all the LCs on both sides, that will be automatically a separation of S, okay, not a separation of S hat, okay, S hat is finally will not be, S hat is actually connected, so there are no separations, so that is what we are going to prove, but S has many separations, okay, so we have to study this one deeply, okay, S is taken as separation A, non-trivial separation, then I am going to claim something, for each C inside C, this C comma 1 will be either inside A or inside B, because A union B is the whole of S, accordingly tau C A and tau C B will be defined, right, which one you do not know, there will depend upon whether this point is inside A or inside B, to treat both of them together, I will just have a notation tau C, okay, there is lot of symmetry here, because A, A B are nothing but separations, A, S of A B is S of B also, assume that tau C is positive, okay, so I am coming back into the step, this step here, tau C is 0, I have something, now suppose tau C is positive, what happened, tau C is always less than 1, that much we know, okay, suppose C 1 is inside A, there are two cases, right, so just for definiteness, suppose C 1 is inside A, then the definition of infimum, it follows that C comma tau C is in the closure of A, right, this closure I am using in wire, inside I cross I, okay, it is a limit point of this set, therefore this inside A bar, on the other hand, for every epsilon positive, you can find a T between tau minus epsilon and tau C, such that C comma T is inside the complement of A, namely B, right, that is because this is infimum, the moment you take something less than infimum, there will be a point which is not in the set A, if it is not in the set A, it must be inside B because A and B is the whole of it, therefore C comma tau C is in the closure of B also, so what we have proved is, if tau C is positive, C comma tau C is in A bar intersection B bar, assuming that C comma 1 is inside A, but the argument is exactly similar, if C comma 1 is inside B, right, therefore in either case C comma tau C is inside A bar intersection B bar, but what is A bar intersection B bar intersection S, that is A bar intersection S and intersect in B bar intersection S, A bar intersection S, remember A is a closed subset of S, so A bar intersection S is A, B bar intersection is B, so this is empty, okay, this is an element of intersection and this is, the intersection is empty, it follows that C tau C is not in S, okay, look at the definition of C tau C, if C is inside Q, then the definition of the second coordinate must be irrational, 0 or 1, we have assumed it is not 0, it is positive, we have shown that it is less than 1, therefore it must be irrational, okay, if at all if it is in S, but it is not in S means now tau Q must be rational, okay, it is not in S, if it is irrational it would have been in S, for all Q inside Q, you can conclude something for C beyond into P also, but that is not important for us, okay, so we have done something now, tau C is 0, what happens we have understood, tau C positive we have understood what happens, okay, now we exploit this one and write for each R inside 0, 1 and a rational, R must be rational, you can define this for irrational, but it is not of our concern, put Q R equal to all those points inside Q such that corresponding tau Q is equal to R, this tau of C is all the time defined now that is the whole idea, okay, I am still working with a given fixed separation A, A separated B that S equal to A bar B, so that is all fixed, so with respect to that take tau Q's R equal to Q belonging to A bar B, belonging to Q such that tau of Q is equal to this rational, okay, I am not interested in the end point, I am interested only in the in between points, we claim that this Q R is nowhere dense subset of C, this is subset of C, this subset of Q after all, okay, now I want to say that it is nowhere dense, for this we can work inside instead of the bottom line C cross 0, you can work it to copy of C namely C cross R, okay, to show that Q R cross R which is a copy of Q R is nowhere dense in C cross R, okay, first we note that Q R cross R closure, this is a singular term of cross, it is the same thing as Q R closure, again you see what we have shown C comma tau C in A bar intersection B bar, this point Q R cross R is precisely C comma tau C for that one, that is the whole point, so tau C equal to R, right, so that is inside A cross A bar intersection B bar, this is what we have shown, okay, therefore Q R cross R closure intersection S which is Q R cross R closure is same thing as Q R closure cross R, okay, intersection with S, that is empty, okay, so because first coordinate is inside Q, the second coordinate is rational R is rational, now if U is a non-empty open subset of C, I want to show that this is nowhere dense, no where dense means what, Q R cross R bar has no interior, that is what I have to show, okay, suppose you take a non-empty open subset of C contained in Q bar, okay, since P is dense subset of C, okay, it follows that there exist P belonging to P such that P comma R is in Q bar cross R, if you have any open subset of C, there will be a point of P, that is all I am using, but if this point is in Q bar, it will be P comma R will be in Q bar cross R, but then this P comma R will be in Q bar cross R intersection S, okay, and that is empty set, so empty set cannot have a point there, therefore Q bar intersects every open set, sorry, the complement of Q bar intersects every open set, this means where no non-empty open set is contained inside Q bar, okay, therefore Q R is nowhere dense subset of C, okay, now we are coming more or more closer, step three, so more topology now, I define another set E such that all points of, all points of the canter set such that tau C equal to 0, okay, we claim that E is a closer subset of C, that is the first step, put A prime equal to all those C such that L C is inside A, B prime equal to all C such that L C is inside B, okay, remember tau C equal to 0 implies L C is inside A or B, right, depending upon C comma 1 is inside A or C comma 1 C inside B, so that is our first step, right, therefore this E is nothing but A prime union B prime, so it suffices to show that A prime and B prime are close, then E will be close, okay, now look at L C bar is nothing but C cross the whole of interval I because it will either have all the irrational points or all the rational points, in either case the closure is the whole of the interval, so it is C cross I for C inside C, also note that for any subset X of I, okay, we have X cross I closure is nothing but X bar cross I, this is a general property actually nothing to do with I, if you take X cross Y closure, okay, Y is in any arbitrary subset, X is a subset of some other thing then it is always X bar cross Y, okay, so that is what I am using the product topology here, therefore look at the union of all L C's where C range is over A prime, namely all those L C's are inside A, so this union is inside A, I am taking the closure, this closure, I can take first L C closure here, okay, then this is a larger subset than this one, so the closure is contained in the closure, I have to take closure, on the other hand these are close subsets, right, L C is contained inside, contained inside L C is same thing as L C here, the closure is contained in this big thing, the favor of this union is contained in the closure here, therefore it is closure is contained here, therefore there is an equality here, so I have to say this one because this union is an arbitrary union, it is not a finite union, so these two are the same, but now L C bars are what they are singleton C cross I, right, where singleton C range is over A prime, right, so I can write this as now A prime cross I bar because each L C bar is A cross singleton A cross I, so this is A prime cross I, the inside subset, so it is closure, but this closure is same thing as A prime closure cross I, alright, moreover each L C by definition where C is inside A prime is inside A, therefore this left hand side union is inside A, therefore its closure is inside A bar, therefore if you start with a point inside A prime bar here, then you should take X say C cross I, we have L C bar correspond to L C bar which will be contained inside A bar and hence this L C will be equal to L C bar intersection S, that is my definition, right, because L C are close subsets of S contained inside A bar intersection S which is same thing as A because A is a close subset of S, so L Cs are contained inside A, okay, that is what we know already, okay, therefore this means that C is inside A prime, C, I started C belong to A prime these L Cs are inside inside A, but now I have taken C inside A closure for that C also we have proved that L C is inside A, okay, therefore A prime bar is inside A prime which means that A prime is closed, all these closures are now, okay, inside C they are they are subsets of C, C cross 0 we can take, similarly argument is similar there is no there is no change here, B prime is also closed, therefore set of point C inside C for which tau C is 0 is a closed subset, we have proved so much of topology on this one, we just do not know whether this is non-empty, for all your discussion this may be empty set, we have not proved that it is non-empty, we have not proved that any of this L Cs is contained inside A or B, they may be all dissected by A and B, okay, having done so much of topology we still do not know that, the next step is precisely that, E is dense in C and hence E is actually put the whole of C, the strange thing is you try to prove it is non-empty just non-empty there is no other way, we are actually going to prove that E is C and this E is the typical way wherever you apply Bayer's category theorem and that is what we are going to do, we are going to apply Bayer's category theorem here, okay, we are not just proving it is non-empty we are actually proving that this is dense subset, okay, being closed it will be equal to the overall, okay, so another notation here just temporary notation, Q intersection 0 1 O upon all rational points let us denote in them by lambda, the only thing that I am concentrating on this one is a countable set, this is going to be indexing set now, it follows that the elements of C are the following, those which are inside P, so I am writing the singleton P, singleton P P inside P then those which are in E, this E may be empty I do not know and those which are indexed by lambda what are those set Q R, take R inside lambda and take the Q R which we have proved nowhere dense subset, okay, this is what the step we have proved that Q R's are nowhere dense subset, okay, yeah, as soon as these R is not in A, okay, not in here that means these things are inside Q, the Q R each of these Q R are nowhere dense subset, so many of them are the countably many, so C is the union of these things, what is this union, countable union because P is countable, this is one single element, one single set and these are countably many sets, so this is a countable union, each singleton P is nowhere dense, right, each Q R is nowhere dense, if E is nowhere dense what happens, you will get a contradiction to Bares category theorem because C is a complete metric space, indeed once you write like this as countable union the stronger statement of Bares category theorem says one of them must be dense, so this is nowhere dense, nowhere dense, this must be dense, okay, so this part turned out to be easy for us but without Bares category theorem it is just impossible to do anyhow anything okay, so to prepare this much we have to apply Bares category theorem we have to do all this, alright, so now we can complete the proof of both B and C very easily, so what is B, B says that it does not satisfy S1, it does not satisfy S1 means given I have to produce two points, some two points, there exist points X and Y which cannot be separated, so what do I do, okay, given any separation S, S, A, B we have the corresponding E is equal to C, what does this imply, this implies that all C inside C, L C is either inside A or inside B, A prime only B prime is C, right, so they are inside A or inside B, they cannot be both because they are disjoint, A and B are disjoint, therefore for any C inside C the point C0 and C1 cannot be separated because C0 and C1 is in the same L C, over, so I have produced plenty of pairs of points which cannot be separated, essentially we have observed that there are vertical separation, but this says that two points lying on the vertical line they cannot be separated, so there is nothing like horizontal separation, okay, there is nothing even slanted separation and so on, only vertical separations are there, plenty of them because C itself is what totally disconnected, so this proves part, it proves the proof of B that S does not satisfy S1, okay, so if you remember like this, this is proof theory, the proof of this topological part, okay, they are difficult to remember but you can reproduce them, there is no problem, alright, now you will see that B, C is also easy, okay, so proof of C, now remember that S hat has one extra thing there, namely the entire line segment, 0, 1, close interval 0, 1, cross 1, right, so that is S hat, suppose S hat has a separation, we want to prove S hat is connected, any non-connected say disconnected space must be having a separation, non-trivial separation, non-trivial separation means what A and B are non-empty, both A and B are closed and they are disjoint and the union is the whole thing, that is all, so I am just recalling what is the means of non-trivial separation, but I cross 1 is a connected set contained inside the left hand side, therefore it must be contained inside A hat or B hat, what does this imply, for definiteness sake let us say I cross 1 is inside A hat, okay, it cannot be on the other side also, put A equal to A hat intersection S and B equal to A hat intersection, B hat intersection S, from S hat come to subspace S, then S itself will be separated, namely S equal to A bar, right, A and B will be close of sets of S and they are disjoint obviously because A hat and B hat are themselves disjoint in S hat, okay, so this S is A union, S A separated B, as soon as we have part B says we have 1 comma C comma 1 is either inside A or inside B, but right now C comma 1 are all inside A by this assumption, you see the entire i-chromoism A hat, so all those points are in A they will be inside A, A hat intersection S is A, so all the points are inside A for all C inside C, which just means that A prime is the whole of C and B prime is empty, right, this means the entire set S is inside A, okay, but A is a subset of A hat, therefore S hat itself is in A hat because S hat is nothing but you know it is a closure of S and this is a close subset, a contradiction to the assumption that S hat equal to A hat bar B hat is non-trivial separation, so we have proved that S hat is connected, okay, now I come to the final construction of the Kuratovsky, Knaster Kuratovsky example, so this is the picture, starting with this is dot, dot, dot, all these is the canter set, okay, then what we did we took lines, vertical lines like this for P inside capital P all the second coordinates were rational, for the in the complement all the second coordinates were irrational, so that was my S, then we added this line to get S hat, but now what we want to do is you just bring all the entire line segment here to a single point here, collapsed it, how do you collapse it, along the horizontal lines, anything in the horizontal lines keeps moving, so this line segment from this 0 to 1 will be moved to from this point to this point depends upon how high it is from what, what y coordinate is, second coordinate is, you just move them like this, so this line has become one single point, okay, in other words if you delete this point and if you delete the entire line here, whatever you have remaining thing here will be the open path will be homeomorphic to a subspace of S, right, namely all the points in the C comma 1 are removed, okay, being a subspace of a totally disconnected space that will be also totally disconnected, alright, so what happens is this A is the point which is apex point of this tent, that is why it is called canter's tent, if you remove that point it becomes a leaky tent, that leaky tent is totally disconnected, why this is connected, this is just obtained by quotient of a set, okay, where we have identified this entire line to a single point, that is all, okay, I have written down the formula to get a set, this I call this as capital K, so how to get this one from a set, so I will show you that one now, okay, here is the example, okay, so we construct final construction of this curly K which is nothing but the quotient of a set wherein I cross 1 is collapsed to a single point, so this is a notation, okay, however since we want to display K as a subspace of R2, we consider the function f from I2 to I2 given by f of t comma S, you see the second coordinate does not change at all, so everything is happening along the horizontal line, horizontal what I am going to do, I take 1 minus S times t plus S by 2, okay, when S is 0 what is this point, it is t comma S, it is identity map, when S is 1 what happens, this will be the 0, this will be 1 by 2, 1 by 2 comma 1, right, so all the point t are going to a single point when S is 1, that means I cross 1 is going to single point, if S is less than 1, this is a bijection, okay, check that, so put kappa equal to f of S, the image, check that image of f is a convex hull of, okay, it is a convex hull of t, okay, image of f under, image of f means the hull of I2, f of I2, what is that, that will be the strangle or based on its I, the base of the strangle will be this, this is the closed interval, then this will be the effect, so this kappa is going to be a subspace of the strangle, same kind of description is there all the time, restrictions on the y coordinate of those points, okay, f is restricted to I cross 0, 1, okay, cross 1 is throw away the line, homeomorphic on to t minus half cross 1, so throw away the line, top line from I cross I, throw away the point from t, that is a homeomorphism, therefore it follows a kappa which is equal to f of S, it is a connected space being the, being the quotient of a connected space and kappa minus half comma 1, being homeomorphic to a subspace of S actually is totally disconnected in the picture above this a, as I put a here which is nothing but 1 by 2 comma 1, okay, so everything is proved, so let us stop here with the final remark that k satisfies neither S0 nor S1, why, because it is connected over, we shall see later that adding an extra point does not destroy S3, therefore we conclude that k minus half 0, this is neither S2 nor S3, alright, let us stop here, next time we will study a little more about 0, I mean S0, S1, etc, introduce what are called as 0 dimensional spaces and so on, slowly we will start our dimension theory, thank you.