 Hello, welcome to NPTEL NOC, an introductory course on points of topology, module 54, groups of homeomorphisms. Remember, this is a part of our compact study of compact open topology. So for the rest of the stock, we will always assume that X is a locally compact house topological space, though we may not mention it again and again. And HX denotes the group of all homeomorphisms of X, self-homeomorphisms. That CO denotes the compact open topology on HX. We will discuss the problem whether this HX with the compact open topology is a topological group or not. It is shown that this is not so in general. In general means just under this hypothesis, namely, X is locally compact house topology. It is also shown that under the additional hypothesis that X is locally connected, the answer is yes. An example to illustrate that local connectivity is not, you know, is necessary, is included. That is not correct. Local connectivity is necessary, is included. The motivation for this discussion is that a positive answer to this problem has applications in bundle theory. We are unable to locate any discussion on this topic in the existing literature. Main question is the following. Does the set HX of all homeomorphisms of a topological space X form a topological group under the compact open topology with the usual compositional law of functions? Whenever you take self-homeomorphisms or self-atomorphisms of any kind, the group law is always a compositional law. Of course, it is natural to demand that X is locally compact house topology because we are dealing with compact open topology. It seems that all working mathematicians who have to use this result assume the truth of it. I too have been assuming this though I have not used it in any of my research works. The point is that the statement is true for a large class of spaces which are interesting to topologists. For instance, if X is compact, we will see an easy proof of the assertion here. X is discrete, you can do it by hand and see that it is also okay. X is a manifold, not easy. X is locally finite CW complex, that is not easy. So, these are the things some sort of proofs seems to be there in the literature but nobody has written down anything. So, whenever people have used it especially for manual holds, apparently they are sure of that the result is true somewhere. So, that is the whole situation. We first notice that the proof of an affirmative answer to the above question in the case when X is compact is quite easy. So, naturally your first attempt to prove the same for a non-compact, locally compact first of space is to go to the one point compactification, do you know use the theorem there and try to bring back to come back. Note that every element f belonging to Hx has a unique extension f hat which is also homeomorphism such that the extra point f hat of star is just the star that pointed infinity whatever you want to call x star is the one point compactification here. Also note that the association f going to f hat can be used to identify the group Hx with a subgroup of Hx star namely those which fix the point at infinity. Therefore, we are led to ask the following question does the compact open topology on Hx coincide with the subspace topology from Hx star start with Hx star give the compact open topology you can take the subspace topology because this is a subset now you can treat this as a subset actually some group, but the question is whether that subspace topology is compact open topology in its own right in on x on Hx. So, it turns out that this question seems to be harder than the origin question indeed one can more or less see that the two questions are equivalent at least this harder in a sense this stronger in a sense you can if this is true then you can get an irreproof of question number this question main question. So, the answer to our main question itself is not always in the affirmative the example that we have is obtained by removing origin from the standard deleted middle one third counter fact in the unit closure interval on the positive side we have the following result let x be a locally compact locally connected half the space and Hx denote the set of all homomorphisms of self homomorphism ok take the compact open topology on Hx then this Hx co is a topological group on the way to prove this theorem we shall see that the result hold for all compact off-door spaces without the extra assumption of local connectivity. So, we do not have to prove that one separately ok. So, let us start proving but there will be many interruptions in between. So, let us hope we can start proving recall that the compact open topology is generated by sub basic open sets of the form is you know angle bracket KU such that all f belonging to Hx such that fk is contained inside you remember this notation we used for all continuous functions when we are dealing with space of all continuous functions here I have restricted myself to only homomorphisms of x to x. Therefore, the same notation I will use the notation is changed that is why I have carefully rewritten it here. Now, this is actually the old notation KU intersection with Hx that means only homomorphisms are taken not just all continuous functions. So, that is what you have to pay attention to this one. In any case inside Hx is subspace of all continuous functions from x to x right. So, this is subspace topology there. Therefore, if you take these basic the sub basic open sets in the bigger topology and intersect them with the subspace in the subset they will be sub basic open subsets for the subspace topology. So, there is the no problem in that one. So, this is the subspace topology from the compact open topology on all continuous functions for a locally compact of the space by b of our earlier theorem namely exponential correspondence theorem. Continuity of the composition map Hx cross Hx to Hx for f comma g to f component g is a consequence of the continuity of the corresponding function evaluate under evaluation map and so on right. You have evaluated the x Hx cross Hx cross x and evaluated f g operative numeric to f g of gx. This is the function this evaluation function is a map into x. So, if this is continuous then that is continuous this could be any domain some domain into the space of continuous function in a function space is continuous if and only if composite with the evaluation map here is continuous that is all we have seen in this is the theorem we have we have used several times now we are going to use that also ok. So, how to verify that this f comma g comma x going to f composite g of x is continuous on this product space Hx cross Hx cross x I have taken to Hx all right. So, to verify this namely this this one fix f g and x such that this composition and operated on x is inside an open subset u ok. Then I have to produce a neighborhood of f cross a neighborhood of g cross a neighborhood of x such that under this function the entire neighborhood goes inside u. So, that is the continuity at this point. So, start choosing an open subset v such that gx is inside v and v bar is compact and f of v bar is contained inside u. So, this I am doing the continuity of f f of gx goes inside u. So, some open subset goes inside u under f that open subset v can be chosen to be compact also. So, closure is compact also because the local compactness. Similarly, once you have chosen this v now f of x goes into v right. Therefore, you will get an open subset w such that x is inside w w bar compact g of w bar inside v. Now, it is straight forward verification to see that v bar cross u. So, this v bar is compact this is open right. So, all function all homeomorphism taking v bar into u that is a neighborhood neighborhood of f because f is one of them. So, this is an open subset this is an open subset this open subset contains g and this is the open subset started with chosen such that w such that x is inside w. So, this is an open subset of f comma g comma x this is open neighborhood. So, if you take you know you take arbitrary you know say f prime g prime w prime with this property the composition will land up inside w is very straight forward to f rest salvation. So, in particular it follows that once you have proved that the composition is is continuous you can restrict it to fix f look at g going to f composite g. So, that is a left multiplication by f fix g. So, fix f sorry fix g now f going to f composite g that is a right multiplication by g. So, both of them will be continuous that is all. So, this is a standard thing. So, left which is left multiplication by f r f is a right multiplication by f. So, both of them are continuous from h x to h x the main thing here is to prove that inverse is also continuous once you prove that it follows that the group laws are continuous. Therefore, the h x is a topological group. So, this is what we have to prove here is an elementary result that we want to use since it is does not seem to be very common let me just take a few minutes to state it clearly and then prove it also. In general result let g be a topological space with a group operation g cross g to g continuous only assumption is the the multiplication is continuous. Suppose further this eta from g to g given by g eta g g inverse that is a inversion map. Suppose this continuous at just at a just at any one point. So, better take at a then eta is continuous on the whole of g this kind of result you must be we must have used in any topological group itself like if you have a continuous homomorphism it any homomorphism which is continuous at identity e comma e will be automatically identity on the whole space on the whole group and so on. Differentiable function if they are group homomorphism differentiability at a single point then it will imply differentiability everywhere that kind of thing. But in all of them you already have a topological group here the kind of result that were a technique that we are using the same. But only continuous function is only multiplication is continuous the eta g is not yet continuous that you cannot assume we are going to prove that this inversion eta is continuous now. So, that is why this is not all that common. So, let me tell you the proof we first note that for each g inside g the left multiplication right multiplication are continuous because of because the multiplication itself is continuous. Since Lf inverse is equal to L of f inverse and Rf inverse is R of f inverse it follows that Lf and Rf each of them is a homomorphism okay first of all they are continuous well same reason L of f inverse is also continuous because f inverse is another element there okay. So, now let f belong to g be any element we want to check the continuity of eta at that point okay. So, first you check the following thing the inversion composite Lf is nothing but Rf inverse of inversion as a map from g to g as a function from g to g. So, this is the commutativity law eta composite Lf when comes it on this side you will be Rf inverse of Lf okay this is very strict for verification this is true in all groups okay. Since Rf inverse is also continuous on the whole of g it follows that Rf composite eta is continuous at e okay this is a homomorphism if you just a continuity of this one implies eta is continuous at e. So, when you take composite another continuous function the composite will be also continuous at e okay. So, this side now eta composite Lf we know is continuous at e but now Lf is a homomorphism therefore it will follow that eta is continuous at Lf of e okay. So, all that you need is to use that Lf is open mapping okay. So, this will be easy indeed you can if you are familiar with taking the sequence you know sequential continuity and so on you can use that also only thing is the space that you have taken you will have to assume first countable the alternative is there you have learned it in this course namely you do not have to assume first continuity what you do you take nets show that use convergence of nets okay. So, that I will need to use an exercise this also just entertaining exercise on the other hand using nets to prove this one it is like using an atom bomb to kill a sparrow. So, let us use this result now namely continuity at one point namely at identity is enough to prove the continuity of the whole thing okay. Therefore, this come back to this special case now eta inversion in Hx to Hx okay I have to prove that it is continuous at the identity element of this group Hx okay. Also note that the inversion is a after all by the very definition it is a inversion usually inversion of inversion is identity eta inverse you know this is a involution okay eta inverse is eta so they are same. So, showing that it is continuous same thing here showing that it is open okay we need to show that sectors of form eta of k u are neighbourhoods of identity of x where k is inside u because neighbourhood of identity means what f of k must go inside identity belongs f of k is k itself it must be inside k must be inside you that is all we have to take. So, k is compact and u is open that is the standard thing for all sub basic open cell. So, we have to show that you know eta of k u is same thing as eta inverse of k u we have to show that it is a neighbourhood of identity because these are neighbourhoods of identity start with provided k is contained inside u k compact and u open okay. So, just do this elementary algebra here f is a point of I mean function in eta of k u means f inverse is inside k u right. So, that means f inverse of k is going inside u and that is the same thing as saying k goes inside f u because f inverse is after f is a homeomorphism if k is goes f u because now k is f is a one one mapping okay it is a bijection complement of k will contain f of the complement but this is same thing as f of the u complement. So, f of u complement is going inside k complement it is same thing as in this notation f belonging to this angle u c comma k c. So, starting with this an open subset you have to show that this is an open subset in the definition of compact open topology such sets are not there you see to allow this as a sub basic open set I need this one to be compact and this one to be open luckily since k is compact k is closed so inverse image the complement is open but why u c must be compact use just an open subset okay. So, that is the one of the main reason why this is why this is a non trivial problem. So, we have to show that this is actually open we do not need to show that we have to just show that this is a neighborhood of identity that is enough okay that will be equal to showing that this open actually eta itself will be a continuous function after that. Therefore, we need to prove that sets of a form angle u c comma k c where u is any open set and k is compact k contain inside u these are neighborhoods of identity of course we need to consider only the case when k is non-empty okay if k is the whole space k c will be empty right. So, if k is empty k c will be the whole space so this will be automatically a whole you know there is no condition on that one so that is not a and u c is non-compact if u c is compact then this again open subset basic open subset the sub basic open subset there is no problem. So, in particular if u c is compact means what u as arbitrary open subset therefore u c is an arbitrary closed subset when it is compact the space itself is compact once a space is compact there is no more work to do because this will be compact subset that is an open subset these are some basic open set set so you are home so that is what I say that we have already now got a proof of the assertion that you know in the case of x is compact okay this is on the way but we want to go ahead without this one so far we have not used local connectivity at all okay just compact order space is good enough okay now we want to use local connectivity okay so here is one more elementary lemma okay so we should state a proof separately instead of using it in the run of the proof let x be a locally compact locally connected half-dough space let k contain inside u subsets of k such that k is compact and u is open then the registration opens of set w of x such that this w is the union of finitely many connected open sets each of which is actually compact also in any case what I want is w bar is compact and k is contained inside w contained inside w bar contained inside u okay the statement is k is compact u is open so in between you can always get this one such that w contains a w bar by regularity but what I want is this w itself is the union of finitely many connected subsets and each of them such that their closure is compact so that w bar will be also compact so that is the extra thing and that extra thing comes from local connectivity okay the rest of the statement is just regularity here let us go to the proof this not very difficult thing for each point inside k because that x belongs to u and u is open and because the space is locally compact and connected locally connected I can choose x belonging to vx contained inside vx bar contained inside u such that this vx is connected vx bar is compact the first you choose such that vx bar is compact inside that you can choose a local local connectivity you can choose another open set which is connected its closure will be contained inside the closure of the original one which is compact so it will be compact that is the way you can prove this one okay now so that is what this is possible because x is locally compact first and then locally connected also since k is compact you can get a finite cover k x1 x2 xn all of them inside k such that k is contained inside vx size i ring to i ring 1 to k I am denoting that as w so that was the assertion there is such a w so this w is the union of these connected subsets each of vx bar is compact therefore w bar is a union of finitely many close up say v bar so that is also compact so all of them is contained inside u over so we will use this one and then let us continue with the proof of theorem so let vx size and w etc be as in the previous lemma just now which we have proved for a notational sake I will put l1 equal to w bar I am starting from some kind of induction not very big induction only number of steps here using regularity of x remember this w bar is compact and contained inside u so I choose this l1 is w bar l1 is contained inside u so in between I will choose another l2 which is also compact which is a neighborhood of this one so l2 interior is contained inside l2 that contains l1 so similarly in between l2 and lu again another one will contract l3 interior contained inside l3 okay so these are all compact subsets now okay l2 and l3 their interiors contain the previous one okay so l1 l2 l3 contained inside u now you take v equal to interior of l3 minus l1 by the way this kind of things we have done several times in proving the para compactness and so on okay whenever you have hostorsness and locally compactness and so on so you are using normality is here strongly regular because things are compact and so on local compactness is extra thing that is all okay so take this v interior of l3 minus this close set l1 so this is open subset and put t equal to just the boundary of this l2 l2 is a compact set throw away the interior of l2 so that is the boundary of l2 so I will just denote it by t okay so here is a picture what we have done so far started with this rectangle here k contained inside this large open set here so in between the first thing was to take a w such that w is union of finitely many connected open subsets such that each closure is compact so those are the vx size okay so this w covers the whole of k and it is contained inside u that is all after that I have chosen the notation w bar to be l so that is a compact set then I choose l2 then I choose l3 okay so that is all picture what is this v this v is this l3 minus l1 interior the other way around l3 interior minus l1 so there is an open subset so did I make you open this here l3 interior minus l1 right and then t is just the boundary of l2 okay so I have drawn nice pictures nice pictures will not be all that nice okay there I have drawn a circle here and on this t cannot need not be connected and all that the only thing is this w is a finite union of connected set okay there is no other way kk is not connected u may not be connected nothing there is no connectivity assumption anywhere else now you take g equal to xi vxi triangle so this is an open set intersect it i rank 1 to k that is an open set basic open set intersect it with further with t comma v t is compact v is open so this is an open set xis are in vxi t is inside v therefore identity map belongs to this intersection okay therefore g this g is a neighborhood of identity in hx now look at the complement of c complement of u this portion okay this portion is contained in the complement of l2 also because l2 is inside u right so complement of u is inside complement of l2 and complement of w contains complement of k because k contains w k is contained in w so I am just reversing De Morgan Lawyer that is all it follows that if you take l2c wc the angle here is contained inside uc kc wc is inside kc so everything coming here will be inside kc also on the other hand this is a smaller set l2c is a larger set if a larger set is going inside here the smaller set uc will be also going inside that under whatever function you take so this angle all functions all homeomorphisms from x to x which take l2c into wc is automatically contained inside this one okay this is satiric result property of this angle function that stops therefore it follows it suffices to prove we want to prove what we want to prove to forget that one that this g we have selected right where was Christian this g right this is an open subset we want to show that this is neighborhood of identity contained in this open subset namely uc kc okay so instead of that we can just show that this is contained inside this set here that is a stronger statement okay so let us prove that g is contained inside l2c wc okay now consider this thing t is a closed subset right so x minus t if you think of this one what is t? t was the boundary of l2 right whatever is remaining is interior of l2 minus the complement of l2 complement of l2 is open interior of l2 is open they are disjoint the union is precisely x minus the boundary of l2 that is t so this is a standard way of getting a separation okay this nothing nothing very special about l2 and you could have taken any compact subset okay and take the interior of that and the boundary of that the boundary of that will separate the interior with the complement that is all for now take any function any homeomorphism in g any homeomorphism we will do for this this particular thing but we are interested in what is happening for elements of g okay so since f is homeomorphism f of this will be another separation what is f of x it is x what is f of t it is f t so x minus f t will be put f of l2 interior and disjoint union with f of l2 complement okay a separation goes to separation under a homeomorphism okay now look at all the xis which are inside v xis okay and v xis are contained inside l2 interior right because they are inside l1 itself so we have f of xi if you take f of that it will be inside f of l2 interior also f t is contained inside v okay in this picture if you remember f t remains here t is here f t will remain here because f is inside g right f is inside the t of f of t is inside v that is the hypothesis here so f t inside g f t inside v and hence does not intersect any v xis because v xis are inside l2 interior instead l1 interior sorry l1 interior so I have subtracted the entire l1 from l2 l3 interior so that is me so they do not intersect v xis okay and hence all the v xis must be inside x minus f t okay because f t does not intersect this one so v xis are x minus f t but v xis are connected so this is a separation therefore v xis must be in one of them okay so v xis is connected on the other hand v xi intersection f2 f of l2 interior is non-empty because f of xi is there right xi v xi f is here you see again x is inside this one so xi f of xi is inside v xi xi is also in v xi but xi does not move out of v xi under f so this f of xis are in the intersection so v xi intersects this one therefore it must be completely contained inside f of l2 interior so it cannot intersect this part okay being connected okay this is true for all I I have never used anything about xi either other than anything about I here special so the entire w which is the union of v xis is contained inside f of l2 interior okay but look at this one if something is contained in the l2 interior its interaction with f of l2 c is empty right therefore what we get is w c the complement of w which contains complement of this the complement of this this first part will contain the this whole thing here and more perhaps in fact the complete complement is being taken in x so it will contain ft also does not matter it contains f of l2 c so what we have proved f of l2 c is inside w c okay so out of that mean f is inside the angle l2 c angle w angle w c that is what we wanted to prove l2 c w c started with an f element here it is here so that is the proof okay so proof is over indeed the next thing namely producing a counter example that is if at all you think this is difficult that will be little more difficult okay so let us come to that so before that I will tell you the history this history is very recent after all it is just about three years old so whatever I remember I have put it here already attempt to find a justification for locally connectedness assumption in the above theorem initially led us naturally to the topology sine curve recall that this topology sine curve is a subspace of r2 is a compact subset of r2 actually which the union of the graph of sine pi by x on the interval open interval 0 to close interval 1 open to close okay you take sine pi by x take the graph together with the closed interval 0 cross minus 1 plus 1 on the y axis okay I am just I cannot go into the full study of sine pi by x now but I am just recalling you this was we we this is a compact set if you take this one with the compact open topology the homeomorphisms of this space that will be automatically topological group because this is compact set so what we want to do is you know destroy the compact compactness by removing a point so do that namely take this very special point 0 comma 1 okay 0 comma 1 or 0 comma minus 1 these are two special points there so remove one of them so remove this one from this up and obtain the topological space x which is now non-compact it is locally compact okay this is subspace of it is also also and it is connected but not locally even locally connected locally it is not locally path connected it is not locally connected either we can then think of the original space as the one point compactification of x okay so I was trying to prove or disprove actually I was trying to prove that in this case there is an assertion namely affirmative answer namely Hx is a topological group but it turns out that understanding the structure of its the group structure of Hx is very important and we don't know it very well yet so I had to abandon this example but this leads to a sub question here namely this attempt is what that is important here going to one point compactification right forget about this particular example try to do it in general okay so that is why we have that the sub question anyway there is matter so the point is that I wanted to say is that this attempt was not successful yet successfully yet I haven't given it up yet of course okay so the next example is you can guess what is it the canter set so canter set that did help okay so let us study the canter set now so what I am going to do canter set is also compact remember that so I want to destroy the compactness by removing a convenient point you can perhaps do it with any point but most convenient is point remove the point 0 okay so recall that canter set was obtained this is one third middle one deleted middle one third canter set namely for each integer i 1 2 3 whatever i divided by 3n to i divided by i plus 1 divided by 3n okay you keep deleting these open interval first you delete 1 by 3 and 2 by 3 1 by 3 to 2 by 3 and then you 1 by 9 to 2 by 9 4 by 9 to 5 by 9 and so on you keep deleting them okay that is the that is a construction of the canter set okay so if you take some some open interval of this length 1 by 3 power n and next one 1 plus 1 k plus 1 3 power i i i by 3 power i plus 1 3 power n of this of this length 1 by 3 power n here this may be empty because this is already in the deleted part if it is not empty inside that you will get a carbon copy of the canter set again a subspace of a which is of this nature provided non-empty depending upon value will be called a canter subset it is again carbon copy of the canter set which instead of starting with 0 1 I could have started this interval and then did the same procedure for that interval that is all so we shall use the following fact about a canter subset so canter subset I am denoting by a so each a each canter subset is homeomorphic to the entire canter set c through an affine linear map endpoints you take and take an affine linear map which will send the endpoints to endpoints that is all automatically it will give you a homeomorphism of the canter set each a is clop on in c open as well as closed okay if j is an open interval non-empty open interval and a intersection j is non-empty then a intersection j inside that a intersection j contains a canter subset so that may be a prime or a 1 or b 1 whatever each open interval okay that is the property of canter set it is repeated it is copied inside every canter subset also that is what I wanted to suggest finally this may need a little more proof similar to our lemma about local connectivity and so on but see here this is a different kind of game but proof is not difficult take any canter subset contained in union finite union li okay of some open finite covering by canter subset okay it is not an open covering it is clop on covering canter subsets are clop on you know that it is not just open covering okay these are the finite may finitely many compact subsets of li that is enough a is a canter subset these are the best comp any compact subsets okay there may be some points also here and so on some intervals also I can take I can take compact subsets of r to begin with so a is contained in the union and each li is compact if all that around then there exists a clop on subset b of a which is homeomorphic to c i that all that I say b is another canter subset that is all there is a canter subset of b which is canter subset of a such that the property is important b intersection li is non-empty implies b is completely contained inside li see there are finitely many of them look at l1 b intersection l1 non-empty it must be inside that completely if it is empty it is fine so some of them may not intersect the moment they intersect they will completely contain b so that is the meaning of this one okay for this is true for all the r so some of them contain completely some of them may not contain that is all none of them may contain that is not possible because union is inside l r the whole thing is l r right if all of them are disjoint that is not possible okay so this is the assertion here so this will be used in a very peculiar way you will see that namely so now I take x to be the canter set minus 0 we claim that the inversion map is not continuous okay so it cannot be a topological group under the compact open deposit as seen before it is enough to produce a compact subset k and then open subset u such that this eta k u which is u c k c this we have seen already is not a neighborhood of identity over if it is a neighborhood of identity if you have to all the time then you know that eta would be continuous so one single example of this nature will show you that it is not continuous so all that I have to choose is start the compact subset k and then open subset containing that look at u complement and k complement show that this langer u complement k complement that is of course it is it is a you know it contains identity because k is contained inside you know u complement is contained okay the identity is there but this is not a neighborhood of identity no basic sub basic open this intersection of sub basic open set no basic open set containing identity will be contained inside this one so this is the claim okay okay so what is the choice I have to choose some k and something right one single choice such that this happens is over so what is the choice our choice is k to be this two-third to one intersection three so this is a another cantor subset okay first zero to one-third contains zero which I have thrown away this is two-third to one that part I am taking so that is my k which is a open set so I take that it is a rescue it is compact it is open okay so I can take k equal to u so in this one I am shown that this kc uc kc is not open that is what I will show not neighborhood of identity so for brevity we will just put c0 equal to k complement which is x minus k remember where I am taking the complement here now in the space x okay we now show that this c0 c0 the c0 is k k is also u okay so I what I have what you see here c0 c0 it is not a neighborhood of identity okay the statement is obvious so you so far the only notations and all this I have done so now I have to prove this one okay now the proof starts let k1 k to kr be any finite collection of compact subsets of x let u i be open subset containing k i's one more k0 you take namely k0 equal to u0 equal to c intersection two-third one okay you include this set also we claim that the intersection I range from one to r k i u i this is obviously you know candidate if if this contains identity and this is an open subset all open any open subset which contains will contain an open subset of this nature because these are basic open subset and this is most general I whenever you have chosen like this this subset is not contained in c0 c0 is what I have to show even if one such thing is contained in then I am finished I cannot my claim is wrong so for every k i is chosen like this finitely many compact sets and open subsets like this if I want to show don't worry about k0 why you should have taken the k0 you know I want to show that k i ui itself is not contained inside even after taking the intersection it is not contained in even the smaller set is not contained in larger set will not be contained so that's why I can take this extra thing also this is a special one I have taken but these are general things okay so that will show you that c0 c0 is not a neighborhood of identity so how to construct you have to construct a special function here now which is a homeomorphism of this x c minus 0 which belongs to this one but not here that's all the idea is to construct a homeomorphism from c to c itself shuffling suitable canter sets such that 0 goes to 0 so that if you throw away 0 it is still a homeomorphism of x that's all and this f is on the LHS here but not on the RHS okay now the construction starts look at these finitely many k1 k2 kr which are compact subsets of x okay 0 is not a point of x it is inside the canter set c right so this distance is positive this is the usual metric you know usual distance in R nothing more than that okay this distance is positive choose n such that 1 by 3 power n is less than d okay so that 0 to 1 by 3 power n that interval does not intersect any of this k i that's all okay a small interval around 0 is taken of course 0 will be thrown out afterwards that interval is not intersected by none of this k i that is the important step here note that this k0 whatever you have taken k0 is this this is one of the canter sets right it is a canter subset the k0 is a canter set by above remark remark 4 I told you k0 I take now a equal to k0 and union i rank 1 to r k i so this is not a is that k0 union all these things okay so then a is contained inside that okay a I apply a equal to k0 okay I take a equal to k0 and k0 is i rank 1 to r k i k k0 is there k1 k2 kr may not cover k0 so I am taking k0 also but what are the they are compact subsets so I can apply to the a a a is taken as k0 but I apply this remark 4 there exist a clopon subset b okay which is a copy of canter set such that this b is contained inside k i i rank 1 2 3 up to s and the rest of them it does not intersect this s may be 0 whatever s it will not be 0 because s will have to be contained in a k0 okay so that is the one of the for example suppose k i's were all disjoint from k0 then I can take k equal to this this b equal to k0 itself over okay so b is inside k i so I have put i equal to 0 here very precisely but 1 2 3 up to s they are shuffled k0 is a special one all k1 k2 kr arbitrary you can label them whichever way you like you don't know what I am taking the first s of the s of them okay and labeling them as 1 2 3 up to s s of them what is the property of that they all contain b that's all this s may be just 0 also I don't know this only 0 part okay that matter the rest of them don't intersect k i at all so q suggesting after that's why after re-indexing the set k1 k2 kr now let us do some some more in a bifurcation the whole idea is the subsets of this canter subsets can be written you know disjoint union of so many copies of themselves and so on that is the whole idea this fractality of the canter set is used to be very nicely take c1 equal to 0 to 1 3rd raised to n intersection c okay so this part does not intersect or none of the k i that's what we have chosen this one right this one one third raised to n that has been chosen that way so look at that one call it as c1 you write c1 itself as disjoint union of two canter subsets okay you don't you don't have to worry just write it as disjoint union of two canter subsets because this is canter set how it's how it's the next thing is how it's constructed from this interval you remove the middle one third so you get two of them so take those intersections so c1 1 and c1 2 these can always do that okay whichever way you like doesn't matter now you write k i equal to k i prime union with b because these k i's contain b right what is b b is a canter set therefore it's a clop on okay so I can again write k i as disjoint union of k i a primes which will be also clop on okay because b b is clop on and I am taking the complement here k i a primes but what happens to other things don't have to worry because k i's don't intersect that part okay finally you have b itself is subset of k naught right it's subset of k naught k1 k r so that b itself you write as a b1 union b2 so all these b1 b2 b and c1 c1 1 c1 2 they are all canter subsets so disjoint union of non-empty clop on set okay now we assume that 0 is in the closure of c1 1 remember 0 must be I have taken the whole thing right so 0 is inside c1 1 part you can take so I have actually I should take open interval so c1 it is in one of them so it is in the in the part of c1 that's all I want to say okay 0 is in the c1 1 part okay note that c1 1 c1 2 b1 b2 are all canter subsets and all are clop on sets that's I am repeating here that's all now look at the union of all these 1 2 3 4 that is such a clop on set inside c so the entire c can be written as c1 1 c1 2 b1 b2 disjoint union disjoint union with another L which is also a clop on set alright now the major work is over now I can define my the homeomorphism by merely shuffling these by merely shuffling these things so one thing you have to notice is this L is a complement of all this will contain all the complement of b inside Ki a Ki prime and all those things which does b i b does not intersect namely Ki varying from 1 to s s plus 1 to r so L will contain all of them okay L will be many more other things also all rest of the c is there so now define f from c to c as follows first c1 1 c1 both c1 1 is a subset of c1 right but both of them are canter subsets so take a linear homeomorphism which is order preserving and both of them contain 0 so make 0 go to 0 you could have taken the the other way around so order preserving will automatically imply that f0 okay if I say f0 0 the rest of them has to be order preserving because this is homeomorphism okay so let f from c1 2 to b2 be the order preserving homeomorph, linear homeomorphism both of them are canter sets there is a homeomorphism all that I am assuming so they are all disjoint so I am totally independent in defining these homeomorphisms so f1 this one f1 this one and so on okay then f from L to L being the identity map last b to b1 see b1 is a subset now c1 was subset and I went to larger set here now I have to cover this part so I will take b to b1 be the order preserving linear homeomorphism so since I have covered the entire of c okay because they are all disjointing and they are going to disjointing the image and the domain and image are covered completely so f is a homeomorphism is a continuous bijection and this is a homeomorphism f0 is 0 therefore if you throw away 0 f restricted homeomorphism from x-ray okay now look at three this identity on the on this L and four which says b equal to b1 so these two are important to follow that this f takes Ki to Ki okay these Ki's are fine because where are they they are inside L so that that's identity what about these Ki's these Ki primes are again inside L only the b part you have to see all right the b part is going inside b1 no so this Ki is going inside Ki I am not claimed that f is a homeomorphism here no f of Ki goes into Ki for each I okay and these are contained inside because Ki's are contained inside UI so each Ki Ki is contained is a Ki UI so this intersection is contained inside so f is here so that was the first thing we have to see now finally I have to see that f is not in C0 C0 since f sent C12 which is inside b2 to all the way inside 231 okay remember so this b was subset of 231 here so C12 is somewhere away from this one right so it follows that f is not in C0 C0 okay what is C0 remember C0 was the complement of let me just show you what was C0 to begin with this is C0 is a composite short notation for complement of K okay and my K itself is 231 intersection C right so it has gone out of the complement so it is not going inside C0 at all so therefore this f is does not take C0 inside C0 so the proof is over namely canter set the group of homeomorphisms is not a topological group so there are a couple of home questions that automatically rise in this context namely find a criterion for Hx under co under compact upon topology to be a topological group where you do not try to put anything other than x equal to locally compact order of space the criterion means what suppose this is locally then x must be this if x is something something then this is a that is one way that I have given you namely locally connected locally compact top of this right equivalently you can take one point compactification try to solve the subspace prop is the subspace on Hx coming from Hx star whether that is you know the induced that is the the subspace topology whether that is compact upon topology okay or you can try to solve this problem in the case of sin sin 1 by x and so on interesting other interesting cases like that so there are quite a few problems versus one well that is it so so I told you that these these things were done a few very you know recently about three years back so during a conference during a workshop in characteristic classes in which Shamik Paul attended it and we were sharing a room for some time and that time I was discussing with him then I am thankful for Alain's you know Alain Hatcher I asked him whether he knows he says he doesn't know but maybe he looked it here and so on then I asked Delin Salivan and so on the most important of all my initial attempts were discussed with permission so he said look here you have to be careful and so on permission was he's painfully went through the initial versions with so many typos and so many missiles there but he he saw through the whole thing and made it you know where I should be careful and so on gave up some warnings so I should thank all of them of course I thank you for listening to this one also it's a good opportunity of presenting this one okay so thank you we will meet next time