 Welcome to module 37, last time we introduced the notion of local connectivity and local path connectivity and proved a few results there. So today we will examine a number of illustrative examples, either they are concrete examples or they are positive examples and so on. The first example is what you have already met, okay, it is very easy to state which will be locally connected but not locally path connected, this is very easy to state namely the co-finite topology on say take set of integers or natural numbers and any countably infinite set, okay, countability is important, infinite also important, okay. If you do not take infinite, finite set co-finite topology with discrete topology, discrete topology with more than one point is neither connected nor path connected but surprisingly because points are open this is both locally path connected as well as path connected, locally path connected so I am not interested in discrete topology, alright. So the first thing is every non-empty open set will intersect every other non-empty open set in the co-finite topology because complements are finite, right. So a finite set cannot contain another set which is complement is finite that is not possible when the whole thing is infinite. So any two open sets intersect therefore there is no separation, okay that means that the space is connected similarly it is locally connected also because every open set will be also the same property you cannot separate it out, you take any open set that itself will serve as a connected neighborhood inside that you have to take connect it is already connected, right. The subspace topology on a open set is again the co-finite topology only and it is uncountable sorry it is countable and infinite therefore the space is locally connected and connected. However we claim that every continuous function omega from closed interval 01 into the space is a constant once we prove that it follows that two distinct points cannot be connected by a path, okay. So we are going to prove strongly that the path connected components of the space are singletons, okay strongly path disconnected in that sense, okay. So let us see how we prove this one. Suppose you have a path which is non-constant that means what the image should contain at least two points, okay. Every point in X is a closed point singleton X is closed why because the complement is open that is all, okay. You could follow that the closed interval 01 is a countable, distinct union of proper closed sets what are those closed sets just take omega inverse of singletons. Closed set singletons are closed, omega is continuous so omega inverse is closed subset. As X range is over its image of omega Y must be image of omega contained inside X take the disjoint union these are the fibers after all so they are disjoint, right. Disjoint union each of them is closed what is Y, Y is omega of 01 the image the points are varying over Y, okay there are at least two of them, okay 01 has been written as disjoint union of closed sets, okay. One thing is clear namely this Y cannot be finite Y if this Y is finite this will give you a partition of 01 but 01 is connected so there is no partition, okay there is no separation this will give you a separation so that is not possible, alright because once if there are finitely many you take one of them everything is closed the finite union of other things are also closed so you can write a disjoint union of two non-empty closed subsets then that is over but if it is infinite then there is no contradiction, okay so 01 may be connected but you can always write it as disjoint union of singleton sets for example singleton sets are closed so that is not a contradiction, okay. So what we are going to prove is somehow I want to get a contradiction to such a description the first thing is I notice Y has to be infinite but what I am going to prove is the following a general statement no closed interval a, b less than b is a countable disjoint union of proper closed sets Y being a subset of X X being countable this Y will be countable okay therefore immediately this description is not possible is a contradiction when I take a, b to be the interval 01 so this is a general statement I am going to prove in this way in this one I am assuming that countable means finite also which we have seen that finiteness is not possible that is easy because all intervals are connected so the point is now countable is not possible is what we have to see, okay. So here we will have to use Bayer's category theorem in a clever way so that gives you an opportunity to use Bayer's category theorem. So watch out suppose you have a, b equal to disjoint union of f n's where each f n is a closed proper subset of a, b okay you see if f n's are empty and one of them is a, b then there is no contradiction so they must be proper that is important proper means what non-empty as well as not the whole space take d n equal to the boundary of f n by this I mean the boundary points of f n inside a, b they are all subspecies of a, b now the usual topology of a, b take d n as boundary of f n and take this d equal to union of all these d n's okay note that since a, b is connected and each f n is a closed subset of k, b no f n is open if f n is open what happens if one of them f n is open what happens it is also closed I have assumed right each of f n is closed to begin with so proper nonempty closed and open subset will contradict so since they are closed they are not open it is a consequence okay so when you take interior, interior is a strictly smaller subset than f n when you throw away the interior from f n what you get is the boundary because f n's are already closed I do not have to take the closure remember what is the boundary point boundary points are closure minus the interior so closure is f n itself interior is empty sorry interior is not the whole space so interior may not be empty but you throw away the interior d n's boundary of f n's are non-empty, boundary of f n's would have been empty if f n is open also therefore each d n is non-empty boundary of a subset is always a closed subset of the set the set is closed so these are all closed subsets of it okay d n's are non-empty closed subsets okay it also follows that d n is infinite but we are not actually using this fact okay d n's are infinite means what each of them I have proved that they are non-empty and to begin with I have taken these things are infinite we should this is finite then we already know that the case is over okay but this fact we are not going to use okay right now our argument will include this also anyway so each d n is non-empty closed subset right but now we have lost because these d n's do not cover a b right the f n's are covering a b now I have taken smaller subsets so what do I do with them take d which is equal to union of all d n's right how do you get d n's by deleting the interiors of each f n right therefore this a b minus interior of all f n's is precisely equal to d okay therefore this is a closed subset of a b therefore it is a complete metric space now using Baier's category theorem we will arrive at a contradiction how do we use Baier's category theorem namely if we show that each d n has empty interior inside d okay then the proof will be over this is very important because d n's are boundary of f n I have removed the all the interior points there so you may say it is over no no these interior points were inside a b now I have a different subspace we have this subspace namely d I am applying the Baier's category theorem to this complete metric space d so I have to show that each d n has empty interior inside d okay then the Baier's category theorem says that countable union of such things again no evidence they cannot be the whole of d okay so that is the end of the proof right so let us go ahead how to show that d n has empty interior okay take any point in d n show that no by the way what is the topology topology it is coming from r right from a b everything is substitutes of r right so take a open interval around a point intersect it with d if it is contained inside d n that will be the interior okay take an interval intersect it with d show that that is not contained inside d n then it will show if you follow that interior of d n is empty so this should be have to do for every point of d n what is d n boundary of f n take a point and take an interval around that by the definition of boundary okay j must boundary now I am using f n point of d n is a point of f n also but it is a boundary point of f n first you use that property that d n is a boundary of f n okay by the definition of boundary j must intersect both f n and complement of f n right so it intersects the complement of f n where inside the the closed interval a b okay hence must contain some point of some other f m because the entire a b is written as disjoint union of f n's if a subset is not contained inside one particular f n it must intersect some other f m m not equal to n okay now suppose j does not intersect the d m boundary of f m d m is boundary of f m okay that means that it intersects the interior of f m all right it intersects f m it doesn't intersect the boundary so what is left out it must be interior of f m but then what happens interior of f m this interior is taken inside inside the closed interval a b remember that interior of f m is an open set complement of f m is also an open set if you take j intersection this and j intersection that that will be the whole of j intersection a b I have not written a b here because I am taking the interval j itself inside the interval a b okay so therefore this will this will give you a separation of j and that is a contradiction therefore this j I said j must intersect the d m the boundary of m okay so what happened I started with an interval around a point in d n that intersects d m m not equal to n but these are disjoint sets right therefore this j is not contained inside d n so there is a point d after all all of d m is also contained inside d so j intersection d there is a point here it is not inside d n this shows that d n has empty interior inside d and that proves that no closed interval can be written as a countable union of disjoint you know closed subsets proper closed subsets in particular the co-finite topology on a countable set is what happened is not path connected every path component is a single term let me give you some other examples now this is called the broom space let me show you the picture here in the next plane so I start with a closed interval so let us call this as 0 to 1 okay here I take the line segment between 0 0 and 1 comma 0 in the plane so this is 1 by 2 sorry this is 1 comma 0 what 0 you know it is 1 comma 1 this is 1 comma 0 so 0 0 and this is 1 comma 1 this is 1 by 1 comma 1 by 2 this is 1 comma 1 by 3 this is 1 comma 1 by 4 and so on dot dot dot dot dot so 1 comma 1 by n approaching 1 comma 0 here all those points join to 0 0 so it looks like a broom this is the broom space this is the closed subset of r cross r it is closed and bounded also okay it whole thing is contained inside the square i cross i clearly it is star shaped at this point right take any point there is a line segment to this point therefore this is path connected therefore this is connected so why this example is there for the reason that you know take a point here on the x axis 0 to 1 so let us take this itself namely 1 comma 0 if you take any neighborhood of that take a ball around that and intersect with this one you will have lots of these line segment disjoint sets no matter how small this ball you are taking okay it is going to be intersection with this space will be disconnected infinitely many line segments will be there along with a small line segment on the x axis so at this point it is not locally connected or locally path connected the same thing holds for all the points here except the point 0 0 at 0 0 it is clear if you take any neighborhood then you can take a small ball around that then it will be star shaped again so there is no problem okay so that is a broom space here so it is path connected it is not locally path connected at any point actually on the x axis except 0 0 at least 1 0 it is easy all right so let us go to slightly more complicatedness namely what i do i take a small copy of the broom space here again i have not drawn all these lines here they are there inside okay so this is a broom space b so instead of 0 to 1 i have made it 0 to half here okay this is 0 0 this is half 0 this is 3 4 0 and so on take half of each time take the side also like this also make it half keep making half half like that so this will go add infinity term infinitely many of these things will be there okay along with this just point will be there last one there is nothing there left out if you come slightly out of that point you will have lots of brooms here infinitely many brooms okay so this is again the whole thing is contained inside i cross i okay and it is a close subspace remember i have not drawn these there are from 1 by 2 1 this is 1 this is 1 by 2 comma 1 by 2 you can join them you have joined them all those things are there so this is iterated broom space so here i have defined it denote the broom space defined above by b in the first one scale it down by a factor of 1 by 2 power n and shift it at the point 2 power n minus 1 minus 1 by 2 power n minus 1 comma 0 so all these points approaching 1 comma 0 to obtain b n let b twiddle be the point 0 1 comma 0 has to be there okay you take that one union all the b n's okay now show that every open connected subset of b twiddle which contains 1 comma 0 contains the entire segment 0 1 comma 0 okay the entire line segment the x axis x axis part however show that every neighborhood of 1 comma 0 contains a connected neighborhood and hence b twiddle is weakly locally connected at the point 1 comma 0 but not locally connected at the point 1 comma 0 if you want an open connected set you will have a problem if you do not want open sub connected set just a connected subset they are there this is the whole idea so let us try to see this one at least some part after all I have left this as an exercise to you but let me just explain so first part is to show that every open connected subset b twiddle which contains 1 comma 0 contains the whole line segment therefore it cannot be arbitrary small that is all you cannot have a connected open subset which arbitrary small if you show that one so how do you show that go to this picture I want to take an open connected subset so what you have to do take a open ball around this one which you have taken open ball around this one right intersecting this space as soon as you take an open ball around this one infinitely many of these triangles or these booms are already contained inside this one okay nevertheless what happens because it is an open thing okay suppose you have come up till here there will be always a line from here to here half wing so this is precisely what I have at some point okay suppose this point is there then a small neighborhood of that point must be there in your open set then all the line segments should be there so in order that those line segments are connected you have to go to the base of this you have to go to the base of that broom that means suppose once you have suppose this one is 2 power n minus 1 minus 1 divided by 2 power n minus 1 somewhere here this point is there means there is small neighborhood around that one is there then there will be line from here to this point therefore this point is there okay so you just keep that this whole line segment will be there because it is connected there is no other way to connect this point and this point now once this point is there a small neighborhood around that will be there because I am assuming openness open connectedness then there will be some line here some point here and the line all the way up to this one so this way you will end up here this entire line segment will have to be there in any connected open subset containing any of these points in particular 1.0 okay you will have to keep backwards forward you do not have to worry so if we take a neighborhood here you do not go forward but you go all the way here that is so for 1.0 the whole line segment will be there all right so connected open set is a problem on the other hand take a point here then 1.0 take a neighborhood now do not go backwards look at wherever you have stopped up till here okay that will be connected neighborhood that is not an open set you just take the broom only do not take all these points up to broom you take okay everything forward so that will be connected never it is a neighborhood because a smaller open subset will be there a smaller open subset will be there which is not connected but disconnected and it is contained in the original neighborhood therefore this is low weakly locally connected at this point okay and not locally connected all right I come back to the topology sine curve now it will be very easy but we will have to just sum it up the topology sine curve we see that it is connected but not connected that we have seen now take any point p on the y axis part okay x coordinate 0 y coordinate is some point p and take epsilon to be less than half just for being safe then the ball B epsilon p intersect with the topology sine curve x consists of infinitely many disjoint arcs on the graph for the same reason that we have discussed for the broom space what happens is the topology sine curve is not connected at any point locally connected at any point on the y axis okay let us have a picture of this one first here so on the y axis here to take any ball around here it should not ball should not be such that it contains the the line segment here and line segment here at at y equal to 1 and y equal to minus 1 so I take the ball of say epsilon epsilon by 2 sorry 1 by 2 less than epsilon 1 by less than 1 by 2 take a ball here let us magnify it okay so magnify it and look at it what you look at it this may be the your point and that means epsilon ball you will let all these curves okay part of the curves what curve sine pi by x curve going and going like this I have drawn this one thick line this is not a thick line this is just a line there are infinitely many lines like this infinitely many arcs like this coming that's why it looks thick that's all if you may if you again magnify this taking a small ball here it will have the same picture again there will be infinitely many components there so this is locally not even connected so it is not locally path connected either for every point on the x y axis so it is similar to the broom space there okay but it has this property also is what we wanted to tell you that thus let us consolidate what are the things that we have done for topology sine curve it is neither locally connected nor locally path connected just now we saw this topology sine curve serves as an example or you can say counter example to show that connectedness does not imply path connectedness local path connectedness or even local connectedness because it's a connected space but it is neither of any of these things closure of a path connected set need not be path connected because the sine part sine graph graph of sine sine x part is connected part its closure is the whole space and that is not path connected so finally analog of 3.34 is not valid for path connectedness what was nl 3.4 3.4 that that if the bottom space is connected the fibers are connected of a quotient map then the top space is connected so if you replace path connected everywhere if you if you connectedness is replaced by path connectivity the theorem as such is false so that is the meaning of this one so all these things we have seen okay now I will come to another important part here maybe that is the last part for today how to use connectivity positively to derive some some interesting results okay so so here is something look at the the intersection of I mean union of the two axis x axis and y axis if you are bothered about the two big and so on cut it off mod x less than equal to y mod y less than equal to 1 mod x less than equal to 1 mod y less than equal to 1 okay so this is just x axis and y axis minus 1 to plus 1 minus 1 to plus 1 you can feel that this space is not homeomorphic to an interval you know this also compact matrix space you also compact matrix space so whatever topological property that you have seen so far will not be of any use other than connectivity both of them are connected also both of them are path connected okay so what is that how to use connectivity to show that it is not homeomorphic to a closed interval minus 1 plus 1 minus 1 plus 1 I take because yes to mod you cannot take any closed interval they are all homeomorphic to each other but they are not homeomorphic to this space x okay so how to see that suppose you have a homeomorphism from here to here now take a point here take its image we may be zero one point remove both of them from from their spaces namely from the domain you remove this point co-domain you wish me its its image remove that then the homeomorphism whatever call let us call restrict to the subspaces that will be homeomorphism again a homeomorphism from x to i restricts to a homeomorphism from a to f a for any subspace f a okay so that is easy to see that is what I am going to use here my a will be minus 1 plus 1 minus 0 then that a we know is disconnected as two components right right if I go here will it always have two components that is the point right so what I want to do immediately I see one way nice way of looking at it namely remove 0 comma 0 from here the point of intersection of the two lines immediately you see that there are four path connected components for the complement so take a homeomorphism this way other way around f x2 minus 1 plus 1 remove the origin here remove the image of that here image of that maybe any point I don't know which point I am just remove one point from minus 1 to plus 1 what do I get if the point I removed is the end points then the space is still connected path connected if it happens to be some somewhere between namely minus 1 less than f x less than 1 then it will have two components but here I have four components a homeomorphism has to induce what by ejectivity of the connected components by ejectivity of path connected components so what is wrong namely our assumption is wrong that there is a homeomorphism from x to minus 1 plus 1 okay so that is the gist of this one x minus 0 0 to minus 1 plus 1 minus f of 0 0 that will be homeomorphism this has four components this has at most two components some case it may have one component so that is a contradiction okay now I have listed the number of exercises one of the exercise here x 3.55 for example is directly from this one on the last example that I gave you what did you say take take English capital letter okay as subspace is of your plane namely r cross m n z look at all this I claim that they are all homeomorphic to each other in fact they are all homeomorphic to an interval easy to see that right you can straighten out these things that is the meaning of after topology homeomorphism but none of them is homeomorphic to the letter o can you see that why if there is a homeomorphism here you remove a point here from here okay let me point from here what happens that is still connected right but here if the point has to be removed only from the interior so what you should do start with the homeomorphism here to here take an interior point here remove it there will be two components no matter what point you remove here it has only one component from the circle from the shape oh the shape oh you can think of a circle it is homeomorph circle remove one point is still connected so none of them is homeomorphism so this is another example I am giving you here so using this idea what I am telling you is classify all letters that 20 26 letters up to homeomorphism okay enjoy this exercise there are other things which you can have a look at it okay so this exercise for example is about matrix spaces and part of it most of them are from linear algebra and little bit of analysis but last one is gln son etc whatever you have met last time they are all locally path connected spaces the previous exercise told you that they are connected therefore they will be path connected so the exercise here is to show that they are locally path connected the challenge so all these exercise earlier exercises will help you to solve that thank you so let us close it here