 Welcome to module 36, Local Connectivity. Last time we studied path connectivity, connectivity, components and also studied one important example of topology sine curve which illustrates that connectivity need not imply path connectivity in particular. Intervals are connectivity, intervals is connected. In fact, the connectivity was thought to explain the intermediate value theorem of an interval and that is the key for path connectivity implying connectivity. So, though we showed that the other way around is not true, there is very much close relation between them. So, now let us look at what happens inside Rn instead of just R. R is too simplistic for further investigation, for deeper analysis of the whole thing. That is why we are going to Rn and get some experience there and then study you know introduce more general concepts and so on. That is the whole idea. So, here is a result on Rn. Take any non-empty open set, you can take empty set but that is not going to make any difference. Non-empty open subset of Rn, it is connected if and only if it is path connected. So, here is a strong result. Connectivity need not imply path connectivity in general but here is a namely open subset of Rn. Connectivity is the same thing as path connectivity. It is not just for intervals, any open subset. If it is connected, it is path connected. One way is we have seen already, path connectivity implies connected. So, we have to see only the other way around here. So, start with an open subset A. We used to show that given any two points of A, there is a path joining them completely inside A, not inside Rn. Rn is no joke, it is very straightforward because whole Rn is a vector space or it is a convex set. Now, A is not given to be a convex set. A is only open set with QL. So, this follows if we show that all points of A can be joined to one chosen point Z0. That Z0 is not very specific. Any Z0 you choose. First of all, any two points can be joined means any point can be joined to Z0 but that is enough we have seen earlier. So, let us form a subset of A, all Z belonging to A such that Z can be joined to Z0 by a path inside A. What we wanted to show is that it is the whole of A. Starting with an open set and fixing a point Z0, we make this subset. But this subset which I have denoted by U, what we want to show is this U is the whole of A. Now, the method of proof is going to be like a principle here. How to use connectivity to show a lot of other results. So, this is the starting of this one which you have not done so far. So, observe this method. We shall show that U is a non-empty open set such that its complement is also open. Then because A is because you start with A connected then you want to show that this U is the whole of A. Because A is connected it will imply that complement of U cannot be non-empty it has to be empty. Otherwise U and U complement will make a separation of A. So, U must be the whole of A. So, what I have to show A this U is non-empty this is open its complement is also open or equivalently you can show that U and its complements are both closed either way. So, we will show that both of them are open. The first thing we observe is Z naught itself is inside U. Because Z naught can be connected to Z naught by the constant look. Therefore, U is non-empty that is easy. Now, suppose Z is inside U then I have to show a small ball open ball around Z is contained inside U. So, that is enough to show that U is open. So, fix a path gamma from A to inside A from Z naught to Z. That is the meaning of Z belongs to you there must be a path. So, you fix it. So, there is a path from Z naught to Z naught now completely inside A. But now A is an open set Z is inside U, but Z is inside A also by definition U is a subset of A. Therefore, you can find a positive delta such that the open disc of radius delta around Z is completely contained inside A. But then every point of W every point W inside this open ball open disc can be certainly joined to Z because center all that you have to do is the line segment Z W. The line segment Z W is completely contained inside B delta Z. So, it is contained inside A. So, gamma comes from Z naught to Z take this line segment that will be a path from Z naught to W. Therefore, W is inside U that just means that B delta of Z the whole of B delta Z is inside U. So, one part is over that U is open. This argument is very similar. It will also show that U is closed or complement of U is open. So, let us just get through it. Suppose, Z is not a point of U. That means it is not complement. Then no point that this B delta Z is the same choice you can do that namely it is contained inside A. Then no point of W no point W of B delta Z could have been joined to whole of Z naught inside A because if that is the case then Z naught to W I can always take the line segment. So, if there is a path from Z naught to W any one point then we could have joined all of them. That means that none of the points in B delta Z are inside U which is same thing as B delta Z is contained in the complement. So, the complement is also open. So, that completes the proof that U is equal to whole of A therefore A is path connected. So, I repeat this process. Suppose you want to show something holds for a whole some subset which you have defined. Then you make a subset of that set which is those points for which the property holds. Luckily if you can show that it is first of all non-empty and open and complement is also open then you have finished the proof that that subset must be the whole of A. So, this is a principle. This is like a meta theorem. It can be used several times. It has been used several times in topology. So, here what we have used what was the key? The key is that the basic open subsets namely the balls they are path connected. Any two points that can be joined the balls are actually convex. So, that is what we have used. As usual what matters for us is that the connectivity of the ball path connectivity of the ball. So, now you see if in a topological space basic open subsets are path connected then the same proof will work namely any connected open set will be path connected. That is our next theorem. Before that perhaps we should make this property as a you know you can name it and that is what is called as local path connectivity. Since we are going to study both connectivity and local path connectivity and path connectivity together maybe you should do the same thing for connectivity also. So, we will have two different definitions here. Start with a topological space we say x is locally connected at a point is again point wise. If for every neighborhood u of x there is a connected open neighborhood of x inside x such that that v is contained inside u. Start with any neighborhood you must be able to get a smaller open neighborhood of the point which is connected that is called locally connected at the point x. For one open one neighborhood if you find it that is not enough for every neighborhood you must be able to find this such a thing smaller one. So, that is the local property. No matter how small or how big you chose the original neighborhood inside that you must be able to find an open connected neighborhood. If you replace connectivity by path connectivity what you get is locally path connected at that point. Just like continuity and other things if such a thing is true for all the points of x then we say the space is locally connected or locally path connected as the case may be. I repeat suppose the space x has the property that for every point x it is locally connected then x is called locally connected. Same thing if at every point it is locally path connected then we say x itself is locally path connected. Okay now as anticipated this was only a name we have given what theorem we have immediately this theorem which is exactly the carbon copy of whatever we did for R n. A space is locally path connected. Locally path connected first of all you find only every path component of every open set is open okay. So, I am going towards characterization of locally path connectivity or locally connectivity. First let us do it for connectivity drop out this path locally connected if and only if every component of every open set is open okay. What have to do? Take an open set look at its connected components all of them are open. How do you show that a connected component is open? Take a point there then you must produce an open subset contained inside that but local connectivity says that at each their point given an any open set he started with an open set remember that. Given any point there is a connected open subset therefore that connected open subset is contained in the component because component is the largest one. So every point has a connected subset which is open right. So, the union of all these open sets with the connected component. So, the each connected component of an open set is also open. The converse is obvious because these connected components themselves you can take it as as neighborhoods okay. Start with any open set. I want to find a neighborhood which is connected and contained in that open set. Take the connected component over okay. Exactly same way the path connectivity also works because now the property that I have used is there is no difference between the conditions okay. Just remove just put the connectivity wherever we have used path connectivity that's all. So, once again I repeat as can be seen immediately it follows that every open subset of a Euclidean space is locally path connected that is property we have used. Now you have given a name clearly locally path connected implies locally connected just like path connectivity implies connectivity. The above theorem in fact contains a proof of the fact that every locally path connected and connected space is path connected okay. Directly I could have proved by using the carbon copy of the proof of a theorem that we proved for Rn but here is a shortcut because I already proved a stronger result here okay. So, let us see how this proves that. You start with a connected space which is locally path connected okay. Therefore it is locally connected. Therefore its components are open but it is a connected space already okay right. So, how do you use that it is locally path connected if all components are open all components are closed also in a partition if the partition members are open they are closed also if they are closed you cannot say they are open because what suppose you are taking one of the members look at all the other members each of them is open. So, union is open therefore this set is closed but A is connected if the connected components are open okay you cannot have more than one component okay. So, that is the proof here understand. So, by this by the way this just doesn't it is stronger than just saying that locally path connected and connected implies path connected. So, okay this is stronger thing every open subset of a path connected space okay has the property that all the components are open similarly for connectivity. So, these two two different statements but use the prefab that path connectivity local or otherwise implies connectivity local or otherwise. So, all that I have to do. So, this is a very strong thing. So, remember this one properly okay if we drop the condition open on we in the definition in the definition of a local connectivity or local path connectivity I insisted let us go back again once again every neighborhood of x there is a connected path connected whatever open neighborhood of b which is connected. So, suppose I drop this openness here every neighbor has a smaller neighborhood is connected that is obviously a weaker form of the local connectivity. So, we will call it as weekly local connected similarly you can make weekly locally path connected whether it is really weak or not that needs to be verified sometimes a weaker condition weakly looking condition may be equivalent. So, this is what I repeat here this is what I have done you drop out the openness of way in the definition of definition given above we get perfectly valid definition of locally connectedness why I am giving this one is some some authors may give you this as a definition but then we be careful our definition is stronger. So, this is called weakly locally connected there are spaces which have points at which the space is weakly locally connected but not locally connected once I give you this example then you know that this concept is really weaker than the original local connectivity. Okay, so we will come to those examples however it turns out that if a space is weakly locally connected at all of its points then it is locally connected in our sense. So, this is why many authors use this condition because they are not defining it to point wise they they are only interested in local connectivity as a global thing so they define it everywhere so they are overlooking this one that is why they are how happy they are not bothered about okay that is all. So, however it turns out that if a space is weakly locally connected all its points then it is locally connected okay this theorem use it cleverly it is not very difficult okay to prove that if something is weakly locally connected at every point then it is actually locally connected all right okay you have to work out that when you have to think about how to prove that it is not difficult only then you understand the difference here see you have a neighborhood open set you have a smaller neighborhood which is connected that neighborhood may not be open so to get an open set you may have to get take a smaller one a smaller one may not be connected you see you take a connected space not all sub spaces are connected otherwise the whole thing would have been you know just you global connectivity how enough okay when you go to smaller set or bigger set connectivity may fail all right so that is why all these things have to be carefully first of all understood the definition should be carefully understood in several topological problems about the plane r cross r local connectivity or absence of it plays a very crucial role we won't have time to study all that especially in the dynamics of sea you know there is a lot of what Julia says Mandelbrom says and so on okay in that these local connectivity and local path connectivity are important notions connected components of a space are closed subsets of the space which we have seen once you take a connected space connected subset its closure is also connected therefore components must be closed further if exists locally connected then connected components are also open this what we have just seen similarly path connected components of a locally path connected space are all open and hence they are closed but I am not saying that without path local path connectivity take any space and take a path connected component it may not be closed so you have to be careful about that so we have we have already seen an example of this in the topology sine curve so let us just go through the proof of this one we have seen that each connected component c of x is closed right take connected component take the closure of that that is also connected but it is larger no it should not be larger because c is a connected component so therefore c equal to c bar now assume that x is locally connected then for each x belong to c you can find a neighborhood open neighborhood which is connected therefore every point has a open set contained inside c right actually why it should be contained inside c every point you have we have seen that I am proving it here again c union u will be connected right but c is contained in this union but c is maximal so they are equal so you have already proved it but I have I have just put the proof here again the same thing works for path connectivity also if c is path connected and u is path connected and the intersection has a one point there then the whole thing is also path connected so that that part is the same thing locally path connected or locally connected implies the components are open okay so next time we will see more and more examples especially to bring out the difference between local path connectivity and local connectivity thank you