 So, welcome to this lecture. So, in the previous lecture we had introduced the notion of path connectedness and we define an equivalence relation on x and we ended the lecture by stating this proposition. So, let us quickly see a proof of this proposition it is really easy. So, first we want to show that every path connected space of x is contained in some x i right. So, let t be a path connected subspace of x right and so, let x be in t then x belongs to x i for some i right. Now, we clean that t is contained in x i right to show that t is contained in x i it suffices to show that. So, for any it suffices to show for any t in t there is a path x joining x and t right, but as t is path connected this implies that there is a continuous path t says that gamma 0 is equal to x and gamma 1 is equal to t. But then we can just take the inclusion z x which is continuous and therefore, the composite of these is going to be continuous. So, this implies that t and x are in the same. So, this implies that t contains is contained x x i. So, this proves one. So, we have shown that every path connected subspace of x is contained in some x i and proof 2 this also obvious right. So, let x be in x i. So, we need to show that any 2 points in x i can be joined by a path. So, this is clear from ok. So, since x if x and y are any 2 points in x i then by the definition of yeah. So, this is clear from the definition of path components. So, both these together prove that each x i is a maximal path connected subspace of x ok. So, recall that we are a similar proposition when we were talking about connected components, but that proposition had third point which said that every connected component is closed that is not true of a path components. So, let us see a counter example. So, a space which is, but not path connected. So, first consider this. So, let us make a picture of this space. So, this is 1 and let us take half and this is 1 by 3, this is 1 by 4 and this is 1 by n right. So, at each of these 1 by n's we make a straight line of let us say length 1 ok. So, we remove the origin right. So, let us just look at what the space C is. C is a subset of R 2 right and it consists of points 1 by n comma y. So, in the x coordinate we have 1 by n and y where n is a natural number and y is in the interval 0 1 this. So, this will contain all these lines. So, then we also need to take this region. So, this is we delete the origin, but we take the rest. So, let me just write what this is this is the points 0 x coordinate is 0 and y coordinate is can be any y where y is in the half open interval 0 1 ok. So, in particular this point p is 0 comma 1 ok and then we have to take the. So, this is this line and we also have to take the x axis. So, that is or part of the x axis which is union x comma 0. So, the y coordinate is 0 over here over here and where x we have to remove the origin and we can take any x ok. So, this region in black. So, let us say I just of the same height 1. So, this is our C ok. So, this is our. So, let us see the subspace topology from R 2 right. So, R 2 has the metric the standard topology which is the same as the 1 given by the standard Euclidean metric. Now, let gamma from 0 1 to C be a path such that gamma of 0 is equal to this point 0 comma 1 which is this point p. So, then the claim is the image of gamma is completely contained inside. So, what is y? Why is this subspace that is why is this line? Why is this line over here? So, the starting point of gamma is this point p and the claim is if we take any continuous map from 0 1 to C which starts at this point p then it cannot go outside the subspace y. So, let us prove the claim. So, let us assume. So, let us assume this is not true right and that the image that is the image of gamma moves out of. So, note that y is a close subspace of C right. In fact, so C is contained in R 2 and it has a subspace topology from R 2 we have the projection to y projection to ok. So, this is the second projection projection to the second coordinate and the subspace y is exactly equal to. So, let us say this is i. So, i compose p 2. So, p 2 is continuous the projection maps are continuous and the inclusion is continuous because C has a subspace topology and therefore, the composite is continuous and y is exactly the inverse image of 0 in C right. So, therefore, y is a close subspace of C right. So, now, we define this set S to be equal to is those x in 0 comma 1 such that the image of 0 comma x is completely contained inside y. So, let t naught be the supremum over. So, S is a subspace of before you write that S is contained in 0 1 right and clearly S is non-empty as 0 belongs to S because gamma of 0 comma this interval which is just gamma of 0 which is equal to p this belongs to y ok. So, then let t naught be the supremum over all S in S ok. So, we claim that we first claim that gamma of t naught is in y. So, why is that? So, since t naught is the supremum. So, this implies there is a sequence t n's in S such that t n's converge to t naught right. Now, this as gamma is continuous this implies gamma of t n converges to gamma of t naught, but note that as t n belongs to S gamma of 0 t n is contained in y which in particular implies that gamma of t n belongs to y right and since y is closed every time we have a sequence of points inside y which converges some point. So, it will mean that limit point is contained in y. So, this implies that gamma of t naught is in y ok. So, if t naught is equal to 1 then what is this mean? Then this implies that gamma of 0 comma 1 is contained in y which is what we wanted to prove. So, then we are done. So, let us assume that this is not the case right. So, let us assume that t naught is strictly less than 1. So, we have this interval 0 1 and t naught is somewhere over here ok and copy this this is our gamma right. So, gamma t naught is contained in y right. So, this is to this point is gamma t naught right. We just showed that gamma of t naught is contained in y. So, this is our y. So, we take a small neighborhood around gamma of t naught such that u misses the u does not contain the origin. So, we take us. So, u is let us say a small neighborhood in r 2. So, we could for instance take a small square like this. So, then gamma inverse. So, gamma is a map from. So, gamma is a map from 0 1 to c right which is contained in r 2. So, let us just view gamma as a map from the gamma as a map from 0 1 to r 2 is therefore, continuous because c has a surface topology right and therefore, the inclusion is continuous. So, we can just take. So, gamma inverse of u let us viewing gamma as a map from 0 1 to r 2 is an open subset open in 0 1 right and so it contains the interval t naught comma t naught plus epsilon for some epsilon positive. So, now note that as t naught is the supremum of elements in s the image gamma of 0 comma t naught this half open interval is contained in y. Moreover, we have proved that gamma of t naught is also in y. So, this implies that image of this closed interval is contained in y, but now if we take any delta. So, if delta is strictly greater than t naught then gamma of 0 comma delta cannot be contained in y otherwise or else t naught will not be the supremum of s right. So, thus there exists delta in this half open interval in this open interval such that gamma of delta does not belong to y. So, now we restrict gamma to this t naught comma delta right. So, this restriction is mapping to c, but the image actually lands inside u intersection c which is contained inside yeah. So, therefore, we get that t naught comma delta to u intersection c is continuous. So, now, let us make a picture of u intersection c. So, u intersection c it looks something like this. So, this is u this is gamma of t naught and there are all these lines at 1 upon n which are converging at x equal to 1 upon n which are converging to x equal to 0 ok. Gamma of delta is somewhere over here as. So, gamma of delta is not in y. So, it lands outside and it also lands inside u. So, therefore, it is of it is of the type 1 upon n comma y for some right, but now we can find two open subsets non-empty open subsets in u intersection v. So, let us say this is v 1 and let us say this is v 2 this v 1 right. So, then so u intersection c we can write it as v 1 intersected c this joint union v 2 intersected c ok. So, now gamma from t naught comma delta. So, this contains gamma of t naught and this contains gamma of delta. So, gamma is from 2 u intersection c is continuous. So, this implies that t naught comma delta can be written as a disjoint union of gamma inverse v 1 intersected c inverse of v 2 intersected c right, but this is the contradiction as t naught comma delta is connected. So, this could contain t naught and this would contain delta right. So, that is not possible. So, thus t naught has to be equal to 1 and this implies gamma of this entire interval is contained in y right. So, what does this mean? So, this means that if we take a path which starts at this point y then it has to at this point p then it has to remain inside y it cannot move outside y right. So, therefore, so if this point is if I take this point this is 1 comma 1 right. So, this shows that so this shows that there is no continuous path from 0 1 to c which joins these points 0 1 and 1 comma 1 ok and so right. So, this implies that c is naught. So, on the other hand so it is clear that c minus y is path connected. Why is that? Because if we take any point in c minus y. So, we have deleted this we have deleted y. So, any point any two points will look like. So, two points will look over here. So, we can first come from here to the x axis over here this point and then we can travel to this point and then we can go up. So, we cannot do this if our point is on y because the origin has been left out. So, we cannot pass through the origin. So, it is clear that c minus y is path connected. So, this implies that c minus y is connected and so this implies that the closure of c minus y in c is also connected right. Because if we take any subspace a then its closure is also connected yeah, but the closure of c minus y c minus y in c is exactly is all of c. Because if we take any point on y we can find a sequence with the same y coordinate and this sequence we can find it which converges to y which converges to this point q. So, this shows that c has two path components namely y and c minus y and just one connected component this one connected component c right. And this also shows that since c minus y is not closed in c this implies path components need not be closed ok. So, we will end here. Thank you.