 So, in the previous lecture we introduced a notion of connectedness and one of the most basic spaces is the interval are the closed intervals a comma b inside r. So, today we will prove that these are connected. So, let us just begin. So, proposition 0 1 is connected. So, proof. So, let us assume that 0 1 is not connected right. So, then there exists non-empty open sets u and b such that these are also disjoint 0 1 is the disjoint union u disjoint union v right ok. So, we may assume. So, one of these contains 0. So, we may assume that it is in u 0 belongs to u. So, let us consider the set s this equal to those x in 0 1 such that this entire interval 0 x is continued right. So, if we were to make a picture. So, this is our interval 0 1 or u may be some combination of open sets and v is also some open subset. So, z u and v are disjoint right. So, we are assuming that. So, let us say u is it could be a complicated open set we do not know what it looks like, but it contains 0 right. So, this x this s is defined as the set of all those x is such that this entire closed interval 0 x is continued u ok. So, clearly s is non-empty because 0 is in s right because as the closed interval 0 comma 0 this just equal to singled in 0 this is definitely in u. Now, let a be the supremum. So, we take the supremum over all elements of a yeah. So, we claim that a is in u. So, by the definition of supremum right by the definition of supremum there is a sequence of x n's in x such that x n's converge to x x n's converge to a sorry. So, let us call this a n. So, now as u as v is open as v is open in this interval 0 1 and u is the complement of v this implies that u is closed in 0 1 right. And since u is closed and each a n belongs to u each a n belongs to u because a n is in s implies 0 comma a n is subset of u in particular a n also belongs to u right. So, therefore, and a n's converge to. So, u is closed a n belongs to u and a n's converge to a these together imply that a also belongs to u. So, a we have proved that a is in u. So, now note that as a n is in s this implies this interval 0 a n is u by the definition which implies when we take union n written equal to 1 0 a n this is also continue ok. So, now observe that. So, if we make a picture. So, this is a point a and we have the sequence of a n's a 1 a 2 they do not have to increase necessarily, but they will converge to a they are going to get closer and closer to a. So, now if you take any point y which is strictly less than a right. So, then we can choose an epsilon neighborhood around a which does not contain y right and for all n sufficiently large the a n's will be contained in this epsilon neighborhood. So, therefore, this implies that for any y in 0 comma a any y which is greater than equal to 0 and strictly less than a we will have that y belongs to 0 comma a n for n sufficiently large. So, this shows that 0 comma a this half open interval is contained in u and plus we have proved that a is as we have already proved that a is in u right. So, this implies that the interval 0 comma a this closed interval is contained. So, now from now we claim that a has to be equal to 1 if a is not equal to 1. So, if a is strictly less than 1 right. So, once again we have 0 we have 1 here and let us say a is somewhere over here then there is epsilon positive such that the interval a minus epsilon comma a plus epsilon intersected with 0 1. So, a minus epsilon intersected with 0 1 this is contained in u ok. So, we can choose a plus epsilon such that a plus epsilon is strictly less than 1 we can do this as a is strictly less than 1. So, this implies that a comma this this half open interval is contained in u which implies that 0 comma a we already know is contained in u. So, therefore, the unit epsilon is contained in u. So, which implies that 0 comma a plus epsilon is contained in u which implies that. So, this is epsilon by 2 this interval a plus epsilon by 2 is completely contained in u ok. So, but this contradicts the fact that. So, this implies that a plus epsilon by 2 is in S right, but this contradicts the fact that a is the supremum of elements in S. So, therefore, thus a is forced to be 1 right in particular. So, this implies that this entire interval 0 1 is contained in u which contradicts non-emptiness right thus 0 comma 1 cannot be disconnected. So, this implies that 0 comma 1 is connected. So, this completes the proof. So, let us make a remark. So, a slight modification of the above proof shows that when a and b are two real numbers such that a is strictly less than b, then the closed interval a comma b is connected. So, we just have to imitate the proof and that is easy. So, this is left as an exercise ok, but we can do this in a simpler way which we will just which we will see immediately. So, non-immediately a little later. So, as a corollary of this let us show that R with the standard topology is connected. So, proof then we can write R as the disjoint in a two open sets where u and v disjoint non-empty open subsets. So, we can choose any a in u and b in b and without loss of generality may assume a is strictly less. So, then intersecting this relation R is equal to u disjoint union v with the interval a b this implies a b intersected with u disjoint union a b intersected with v. So, as so both sets are non-empty as a is contained in a b intersection u and b is contained in a b intersection b right. So, this gives this shows that a b is disconnected the interval a b is disconnected which is a contradiction right because both these are open subsets in the subspace topology on a b and here a b has a subspace topology. So, therefore, we have written the interval a b as a disjoint union of two non-empty open subsets. So, which contradicts the fact that a b is disconnected. So, as a corollary of this so, let us try to understand what are all the connected subsets of R. So, let y be a non-empty connected subspace of R. So, then y is one of the following right. So, basically it is an interval. So, so a comma b are real numbers with a strictly less and b either this or a comma infinity. So, let us prove this. So, the idea is simple. So, we just let we let a to be we define a to be infimum y in y y and b to be supremum right. So, here we have a real line and our y is some subset. So, we take the infimum of infimum and supremum of all elements in y right. So, then a is less than equal to b. So, note that a and b they are allowed to be plus minus infinity. So, if a is equal to b so, this clearly implies that y is just the single sense that set a right. So, I leave this as an exercise right and clearly in this case and equal to. So, if a is strictly less and b then we claim that the open interval a comma b is contained in y. So, if not then there exists c which is contained in this open interval such that c does not belong to y right. So, a is over here b is over here and there is some c such that c is not in y, but then this will imply that we can write y as y intersection minus infinity comma c is joint union y intersection c comma infinity. So, let us check that this is non-empty right. So, as a is the infimum of elements in y there is a sequence converging sequence in y which converges to a and since a is less than c. So, there will be elements of the sequence in the interval elements in the sequence which are strictly less than c. So, this shows that this is non-empty and similarly this is non-empty right because b is the supremum of elements in y. So, there will be a sequence of elements converging to b and since b is strictly greater than c. So, almost all members of the sequence will be greater than c right. So, thus this gives a contradiction. So, this contradicts the assumption because we have written y as a disjoint union of non-empty open subsets. So, thus the interval a comma b is contained in right y has the subspace topology and in the subspace topology both these sets are open and also notice that y is contained in a comma b right as a is equal to infimum of elements in y y and b is equal to supremum right thus we have a comma b is contained in y is contained. So, this part is relevant only if a and b are finite and from this. So, from this we can easily conclude ok. So, this is left as an exercise. This result it gives us a description of the of the connected subsets of the real line ok. So, next we want to see a very useful proposition which talks about how the interaction of connectedness and continuous maps. So, proposition let f from x to y be a continuous map. So, assume x is connected. So, then f of x which has a subspace topology from y is connected ok. So, the proof is easy if if not then there exists open subsets u and v such that f of x is the disjoint union of f of x intersected u ok. So, these are open subsets in y u and v are open subsets in y. So, let me emphasize that u intersection v need not be empty. So, what is given we are assuming that f of x is disconnected which means there are two open subsets in f of x, but every open subset of f of x is of the type f of x intersection intersected with u. So, therefore, we can we can only say that f of x intersected with u and f of x intersected with v these two open subsets of f of x are disjoint. So, this is just a word of caution ok. So, and both these f of x intersection u and f of x intersected with v are non empty ok. So, note that this implies that we can easily check that x is equal to f inverse of u disjoint union f inverse of v. So, this is an easy check. So, moreover as f of x intersected u is non empty this implies f inverse u is non empty and similarly f inverse v is non empty. So, as f is continuous both f inverse u and f inverse v are continue are open yeah. So, thus we have written x as disjoint union of non empty subsets, but this contradicts the connectedness. So, thus f of x is connected in the subspace. So, we will see several useful applications of this proposition that we prove in. So, we will end this lecture here.