 So, in the previous lecture we had introduced metric spaces and the last result we had proved as the following lemma. So, we introduced metric spaces and we saw how given a metric space we can put a topology on it and we proved the following lemma we proved that if a is a subset of x which is a metric space. So, then x belongs to a closure if an relief there is a sequence x n's in a with x n's converging to x. So, using this lemma we will give a convenient criterion for a subset in a metric space to be closed. So, let x be a metric space a contain x be a subset then a is closed if an relief it has the following property. So, if x n's is a sequence in a and x n's converge to x then x is also. So, let us prove this lemma. So, recall we had proved that a is closed in a topological space if an relief a is equal to a closure right. So, we will use this criterion yeah. So, first let us assume that. So, it is enough to show that a is equal to a closure if an relief x n belongs to a and x n converges to x then x belongs to a. So, first assume that a is equal to a closure right. So, then we have to show that the set a it has this property. So, let us see this. So, if x n's are in a and x n converges to x then by the lemma we proved in the previous lecture x belongs to a closure right and we are assuming that a is equal to a closure. So, this implies that x belongs to a thus x has this property. So, conversely suppose x has suppose a has this property this property has a property that every time we have x n's in a and x n converge to x then x is in a right. So, we assume that a has this property and we want to show that a is equal to a closure. So, let x be an element of a closure it is obvious that a is contained in a closure. So, we just have to prove the converse. So, we take an element x in a closure. So, by this lemma by the lemma by this lemma that we proved in the previous class there is a sequence x n's in a such that x n's converge to x right. So, by this property that a has every time we have such a sequence then x belongs to a right because of this property. So, we have a sequence x n's in a and x n converge to x. So, we have x belongs to a yeah. So, this implies that. So, this implies that x belongs to a thus a is equal to a closure right. So, thus a is closed. So, to check that a subset is closed we just need to check that every time a sequence of points in that subset a x n's x n's converge to x then we just have to check that a this point x is in is in a to check the continuity of a map between metric spaces. So, that is this theorem. So, let x and y be metric spaces f from x to y be a map of sets ok. So, then f is continuous if and only if for every sequence x n's converging to x we have f of x n converge to f of x. So, this sequence is in x and this sequence is in y right. So, this gives us a way to check continuity of maps between metric spaces right. So, let us prove this. So, let us assume that we first assume that x f is continuous. So, for every and let x n converging to x be a sequence for epsilon positive consider the subset the open subset the epsilon ball around f of x ok. So, so here we have x and this is f here we have y and here we have this f of x. So, we just take this open ball around f of x of radius of silent and we look at its inverse image right. So, since f is continuous. So, its inverse image may be some open subset like this implies that f inverse of this open ball is open in x and moreover since x is in this open subset right. So, x is somewhere over here and this is an open subset which contains x. So, therefore, there is a small delta a ball of radius delta around x which is completely contained in this open subset. So, this implies that there is delta positive such that this ball of radius delta around x is completely contained inside f inverse of. So, as x n's converge to x by the definition of convergence right there exists n such that for all n greater than equal to n these x n's are in this ball. So, we have this sequence of x n's which converge to x. So, after finitely many all these x n's are going to be in this open ball b delta are going to be in this delta neighborhood of x right. So, this implies that which is contained in this subset right. So, this implies that for all n greater than equal to n f of x n belongs to E epsilon and by the definition. So, by definition by definition of convergence this implies that f of x n converges to f of x. So, this proves one part of the lemma ok. So, now let us prove the converse. So, to prove the converse it suffices to show that. So, to prove the converse means we are given that f satisfies a particular property it takes convergent sequences to convergent sequences and we have to show that f is continuous. So, to show that f is continuous it suffices to show that the inverse image of a closed subset is closed z contained in y is closed. And to show that this is closed we will use this lemma that we just proved right or not this lemma. So, we will so let us see. So, to show that f inverse z is closed we will use we will show that f inverse z is equal to its closure. So, let us pick. So, let x be an element in f inverse z closure right. So, this implies that there is a sequence x n's converging to x where x n's is in f inverse of z right. So, x n is in f inverse of z implies that f of x n's is in z. So, moreover since f takes convergent sequences to convergent sequences that is the property that f has if x n converges to x then f of x n converges to f of x right as x n converges to x this implies f of x n converges to f of x right. Now, z. So, as z is closed this implies that f of x belongs to z z is closed f of x n is in z and f of x n converges to x this implies that f of x belongs to z right. So, thus x belongs to f inverse z. So, we started with the point x in f inverse z closure and we proved that x belongs to f inverse z. So, this implies that f inverse z is equal to f inverse z closure right. So, this implies that f inverse z is closed which implies that f is continuous ok. So, that kind of brings us to an end of the second part of this course. So, in the first part we introduce some the notion of a topological space and basic examples. In the second part we introduce continuous maps between topological spaces and we studied their properties. In the second part we also saw large we also defined metric spaces which gives us a large collection of topological spaces and large and most important collection of topological spaces. But now we will. So, now we are coming to the third part of this course. So, in the third in this part we will introduce topological properties of topological spaces right. So, in the first property we are going to introduce is that of connectedness. So, we ok. So, we recall two spaces two topological spaces if there is a bijective continuous map f from x to y such that f inverse is also continuous right. So, two topological spaces are the same. So, the word same is in quotes right same as in as far as continuous maps are concerned they cannot be differentiated if if they are homeomorphic. So, it is natural to ask it is natural to try and classify topological spaces right. For instance we can take the interval 0 1 and we can take the disjoint interval 0 1 I mean the this topological space which is the disjoint in the union of two intervals right and we can ask are these homeomorphic. So, if we can find some topological property which 0 1 has and this disjoint in of two intervals does not have then it would mean that these two cannot be homeomorphic right. So, we will we will introduce this notion of connectedness now which will help us distinguish between two topological spaces ok. So, in this lecture we shall introduce the notion ok. So, definition let x be a topological space suppose there are non empty open subsets u and v. So, they are non empty both of them that is important such that u in. So, u intersection v is empty they are disjoint and x is the union of these. So, then we say x is disconnected otherwise if x is not disconnected then we say x is connected ok. So, immediately. So, we have this property we have defined this property of topological spaces. So, we can ask the following questions what we can ask is r connected. So, r is with the standard topology what about r n again with the standard topology what are the connected subspaces of r is there a nice way to describe these right or for that matter of r n. So, 3 if x and y are connected can we say something. So, one of the topological spaces we constructed out of x and y is the product about the connectedness of x cross y. Then we saw various examples right connectedness of g l n r s l n r s o n g l n c s l n c and so on u n right what can we say about the connectedness of these objects right. So, in order to answer these questions we will need to develop some results and tools which we can use here. So, ok. So, let us begin that. So, the first proposition we are going to prove is let u contained in x be dense. So, if u is connected then x is connected. So, let us prove this. So, let us assume that x is disconnected and arrive at a contradiction. So, as x is disconnected there are non-empty open sets u 1 and u 2 such that which are disjoint such that we can write x as a disjoint union x is a disjoint union of u 1 and u 2 right. So, now we simply intersect both sides with u. So, intersect with u to get u is equal to u intersection u 1 disjoint union u intersection u 2. So, as u is dense in x. So, as u is dense in x this implies that u intersection u 1 is non-empty and u intersection u 2 is non-empty right, but this shows that u is disconnected which is a contradiction. So, this implies that our hypothesis is wrong. So, that means x is connected. So, as a corollary we have let a contain in x be a subspace if a is connected then the closure is connected right proof recall that. So, when we talk about a and a closure all these are subsets of x and we are always giving it the giving these the subspace topology. So, recall that we have proved a is dense yeah. So, applying the previous proposition implies that a closure is connected ok. So, we will end this lecture here.