 So, welcome to the next lecture. So, in this lecture we shall introduce the class of, we shall introduce the notion of metric spaces. So, metric spaces provides perhaps the most important class of the logical spaces. So, and so after we introduce metric spaces we will also see I mean we have talked about open subsets, closed subsets, continuous maps, we will see we will give descriptions of these in terms of metric, okay. So, let us begin by recalling the definition of a metric space. So, we recall the definition, let x be a set that d from x cross x to the real numbers which are greater than equal to 0 be a function map which satisfies the following three conditions d of x comma y is equal to 0 if and only if x is equal to y d of x comma y is equal to d of y comma x and d of x comma z is less than equal to d of x comma y plus d of y comma z for all x comma y comma z. This is for all x comma y and x. So, this is called the triangle inequality. So, the d is for distance and we know that if we have a triangle. So, this x this is z and this is y then the distance between x and z is less than equal to the distance between x and y and y and z. So, if these three properties are satisfied then so then we say that then the pair x comma d is called the metric space. The map d is called the metric on x or the distance function. So, we just need any function d which satisfies these three conditions. So, given any set x, there could be lots of metrics on x, but the most common example of a metric space we encounter of metric space we encounter is R n with the Euclidean metric. So, that is if we have two vectors in R n. So, x is x 1 to x n and similarly y is y 1 up to y n. So, then t of x comma y is defined to be the square root of the sum of squares of the difference of x and y. So, let us check that this defines a metric. So, let us check that this defines a metric. So, note that the first two conditions are the first two conditions are trivial to check. The only non-trivial condition is the triangle inequality. So, which we will prove now here as a proposition. So, this d of x comma y we shall write this as norm of x minus y, the two norm. So, we will use this notation instead of writing d of x comma y. So, what we need to prove is that with this in this notation that the norm of x minus y is less than equal to or alternatively I mean we can say, we can define let me just. So, if for a vector x in R n we define the two norm of x as summation x i square the square root of this. So, then clearly then I do not need to write this as a definition. So, then clearly d of x comma y is equal to the two norm of x minus y. And check that this defines a metric to check that this defines a metric, we need the following proposition that is precisely the following proposition the norm of x minus y is less than equal to for all. So, let us prove this. So, it is enough to check that and this left as an easy exercise that it is enough to check this for all. So, in order to do this. So, consider the map this inner product defined as. So, we first claim that this absolute value of this quantity is less than equal to. So, let us first prove this claim and to see this. So, let us write let us define w to be equal to y minus t x where t is a real number. And now let us compute So, this inner product of w with w and this is equal to this is bilinear right this is bilinear in both. So, what that means is if we fix x that is fix if we fix x then x comma lambda y 1 plus y 2 is equal to or if we fix x then it is linear in y right. And similarly if we fix y then linear in x right. So, using the bilinearity we get that this is equal to minus 2 t 2 x right. And it is also symmetric bilinear and plus symmetric right because clearly for symmetry x comma y is clearly equal to y comma x. So, using the fact that this bilinear and symmetric we get this about this is precisely equal to t square minus 2 t. So, for a fixed x and y we view this as a quadratic in t right. There is a degree to polynomial in t and as this and as it only takes positive value greater values greater than equal to 0 it follows that the discriminant is less than equal to 0 it follows that the discriminant is less than equal to 0 right. But what is the discriminant? So, the discriminant is 4 is less than equal to 4 into which proves our claim right this is the claim which we want to So, using this claim we will now prove the proposition. So, note that. So, we have x minus y o square is equal to this is equal to and this is less than equal to yeah. So, over here we have used the claim. So, oh sorry we have not used the claim yet this is less we are just taking absolute value this is less than equal to now here we have used the claim. So, this implies it and note that the norm of y is same as norm of minus y. So, therefore, we can just replace y by minus y and we get. So, thus so this shows that this R n comma d is a metric space. So, this proves triangle inequality holds and therefore, this is a metric space. So, this is the main example of metric space. So, now given a metric space how can we use the metric to define a topology on it. So, let us see that. So, given a metric space x we can use the metric to define a topology for that for x in x and an epsilon positive define the open ball. Of radius epsilon around x as follows ok. So, b epsilon x is defined to be those y in x says that the distance of x from y is strictly less than epsilon ok. And we consider the collection there is something which we are familiar with b contained in the power sort of x defined as. So, b is the collection of all these open balls around various points of x x is in x and epsilon is positive. So, it is easily check that. So, it is easily check b satisfies the two conditions for being a basis. So, thus b defines a topology tau on x which has b as basis right. And so, therefore, in other words. So, thus a subset u contained in x is open in this topology if and only if it satisfies the problem following property for every u in x or sorry for every x in u there exists an epsilon positive such that this open ball of radius epsilon around x is completely contained inside. So, this is very similar to the examples that we saw in the beginning of the course that is R n. So, if we have a topological space x over here. So, a set is open if and only if given any point x we can find a ball of radius epsilon around x which is completely contained inside. So, the epsilon of course depends on x. So, it is trivial to check and this is left as an exercise that the topology defined on R n using the Euclidean metric. Is the standard topology and this check is left as an exercise. So, now we want to understand what it means to be in the closure in terms in a metric space. So, in terms of the metric. So, definition let x be a metric space let x n for n greater than equal to 1 be a collection of points of x of points of x. So, we say x n converges to x if for every epsilon positive there is an n sufficiently large such that for all let us say n naught and this n naught depends on epsilon depends on epsilon. So, is that for all n greater than equal to n naught the x n's are contained in the epsilon ball around x. So, this x n. So, I should write x n converges to we say x n converges to x right and it is often written as x n converges to x ok. So, roughly if we have x x is here and we have sequence of x n's. So, we want these this entire the sequence to get closer and closer to x as our n tends to infinity right. So, or precisely what we want is here no matter which no matter how small we take epsilon when we take the epsilon ball around x after finitely many n's all the x n should be in that epsilon ball ok. So, let us prove this lemma let x be a matrix space and let a contain in x be a subset then x is in a closure if and only if there is a sequence x n converging to x and x n belongs to a. So, x is in the closure of a if and only if there is a sequence in a which converges to x right. So, let us take a simple example. So, if we take let us say this region. So, then given any point on the boundary given any point of the boundary we can find a sequence inside this red region this region x square plus y square strictly less than 1 right which converges to this point of the boundary. But if we take some point outside this boundary then we will not be able to find a sequence. So, then we can find a small neighborhood around this point outside which does not beat the set a. So, therefore, we want no matter which sequence we take inside a yeah it will not satisfy the definition of converging to x if we take a point outside this circle x square plus y square is equal to 1. So, that is a that is a small example of what is happening. So, let us see a proof. So, let us assume that suppose x belongs to a closure right. So, let us say x is over here right. So, then by the definition of closure every open subset u containing x meets right. So, take u to be b 1 upon n x. So, we take u n to be this collection right and we choose. So, each u n it meets a because x is in the closure right. So, choose any x n in b 1 upon n this epsilon this 1 upon n neighborhood of x. So, we are taking this we are taking smaller and smaller neighborhoods right. And inside each neighborhood we are choosing each of these neighborhoods it meets this red region and we choose one point in the intersection a yeah. So, then we claim that x is equal to x n converge to x right. So, to prove this what we need to show. So, we need to show that given any epsilon positive there exist some n naught very large. So, that such that for n greater than equal to n naught we have x n belongs to e epsilon x right. So, let us show this. So, note that first choose. So, we can choose n naught sufficiently large so that 1 upon n naught is strictly less than epsilon right. So, then we claim that n naught we have b 1 upon n x is obviously contained in b 1 upon n naught x which is contained in b epsilon x right. And moreover this these balls these are nested and so on right. So, from this and x n is here x n plus 1 is here and so on right. So, all this is contained in b. So, therefore, this shows that. So, this implies for all n greater than equal to n naught x n belongs to b epsilon x which implies that x n converge to x ok. So, conversely suppose so in the first part of the proof we showed that if x is in a closure then we can find a sequence of x n's in a. So, that x n's converge to x yeah. So, and conversely so suppose x n's is a sequence in a such that x n's converge to x then we need to show that x is in a closure right. So, once again we will use the definition. So, let u containing x be an open subset right. So, by the definition of the topology on x which is given by the metric right. So, then there exists an epsilon positive such that b epsilon x is completely contained inside u. So, u is some set open set which contains and let us say x is here. So, we can find an epsilon. So, z b epsilon x is completely contained inside u right. And so as x n converge to x. So, this implies that for all n greater than equal to n naught there exists n naught such that for all n greater than equal to n naught x n belong to b epsilon x right. So, this implies that x n belongs to b epsilon x intersected a which is containing u intersection right. So, thus u intersection a is non empty right. So, thus x belongs to a closure. So, we have given a nice. So, in case of metric spaces we have given a nice and intuitive criterion for what it means for a point to be in the closure of a set right. So, right. So, once again we return to our earlier example. So, the closure of this set we are taking this open region a is equal to x square plus y square strictly less than 1 right. So, the closure of this is exactly the set a closure is equal to x square plus y square is less than equal to 1 ok. So, we will end this lecture here.