 So, in the previous lecture, we had we had ended the previous lecture with this exercise. So, let us just do this exercise. So, what do we have? Let us make a picture of this. So, let us say we have R n. So, this in our case let us just take R 2 and we fix this point a comma b and we want to show that the complement of this point. So, we are leaving out this point. So, we want to show that the complement is an open subset right. So, let us just take any point x comma y right and then. So, how will you prove this? So, let x bar m in R n the complement of this right. What this simply means is that x bar is not equal to a bar which implies that there is some i such that x i is not equal to a i right. So, let epsilon be the absolute value of x i minus a i right and this is positive yeah. So, we claim that s epsilon by 2 of x is completely contained inside minus this a 1 up to a right. So, in other words we are taking this distance epsilon this epsilon and we are taking a square let us just to be safe let us just take epsilon by 4 right. So, we are taking this s epsilon to the by 4 around this point x comma y. So, let us prove this. So, suppose if y equal to y 1 up to y m belongs to s epsilon by 4 x. So, then this implies that mod x i minus y i is strictly less than epsilon by 4, but then this implies that yeah. So, this implies that y i it cannot be equal to a i right because x i minus a i is equal to epsilon yeah. So, therefore, this implies that right. So, this proves that this thing inside in the complement given any point in the complement we have found a basic open subset which contains that point and it is completely contained inside the complement. So, therefore, the complement is open. So, this implies that oops I am sorry is a open subset is an open subset. So, this implies that a is an open subset which implies by the definition of closed subset the singleton is a closed subset ok. So, with this let us begin the next lecture. So, in this lecture we will define the closure of a subset yeah. So, let us first write the definition and then we will see what this means by means of some example. So, let a contained in x be a subset. So, it can be any subset does not have to be open or closed yeah. So, the closure of a is defined as follows. So, a closure is equal to those points in x such that for every open set or rather if u is an open set is an open set containing x then u meets a, u intersection a is non-empty ok. Let us try to understand this by means of an example. So, the simplest examples are on the real line. So, let us take the real line with the standard topology and we can take the interval 0 1. Let us take this interval 0 1 and let us see what its closure is. So, suppose we take any x which is not 0 or 1 let us say x is strictly less than 0. So, then we can find a small neighborhood around x which does not meet this interval a. So, a is equal to this open interval 0 1 right. Similarly, if we take any. So, this implies that. So, this implies that if x is strictly less than 0 then x does not belong to a closure yeah. Because the closure requires that every neighborhood every open subset which contains x should meet our set a right. And similarly, if I take any x which is strictly greater than 1 then also we can find some epsilon small enough such that the epsilon neighborhood does not meet a. So, once again this implies that if x is strictly greater than 1 then x does not belong to a closure. And let us make a remark maybe I should have made the remark here itself. It is completely obvious it is clear that a closure a is contained in a closure ok. So, let me just highlight that. So, therefore, a closure. So, in a now in this example. So, in this example a is contained in a closure and a closure has to be contained in this closed interval 0 1 right. That is what we have seen because if we take anything which is strictly less than 0 then it is not in the closure. If we take anything which is strictly greater than 1 then also it is not in the closure the only possible points which are which could be in the closure are 0 and 1. So, let us see if they satisfy the definition of a closure of being in the closure. So, if we take 0 then no matter which neighborhood of 0 we take if we take any open subset. So, if u is an open subset which contains 0 right. Then by the definition of the topology then there exists an epsilon positive such that d epsilon 0 the ball of neighborhood radius 0 epsilon around 0 is contained in u right. But clearly epsilon by 2 belongs to this ball and epsilon by 2 is contained in 0 1 right. So, this implies that u intersection a is non empty. So, therefore, 0 is contained in a closure and similarly one can check easily that 1 is contained in a closure. So, therefore, a closure in this case is precisely the closed interval 0 1 we can give a slightly more complicated example. So, let us give an example in R 2. So, we can take this region right. So, this region is a is equal to those x comma y in R 2 such that x square plus y square is strictly less than 1 and I would not write the details, but you can check that if we take any point x comma y a comma b let us say. So, that a square plus b square is strictly greater than 1 then we can find a small neighborhood which does not meet a there exists epsilon positive such that s epsilon a comma b intersection a is empty right. And similarly we can check that if a square plus b square is equal to 1 then for every epsilon s epsilon a comma b intersection a is non empty right. So, this will show that a closure is exactly the set x comma y in R 2 such that x square plus y square is less than or equal to 1. Note that a is obviously containing a closure. So, a closure is just adding the boundary this boundary circle ok. So, as we proceed we will get more familiarized with this notion. So, for now let us prove a lemma. So, although we will write a proof it is good to keep a picture in mind while we prove these statements. So, let a contain an x like for instance a picture in R 2 will be good enough be a subset a closure is closed in a is closed in x. So, that is why the word so, this justifies ok. So, let us prove this. So, it suffices to show x minus a closure is open right the definition of closed subset was the complement should be open. So, that is what we are going to show that the complement is open. So, let us pick. So, let x be an element in x minus a closure right. So, that is what this means that x does not belong to a closure, but then by the definition. So, then by definition there exists an open set u containing x u contains x or which contains x which contains x and u intersection a is empty right, but it follows that if y is any point in u right. So, then y has an open subset which is u itself which does not meet a yeah has an open subset namely u which does not meet a. So, thus y is not contained in a closure right. So, therefore, we have proved that. So, this implies that u is completely contained inside x minus a closure. So, thus for every x x in x minus a closure we have found an open set u sub x which is which contains x u sub x is contain x minus a closure right. So, and so therefore. So, this implies that this x minus a closure we can write it as union x minus a closure is u sub x right and each of these is open and an arbitrary union of open sets is open. So, this implies that x minus a closure is open ok. So, using. So, yeah. So, roughly this says that if we take any point here we can find this small neighborhood which does not meet the closure and similarly if we take any point here and the complement which is this open region in red is open ok. So, let us prove this proposition the next proposition. Let us use this lemma to prove the next proposition a set a is closed if and only if a is equal to a bar or a closure right. So, let us prove this. So, recall let us just recall the definition of this a bar. So, x is in a bar if it has the property that given any open subset u which contains x it should meet a. So, let us prove this. So, let us assume first assume that a is closed. So, we need to show that a is equal to a closure. So, since a is contained in a closure we already know this is obvious we have to observe this in the remark which followed the definition it is enough to show that a closure is contained in u. So, taking complements. So, we should prove this to show this it suffices to show that x minus a closure contains x minus a and which is what we are going to prove. So, let us take let x belong to x minus a. So, as a is closed it follows from the definition that x minus this implies x minus a is open. So, let us denote this open subset by u then. So, thus there is an open subset which contains x and such that u intersection a is empty right u is defined to be the complement of a. So, therefore, u intersection a is empty right. So, thus x does not belong to a closure a bar if you like yeah by the definition of this a closure. So, this implies that x belongs to x minus a. So, thus we have proved that we have proved this and therefore, we have proved this right. So, this implies that a is equal to a bar that proves one direction of the proposition. So, for the other. So, next let us assume a is equal to a bar. So, then by the previous lemma. So, what is the previous lemma say? The previous lemma said that a bar is closed in x right a bar is closed in x and since a is equal to a bar. So, thus a is closed in x which is exactly what we wanted to prove right we wanted to show that a is closed. So, this completes the proof of the proposition. So, as a corollary as an easy corollary. So, let b be a subset of x. So, then b closure closure yeah. So, closure of the closure is equal to b closure. So, if we take closure then taking closure again makes no difference. So, let us see how to prove this. So, proof. So, from the lemma. So, the lemma says that this lemma over here it says that no matter which subset we take the closure is always closed in x right. So, applying this lemma from the lemma we get that b closure is a closed subset and this proposition yeah the above proposition says that b closure is closed implies b closure is equal to b closure closure closure implies that b closure is equal to b closure closure. So, applying the previous proposition a is equal to equal. So, we will end this lecture by with two exercises with two easy exercises. Let a contained in b b subsets then a closure is contained inside b closure that is the first exercise and the second exercises. So, let z contained in x be a closed subsets and let a be a subset of x which is contained in z yeah. So, then a closure is contained in that ok. So, let me make a remark. So, exercise 2 says that a closure is closed in x. So, exercise 2 and the above lemma imply that a closure is the smallest closed subset containing. So, we will end here.