 So, welcome to this lecture. So, in the previous lecture we define continuous maps and we had entered the lecture by this lemma which I had left as an exercise. So, let us just prove this lemma perhaps some of you are able to do it. So, let x and y be topological spaces let f from x to y be a map of sets and we want to check that f is continuous. So, let b be a basis for the topology on y then if f inverse v is open for every v in v then f is continuous. So, this lemma gives us a convenient criterion to check whether a function is continuous in terms of checking only for basic open sets. So, prove. So, let u contain in y be an open set then for each x in u there exists a basic open set v sub x such that x belongs to v sub x and v sub x is contained in u. Therefore, this shows that we can write u as a union over all the x's these v sub x. So, then it is a straightforward checking set theory that f inverse u is equal to union f inverse of v sub x each of these is open as this v sub x belongs to b and by hypothesis assumption is that f inverse v sub x is open and as arbitrary union of open sets is open this implies f inverse u is open and since this happens for every open set u thus f is continuous. So, next we want to show that the space for continuous functions which are real valued or complex valued maps are is a nice has some nice algebraic structures. So, to do that we are going to begin with the following theorem. So, we have the following two. So, before that consider with the standard topology and there are two natural maps which we can define. So, then the following maps. So, the first one is the addition map this is given by a of x comma y x plus y and the second is the multiplication map this given by m of x comma y is the multiplication. So, both these are continuous. So, we will use the above lemma to prove this theorem. So, proof by the above lemma. So, it suffices to show a inverse of b epsilon z. So, let us say z is a real number and epsilon is positive or the standard topology on R has this as basic open sets and we have to show that these the inverse images of these are open. So, in order to do that that x comma y be in order to and recall that we had defined this as epsilon x comma y to be those x prime comma y prime in R 2. So, is that x prime minus x is less than epsilon and y prime minus y is less than epsilon. So, note that if x prime comma y prime is in this set then the absolute value of minus x plus y is less than equal to x minus x prime plus y minus y prime which is less than equal to epsilon strictly less than epsilon. So, this implies that the absolute value of a x prime comma y prime minus a x comma y is strictly less than 2 epsilon for x prime comma y prime in S epsilon x comma y. So, this shows that S epsilon by 2 of x comma y is contained in a inverse of b epsilon x plus y. So, let us check why this happens to see this we need to show that if x prime comma y prime is over here then a of x prime comma y prime belongs to b epsilon x plus y, but we have checked that if x prime comma y prime is here then a of x prime comma y prime minus a of x comma y is strictly less than 2 times epsilon by 2 which is epsilon, but this a of x comma y is precisely x plus y. So, this implies that a x prime comma y prime belongs to b epsilon x plus y. So, we have simply rephrased the above computation in this statement. So, let us go ahead. So, let now using this we will prove that in the inverse image of all basic open sets are open b in R be a basic open subset. So, suppose x comma y is in a inverse of b epsilon z this implies that x plus y belongs to b epsilon z. So, this b epsilon z is b of z sorry and this is the epsilon ball around z and x plus y is somewhere over here. So, this implies we can find epsilon prime positive such that this b epsilon prime of x plus y is completely contained in b epsilon z right and from the above computation the above computation it follows that s epsilon prime by 2 is contained in a inverse of b epsilon x plus y which is contained in a inverse of b epsilon z. I should write s epsilon by 2 of x comma y epsilon prime by 2 of x comma y right. So, what this shows is that so thus for every x comma y in a inverse b epsilon z there exists some epsilon prime positive such that s epsilon prime by 2 of x comma y is contained in a inverse of b epsilon. So, therefore, by the definition of the standard to polygenar 2 thus inverse of b epsilon z is open in R. So, this shows that the addition map. Similarly, let us show that the multiplication map is continuous. So, we prove the second part now. Once again we start with a small computation let us start with a delta which is positive and less than 1. So, and suppose that prime comma y prime belongs to s delta x y. So, this is in R 2 ok. So, then the absolute value of x prime y prime minus x y this is equal to x prime y prime minus x y prime plus x y prime minus x y. This is less than equal to x prime minus x into y prime plus minus y which is strictly less than x prime minus x into mod y plus delta. This because since y prime minus y is less than delta. So, this implies that mod y is mod y prime is less than mod y plus delta plus mod x. So, this is strictly less than right now this is also less than delta this quantity over here. So, we use that and this quantity is less than delta. So, plus is equal to delta into mod y plus mod x plus delta which is strictly less than delta into mod y plus mod x plus 1 ok. So, this computation shows that. So, we can this refreeze this computation as when we apply m on s delta x comma y this is going to be contained in b of delta into mod y plus mod x plus 1 of x y. So, once again let us check why this shows this this because if x prime comma y prime is in this set s delta x comma y then m of x prime comma y prime which is equal to x prime y prime. We have just now check that over here is belongs to of x y. So, sorry there is a type over here. So, now let us so let ok. So, we fix any epsilon between 0 and 1 and define delta to be equal to epsilon by mod x plus mod y plus 1. So, then clearly delta also lies between 0 and 1. So, as a result what happens is so this implies that m of s epsilon by mod x plus mod y plus 1 x comma y is contained in b of epsilon x y or which is same as saying that s of epsilon by mod x plus mod y plus 1 of x y is contained in m inverse of b epsilon x prime. So, now with this computation in mind. So, suppose x comma y belongs to b m inverse of b epsilon z where z belongs to r and epsilon is positive. Then there is an epsilon prime which is between 0 and 1. So, this is r z and this is epsilon and x y is somewhere over here. So, we can always choose epsilon prime and further assume that epsilon prime lies between 0 and 1 such that this ball of really epsilon prime around x y is contained in this ball of really epsilon around z. So, from the above we conclude that s epsilon prime by mod x plus mod y plus 1 of x y is contained in m inverse of b epsilon prime x y which is contained in m inverse of b epsilon z right. So, once again given so therefore, given any point x y. So, given any point x y in m inverse b epsilon z we have found this positive quantity such that this basic open set around x y is completely contained m inverse of b epsilon z. This implies that m inverse of b epsilon z is open. So, this shows that the multiplication map is also continuous. So, this completes. So, next we prove a similar theorem, but about r minus 0. So, let r star be the set we move 0 from the real line. So, with the subspace topology. So, and consider there is a very nice map on this set. So, consider the map there is a natural map given by f of x is equal to 1 upon x and the claim is this map is continuous. So, let us prove this claim once again. So, x ok. So, let us begin with an x in r star. So, one checks easily if epsilon is strictly less than mod x then b this basic open set is completely contained in r star. That is obvious because this is let us say this is 0. So, this is r star we have to move 0 and if we take any x and we choose an epsilon which is less than mod x then this neighborhood this interval around x will be completely contained in r star and the thing is. So, it is easy to check the collection b epsilon x where x belongs to r r star epsilon is strictly less than mod x is a basis for the subspace topology. So, therefore, we will check that. So, thus it suffices to check f inverse of b epsilon x is open when epsilon is strictly less than mod x. So, in this situation so, what do we have? We have we have removed 0 and let us say r x is somewhere over here. So, we take this neighborhood epsilon x minus epsilon and x plus epsilon then one easily checks that this set f inverse of this interval x minus epsilon comma x plus epsilon is precisely equal to the interval 1 upon x plus epsilon which is open in r star in the subspace topology. So, this proves that f is continuous. So, we will end this lecture here.