 So, let us begin the next lecture. So, in the previous lecture we repeatedly used this result or remark we used the following result which I had left as an exercise, but let us just see how to do this. So, let us write as a proposition the standard topology and product topology on R n are equal. So, what do I mean by product topology on R n? So, R n is equal to this product n times and we give each factor the standard topology and then we take the product topology. So, proof. So, recall that we had proved that if x is a topological space is a set with two topologies tau 1 and tau 2, B i is a basis for tau i then if B 1 is contained in tau 2 implies tau 1 is contained. So, and this we had used this result to prove that topologies are equal. So, we will use this once again. So, let tau 1 the standard topology on R n and let tau 2 denote. So, right. So, a basis for tau 1 is given by sets of the form S epsilon a comma well x here. So, here x is a point in R n. So, x is equal to x 1, x 2 and this set S epsilon x recall this was those points y in R n. So, it is that mod x i minus y i is less than epsilon. But this set is precisely equal to the product of B epsilon x i, where B epsilon x i is this x i and this is the interval x i minus epsilon and x i plus epsilon. And each B epsilon x i is open in R and therefore, this is an element of the basis for the product topology. So, since belongs to B 2 this is equal to the basis for the product topology. So, this implies that this B 1 which consists of these sets S epsilon is contained in B 2 which is contained in tau 2. So, this shows that tau 1 is contained in tau 2. Now similarly let us prove that tau 2 is contained in tau 1. So, for that we will show that. So, next we will show that B 2 is contained in tau 1. So, what is the typical element of B 2 look like? So, an element of B 2 looks like product u i i equal to 1 where u i's are open. So, let x be a point in this product u i. So, then for each i there exists epsilon i such that this B epsilon i x i is completely contained in u i. So, we choose epsilon equal to minimum of all these i's among these epsilon i's. So, then this implies that B epsilon x i is contained in u i for all i which implies that product B epsilon x i is contained in this product of u i for all i right. But that implies that S epsilon x is contained in because this set is equal to this S epsilon x. So, therefore what this shows is that. So, if n is equal to 2 right. So, this is my u 1 and this is my u 2 and we have taken any point and we have found S epsilon around it. For each x in u 1 cross u 2 there is an epsilon such that this open square of side length 2 epsilon is completely contained inside u 1 and u 2. So, this implies that. So, taking union over all x. So, we get that this product u i is equal to this x and S epsilon x right. Of course, the epsilon is depends on x epsilon varies with x right. So, this implies that. So, since we have this set product of u i's and in the standard we have we can write it as a union of basic open sets in the standard topology and arbitrary unions of open sets are open. So, this implies that this product of 1 to n is open in the standard topology that is tau 1 right. Because we have written it as a union of open subsets in tau 1 right. So, thus this shows that b 2 the basis is continued in tau 1 which implies that. So, this shows that. So, this shows that the standard topology and the product topology on R n of D. So, this is a result which we had repeatedly used in the previous lecture. So, now let us continue. So, next we are going to see an example. So, in this example we will give a set theoretic description of a map and the exercise is to show that that map is continuous. So, in order to show that that map is continuous we will use what we have learnt earlier. So, let us first describe the map set theoretically. So, let h contained in R n be the hyper plane h is equal to x 1 x 2 up to x n minus 1 and the last coordinate is 0 well I can write as right points of this point this type inside R n. And similarly let h prime be the hyper plane. So, here the last coordinate the first n minus 1 can be anything, but the last one we want is 1 ok. So, let us we can make this both of these let me make h prime and read maybe. So, this is h prime and this is h. So, we are interested in considering R n minus h prime right. So, we will define a map. So, we will define which we denote by phi from R n minus h prime to h. So, what is the description of this map? So, we take any point x in R n minus h prime and we take this point p this point p is. So, this point is p and this is R x right. Now, we join x and p by a straight line and we extend this straight line till it means h yeah. So, this x and so, so the description in words is let x be in. So, let me just write this. So, let x be in R n minus h prime join x and this fixed point p which is defined to be 0 and let q in h be the point where this line meets h. So, there will be a unique such point and. So, we define the map as define phi of x to be equal to q ok. So, the claim we want to make is let R n minus h prime and h have the subspace to poly from R n ok. So, then phi is continuous. So, before we prove the claim let us make a remark over here the subspace to it is it can be easily checked. So, note that h is isomorphic to R n h is the inclusion image of the inclusion i from R n minus 1 to R n right. So, what is the map i x 1 up to x n minus 1 it gets map to x n x 1 up to x n minus 1 comma 0 right and this map is an inclusion and the standard topology on R n minus 1 agrees with the subspace topology from R n yeah and a similar remark holds for h prime ok. So, having made this remark let us prove our claim. So, the idea is to first describe phi in terms of coordinates and then see that each of the coordinate function is continuous yeah. So, let us just ok. So, that is like the brief idea or the main point of the idea. So, let us see how to prove this claim. So, the image of h the image of phi lands in h and h has a subspace topology and. So, recall what we have proved. So, we have R n minus h prime. So, we can view phi as a map to R n right, but the image lands inside h this h and we have this inclusion and since h is given the subspace topology. So, to show that. So, this is actually well this phi and let us call this yeah this i compose phi right. So, to show that. So, to show that phi is continuous it is enough to show it is enough to view phi as a map from R n minus h prime to R n and show that this map is continuous ok. So, we that is exactly what we are going to do now. So, we will now compute a formula for this map in terms of coordinates. So, let us do that. So, we are going to make this map precise. So, let x 1 up to x n be in R n minus h prime right. Every point in the line joining x and p is of the form x plus t times p minus x, where t is in R right. So, that is clear because we have our x and we have this direction p minus x or we can take x minus p we have this direction this is the direction of p minus x. So, all points on this line are given by moving along this direction for different values of right. So, a general point looks like on this line as this as these coordinates this is 1 minus x n which is equal to 1 minus t into x 1 1 minus t into x n minus 1 and the last coordinate is x n plus t into 1 minus x n that is right. So, this is a general point on this line has this expression. So, we are looking for the point which lies on h right. So, we have to set the last coordinate to be 0 yeah this point is on h if an relief x n plus t times 1 minus x n is equal to 0 that is if an relief t is equal to x n upon x n minus 1 yeah and note that this is well defined. So, this t which we will denote by t naught is well defined is well defined since x n is not equal to 1 as x does not belong to h right. So, we have this t naught is equal to x n upon x n minus 1. So, let us compute what is q want to be. So, q is going to be. So, this implies that phi of x is equal to. So, maybe I can write phi of x 1 to x n is equal to 1 minus t into x 1. So, let us compute what is 1 minus t naught is minus 1 upon x n minus 1 right. So, this is equal to minus x 1 upon x n minus 1 minus x 2 upon x n minus 1 x n minus 1 upon x n minus 1 and 0. So, this means that. So, we show that. So, therefore, to show that phi is continuous it is enough to show that each of the coordinate functions are continuous because R n now has the yeah well the product topology on R n is the same is the same as a standard topology. So, it is enough to show suffices to show that phi i yeah phi i. So, this is from R n minus h prime to R n and then projection on to the i th coordinate to R. But, what is this? So, if x 1 up to x n it maps to minus x i upon x n minus 1 right. If i is strictly less than n and x 1 up to x n maps to 0 if i is equal to ok. So, we want to show that phi is continuous. So, phi is from R n minus h prime to R n right. So, here the standard topology is same as a product topology and therefore, it is enough to show that this continuous when R n has the product topology and to check that phi is continuous in the product topology we just have to check that each of the coordinate functions is continuous and that is exactly what we are going to do right. So, each of the coordinate functions is given by minus x i upon x n minus 1. So, that is when i is less than 1 yeah, but now note that minus x i. So, yeah so from R n minus h prime to R we have these 2 functions minus x i this is. So, this is just the projection R into R we have x this is just the ith projection this is minus of the ith projection yeah. So, if so this minus minus x i upon x n minus 1 this is equal to the constant function minus 1 times the projection map times the map 1 upon x n minus 1. So, x n is a continuous function x n minus 1 is a continuous function on R n and x n minus 1 never vanishes on R n minus h prime. So, therefore, 1 upon x n minus 1 is a continuous function on R n minus h prime yeah. Therefore, all the product of all these is continuous all these or these 3 these 3 are continuous on R n minus h prime. So, the last one is continuous because x n minus 1 never vanish does not vanish as x n minus 1 does not vanish right. And so, their product is continuous. So, and of course, the last coordinate function is just the constant function and it is an easy check that the constant functions are continuous right. So, this implies that. So, this implies that phi from R n minus h prime to R n is continuous and. So, also phi from R n minus h prime to h is continuous ok. So, this completes the proof. So, now notice that ok. So, let us make some remarks. So, remarks So, we could have written phi directly as phi from R n minus h prime to R n minus 1 right as x 1 up to x n maps to minus x 1 upon x n minus 1 minus x n minus 1 upon x n minus 1 right. And this is clearly a continuous map due to the same reasons as above and what we can do is we can restrict this map to the sphere minus point. So, let us just see what is happening. So, let us just copy this diagram over here and we paste it here ok. So, when we make the sphere the unit sphere. So, the unit sphere meets h prime exactly at this point p ok. So, s n intersected with R n minus h prime is exactly s n minus this point p this is an easy check which I will leave to you. So, then so, we have s n minus p is subset of R n minus h prime and here we have this map phi to R n minus 1 ok. So, sorry I should write s n minus 1 is s n minus 1 ok. So, if we give s n the subspace topology or s n minus 1 the subspace topology then this phi restricted to s n minus 1 minus this point p from s n minus 1 minus p to R n minus 1 is continuous because restriction of a continuous map to a subspace is continuous. Here note that for that to happen it is important that the subset is given the subspace topology here ok. So, here are some exercises. So, the first exercises show that phi restricted to s n minus 1 minus this point p R n minus 1 is a bijective map of sets ok. So, exercise 2 is so, let us psi from R n minus 1 to s n minus 1 minus p denote the inverse. So, apriori is just a set theoretic map set theoretic inverse of this map ok. So, then show that psi is continuous. So, here is a hint to do this exercise. So, we have psi from R n minus 1 to s n minus 1 minus p right s n this is included inside R n minus h prime and s n minus p has a subspace topology from R n minus h prime which in turn has a subspace topology from R n right. So, therefore, to show that psi is continuous it is enough to show that this composite is continuous yeah. So, therefore, if you can compute the coordinates of this composite function and show that each coordinate is continuous. So, thus enough to compute coordinate functions of this composite and show those are continuous ok. So, that is exercise 2 and that is also a hint to this exercise and that is all leads us to a definition let f from x to y be a bijective continuous map. Let g denote its set theoretic inverse which exist since f is bijective right. So, if g is also continuous then show that I am sorry nothing to show then f is called the homeomorphism. So, ok and I like to give one more exercise which is very easy let f from x to y be a bijective continuous map show that f is a homeomorphism if and only if for every u open in x f of u is open in y that is part a of the exercise and part b of the exercises if f is a homeomorphism then show that its inverse g is also homeomorphism. So, we will end here.