 So, let us return to the example of subspace topology and now we will consider a slightly more complicated example. So, let X be R with the standard topology. So, this is example 4. So, this is lecture 5 and consider the inclusion map. So, I from R to R 2 which sends X to X comma 0. So, we are just embedding the real line into R 2 as the X axis. So, this is 0. So, point X is going to X comma 0. So, let S denote the standard topology on the real line and let tau denote the standard topology on R 2. So, let Y be equal to image of I. So, Y simply the X axis and we can put. So, we can identify the real line with Y and we can put the subspace topology on Y. So, then so what this means is. So, tau sub Y is equal to U intersection R where U is an open subset in tau. So, here we can take any open subset and we can intersect it with the X axis. So, which means over here we will get the union of these intervals and this map is I. So, this U intersection R we can also view it as I inverse of U. So, the claim is. So, the claim we claim that that S and this tau sub Y are equal. So, these are the same topologies on the real line. So, let us prove this. So, instead of proving this directly we will prove two lemmas which will be helpful and they will be helpful in other context as well. So, lemma. So, let X be a set and suppose there are two topologies on X. Say tau 1 and tau 2. So, let B 1 let B I be a basis for tau I equal to 1 comma 2. So, we are given two topologies on X exactly as in this previous example we have two topologies S one is S and the other which is a standard topology and the other is tau Y which is coming as the sub space topology from the standard topology on R 2. And we have B 1 is a basis for tau 1 and B 2 is a basis for tau 2 and if B 1 is contained in tau 2 then tau 1 is contained. So, if the basis for tau 1 is contained in tau 2 then all of tau 1 is contained in tau that is a content of this lemma. So, let us prove this. So, let U be. So, in a subset of X which is open in tau 1 or which is simply an element of tau 1. So, we will show that U is open. So, since B 1 is a basis for tau 1 for every X in U there exists a W sub X which is in B 1 such that this W sub X is also a subset of U. So, as a result thus we may write. So, we can write U as the union of ok. So, let I be the collection of such W sub X for every X in U. So, every X in U we choose one such W X and we take I to be the collection of all these W X's. So, then clearly U is equal to the union over these elements of I's over the elements of I this W sub X. So, this inclusion is clear as each W sub X is contained in U and therefore, the arbitrary union is contained U and this inclusion is clear because given any X in U it is in W X. So, now since each W X is in B 1 and B 1 is a subset of tau 2 this implies that W X is also in tau 2 that is W X is open and since tau 2 is a topology and an arbitrary union of open sets is open. So, since this is by condition 3 in the definition of topology arbitrary unions of open sets or simply unions of open sets are open. So, this implies that this union over I W sub X this is also in tau which implies that U is in tau 2. Therefore, we have proved that given any U in tau 1 it is in tau 2. So, thus given any U in tau 1 U belongs to tau 2 this implies that tau 1 is contained which completes the proof of lemma. So, as a corollary if with same notation as in the lemma if B 1 is contained in tau 2 and B 2 is contained in tau 1 then we get tau 1 is equal to tau 2. So, we will use this corollary and we need one more lemma the proof of which is easy and left as an exercise. Let X be a set with a topology tau let Y contain in X be a subset and let tau sub Y denote the subspace topology. So, if B is a basis for tau then there is a obvious candidate for a basis for Y B sub Y is defined to be U intersection Y where U belongs to B is a basis for tau sub Y and the proof left as an exercise. So, using these two lemmas this more precisely this corollary and this lemma we will show that. So, proposition we will prove our claim that let S and tau sub Y be as described above then S is equal to tau sub Y. So, recall what we are doing S is the standard topology on the real line and tau sub Y is the topology on the real line defined as follows. We can embed the real line into R 2 as the X axis and we can take the subspace topology on the X axis from the standard topology on R 2 and we have to show that both these topologies are equal. So, proof. So, recall that tau is the standard topology and this had as basis the collection S epsilon A comma B is equal to those points X comma Y in R 2 such that absolute value of A minus X is less than epsilon and absolute value of B minus Y is less than epsilon. So, then B sub Y defined over here is the is equal to the intersection intersected Y for S epsilon A comma B in B. So, on the other hand this is a basis for the subspace topology on the real line. So, we have the real line this is map I. So, on the other hand let C be the collection of B epsilon X. So, here X belongs to R and epsilon is positive union the empty set. So, here this B epsilon X is contained in R and recall that B epsilon X is the interval X minus epsilon comma X plus epsilon this is a subset of the real line. So, then C is a basis for the standard topology on R which is S. So, we want to show S is equal to tau Y and by the corollary it is enough to show B sub Y is equal to C. So, C is a basis for S and B sub Y is a basis for tau sub Y we want to show that S is equal to tau sub Y and therefore, we will apply this corollary. So, let us check that B sub Y is indeed equal to C. So, let us look at what the elements of B sub Y look like. So, given if A comma B is here then it may happen that S epsilon intersected with Y, Y is the X axis this is S epsilon A comma B. It may happen that S epsilon A comma B intersected with Y is the empty set here that is one possibility the other possibility is A comma B is here and then S epsilon intersected with the Y axis is going to be B epsilon. So, that is the only there are only two possibilities here. So, note that B sub Y or rather for every S epsilon A comma B in B sub Y either intersected Y is the empty set or S epsilon A comma B intersected with Y is exactly B epsilon A. So, we are identifying Y we are identifying the X axis with we are identifying the X axis with the real line using this embedding I and because of that. So, this S epsilon A comma B intersected with Y is going to be identified with this B epsilon A over here. So, therefore, so thus it is clear that this B sub Y is contained in this collection C because C contains B epsilon X for all X and R and conversely if we take some element B epsilon X in C then we can take we have X we can take the epsilon square around X with X comma 0. So, what I am saying is S epsilon X comma 0 intersected with Y is equal to B epsilon X yeah. So, thus C is also and of course, the empty set is contained in B epsilon because we can just take some S epsilon over here and when we intersected Y we get the empty set. So, every element the only elements of C is either the empty set which is contains B B sub Y or it is of the type B epsilon X which we have just shown is containing B sub Y. So, therefore, we have proved both inclusions which implies that thus this both these bases are equal which implies that the subspace topology is equal to the standard topology or not. So, thus we have defined the so, in the same way. So, let me just so, in the same way in the same way we may embed R 2 into R n as a hyperplane. So, I can take X 1 X 2 and this maps to X 1 X 2 0 and in the same way one can show that and one may show that the standard topology in R 2 is the same as the subspace topology. So, let us call this I on I of R 2. So, I is an inclusion so, using I we can identify R 2 with its image and using that identification we can take the subspace topology on I of R 2 and transfer it to R 2. So, and then R 2 now has two topologies and both these topologies are the same. So, we have seen the subspace topology and next we want to define product topology. So, this ends our discussion on subspace topology. So, the next thing we want to define is a topology on the product of two sets each of which have a topology. So, but before that we need so, next let me just write down suppose X 1 and X 2 are two topological spaces then we next want to define the topology on X 1 cross X 2. So, there is a natural way to do this and let us see how do it. So, let us first prove a lemma let X and Y be topological spaces and let B contained in the power set of X cross Y be the collection of subsets defined as. So, B is defined to be U cross V where U is in the topology on X is open in the topology on X and V is open in the topology on Y. So, when we say X and Y are topological spaces it means that there are topologies tau which we denote tau sub X and tau sub Y. So, this is the definition of B and what we want to show is. So, the content of the lemma is then B satisfies the two conditions in the proposition on generating topologies. So, I am referring to this proposition here oops that was in the previous lecture sorry. So, in the previous lecture we prove this proposition. So, let me just recall here. So, recall the two conditions. So, let us prove recall the two conditions we need to check where the following a when we take the union of all W in B then we get the entire set. So, in this case that is X cross Y and the second condition is suppose W 1 and W 2 are in B and X is an element in the intersection then there is a W in B such that X is in W and W is contained in W 1 intersection W 2. So, if these two conditions were satisfied then we saw that this collection B it defined the topology which we denoted tau sub B. So, let us just check that these two conditions are going to be satisfied. So, proof oops sorry. So, first note that since X belongs to tau sub X and Y belongs to tau sub Y this implies from the definition of B X cross Y belongs to B. So, therefore, clearly this union of this W W it contains X cross Y and of course, it is a subset of X cross Y. So, thus therefore, this thing is actually equal to X cross Y. So, this proves A this proves that A is indeed true. Let us check the second condition. So, for B. So, we are given W 1 and W 2 in B. So, W 1 comma W 2 are in B this implies from the definition of B recall that B is defined as above. So, W 1 is equal to U 1 cross V 1 where U 1 and let me just write and W 2 is equal to U 2 cross V 2 where U i's are in tau X and V i are in tau Y. So, let us take a point. So, a point in W 1 intersection W 2. So, looks like A comma B which is there in U 1 intersection sorry cross V 1 intersection U 2 cross V 2. Now, this implies that A belongs to U 1 and U 2 and B belongs to V 1 and V 2 which implies that A belongs to U 1 intersection U 2 and B belongs to V 1 intersection V 2. Now, since tau X and tau Y are topologies note that U 1 intersection U 2 is in tau X and V 1 intersection V 2 is in tau Y right. So, thus we have. So, we have A comma B this is in U 1 intersection V U 2 cross V 1 intersection V 2 which is an element of B right because this is in tau X and this is in tau Y and this is in turn contained in U i ok. This is contained in U i cross V i for i equal to 1 comma 2 yeah. In other words A comma B is contained in U 1 intersection U 2 cross V 1 intersection V 2 which is contained in U 1 intersection U 1 cross but this is W 1 and this is W 2 right. So, therefore, we can take this set as W. So, given any A comma B in the intersection we have found a W in B. So, is that A comma B is in W and W is contained in the intersection. So, this proves that B satisfies the two conditions to generate a topology and. So, generates a topology which we denote tau B which has B as a basis. So, this topology. So, this ends the proof of this lemma. So, this completes the proof of the lemma and let us just define product topology. Let X and Y be topological spaces then the topology defined on X cross Y in the previous lemma is called the product topology on X cross Y. So, a basis for this topology is given by sets U cross V where U belongs to tau X and V belongs to tau Y. So, we will end this lecture here.