 Welcome to module 24 of point set topology part 1. So after discussing some kind of general methods of obtaining new topologies with basis and some basis and several examples of that type, we will continue one more very important example again. Consider the Euclidean space Rn with the usual topology. We know that it is induced by several equivalent norms and thereby by linear matrix which we denote by d1, d2, etc. d infinity, out of which the so called round metric, Euclidean round metric given by d2 of xy equal to square root of summation of xi minus yi squared, this being the central which is called a usual topology. All of them give the same topology is a God gift for us. Let us have a look at the picture for this one just to recall you, just to feel good about it. So this was the topmost one here was the d infinity ball of radius 1. So these were some dps. This was this circle was d2 and this was d1. So like this there was such a relation and we have discussed this one quite in depth. Now let us concentrate only on the square and the circle, the square corresponding to d infinity and the circle corresponding to the d2 norm. Just again I recall this one namely this picture tells you that inside the circle you have squares, inside the square you have circles, circles, squares, squares and so on. This is centered all at the say one single center that is not even necessary. You can take any point here, you can put a inscribed circle and from there inscribed center and so on. At every point it is true. So what does this mean? If you want to control some phenomena by the circle, control it by the square automatically it will get controlled by the circle. Control it by square it will be the circle and so on. And vice versa if you want to do this one you can go to control this one. So this is what we have used in several times in elementary mathematics like in complex analysis we said a complex valued function is sequences convergent if and only if both the sequences of real parts and imaginary parts converges if and only if. So this is what is going to happen. What we have what is this? If you look at all the all the disks open disks centered at various points that forms a base for the usual topology. On the other hand the same thing is true for all squares centered at various points open squares. So both of them give you a same topology, both of them are bases. So in terms of our modern new terminology these things are bases for the same topology. So I repeat B1 be the set of all open disks with respect to the D2 metric we have seen that this forms a base for the virtual topology. On the other end the B2 of all open squares also forms a base for the topology these are you can call them as unit disks in R infinity or in D infinity topology. Indeed the above picture tells you that the topology is the same. Same means what equal to the usual topology. So here is a general theorem first of all I want to say suppose B i i equal to 1 and 2 be any basis for x i and tau i for i equal to 1 and 2. Then the collection B1 join B2 read it as B1 join B2 collection of all members of subspaces of x1 cross x2 I am taking 1 B1 first part is B coming from curly B1 and second comes from curly B2 B i such that B i to take their characterization product. So this will be a subspace of subset of x1 cross x2. Some author just write B1 cross B2 here and that can cause huge confusion. So I have made this some funny notation you read it as B1 join B2. This is a base for a unique topology tau on x1 cross x2. This is a general statement now motivated by our old you know this observation take x i is to be R in both the cases tau i will be given by open intervals open interval cross open interval is also a basis precisely what I wanted to what I got the lesson from here. So that is what I am generalizing it here now. The claim is that this will be a base for a topology. So you have to verify those two conditions B1 and B2. Moreover look at the family tau 1 cross tau 1 join tau 2 which is same thing as again by definition I am repeating it here U1 cross U2 where UIs are in the tau i. This will be also a base for tau the same tau in particular if you choose different bases they do not give you different tau because this tau 1 cross tau 2 is the same okay for all of them this is also a base. So this topology tau is independent of what basis you choose because it is equal to base topology generated by this tau 1 cross tau 1 joint out okay. So I have to do two things here first of all this B1 join B2 I have to show is a base and then I have to show that tau 1 tau tau is also the same base that is all I have to show okay two things I have to show. So given X1 X2 you know X1 cross X2 because BI's have this property B1 and B2 have the property I can choose you know BI inside curly BI such that XI inside BI then X1 comma X2 will be inside B1 cross B2 that is an element of B1 join B2. So condition B1 is verified. Similarly take two members B1 cross B2 B1 prime cross B2 prime inside B1 join B2. Suppose this point X1 X2 is in the intersection what is the intersection of this product it is nothing but the first coordinate must be always in B1 as well as in B1 prime the second coordinate must be in B2 as well as in B2 prime which is same as B1 intersection B2 prime cross B2 intersection B2 prime this is pure set theory okay. Then from property B2 for each BI we will get some AI inside BI such that X is inside AI inside the intersection BI intersection BI prime for AI put one end. Then A1 cross A2 will be inside B1 join B2 because A1 A1 is in B1 A1 is in B1 and A2 is in B2 right. So we have point X1 X2 belonging to A1 cross A2 contained in the intersection of these two products and that is a member of B1. So this verifies B2 okay. Therefore B1 cross B2 is a base for it Apollo X1 cross X2 which we shall denote by tau. Now I have to show that this tau is the same thing as the base the topology generated by tau 1 join tau 2. The tau 1 join tau 2 is also a base the proof is exactly the same instead of B1 prime B1 curly B1 curly B2 we write tau 1 P2 all these things are true for same for tau 1 join tau 2 also because tau 1 and tau 2 are also basis for for their own topology their topologies any topology is also a base for itself okay the same argument to show that it forms a base. So if you write tau prime as the corresponding topology generated by this family okay I have to show that this tau prime is equal to tau. But now the base B1 join B2 is inside this one therefore tau will be inside tau prime one part is obvious. Now I have to show the tau prime is inside tau okay to see the other way inclusion it suffices to show that this family the base itself is inside tau we have observed that one then the minimal topology generated the smallest topology containing that will be also inside tau okay. So I am going to show that these families contained inside tau is the same thing as taking one open U1 here and open U2 here and take the product that must be in this topology. So let U1 belong to tau 1 U2 belong to tau 2 okay means what U1 is union of B1Js where all these B1Js are in curly B1. Similarly U2 is union of B2Ks where all the B2Ks are in this curly B2 because they are the basis for tau i respectively okay. Then U1 cross U2 it is the product of the unions which is the same union of the products you have to take product of one element one member here with another member one member another member all all combinations you have to take that is summed over both G and K completely free but this is a union and each of them is inside tau alright so union is not okay. So what we have proved so far is that if you fix basis for tau 1 and tau 2 for topologies and then take the corresponding basis you know B1 join B2 you get a topology and then that topology is independent of what basis you have chosen okay. Moreover we have got an explanation how the points of how the members of this topology tau will look like on the product space X1 cross X2. The topology on X1 cross X2 obtained in the previous theorem this way is called the box topology okay. So in dimension 3 it is actually it looks like a box that is why it is called box topology higher higher dimensions also okay. Once you have got it for X1 cross X2 you can imitate it for X1 cross X2 cross X3 you can iterate it any number of times. So for finitely many products also this will be very very defined what will be the basis take basis for each of them take U1 cross U2 cross U3 cross UN. If you don't have basis take all open subsets in each of this tau 1 take that one that itself will not be topology but it will give you the same box topology as a basis. There are much more open subsets inside the product that is all you have to know okay. So you can take product of BJs J range from J this I am doing for now not only finite in finitely many also any family XJ A belong to J each of them is given a topology and then each topology has a basis BJ then you can take these product of BJs J range over J where BJ range for BJs take all such elements that will be a base for a topology and that topology is box topology on the product the proofs etc it will be the same there is no saturated problem here at all. So I have already told you I have we have motivated I am just repeating it. The above discussion in the example 2.2 I am most saying that the box topology and R2 is the same thing as usual topology given by the Euclidean metric. We should be noted that when we are dealing with an infinite family of topologies there is something called product topology which is somewhat different from box topology. Though we say the set is calculation product we are careful enough to call this box topology. Later on we will introduce this what is exactly product topology. Okay you have to wait a little bit for that. See the two things coincide is not a consequence of this one. We have we had observed that in RN and so on and we pass on to just the maximal topology we imitate maximal topology because the round thing cannot be mutated unless you have a metric in the general cases there is no way of taking squares square roots and summation and so on there is no way. Okay the round thing cannot be mutated. So that is why we are we are thankful to this maximal topology which can be mutated. Okay so that is all for today I have few exercises I will just go through these things not the solutions of course. The first thing you have to observe is that in this box topology the projection maps x1 comma x2 going to x1 or x1 comma x2 going to x2 they are they are denoted by pi 1 and pi 2 these are continuous. So I am writing for 2 as soon as for 2 verify it is verified for n of them but the same proof here works for all the cases namely infinite infinite product also. Next thing is again similar to that take any space z and a function into the product. Okay a function is completely determined by looking at the various coordinate projections coordinate maps namely what are they f1 f2 any map f z2 x1 cross x2 is given by what is happening it is first coordinate what is happening second coordinate. So f1 x comma f2 x so both of them must be continuous if and only if the function is continuous. Okay so verification this one is very straightforward that we once you do that you will get more familiar with the what is happening. Remember to verify that something is continuous you take a base here and check that inverse image of that one is open here. Okay here you do not have you have to be a little more because here is the original thing is there so the base new base has come here so this you have to do directly. Now here is another example of a base coming from you know topology coming from a base by declaring a base this is not a nothing to do with box topology though. Take the polynomial ring in n variables over r or c whatever let us take it c for a while here okay x1 x2 xn okay you can add two polynomials you can multiply the two polynomials you can scalar multiply them and so on this is a actually a vector space as well as a ring so such things are called algebras over this field k kc or c okay for each f in a you can assign a subspace subset of c power n what is that namely all those points x1 x2 xn varying the function f the polynomial f does not vanish I have written out uf earlier I had discussed those which are actually zero so it is a complement of the zero set okay just now we know that if I take the Euclidean topology or the product topology here because each polynomial is continuous these things are open in the product topology okay Euclidean topology but now we are not going to take the Euclidean topology or the product topology here what we are going to take is show that this collection of you know these uf's where f raise over all elements of way forms a base for an topology in cn the important thing here is to verify that uf intersection ug is again a member here remember this just means that this family is closed under finite intersection it is a strong property so once I have told you that you do not there is nothing to verify you have to quote that theorem that is all to say that this is a base so this topology which gives you the topology given by this base is called the cherished topology okay so this is sacrosanct used in algebraic geometry all the time all right the special case n equal to 1 you know this cherished topology is the same as the co-finite topology remember what is a co-finite topology a set is open if it is only its complement is finite so verify this it is very easy n equal to 1 case only n equal to 2 3 etc it will not be co-finite topology okay so here is another interesting topology on r itself look at all half open intervals a comma b where a is closed the left side is closed okay look at all these I will call it as l the collection l this l is a base for a topology on r what you have to verify you have to verify b1 and b2 remember that okay union of all member is equal to whole of r is obvious given any a I can I can always take something smaller than a and then write say a prime comma b so so I will think really cover the problem intersection of two such things you have to see what happens that is it okay so we shall denote it by l itself okay since we are not going to emphasize on these notations too much because notation will may change from author to author I do not want to float one more notation here okay so we will call it semi interval topology more elaborately we could call it left closed right open interval topology if I do this one you will not find it in any literature in literature it is called semi interval topology the semi interval topology can make sense for b closed and a open also that will be entirely different topology if you care you you can say that it is similar but it is it is different topology on r open subsets with respect to that will not be open with resistance on okay instead you could have taken intervals of this form also and we would have got a space which is homeomorphic to l means what it is a topology here there is a topology here if you take x going to minus x something of that nature reversing reflection that will take you the closed interval this way will be image will be a closed until that way the homeomorphism but they are not all the same thing show that this topology l okay is finer than the usual topology on r so it has more open sets all open subsets of r they are there last question is here does it come from any metric keep trying okay before the before the course ends we will get a solution no problem okay and then there is another one interesting one here on the set of integers let us define a topology as follows this I am not introducing it for as an example of a topology this I am introducing as a entertaining application of very elementary topology to produce a very very ancient theorem in number theory okay so what is that this is due to frustansberg frustansberg was just a graduate student when he he published this one in 1955 okay so let us introduce a topology here a b a not equal to 0 look at the arithmetic sequence n a plus b a n range over all the integers so let us denote it by s of a b this is a sequence arithmetic sequence okay n times a plus b and belong to 0 n n n n belonging to the integer let b be the collection of all arithmetic sequences s a b for each a and b there is one that b is the collection and by the very notation what we are going to do this b is a base for a topology let us call this topology f topology in honor of frustansberg okay show that each member of b is closed in f topology this is strange thing right basis basis elements are open but here they are closed also okay I do not have to show that they are open this is by very definition b is contained inside tau show that no set is no finite set is open in f topology so this is not so strange finite sets are not open inside are also okay now comes show that z minus minus one plus one it throw away these two integers minus one plus one okay then you can write the entire z minus that as union of arithmetic sequences where p is a prime you do not have to worry about a b's p is a prime and b is 0 see I have chosen s a b right so a is may have taken as prime number b is 0 look at only those arithmetic sequences that minus minus one plus one is union of these things okay this elementary observation is as soon as I have done that you can conclude that the set of primes inside z inside actually a natural number is infinite okay so this is entertaining this is nothing deeper here but for a student to have found out this is a wonderful thing so it appeared in ams notices so let us stop here and do next time next time we will do another important thing subspaces thank you