 to compare the product topology and box topology on the Cartesian product pi x alpha, alpha belongs to J, we started with the following example namely it take our space r power omega this is nothing but r cross r cross r dot dot that is we have each xn say equal to r n 123 natural numbers then r power omega is nothing but pi xn n belongs to our index set is set of natural numbers and we have defined a map A f from the domain is r with usual topology to r power omega let us see what will happen if we consider the box topology on this product define this function as f of t equal to t, t, t dot dot this is an element in r power omega let us see what will happen to this function whether this function is continuous if we give if we assign the box topology on r power omega so here it is interesting to note the sets of the type a minus 1, 1 cross minus half, half cross etc open interval minus 1 by n for the n coordinate space this is nothing but this is pi unn varies over capital N such case we also write pi unn varies from 1 to infinity where each un equal to the open interval minus 1 by n to 1 by n which is an open set in our hence this let us say u an open set in the box topology. So let us recall that is box topology mean if we call it script beta suffix b set of all pi u alpha belongs to j where each u alpha is open in x alpha. So we are given is a collection this is a collection of topological spaces x alpha tau alpha then this will form a basis for a topology on the Cartesian product that topology we call it box topology. So here our index at j is set of natural number and for each n belongs to capital N un is open interval minus 1 by n to plus 1 by n. So hence this u equal to pi un n 1 to infinity that is same as this open interval minus 1 comma 1 cross minus 1 by 2 comma 1 by 2 cross etc. So this is an open set in with respect to box topology I mean if I did a basic open set but this need not be a this we cannot say it is a open set in the product topology. What is the basic open in the product topology mean x of finitely many the remaining u alpha should be equal to the whole space x alpha that mean here the whole space is r. So if we consider the box topology this is an open set in r power omega with box topology that is this u belongs to tau suffix b. Now what will happen to the inverse image if it is a open set mean we want that our this is an open set f is continuous in inverse image of every open set is open. So now consider this particular open set what is inverse image f inverse of u by definition that is an element in r. So we have u open here then the inverse image is set of all t belongs to our domain r such that the image f of t should belongs to u that is same as set of all t belongs to r such that by our definition f of t is t, t dot dot this belongs to u u is nothing but pi u n is minus 1 by n, 1 by n n belongs to n or we can write n varies from 1 to infinity this implies hence t belongs to here mean this will implies that is set of all t belongs to r such that t belongs to see each image. So belongs to minus 1 by n, 1 by n here all are same so t belongs to open interval minus 1 by n, 1 by n this is true for every natural number for all n in capital N. So that gives this is that is t belongs to here, t belongs to r such that modulus of t I mean r this modulus of t is less than 1 by n, this is for all n and this is possible only when this imply t is 0 this such a t that means this set is nothing but singleton 0 so which is not an open set in r with useful topology. In fact no countable set will be open because some subset of r is open means each point is open is an interior point in fact 0 is the only point it will not contain any neighborhood any open set containing 0 will be an interval containing 0 and contained in that set but that set is singleton 0 which is not an open set. So inverse image of this particular open set u is not open in r with useful topology and this imply that is this is not open in r and hence this gives this implies here r with useful topology to r power omega with box topology cannot be continuous is not continuous. In fact that is the reason is we allow the even component very small open set when n is large this is hence to in fact this approaches to 0 may not still contain small intervals. So with respect to box topology even such a nice function is not continuous now let us see what will happen let us see what will happen when we consider when the same function when f is r with useful topology to r power omega now we change here our topology is product topology. So now in fact not only the domain is particular r and here r cross r cross etc now in place of that take any topological space y. So take any topological space y and a collection of topological space x alpha so y any topological space say y tau topological space this is given x alpha tau alpha alpha belongs to any arbitrary index set a collection of a collection of topological this I am writing for topological spaces. Now consider a function f from the domain y to the product Cartesian product ix alpha alpha belongs to the index set j and what will happen if we consider the topology tau on y here we consider the product topology. So we will use the another notation here let us say tau prime y tau prime and this tau product topology product topology on our x that is the Cartesian product sometime we just write pi x alpha and the index set is understood. Now we already know that what is the for fixed for each fixed beta belongs to j we have this p suffix beta this is the beta coordinate function that is defined from pi x alpha alpha belongs to the index set j to the component corresponding x beta this we have seen that this defined as p beta of an element here is f that images f at beta normally this we also write x suffix beta that belongs to capital X suffix beta. What is the relation between now this f now we have we have f from y to pi x alpha alpha belongs to j. Hence for each alpha belongs to say that for each alpha for each alpha and for here we fix y in y for y in y what is this f of y the image this belongs to the Cartesian product f y belongs to pi x alpha alpha belongs to j that mean f y itself a function any element here mean that is f y is a function from this imply this f y is a function from j to union x alpha alpha belongs to j. So this we call it this is denoted by this imply this f y if we write this as x each element here depends upon f we call it f alpha at y this belongs to where I mean this is essentially for f y at alpha. So that is an element in x alpha so where f alpha y belongs to the correct finding x alpha. So that is for if we fix an alpha that is what fix alpha then we have a function f alpha from y to capital X alpha which is defined as given by. So f alpha from y to x alpha which is defined as defined as f alpha at y equal to this x alpha if we write the notation f alpha y this f y if we write f y equal to x of x alpha alpha belongs to j that mean x alpha this element this is nothing but f y at alpha equal to x alpha then this will be x alpha that is we have for each alpha we have a function f alpha from y to x alpha. So f is so in this sense we say that that is f can be written as f alpha alpha belongs to j. So now if we take the like when we take the co f we have a function f and then take the p suffix alpha the composite function p suffix alpha composition alpha composition f for fixed alpha belongs to j take y in the domain small y belongs to capital Y. So for y in y this is nothing but p alpha of f y f y is we have for again we will change here when we fix let us say beta belongs to j then beta composition f at y equal to beta composition f of y f of y is x alpha now the beta coordinate this will be x beta. So this is nothing but our this is what our f suffix beta at y correct so that is this is true for all y for all y in the domain capital Y. So the domain of this function beta suffix p suffix beta composition f and this component function f suffix beta for both function domain is y and the image p suffix beta composition f at y is same as f suffix beta at y this implies these two functions are same. So hence what is the result is so result now it is in a way trivial if you assume that f is continuous and we know by the definition each projection function p suffix beta is continuous and it is easy to see that composition of these two function is continuous. So f continuous f from y to pi x alpha continuous here we have the product topology here the given topology tau. So this we have called tau prime here tau. So then this is f continuous implies each function here we have denote each f suffix beta is continuous. Now in fact see what will happen the converse suppose each coordinate function f suffix beta is continuous will it imply the f is the given function f is continuous. So here we use the fact that how we have defined the product topology. So if we again recall how we define the product topology is union of s alpha alpha belongs to j where s alpha equal to set of all p alpha inverse of u alpha such that u alpha belongs to tau alpha and we call this as union of such as alpha over script s this will form a sub basis and the topology the smallest topology containing script s is our product topology. So and also we have seen that a function f from a topological space x to any other topological space x prime is continuous if and only if inverse image of sub basic open set. Suppose we know that there is the topology on x prime is generated by a sub basis then it is enough to prove that for each sub basic open set in x prime tau prime that inverse image f inverse of u is open in x with the given topology. So let us take any sub basic open set if we take a sub basic open set u let u belongs to script s then u will belongs to script s alpha for at least one or some alpha belongs to index set j. But what is our s alpha is p alpha inverse of u alpha for some u alpha where u alpha belongs to tau alpha this will imply u belongs s alpha mean again there will x is some u alpha belongs to tau alpha such that this u is nothing but p alpha inverse of u alpha this is our sub basic open set. Now so what is enough to prove is inverse image of this sub basic open set is open in our domain. So now take consider the inverse image of our function f now f inverse of this sub basic open set p alpha inverse of u alpha correct this is same as can prove that this is same as p suffix alpha composition f see that is here we use p alpha composition f the whole inverse this is easy to prove that this is nothing but f inverse composition p alpha inverse. So hence here the reverse f inverse of composition p alpha inverse this is same as p alpha composition f whole inverse of u alpha correct but this we have seen that this is nothing but our f alpha inverse p alpha composition f is f alpha hence that inverse of u alpha. So what is u alpha u alpha is open in x alpha and we have if hence our assumption is suppose each f alpha is continuous. So then that will imply f alpha inverse of u alpha is open in y. So that will prove that the function f is continuous see so note that it is not that we do not know when we start with a sub basic open set depends upon u we do not know u belongs to which s alpha it is not that s alpha is fixed this varies or it depends upon that u. So hence it is essential to have not just one particular f alpha each f alpha is continuous that is what we have proved is this is continuous if and only if each f beta is continuous that is inverse also. If f alpha from y to now we write only i okay if we write here alpha we can change the notation here x beta beta belongs to j this is of there are variables only here the product topology. So this is here we have f from y to this quotation product and f at y is f alpha or if we write this f beta of y beta belongs to j or we write f alpha of y alpha belongs to j and each f alpha is continuous imply f is continuous that is what each f alpha from y to x alpha continuous implies our f is continuous. So this what in the previous example we have consider a particular type of function f see so let us go back to what is that function we discussed was we have our y our y in that example is r our index at j is set up natural numbers and y alpha this x alpha for alpha belongs to n. So that case we use the notation n for n belongs to n xn is r and this pi xn n belongs to n we call it r power omega omega for the natural numbers. So with here and we define f from r to r omega is defined as f at t equal to now let us say this is an element that is what f1 of t f2 of t dot dot fn of t dot dot so here what is our f1 of t our function there here we have taken particularly f1 of t is t identity function f2 of t is t and so on fn of t is t dot dot so that is our function each component function is identity function fn of t is that is here fn of t equal to t that is we what is fn from r to r the identity function which is continuous hence from this result each fn is continuous will imply f is continuous not only this way even if we instead of this function we can take f1 of t let us say t say t square t power n dot dot etc. So each component is continuous then that f will be continuous even such a symbol function is not continuous if we consider the box topology on r power omega hence we observe that they are totally different so that is the product topology is always a sub collection but they are not equal so in fact you can this open set so you leave it as an exercise so this check whether this interval by open interval minus 1 by n this what we have used this is a basic open set with respect to box topology check whether is this pi on open set in r power omega with tau is product topology just enough product topology on this phase so this in many results see now here afterwards so you give any topological space so we can have a collection so we have seen for example this property called that is called host of topological space host of topological space so right now we have seen this definition so that is a topological space x is host of mean you give any pair x y x not equal to y then that can be separated by open set we will get some we can get an open set u containing x and open set v containing y so that this intersection is empty so then such a space we call it host of topological space so now what will happen now take a collection of host of topological spaces so we are given I mean we call it as a theorem that is product of host of topological space is host of see later we will see what will happen to product of next we will see when you say a topological space is connected that is what we have seen connected mean if we we should find the the whole space x and the empty sets these are all always open and these are the only sets which are both open and closed such a topological space is called connected topological space so if similarly we define compact separable and different type of topological spaces then we will ask the question you are given a collection of topological space having certain properties like here each x alpha top of alpha is host of then what will happen to the product space so same question we can ask what will happen with respect to box topology so some are comparatively easier easy to answer but some may be mean highly difficult rather to know the answer so here just to illustrate how to use the product topology or the box topology we will give this following result later we will see other interesting result so if let x alpha top of alpha alpha belongs to some index at j be a collection of host of topological collection of host of topological spaces topological spaces then and tau be the product topology and tau be the product topology product topology on x is pi x alpha alpha belongs to j tau be the product topology on this then each x alpha is host of implies that mean then this x with respect to tau is a host of space this is also a host of topological mean we simply say that host of space so we can ask what will happen to whenever we introduce a new concept the natural question is whether that concept or properties valid or too far on product space see here what will happen if we consider the box topology here it is trivial see what we have f not equal to g that is in the in our Cartesian product pi x alpha that then using that we got an open set u and a open set v belongs to our box topology with respect to we have got already with respect to product topology but what is the relation between product topology and box topologies this is weaker that is sub collection of tau suffix beta the box topology so what is the notation u v they are open with respect to box product topology see this we have proved already but u v belongs to tau imply u v belongs to tau suffix b with respect to box topology also hence we have got open set in the box topology having the same property that will imply even this result is true even in the with respect to box topology so that means product of house top spaces house top even in box topological spaces so as a symbol exercise see this is easy to see result that is we have that is if a 1 a 2 or that is we have x 1 tau 1 consider only two topological space x 2 tau 2 topological spaces consider a 1 subset of x 1 a 2 subset of x 2 then what will happen to a 1 x a 2 and take the closure and a 1 closure and a 2 closure now ask this question whether this closures are same your time or so yeah let us see that whether these two sets are same thing will stop and we will continue in the next class.