 In this lecture we will mainly discuss the notion of continuity between topological spaces that is we have x a topological space y a topological space x with a topology tau let us say y with a topological space topology say tau prime and we have a function f from x to y f is function from x to y. First let us recall how we define this notion called continuity for real line and more generally for metric spaces. We know that suppose capital A will assume that a non empty subset of r and f a function from a to r. Now it is fix x not belongs to the domain note the domain is a subset of r need not be a equal to r in particular it may be equal to r in general it can be any subset non empty subset of r. Then this function f is said to be said to be continuous at x not said to be continuous at x not if and only if it satisfies the following conditions namely you take any epsilon first for each let me every epsilon fix any positive number epsilon for epsilon greater than 0 there exist a delta greater than 0 satisfying the following condition namely such that x is in the domain x is in a modulus of x minus x not less than delta that implies modulus of f x minus f x not this is less than epsilon that is this is called distance between x and x not x is within delta distance from x not then f distance between f x and f x not is less than epsilon. So geometrically what do you mean by this is that is x so this is same as telling that so we cannot minus delta less than x minus x not less than delta this is if and only if this then if you add x not both side throughout x not minus delta after adding this x not will get cancel less than x less than x not plus delta. So that is we have x not consider the point x not minus delta and x not plus delta this that is the open interval x not minus delta to x not plus delta that is in fact a basic open set containing x not but now our domain is a that is what we call it as a space of r then you have to restrict that mean this have not just for any x will take this x should be the neighbor root of x but now over we are going to restrict our neighbor root only to a. So you have to take this intersection with a x is here so it implies again here what do you mean by this that is x that is x belongs to open interval x not minus delta x not plus delta intersection a then that should implies again here we will get minus epsilon less than fx minus fx not less than epsilon so your fx not both side we get fx not less than fx not minus epsilon we are going to add plus fx not this will go so plus fx not that is here this is our sender fx not here radius is epsilon fx not minus epsilon fx not plus epsilon that is whenever x is in a and x is in this neighbor root then that will imply our effect should be within this neighbor root then we say that f is continuous at x not that is if you have for example nice this function f this is our effect x not so then we fix the neighbor root of fx not then we should able to find a neighbor root u at x not in the domain so is that whenever x is in then say together you call it say this is neighbor root of x not in our space that is a implies fx fx belongs to v where in this case v is neighbor root of fx not that is fx not minus epsilon to fx not plus epsilon that is the open interval so keeping this in mind let us define what do we mean by a function f from x to y is continuous at a point x not exactly the same definition f from x to y and we are fixing a x not in the domain then this is our definition then f is f is said to be continuous at x not if only if it satisfies the following condition namely here we fix a neighbor root of x not fx not that we call it v so then we should able to find a delta mean we should able to find a neighbor root of x not that neighbor root is say that x is in u implies f of x is in v that is continuous at x not if and only for each open set in fact enough to consider what is called basic open set so for each open set say v containing fx not there for each open set v containing fx not there exist on open set this is for each open set v in our code of mind for each open set v in y containing fx not there exist an open set u containing x not that means it is a neighbor root of x not in x such that satisfying this condition x is in u implies the image fx is in v that is mean fx is in f of u that and f of u that says that f of u is subset of v for any neighbor root v of x not you should get a neighbor root u of x not satisfying this condition namely f of u is subset of v then we say that it is continuous at x not if this function is continuous at every point x in the domain domain is in this case x then we say that f is a continuous function. Now it is easy to prove that you leave it as a exercise prove that a function f from x to y is a continuous function that mean is continuous that mean continuous at every point of the domain is continuous if and only if you take any open set for every for each v belongs to tau prime that mean v is any open set in y for each v in tau prime this is called inverse image of the set v which is defined as those element x in our domain such that the image fx is in v this is a subset of x and this should belongs to tau that is inverse image of open set is open then from there it is straight forward to prove this. So if we consider in terms of close set so this is same as equivalently that is again exercise f from x but we can just write x to y meaning is there is a topology on y and also a topology on x f from x to y is continuous if and only if inverse image of close set is closed whenever f is closed in y implies f inverse of f is closed in the domain x. So this is in fact start with f here close set f closed then that implies complement is open see this is open this and f is continuous if you assume then this will imply the inverse image f inverse of f complement is open but this is same as this is easy to see that f inverse f whole complement that is open so that will imply this is open in x so that imply f inverse of f is closed in x that is f is continuous mean inverse image of close set is closed similarly we can again easy to prove that again the same idea if this condition is satisfied so whenever f is closed in y imply f inverse of f is closed in x then that will imply f is continuous. Now another more intuitively it is quite natural what you mean by continuous means whenever we have take a concept or we have a set so that is we have a function from x to y here you take any set a then what we mean by f is continuous mean you take x is what we have already seen that what is the closer of a set so all the points of a including then odd with that limit points of a so this we have is a union the collection of all limit points. Now continuous mean whenever take a any set any subset of x you can assume that it is say non empty whenever x is in a closer meaning x is closer x is closer to a that is x belongs to a closer that implies whenever x is closer to a f of x is closer to f of a that is x belongs to a closer implies fx belongs to f of a whole closer this is m is telling x belongs to a closer mean fx belongs to f of a closer this is true for every element so that is this is same thing as telling for every set a f of a closer is a subset of f of a whole closer that is quite natural whenever x is closer to a even we do not have any even no epsilon or delta some way we define the notion called closer that we define using the open set x is closer to a imply fx is closer to f of a now let us prove this result then we will see what do you mean by homeomorphism between topological spaces that is essentially that case both topological spaces are all similar so this will take it quickly we will see so we will we have already this two result inverse image of open set is open inverse or inverse image of a close set is close so continuous if and only these statements are satisfied so we have x tau y tau prime topological spaces it is given that f from x mean with a topological structure tau y tau prime then f is continuous f then this function f is continuous if and only if it satisfied the following condition namely for every a subset of x that been including empty set which is trivial for a subset of x for every a subset of x f of a closer is subset of f of a whole closer of course when a is empty left hand side is empty right hand side is empty so that is nothing to prove so now let us assume that first let us assume that f is continuous f from x to y is continuous then we will have to prove this condition say let us call one so take any non-empty set enough to take a non-empty set so let a be a non-empty subset of our domain x so consider let a a non-empty set subset of x then what we are trying to prove is that is write it so if you know just what is given what is to prove in fact that prove will become easier so to prove f of a closer is a subset of f of a whole closer so take an element in the left hand side so what is an element here is let y belongs to f of a closer then y will be equal to f at an element in x y equal to f of x for some x in for at least one x in a closer fx equal to y x is in a closer now what is to prove is fx this y is in f of a closer to prove we have to prove f this y is in f of a closer we have seen that from the definition it is clear that an element x belongs to any set say closer of a set if and only if for every open set u containing x u intersects in f is non-empty either it may be a point of f or if it is a limit point in any case for every open set u containing x u intersects in f is non-empty this is if and only if so start with a to prove that fx is here so we will start with a open set containing so start with an arbitrary open set so start with an open set say v containing f of x now we have seen that f is a continuous function so in particular so this will imply continuous at every point so v open set containing fx implies there exist an open set u containing x such that f of u is a subset of v so hence by assumption by our assumptions now fv is a continuous function v an open set an open set containing containing fx implies there exist on open set say u containing containing the starting band x such that f of u is contained in v so this we call it 2 so now we got an open set u which contains x and varies x x is in the closer of a say x is closer of a now this now x belongs to a closer u an open set an open set containing containing x that is the neighbor root of x implies u intersection a is non-empty so note that x may not be in a but u intersection a will be non-empty because x is in the closer so this will imply it will have at least one element say take so this is non-empty this imply there exist y belongs to y in the whole space x y is in u intersection a this means what y is in u as well as in a so y is in u y belongs to u so that imply f of y belongs to f of u but we have f of u is a subset of v so f of u is subset of v so one way we have f of y is in v and the other hand y is in a also also y belongs to a implies f of y belongs to f of a so this f of y is in y as well as in f of a so this will imply y belongs to v intersection f of a so we started with a y in f of a closer now and for every open set v containing y we have proved that v intersection f of a is non-empty so this implies this imply the here just we have to be careful about the notation so let us here already we are using say fx equal to y so we say that there exist some y say y prime in u intersection a so that f of y prime is belongs to v and f of y prime belongs to f of a that is we have got that is this imply f of y prime belongs to v intersection f of a so that mean for every open set v containing fx so we started with an open set v containing fx then we have proved that v intersection f of a is non-empty because this element is there so that implies our fx belongs to fx which is equal to y this implies fx belongs to f of a closer f of that is y that is y is in f of a we have started with that is y belongs to f of a closer implies y belongs to f of a full closer that implies f of a closer is a subset of f of a full closer this is the first part so now in fact the other part is quite easier once you assume that now this is what is the converse part is now let us assume that this condition is satisfied converse assume that assume the following that is a for any set a subset of x f of a closer is a subset of f of a full closer the quite natural here is because we have this closer of a set it is natural to prove that inverse image of a close set is close so that will imply it is continuous what is it to prove is f is continuous x to y is continuous so start with a close set in y so let us say f let capital F be a closed set in y what is our climb is f inverse of f should be closed in our domain x so let us call this set is a consider a equal to f inverse of f now by our assumption f of a closer is by assumption yeah consider this set now by our assumption f of a closer this is true for any set f of a whole closer hence what is our a closer is so the right hand side that is that is f of a closer is subset of f of a a equal to f inverse of f so just we are writing f of a closer is subset of f of this correct so but f of this is easy to see that for any function whenever f is a subset of y f of f inverse of f is a subset of f so the closer will be a subset of the closer but what is f f is a close set so which is equal to f for a close set always f closer equal to f now again here what we want is say again we have seen that you want to prove a set is closed what is enough to prove always a will be a subset of a closer so that is what we have to prove they are equal so a closer is subset of a imply a equal to a closer that means that set is closed so now take inverse both sides this imply that is we have we have f of a closer is subset of f this imply f inverse of f of a closer is subset of f inverse of f what is f inverse of f is our a correct so this is our a but again here for any set a in this case a closer is in the domain that set a closer will be a subset of this so we have proved that a closer is a subset of a so that imply a closer subset of a imply a closer equal to a that imply a which is equal to f inverse of f is closed in x so inverse image of close set is close hence f is a continuous function so what we have used here is this is for any sets so you have here can do this exercise for any x and y need not be even any just set x y sets non empty you can assume non empty sets f is a function from x to y then if you take a f in y for f subset of y imply take the pre image that is in x take f then that will be a subset of f this is the first condition and second for a subset of x a subset of x now you have to take f first f of a subset of y then bring it back f inverse then this is superset of a that is a will be a subset of this of course if the map is bijective or 11 and 12 they will be equal so this concepts are very important in topology that is the notion of continuity in fact you see that even in or I mean we have any number of examples from R to R or from a metric space to another metric space here in topology say consider even the nice topology R so let us say we call it either we use U for usual topology or sometime we write S for standard topology so that means let us say x equal to R with usual topology that is this is the topology generated by the collection of all open sets in R so same and we call it y the set is same but now the topology we call it lower limit topology so that is this is the topology generated by this collection set of all closed a open b say that a, b belongs to R a less than b this will generate a topology that topology is this even if you see the set is same x is underlining set x for x R y also R so we can define function define f from R to R so but when we just write this there is no topology so when there is nothing is specified it is normally we take the usual topology but here so what we mean is the domain with our x is this we can just write f is a function from x to y to y R with lower limit topology so define even the identity function define this as the trivial fx equal to x for all x in R question is f a continuous function that the inverse image of open set should be open so what will happen here we know that because in particular every basic open set is also an open set so in particular even take say v close 0 open 1 this is a open set in y because a basic open set an open set then v is on open set in R with lower limit topology so but what is the what is f inverse of a see in this case our function is identity function which is 1 1 and 1 2 the inverse image is set of all I will write it what is definition set of all x in our domain say that fx should belongs to v but what is this is same as set of all x in R say that fx equal to x that is how we have defined x in v that is this is the all those elements in v which are in R that mean v intersection R already v is subset of R so that is same as close 0 open 1 which is not an open set in R with usual topology this is but close 0 open 1 is not on open set in R with useful topology so we have called this so hence this function even the identity function is not a continuous this is not continuous sorry while defining yeah so Rajesh says pointed out some I mean writing mistake here the domain is R with useful topology co-domain is with lower limit topology yeah so our function the identity function is not continuous if we domain is usual topology on R co-domain is lower limit topology see what will happen to the other way that is essentially inverse function R say now if we define say g from R with lower limit topology to R with usual topology in fact all this concept though we define inverse image of open set is open once we know a basis it is enough to prove that inverse image of basic open set is open in fact later we are going to see that still smaller collection inverse image of sub basic open set is open so here now here take any open set arbitrary open set here now we have to prove it so that case any arbitrary open set so then that can be written as union so that we will see first for basic open set again f inverse of a b equal to a b this is open in R with lower limit topology in fact tau with the usual topology is weaker weaker means sub collection of tau here so this will be continuous if the function is just quickly tell what is a homeomorphism so homeomorphism f from x tau to y tau prime is said to be a homeo is said to be a homeomorphism like in any concept in algebra or analysis mean we have this structure is I mean structure on x and y they are similar that case we should have a bijective function which preserves the structure f is homeomorphism if and only if first f is bijective bi mean both 1 1 and 2 bijective 1 1 and 2 second f and f inverse both are we will just write it f from already given f from x to y with the corresponding topologies then f inverse will x is because it is a 1 1 function which is well defined from y to x both are continuous see then almost this two spaces will be similar because whenever now a is open if and only if f of a is open this homeomorphism imply f is bijective and we have a open in x that imply f of a is open in y only a identification the domain when they are in the domain a they go to co-domain the new name is f of a a is open mean f of a is open f of a is open mean a this is if and only if so using this we can see many topological properties will be preserved under homeomorphism see the simple thing for example if we consider even the open interval 0 1 that is our x we consider this as a subspace of r that means we restrict that usual topology to open interval 0 1 and y we call it open interval 1 to infinity again consider this as a subset of x so now we will see as an exercise defined or let f of x is a natural 1 by x see here x is in the open interval here whereas here this is spaces from 0 sorry from 0 and when x is 0 when x tends to 0 this will goes to infinity whether smallest when x is greater than 1 less than 1 it is greater than 1 so it is open 1 to infinity that is what we say that this in fact we can prove that this is a bijective function and f and f inverse are continuous correct so it is a homeomorphism yeah so just we will quickly again some more exercise which we which I should have given in the last lecture just before closing this lecture so in the last lecture we have proved if a topological space is host of x is a host of topological space a subset of x then an element x in x for x in x for x in x x is a limit point of a if and only if for each open set open set u containing x u intersection a is a infinite set in fact what we assumed one way is if it is host of we prove that in a host of space every singleton set is closed and that imply every finite set is closed then we assume that suppose for a for some x in a prime take x in a prime suppose for some open set u containing x this set is finite so that set f is finite then the compliment that mean finite set is closed compliment is open using that we concluded we could prove that we could get a contradiction see that essentially what we have used there is only the every singleton set is closed so that mean this result is also true for a T1 space that is that mean x is T1 then for x in x x belongs to a closer if and only this is an infinite set this is you can easily see also try to give a direct prove for even without using first you do not look at prove that x is in a host of space every singleton is closed then using that we prove that theorem so now directly prove that without using that like that every finite set is closed so let us see in the next lecture thank you very much.