 Welcome to module 25, subspaces, so starting with a topological space we want to define what in the meaning of subspaces of this topological space. Once again we go back to the metric spaces for motivation, given a subset Y of a metric space X comma D, what did we do? We merely took D prime to be the restriction of D on X cross X, that was a metric on Y cross Y, Y you know restricted to Y cross Y, the metric will be on Y. Before that we had a normative space, you take a vector subspace this time and then restrict the norm, that gives you a subspace which is a norm linear subspace. Here what we do want to do? There are no functions here to restrict. On the other hand go back to the topology given by metric, the basis were open balls. The open balls with respect to D prime, they will be inside the subspace Y but the definition will be exactly same and therefore what happens is this Br X D prime set of all Y inside Y such that D prime of X Y less than R, same thing that D of X Y less than R, X and Y are points of Y, that is the difference. But then this is nothing but Br of X D intersection with Y. Only thing is you have to start with X and Y, inside capital Y here. So take a basic element here in the larger topology, larger space intersect it with the subspace subset, those will form a basic elements for subspace topology, the vector subs, the metric subspace topology. If that works fine then that should be the definition in the general case and we will check that, that is precisely what works properly and that gives you a definition here. So Br of Y D prime is Br of Y D intersection with Y for every point inside Y. So take a topological space and a subset Y of X. We make Y into a topological space by taking tau Y equal to all those U such that U is in tau intersected with Y. So I am directly defining this way and I verify that this tau Y itself define, verifies the topology, definition for topology, conditions for topology. Clearly members of tau Y will cover the whole of Y, they will not cover the whole of X because they are all inside Y. And empty set will belong to that, those things are easy. Even if you do not check it, F Y and A U if you check that is enough. Finite intersection of say U1 intersection Y, U2 intersection Y, U3 intersection Y is nothing but U1 intersection U2 intersection U3 intersection Y. But intersection of U1, U2, U3 etc is inside tau, therefore this will be inside tau Y. Similarly for the union. So checking that this is topology is just elementary sectarian. So this is called the subspace topology on Y. So we will have lot of instances wherein we have to use subspace topology. Each time what we do is take an open subset in Y that will look like U intersection Y where U is open ZX. So we have to keep on writing this. So we will not decorate tau and tau Y and so on, we just say Y is a subspace of X and do not mention the two topologies tau and tau Y which are behind the scene. But if there are different topologies then we have to mention this. In the beginning at least we shall mention it till we get used to this concept for a few for a few time till you get the inclusion map is one of the fundamental maps here between a subspace subset and the given set. Luckily this is a continuous map. Indeed more generally take any continuous function from X to Z restricted to Y. The restricted function or original function is continuous today imply the restricted function is continuous. Here X to X identity map is continuous restriction map is what inclusion map. Suppose you have proved the inclusion map is continuous inclusion from Y to X followed by F is nothing but a restriction. Composite of two continuous functions is continuous so that will verify this one. If you verify this one continuous verification of this one follows if you have verified this one that follows. The verification of any one of them is as easy as anything else. Why start with an open set here take the inverse image that is open to come to Y what is F inverse of an open subset here when you take the restriction map. It is just take the full inverse image and then take the intersect it with the subset Y that is all. So that verifies. Now the process of taking this induced apology whatever we have subspace topology subspace topology is also called induced topology is a transitive operation in the following sense. Take a subset Y and another subset of Y let us say Z. Z contains a Y contains a X. Then you can have induced topology on Z as a subspace of Y. You can have induced topology of Z as directly subspace subset of X. So Y is subspace of X, Z is subspace of Y so that is like this is related to this is related to this that is a process. It is same thing as directly Z subspace of X that is the meaning of transitivity. The induced topology on Z by the topology on Y which itself is induced from X the same as that induced directly from X. This is a direct consequence of the fact that I have to finally take a subset of X here A intersected to Z which is same thing as one step A intersection Y first and then taking A intersection Y intersection Z. This is true for all subsets of X in particular when you have open subset you get thus for any Z contained in Y contains like what we have is the tau Z is tau Y Z. The tau restricted to Z the same thing as tau restricted to Y restricted further to Z. In general an open set in a subspace in the subspace Y need not be open in X. However if Y itself is open in X then it is easily seen that every open set in Y is open in X as well. Why because every open set in Y by definition is U intersection Y where U is open in X. If you start with Y as open U intersection Y will be open in X. So that is a simple reason here. Similarly a closed subset in the subspace Y need not be closed subset of X again X again if Y itself is closed then by taking complementations every closed subset of Y will be closed in X also. So closed closed will be closed that is transitivity open open that is open that is also transitivity. So subspace taking subspace is very strong with transitivity relation is here. The last remark implicitly warns us that we need to reexamine various concepts such as closed interior limit points boundary points afresh in the context of subspaces. What is the last remark a subset which is open inside Y in general may not be open inside X. If you want to make it open always then you have to take Y as open but then closed subspace may not be accurate. Now suppose Y is open inside X but now you take a closed subset of Y that may not be open inside. So every time you are taking interior limits boundary and so on you have to be extremely cautious about what is going on. So you have to freshly check whether the old definition for Y itself whether something is working or not. So let us have some elaborate notation here and study the closure properties and interior properties and so on for a Y in the light of this definition subspace to powers. For any subset A of Y let us denote the closure of Y with respect to tau Y everything happening inside tau with CLYA. All that because we are worried that we are having an ambient thing the language of the larger space the country has to be respected here. It is like X is the country and Y is a state inside that. So closure of Y inside A is by definition with respect to everything happening inside Y is tau Y and the same with respect to X namely by the old notation closure of A this will be with respect to X now for the whole thing. So there we are using simpler notation of course we could have closure of A with respect to X also here that is not necessary if there is one more say between Z and so on then we may have to write that Z also here and so on. The ambient the space the original space X the mother space X we will keep the old notation. Similarly interior of Y in A will denote the interior of A with respect to the topology tau Y. If you do not write this Y then it will be a interior of this one as a subset of X in the topology of X. Is the notation clear? This notation will be used only when there is a confusion when there are two different subspaces are involved interior of what where are you taking it? So this is the lemma start with a subset A of Y all of them inside X that is the assumption already. A is closed in the subspace Y if and only if A is equal to B intersection of Y for some closed subset B of X. This was not the definition for us. Okay something is a closed subspace of Y if and only if the complement of A in Y is open in Y from that I want to get this thing namely directly coming from X what I have to do take a closed subset of X intersected with Y just like you have defined open sets you could have defined closed sets also that is the gist of this property one. The second property says the closure of A with respect to Y is just the closure of A intersected with Y if Y is closed okay sorry this is for Y and if Y is closed then the closure of A is actually closure of closure of A with respect to Y is actually closure of A there is no need to take the intersection. In particular a closed subset of a closed subspace is closed Y I am taking in particular suppose Y is closed then when you take A is closed inside Y this closure of Y A will be A okay that means A is equal to closure of A that means that A is closed in X. So this part follows so we have to show this part namely closure of Y A is equal to closure of Y intersection Y closure of A intersection Y and if Y is closed then this is okay let us prove these things by definition A is closed in Y if and only if Y minus A is open in Y right that implies again by definition that there is an open set U in X such that Y minus A is U intersection Y now put B equal to X minus U X minus U is a closed subset of X and all that you have to check is now A is B intersection Y okay so set theoretic thing you have to verify all right B is X minus U okay so something intersection with Y means it must be first of all point of A it is not in Y but not in U but in Y okay second part take X as a closure point of A in the topology Y now given an open subset U of X by definition okay after Y U intersection Y is open now what is the closure point closure point means now X will be inside U intersection Y okay first of all so U intersection Y is a neighborhood of X inside Y right Y X is in the U intersection Y because X is a point of A of Y itself because it is in this topology so X is already inside Y okay so take a point X inside U inside U it will be automatically inside U intersection Y therefore U intersection Y intersection A is non-empty okay why because it is a the closure of this every open subset containing X will intersect A that is the meaning of this but then this intersection is non-empty means U intersection A is non-empty all right therefore X is the closure of A I put Y because I started with X as a point of Y so closure of A with respect to Y is contained inside closure of A intersection Y now I take the other way inclusion start with a point here closure of A intersection Y take any neighborhood V of X in Y okay in Y something is a neighborhood means what by the definition of to Y you have a neighborhood U of X in X such that V is U intersection Y all right now it is closure point in the original topology X therefore U intersection V is non-empty but what is V intersection A it is U intersection Y intersection A right but that is same thing as U intersection A right because A is the subspace of Y so this Y intersection A is just superfluous it is same thing as A therefore X is in the closure of A so we have proved the equality all that you have to do is take the closure of A inside the larger space and then take only points of Y intersect into Y that will give you the relative closure closure of A with respect to the smaller space Y now suppose Y is closed in X A is already inside Y by the way all right so we know that closure of A will be contained inside Y right because closure of A is contained closure of Y but closure of Y is Y therefore closure of A is equal to closure of A intersection Y because it is already inside Y but by definition we now just shown by part 1 is equal to the closure of A with respect to Y okay this is this easier part here we have to completely go through the definitions on both ways okay so the whole idea is if you have done it now then you will remember it then you do not have to do it every time whenever you meet that occasion now similarly for interiors okay X be a typical space Y be a subspace of X if Y is open in X then every open set in Y is open in X also this we have already commented so I have put it here as a reference second part is more important for any subset A of Y interior of Y interior of A with respect to Y contains the intersection of interior of A with Y so here is no equality you see if Y is open then the two are equal okay there is no need to take introduction in particular an open subset of an open subspace is open okay so we have to verify this namely interior of A with respect to Y contains interior of A intersection Y okay actually many times this part the right hand part may be empty but this part may not be empty lots of examples are there like that okay so let us come to this one by the very definition A contained inside Y is open in Y if there is an open subset uses that A equal to U intersection Y since Y is open the intersection is open this is the first part I have repeated it the second part start with a point in the interior of A but this point is already inside Y intersection with Y I have taken so we have a neighborhood open neighborhood view of X such that X inside X such that U is contained inside A okay what is this one interior of A itself is now a neighborhood nothing else see interior of A is an open subset of X but it is contained inside A and X belongs to that means X belongs to U and U is a neighborhood of neighborhood of F right and it is contained there but then U will be also contained inside Y because A is a subspace of X subspace of Y A is a subspace of Y right space subspace of Y that is our exhibition okay if an open subset of X is already contained inside Y that will be automatically open inside Y because I have to intersect it with Y you are getting same U and hence U intersection Y is U is a neighborhood of open neighborhood of X in Y if you have found an open neighborhood of a point that point must be in the interior right so it is in the interior of A with respect to Y so the first part we have proved namely the interior intersection with Y is contained in the interior okay with respect to Y okay there is no equality here now we go to the second part namely suppose Y is an open subset A is contained inside of Y that is given okay so interior of A is contained inside interior of Y by the interior operator as its property but interior of Y is Y right all these things happening inside X therefore interior of A is contained in interior of A intersection Y but this we have proved is contained in interior of A with respect to Y okay interior of A with respect to Y is open in Y so one way of proof this is this part is contained inside here interior of A is open in Y right every interior A with respect to A is open in that okay therefore it follows that interior Y A is open in X itself because Y is open in open in X so it is the first part since it is a subset of A it follows that interior A with respect to Y is contained in the interior of A right after all this is a subset of always is subset of A now we have what is an open subset therefore it is contained inside interior of A because interior of A is the maximum maximal open subset largest open subset contained in A remember that okay so therefore there is an equality here this way equality this is the other way inclusion all right so I have already told you interior how interior can fail to be equality can hope so we will have lots of such examples let us look at r cross 0 contained inside r2 so let us take r2 as X r cross 0 as Y okay now take the usual topology in r2 now the mapping f from r going to r2 given by X comma X comma 0 it is a continuous injection right it is continuous and it is injection with its image what is the image of this map it is the fill Y namely r cross 0 okay so this is how we think of the real line r as a subspace of r2 this is one standard way right so it is it is equal to the image Y with the subspace topology on Y take subspace topology what is subspace topology everything you have to take open subsets in r2 and intersect with r cross 0 then f inverse of f inverse from Y to R is also continuous why because this map is nothing but the restriction of the coordinate function X comma Y going to X so that is the inverse image of this X going to X comma 0 it comes back to X only on X cross 0 so if we have taken the whole space X comma Y you take the first coordinate that is restricted to X cross 0 it will be again that being a restriction of a continuous function this function f inverse is continuous therefore f defines a homeomorphism of r with respect to Y okay the same argument and conclusion hold for 0 cross r also there is nothing very special about the first coordinate being 0 or second coordinate being 0 identical conclusions identical arguments are there all that you have to do is 0 comma Y you can write it so instead of Y you can write instead of X you can write it as Y that is all and then Y going to 0 comma Y more generally with the subspace topology you can take any line given by an equation AX plus BY plus C equal to 0 to make it a nice line as such not a not just a you know the whole space or that I have to assume that this vector AB is not 0 is homeomorphic to r under the mapping depends upon which coordinate A or B is not 0 AB not equal to 0 means either A is not equal to 0 or B not equal to 0 both of them are not equal to 0 that is also allowed but you do not know which one is not 0 so accordingly you have you can choose two different maps both of them may work with A and B are not 0 what is this first coordinate is it is the graph of this function okay Y going Y equal to minus AT plus C by this Y coordinate t 1 t comma this is something FT you have to solve for Y in this equation that is all here you solve for X in terms of Y here so I am instead of Y I am writing T here the T goes to T here second coordinate is Y this is your X coordinate okay so this is elementary in a test standard stuff we are solving for Y in terms of X or X in terms of Y of course that needs either A is not equal to 0 or B not equal to that condition has to be there according as B not equal to 0 or A not equal to 0 you have to notice that all these examples are closed subspaces of R tau okay so therefore if you take a closed subset of R then their image inside this line as a subspace of R tau it will be closed but open no no chance okay so under this I am asking you somewhat vague question here so that I want you to think about this so whole idea here is I have put you know subspaces if they occur with the familiar spaces see this is R cross 0 this is 0 cross R this is any line they are all homeomorphic to R okay they are all homeomorphic to R so this is the first example so now come to second example take X tau be any topological space take any two points A and B distinct points look at the subspace topology on singleton A and singleton B X is topological space okay so singleton A is a subspace subset so there is a topology on it singleton B is a topology on it question is what can you say about these two topological spaces okay separately and together also what can you say okay second now consider special case when X is the Sierpinski space with 0, 1 okay any any A to A comma B will do two distinct points but we have denoted 0 and 1 with 1 being the Sierpinski point which just means that the only open set containing 1 is the whole space 0 singleton 0 is open therefore singleton 1 is closed but singleton 1 is not open that is the meaning of Sierpinski point 1 you know that 0 is open in X whereas 1 is not open okay see here I did not get any subspace which is open subspace they are all closed subspaces and they are all homeomorphic so I want to give an easy example where in subspace is okay 1 is a closed subspace and it is not a closed subset or one is open other is not open what can you say about the topologies was the first question here whatever you have said there combine it with this special case here what you can call okay so come up with some answer and then check it with your tutors okay that is the game you have to do so let us stop here there will be many examples many more new things coming up and so on next time the next time we will have to do union of spaces from smaller things to larger things okay thank you