 Welcome to NPTEL NOC, an introductory course on point set topology part 2. Today we will study some more examples of this S0, S1, S2 and so on. Actually we started same examples last time they were all S2. So, we will continue that here is one of the important example the Hilbert cube. Consider the Hilbert space L2 over a countable family of set points namely you can take the natural number. So, recall that the space L2 consists of square summable sequences of real number and L2 norm there is nothing but take the sum of all the squares and then take the square root. For topological reasons we will have a different look at this one. Indeed we are now going to consider a smaller set here namely let us start with the product of you know infinitely many copies of the interval j which is minus 1 to plus 1. So, let us let us say let us denote it by x, x is j power n. There are many ways to put a metric on it. So, what we will do is we will use this L2 metric L2 norm and then there is a metric L2. So, what we will do is we will take a standard embedding standard means it is convention after all. So, a lot of people are using that especially in analysis. So, what we will do we take a point x here which is xi where each xi is between minus 1 and plus 1 right. So, we will send it to x 1 the second coordinate x 2 will be divided by 2 the third coordinate will be divided by 3 and so on the nth coordinate will be divided by n. So, this is an arbitrary sequence of real number between minus 1 and plus 1. So, this is another sequence. Now, what happens if you take the sum of square that is convergent the sum is convergent. So, this will be an element of L2, x itself is not an element of L2. It is easily checked that this map is a continuous bijection on the space what is that the image is all the coordinates modulus of xi first of all must be less than or equal to 1 I cannot go out of that but actually the ith coordinate should be in modulus less than 1 by i because look at the image here x n is already between minus plus 1 x n divided by n is less than or equal to 1 by n right. So, if you put this condition after multiplying by i it will be still inside minus 1 plus 1. Therefore, that will come will be a point of x. So, that will show you that this is a bijection all right continuity of this one is obvious. So, it is a continuous bijection but actually the way I have defined I am defining the ith coordinate I am just multiplying it by the integer i. So, even the inverse function is also continuous there is no problem which just means that this map p is a homeomorphism of j power n into a subspace of the Hilbert space. That subspace we are denoting by this curly ith and calling it the Hilbert q ok. The notation may be different but it is commonly agreed that this should be the Hilbert q ok. The name is quite standard but not the notation ok. So, now without reference to without reference to h all the time we can just write down the metric which is induced by this a phi on x itself namely given any point x and y inside capital X distance between x and y can be defined to be just take x n minus y n take the square divide by n square take the sum and then take the square ok. So, this is the metric I do not want to write the norm here because x is not a vector space any space which is homeomorphic to j n comma tau d namely what the metric induced topology here that is called a Hilbert q after all we are doing topology for anything which is homeomorphic to this is called a Hilbert q we are not very much particularly interested in the actual metric here but only on the topology and now we know that this topology is nothing but the product topology ok. So, quite often it is convenient to think of this one as just j power n with its this x can be thought of as j power n with its product topology ok this all about convention and notation the subspace q h of h of all points with rational coordinates satisfies s 2. So, this is what we want to say first of all it is a metric space now inside that it satisfies s 2 is what I want to say to see this let j n to j be the ith projection map coordinate projection let p be any point in q h remember q h is again all the coordinates of the point here will be rational ok take any point in q h take an opens of set containing p let v be a basic open set open set containing p that is open neighborhood such that p belongs to v contained inside u after all every open set you know will be containing a basic neighborhood around that point. So, neighborhood around that point this basic neighborhood will be of the form by the very definition of the topology product topology intersection of finitely many pi i inverse of u i where i is u i is an open subset of minus 1 plus 1 the coordinate space ok finitely many of them. So, there will be a maximum n I can include all of them obtain no problem right ok, but here I am going to choose this v such that each u i has this property namely the boundary of u i which will just consist of at most two points depending upon where the point is taken if point is in the extreme interior the boundary of u i will consist of two points both the points must be irrational that means boundary of u i intersection q is empty ok this is done in the interval for each for each i up to 1 2 3 up to n then you take the inverse image of u i under pi i inverse take the finite intersection that is it automatically it will imply that boundary of v intersection q h is empty see this v is a subset of the entire j power n or you can think of h whatever i am taking h intersection with q h is empty because the boundary point at least one coordinate will be irrational because here each of them has boundary irrational ok here you have to use the elementary fact in the product space is namely if you have x cross y and a subset a of x b of y boundary of a cross b is equal to boundary of a cross b union a cross boundary of b use it again and again several times to go into the finite case here finite intersection case you will get this here ok so that is all for each point you can choose a neighborhood such that the boundary is empty means what that neighborhood will define a partition a separation of the point with the complement namely start with u right they are separated out so that is the property s 2 so we have we have verified we have got another example namely subset of all points with all coordinates rational inside the Hilbert q is also s 2 exactly similarly we can show that i h inside h is also you know it is also s 2 what is i h all the coordinates of all the points are irrational ok so you can reverse the rule here that is fine in contrast the subspace q l of all points to coordinates are all rational in the entire Hilbert space l 2 does not satisfy s 2 ok so this is what we want to prove inside the smaller set namely compact set this h this is working but if you go to the entire l 2 so unbounded thing there s 2 is not satisfied ok so we will have examples of spaces which does not satisfy s 2 ok so once again how do you show that something is not s 2 it suffices to prove that the boundary of any bounded open neighborhood of 0 in q l is non-empty ok so at 0 you can prove that you cannot have such a thing you cannot separate 0 by open subsets by close subsets do not contain 0 then over ok so I am going to show that take a bounded open neighborhood of 0 ok then it is a boundary is non-empty inside q l not inside inside a inside a little it is obvious inside q l you have to show it is non-empty ok so in other words you take the entire boundary intersect q l it is non-empty sort of ok so this is done as follows so let me show you the picture here first so this is origin of course my picture cannot contain infinitely many coordinates so you have to think of this one as infinitely many infinitely many infinitely many one inverted time something like that so this is origin this is u is a bounded neighborhood of the origin I start with on the x axis in the first one a 1 axis or x 1 axis I take a point a 1 ok inside u maybe I do not have to take the 0 itself ok take one point be sure in the inductive process I want to do that something so be sure that this a 1 is not far away from the complement of u ok not far away from the complement of u that means what let us say distance between a 1 and the complement complement is a close set let us say it is less than one ok this distance is less than one next I keep this first coordinate a 1 move into the r 2 so keeping the first coordinate a 1 as it is I choose a 2 so that this point is again closer to the boundary so this was a 1 I have chosen so here I will choose it is 1 by 2 this I have 1 here I will choose 1 by 2 so distance between a 1 a 2 and the boundary and the outside of u complement is less than 1 by 2 how do I do that so you can open subset ok so I keep moving inside this arrow ok something all the entire arrow may not be there but there will be some point here as close as keep moving ok having chosen a 1 a 2 a n a n plus 1 will be chosen first a n a n coordinates will remain the same the nth coordinate will chosen so that the distance of this point to the boundary or to the complement is less than 1 by n ok so that is what I am doing here ok we are inductively constructed sequence P n is a 1 0 0 0 a 2 a 1 a 2 0 0 0 that is a 1 a 2 a n 0 0 0 of points inside u such that distance between P n and the complement is less than 1 by n it then follows once I have once I have got this one then follows that the limit sequence namely a 1 a 2 a n ok this P is actually in the boundary of u because all these things are inside u so this will be inside u bar but it is a distance between this P and complement of u is 0 because it is less than 1 by n for all n therefore this will be both in u and u bar and complement of u bar so it is inside boundary ok so how it is done I have already explained it to you ok so let me read that one consider a linear space P 1 set of all x n belong to L 2 such that all nth coordinate beyond that one are 0 that is the x 1 axis you can say clearly P 1 intersect both u and u complement see because u is bounded because u contains the origin so this line has to intersect u because it is u is bounded has to intersect complement so this is the property I am using on that subspace there will be a point inside u very close to u complement as plus f this is what I have to use so pick up P 1 a 1 belong to u such that which tells it less than 1 having chosen P n the same thing you get P n plus 1 it is all x i inside and 2 such that i th coordinates are all 0 beyond n plus 2 ok up to n plus 1 coordinates are there x i must be a i as chosen up to nth ok this subspace just now just a line now see if I fix up all nth coordinate only n plus 1th coordinate is free so that line ok will intersect this point a 1 a 2 a n it go passes through that and it will go out of u c because u is bounded so we pick up x n plus 1 equal to a n plus 1 and take P n plus 1 equal to a n plus 1 such that this distance all that I have to do this one is intersect this line with u and u complement so work inside the real line to get such a point that is all ok once the that coordinate is chosen properly distance between a n up to a n plus 1 and the new coordinate sorry a n and a n plus 1 is own depends only on this point because you have to take a i minus corresponding a i this will be 0 only a n plus 1 square has to be less than 1 by n square that is all ok so that is done so now we will have some theorems ok so what we have done we have shown one example wherein it is not s 2 also it is a matrix space it is an entire Hilbert space so Hilbert space does not satisfy the q l does not satisfy the s 2 property now it is a theorem if x is lindelow then s 2 implies s 3 remember that we have we have assumed that all our spaces are t 1 right from the first lecture on this topic right but just to remind you I have put t 1 in the bracket here ok if x is lindelow then s 2 implies s 3 the proof is somewhat similar to what we have done long back may be theorem 2.5 that a regular lindelow space is normal however for completeness we include the proof here ok because we are not exactly doing normality here nor we are exactly using regularity both are stronger hypothesis s 2 is stronger than regularity s 3 is stronger than normality so let us go through this proof carefully. Take f 1 and f 2 to disjoint closed subsets of x then x is union of their complements because they are disjoint ok therefore given x instead of x either x is instead of f 1 c or x is instead of f 2 c maybe it is in both does not matter one of them will will be there always. Accordingly choose a closed set u x around x such that this u x is equal to x intersects f j is empty accordingly if it is first one the x is inside f 1 c so I can choose u x intersection f 1 empty if x is inside f 2 c also u x intersection f 2 is empty so this is by s 2 the property s 2 assures you that such a closed subset exists ok for each point I have chosen a u x therefore this x is union of all the u x and each of them is open x is lentil of implies we have a countable sub cover here alright now we define a new family of closed subsets remember this is open they are both open as well as closed also ok so how do I do the inductively this kind of steps are same as what we have done improving a regular lentil of implies normality this step is same thing v 1 is u 1 v 2 is whatever you have taken then u v 1 ok you subtract that v 1 so it is u 2 minus v 1 like that v n will be u n minus whatever you have taken earlier namely union of all the iron to 1 to n u i ok I can write here v i is also because v 1 is u r v 1 union v 2 will be union union u 2 and so on so you can do this way this is also correct now what we have is each v i is disjoint from other v i v j so it is a disjoint family each v i is clop on why because being finite union this is also clop on and this is clop on open minus closed is open closed minus open is closed so this will be clop on so each v i's are clop on subset and we have written x as disjoint union of countably many clop on subset more over each v i remember here what is u n minus something so v i is contained inside u i and then v i intersection f i will be 0 because u i intersection f i is 0 for j equal to 1 or 2 accordingly you do not know where they are coming from there are two classes of them right depending upon whether it is a corresponding u x u i corresponds to one of the u x where x is inside f 1 c or inside f 2 c ok the entire family of v i is divided into two families accordingly ok according to whether v i intersects f i ok i equal to j equal to 1 or 2 j v i intersects f j is non-empty now take w 1 to be union of all those v i's such that v i intersects f 1 is non-empty ok and other things are w 2 namely all the v i's v i intersection f 1 is empty so I have taken all the v i's here some of them are here some of them are there and they are disjoint families it follows that x is a union of w 1 and w 2 w 1 intersection w 2 is empty because every v i here and v j here they do not intersect being unions of clop on sets w 1 and w 2 are both open that is enough actually they are clop on also and f 1 will be inside w 1 now ok because f 1 does not intersect any of them but f 1 is contained in the union so it must be inside w 1 f 2 by very definition has to be inside w 1 ok rest of them are w 1 ok so what we have done is writing x as disjoint union of two open set this is a partition this is a separation now right f 1 inside w 1 f 2 inside though that is the property s 3 ok in conclusion we know that a lindelow of t 1 space s 2 implies s 3 but we have already proved s 3 implies s 2 as soon as it is t 1 space so these two are equivalent however even if x is a separable metric space the three of them here earlier one s 0 s 1 and s 2 they are in equivalent ok proof is not all that easy in a t 1 space we have seen that s 2 implies s 1 implies s 0 so we assume it is separable metric space even then you cannot go back in the arrows that is the meaning of this one ok so we will have examples finally so here is one example which is not a metric space ok but it is an easy example start with a countably infinite family infinite set with a discrete topology that is my n tau now take extra points a and b so n disjoint in a and b that is my x ok now I want to put a topology tau hat on this x ok this will consist of all the members of tau along with all subsets y of x with intersect a b and such that x minus y is finite intersect a b means they should contain either a or b or both ok so this is finite but then the complement of that y inside x must be finite ok check that this becomes this definition makes tau hat a topology or x not very difficult which kind of thing we have done several times for each x if x is inside n then it is discrete topology so it is both open and closed in x because all the open subsets in tau they are intersect a way at all so they are all components if a singleton set is both open and closed that estimate must be component right any subject larger than a b is disconnected can you see that take a b itself ok as a subspace of n as a subspace of x what is the topology on itself as a subspace it is discrete space ok therefore singleton a b itself is disconnected therefore anything larger than a b will be also disconnected so this shows that this topology tau hat is totally disconnected all the points are singleton points are components all the components are singletons ok however this is not a t 1 space at all it is not even a t 1 space why because take any neighborhood of a inside x not a b a b if you take a subspace that is discrete but a and b as a elements of tau hat any neighborhood complement a finite any neighborhood of b complement is finite n is infinite so a intersection that u intersection v will be non-empty so that is similar to this co-finite topology therefore this is a not a t 1 space so this a and b are there just to make it non-t 1 space ok but this is totally disconnected space that is the reason we do not want to study such anomalous example that is why we are putting assume t 1 right in the beginning ok so I have repeated it here so the same thing will tell you that of course it is not outside because neighborhoods of n we are not disjoint in particular it does not satisfy sorry sorry this is t 1 but it is not t 2 ok I want a t 1 still it is not good that is the whole idea hence x is not house door that is the point because members of open subsets around a and open around b they always intersect ok if it is not house door it cannot be s 1 s 1 is stronger than house door ok so it is totally disconnected space this is not s 1 next time we will have between the metric space such examples but for that we have to work very hard but this is a very well known example in topology Knauster-Korotowski example ok so that we will study next time.