 Hello, welcome to NPTEL NOC, an introductory course on Point Setupology Part 2, module 40. So far we have prepared ourselves with so called global separation properties for the launch of dimension theory proper. That is what we are going to do today. Recall that a space satisfies S2 if and only if it has a basis consisting of global sets. Guided by the observations that we have in the classical setup, the real line with its usual topology has a fundamental system of neighborhood such that the boundaries are just two points which are 0 dimensional. If you go to R2, it has fundamental system of neighborhood consisting of open disks whose boundaries are circles which are one dimensional and so on. We now make an inductive definition of the dimension. Your key property is this S2. So we begin with dimension minus 1 which is nothing but an empty space. Indeed, only the empty space is defined to be of dimension minus 2. Of course, in all other cases, we start with a non-empty separable metric space just to remind you of that. We say x is 0 dimensional and then we write dimension of x equal to 0. If x has a base consisting of open sets with empty boundaries, you see the same thing as saying that these elements of this base are open sets. It is the same thing as saying that the space x satisfies S2. Empty boundary in terms of our inductive dimension is of dimension minus 1. So this is the key. We will do it like this. It is 0 dimensional if it has a base consisting of boundaries of dimension minus 1. Now suppose n is some positive integer and we have defined what is the meaning of dimension x is less than or equal to n minus 1. We say x has dimension less than or equal to n the next one if it has a base consisting of open sets u such that boundary of u is less than or equal to minus 1. Dimension of boundary of u is less than or equal to minus n minus 1. In that case, we write dimension of x is less than or equal to n. Now that is for less than or equal I have not defined what is this number is not number yet. The entire thing I have defined namely dimension of x less than or equal to n. What is the meaning of that? Namely x has a base consisting of open sets such that the boundaries of each of these open sets is of dimension less than or equal to n minus 1. Now we say x has dimension exactly equal to n. If the dimension first of all should be less than or equal to n minus 1 as defined earlier here and in addition dimension of x equal to is less than or equal to n minus 1 is not true. This I can rephrase as take the least integer n such that dimension of x is less than or equal to n holds. If you take the one lower than that it does not hold. So, this n is the least integer such that dimension of x is less than or equal to n. So, that least integer is called dimension of x. Finally we say x has dimension infinity if for every n dimension of x is less than or equal to n is not true. Now having defined dimension of x as a number I can always say that for every n dimension of x is greater than or equal to n or greater than or equal to n plus 1 greater than or equal to n plus 2 whatever it is for each n it must be greater than or equal to that is the meaning of dimension of x is infinity. Let us take a little time to recapitulate what exactly this definition means just logically. For example, you can easily see that being of dimension less than or equal to n to begin with is a topological property because it has been defined in terms of existence of a base if something is homeomorphic to a space that will also have the corresponding base. What is the property of the base boundaries of this basis have some dimension. What is the boundary under homeomorphism boundary of view will go to boundary of the image of the homeomorphism. So, inductively if f from x to y is a homeomorphism and dimension of x is less than or equal to n we can prove dimension of f x less than or equal to n inductively. So, where does the inductive stock at minus 1 when you have an empty set I need anything homeomorph to empty set is empty set that is all. So, there is nothing more than that. However, under arbitrary continuous function the dimension is not behaved well one would expect that dimension would hold on under a continuous function namely image of a continuous function will be of dimension lower than the domain. That expectation is true only if you qualify this one the function to be more than just continuous function quite often see one function will do that. So, analytic function polynomial function they all do that they will never increase the dimension of course we know that dimension can go down very easily. For example, you can take coordinate projections from r to 0 continuous function constant single thing is 0 dimension r is one dimension that is easy. So, what is weird is that just if you take continuous function you must be knowing the existence of what are called as Piano's curves. The Piano's curves are such that they are continuous functions from a closed interval to the product of two closed intervals to the product of three closed intervals to the product of any number of you know closed intervals. In fact, there are continuous surjective functions from a closed interval into the Hilbert cube itself. We shall not discuss that one here, but we just as a fact we are stating that. Let X be a space of dimension n for some finite dimension and positive. Then it has subspace is of all dimension 0 of course empty set is also there so I can include minus 1 also 0 less than 2 i less than 2 n minus 1 all integers smaller than n there are some spaces and minus 1 also you can include there is no form. How to start with a given point X in X in X there exists a neighborhood u of X such that boundary of u is precisely of dimension n minus 1. For each point there is a neighborhood because the whole has a fundamental system of neighborhood with this property. So boundary of u is dimension n minus 1 so I have got a subspace of dimension n minus 1 but now you apply to a point in the boundary. So this way you keep going down so you get all the dimensions. The above property is false if X is dimension infinity it serves one would expect dimension infinity to have all subspaces of all dimensions I am sorry subspaces of all dimensions right. That is not the case examples of infinite dimensional spaces whose subspace is of finite dimension are all countable. Countable sets we will see we have you see easily that they are 0 dimensional spaces okay or it may be empty also of course. So this is a paper of Hurwig himself so if you are interested in you can look at it and I have given the reference here. Let us now concentrate our attention on 0 dimension for some time slowly we will see we will have to use whatever you have seen into 0 dimension to develop it for higher dimension also. So let us first concentrate on 0 dimension. So the first result is if X is a 0 dimension space then so is every non-zero non-empty subset X prime of X as a subspace topology and the subspace topology okay. So it will prove it very easy being a subspace of a second countable matrix space X first of all X prime is also a second countable matrix space. So it qualifies for definition of dimension but you have already seen that S2 is hereditary okay. So X prime also that is why S2 okay. So that closed chapter namely X prime must be of 0 dimension 0 okay. Of course we have assumed that it is non-empty if we empty of course it is minus. All examples in our previous chapter right in the beginning you can look into example 8.3 they are all 0 dimension except for one of them we have proved all of them that they are 0 dimension in the sense that they satisfy S2. We did not call them 0 dimension at that time in the new definition S2 is the same thing as 0 dimension right. So only thing that we are left to is to prove that example 5 is 0 dimension that it has a base consisting of boundaries of dimension 0. Boundaries are empty or cloap and subsets so boundaries are empty. So let us prepare to prove that one and let us have some more theorems we are more useful than just proving that theorem that proving that example a countable union of 0 dimensional closed subspaces is 0 dimensional. Just now we had an example there you know we just quoted a paper that said every subspace of what is that the theorem here paper here you know every subspace of finite dimension is countable okay. So of course if you assume that these subspaces are 0 dimension countable okay countable then it was fine but that is not necessary each singleton is always 0 he is always closed. Therefore countable union of countable subsets countable set will be automatically of 0 dimension because it is singleton each singleton is closed. So that is the consequence of the theorem now okay if x is 0 dimensional sorry here if x is here. So if x is the countable union of 0 dimensional closed subspaces then it is 0 dimension proof is easy again if x is n rings 1 to infinity cn where each cn is 0 dimensional closed being a subspace of second countable space each cn is second countable. And hence each of them is Lindelof therefore our earlier theorem 8.7 tells us that each cn is s3 remember s2 implies s3 under Lindelof that was a theorem okay but then another theorem says that x satisfies x satisfies s3 because it is countable union of these things okay so that was another theorem I will just quote it here x is a p4 space x is union of it is a countable union of closed sets each satisfying s3 then x also satisfies s3 so that was a theorem. So you see all these background material we have proved that is why our life becomes easy here. So what we have got it here is a countable union of 0 dimensional closed subspaces 0 dimensional of course now it looks easy but we have to use both these theorems all right as a corollary if you have union of just two 0 dimensional subspaces just one of them is closed then the union is 0 dimensional of course it is not direct consequence but how do we use the previous I think countable union of closed sets that is the key. So union of two things one of them is closed all that you have to do is the other one you must be able to write it as countable union of closed subsets that is all okay being 0 dimension all those things will be 0 dimensional also because their subspace is of a 0 dimensional space okay so that is what we will do now start with x equal to union of two closed two subsets one of them is closed both of them are 0 dimensional. So let us assume c1 is closed okay look at x minus c1 that is an open subset of a metric space every open subset of metric space is a countable union of open sets f sigma set okay so write x minus c1 is ironed 1 to 1 to infinity sorry countable union of closed sets ironed 1 to infinity c i prime each c i prime being closed in x okay so you must remember this one we have in the proof of that a metric space is compact we had done lots of metric theory there remember that one of the thing was to write an open set as a union of countable union of closed subsets okay. So I will show you what it was just to recall this was precisely this a n's okay remember this a n's were defined like this okay a n bars will be greater than 0 right if you put equality also here those will be closed subsets and then this a was union of a n's is also union of a n bars so it is a countable union of closed sets so this is how we had we had done that one all right being a subset of c2 because x minus c1 is subset of c2 because x is union of c1 and c2 is 0 dimensional therefore each of these c i is 0 dimensional they are closed so now they are a countable union then we can apply the previous theorem so now we come to the example example 8.6 the fifth one there namely we shall prove that the subspace rnm of rn consisting of points exactly m of whose coordinates are rational is 0 dimensional in fact m equal to 1 and m equal to n we have already proved okay but we are not going to use that we can directly do that here also no problem that is not explicitly used here no problem okay I can allow one less than go to m less than go to n m could be the extreme values also what do I do this is some m between 1 and n right so choose indices i1 less than i2 less than im less than go to n m of them fix them not only that you fix rational numbers r1 r2 rm okay then look at the affine linear subspace which I will denote by l of r1 r2 rm of rn given by these m equations the ikth coordinate this is i1 i2 im so ikth coordinate is equal to this number rk that should be true there are m equations k equal to 1 2 3 okay so that is clearly linear subspace you know shifted shifted at this point xi k equal to rk the coordinate that is all okay so these are affine linear isomorphic to precisely m equations are there inside rn so it is rn minus m okay inside this the subspace l of r1 r2 rm okay of points all of whose other coordinates are irrational is therefore homomorphic to the subspace of rn minus 1 is the same property namely in minus m all the other coordinates are irrational let us solve how many n minus m coordinate irrational we have proved that this is already zero dimensional okay so what we have got is in this subspace if you take only those which have all the all the other coordinates irrational okay then they are zero dimensional okay and they are also automatically closed subspace of that rn minus m that is why they are you know they are zero dimensional as we vary the m tuples r1 etc rm okay you vary these the coordinate rm belonging to q power m okay because remember r1 r2 rm are rational numbers so when you vary you will get all elements of q power m right we get a countable union because q power m is countable countable union which is nothing but the space I am having a notation here q i1 i2 i m which is all x belonging to rn such that x i th coordinate is rational i equal to 100 okay so certain indices were chosen but you have to be free to choose all of them so now you take another union but this time finite union okay when you vary all the indices which satisfy i1 less than i2 less than i m there are finitely many monotonically strictly monotonically increasing sequence of length m okay therefore by the above theorem q i1 i2 i m is zero dimensional and the given say rn m is nothing but union of finite union of these things over all the i1 i2 i m okay so here are some elementary exercises you can try them on your own of course there will be a tf to help you if you do not get it show that a countable product of zero dimension space is zero dimensional next suppose x is zero dimensional show that its wallman compactification is zero dimensional okay this may be a challenging but if you think a little then you will get it let us go to now higher dimension okay so to begin with of course we will have some example the first example as it should be our motivating example namely the real line with the usual topology has dimension one as I have pointed out the fundamental system of neighborhood okay base is for r is all open intervals right the boundary of the open intervals is just two points set which is zero dimension okay so that qualifies r to be dimension one every piecewise smooth curve in any separable banach space has dimension one why because look at look at look at the smooth parts they are open parts right so they have dimension one that is enough for us why that is homeomorphism on the smooth part these are difumorphisms by inverse function theorem or whatever you want to say so in particular it is homeomorphism okay in particular circles parabolas any polygon etc all of them have dimension one in fact countable union of these things are also final dimensional inside r to r 3 and so on because each circles parabolas whatever they are given by equations right so they are close subsets indeed even finite unions of these objects have dimensional countable union you have to be careful okay you have to assume that they are all close subsets that is all in example 8.6 that 8 we have seen that the subspace ql of all points in the Hilbert space l2 of n this is nothing but the little l2 of square summable sequences right so we take ql to be the subspace points with coordinates all rational okay so we have seen that this has dimension positive okay we shall now show that it is actually dimension less than equal to one and hence dimension equal to one okay so by homogeneity homogeneity is what any point can be moved to any other point by a homeomorphism inside l2 of n therefore you can do instead of proving it every point of finding a base and so on you can do it at one single point so you can do it at the origin so what we shall do you have shown we will show that origin has a fundamental system of neighborhoods with boundary of w equal to 0 dimension of 0 for the dimension of boundary 0 okay dimension 0 of the boundary will mean that the space at dimension 1 that is what we want to show all right so if we have fundamental system of neighborhoods which are with their coupon sites then it would have been dimension the space itself will be dimension 0 right so I want to show that these things have dimension boundaries are dimension 0 non-emptive of all they should be right so for 0 less than or less than 1 let us have this notation sr is all x belong to l2 just that the norm of x is equal to r which is nothing but summation xi square is equal to 1 the summation could be infinite actually most of the time it is infinite so it is a convergent series some some converges to r square root of summation xi square square root is equal to r that is the meaning of this one take q of sr to be sr intersection ql namely all such points with each xi being rational it suffices to show that q of sr is 0 dimensional why because sr is the boundary of you know a system of neighborhoods which forms a base for the space lq here for l2 here okay so this is a local base at 0 its boundaries sr intersection with ql we are interested in so I will show that this is 0 dimensional for this we shall identify q of sr with a subspace of qh where h is our hillward q and qha all those points with all the all the coordinates rationally this we have 0 dimensional just before this example 8.6 only before this 8 one this 6 one so let me show you that one so this was the example here right we showed that qh here is jn or q of jn remember qh was homomorphic to this j that is the product apology and this tau d was the metric how we have pulled it out and so on so this qh is 0 dimension is what we have shown okay so I have to go back here now okay so we are going to show that this q of sr is a homomorphic space of space of q of h okay for this we use instead of h we use this j power n namely minus 1 to plus 1 closed interval taken n this natural number that many times product of countably many copies of minus 1 plus 1 okay so consider the map eta of sr to jn which is just the identity map remember this is the bar of radius r r is 0 between 0 and 1 positive but less than 1 therefore each coordinate here of a point s point x in sr is between minus 1 and plus 1 okay it is in minus rn plus r actually so it goes inside this one its identity so it is here there is no problem okay indeed it is continuous why because any function into the product space is continuous if and only if the coordinate functions are continuous in sr in l2 if you take just the ith coordinate that is a continuous function alright so these are continuous functions so identity map we can call it as inclusion map it is not a far away from identity inclusion map is continuous okay you should show that it is an embedding of sr into j power n okay embedding means what it is the homomorphism onto the image that is all I have to show that now what is the image eta of if you take rational numbers here you can rational coordinate it will go into the rational coordinates because this identity map okay and that that is now our ql and that is zero dimensional we will be done right so we have to show that this is an embedding a continuous injection when is it an embedding you should either show that it is closed or it is open closed map it is equivalent to open map because it is already a one-one mapping okay so that is all I have to so let us try to show that it is a closed mapping alright it is supposed to show that it has a closed map this is equivalent to the following statement so take a close subset inside sr its image is closed inside l2 that is what I have to show inside j power n this time the model has been you know we have cut down from l2 to we took the Hilbert cube and then we are studying j power n okay so inside j power n it should be closed that is what I have to show it is the same thing as taking a sequence inside the image okay and then take that sequence such that each coordinate function is convergent coordinate sequence is convergent that is the meaning of a sequence inside the product space convergent alright then I have to show that the sequence is convergent in sr okay so the correct statement I repeat given a sequence xn in sr and a point x in sr such that the coordinate sequence xn i converges to xi okay so all these things are if you take the eta eta xi is equal to eta eta xn of i converges to eta x of i okay so x must be inside sr then only I take eta x alright so if this converges to this one for each i it should imply xn converges to x inside sr now I have to use the topology of l2 here okay we hope you have seen the proof of this statement is in blue color somewhere else okay so if not you can try it I have given you a hint here Cauchy square of course if you don't get it we will explain it to you alright I think today it is enough so tomorrow I will continue with the study of higher dimension spaces thank you