 Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2. Today we will have the last section on the dimension theory, module 44, local separation to global separation. Recall that our study of dimension theory began actually in the previous chapter with a discussion of separation properties which we have named S0 to S3. Keeping in mind that we are going to restrict the class of the colloidal spaces for those which are separable and matrizable and in particular T1, we have pointed out earlier that S1, S2 and S3 are some stronger forms of house dwarfness, regularity and normality respectively. We may call this itself the first step in the passage from local to global. Moreover, the passage from SI to S3, S1 to S3, there is one S2 in between, itself can be viewed as a passage from atom to mass, atomic to aggregate or whatever you want to say, mini-school to largerness and so on. A kind of local to global that is also you can say is the passage from local to global. Recall that one of the first few results that we proved in the previous chapter is that Lindelow plus S2 implies S3. So that made S2 our central object of study. Next in this chapter we adopted S2 to represent zero dimensionality and then inductively obtained its higher dimensional versions to define higher dimension manifolds. Then came the theorem 9.13 and 9.15 which you may term as another step toward globalization. Let me just show you these steps. This was the theorem that if we have subspace of matrizable space, separable matrizable space then x as dimension less than or equal to n if and only if given any closed subset C of x and a point outside there is a closed subset D of x such that dimension of D is less than or equal to n minus 1 and x minus D is equal to a separated b and we are both open and closed subset disjoint with p inside a and c inside b. So we have pointed out that in the case of n equal to 0 dimension of D is minus 1 means D is empty set this is precisely the condition S2. So similarly 9.15 we can also similar things I just wanted to show you one of limit here. If x is a matrizable space and x prime is a subspace of that dimension of x prime is less than or equal to n if and only if for every point b inside x now I am putting condition on points of x all together I am getting condition like that. Condition is dimension of x prime is equal to n what is the relation between points inside x and the subspace will dimension less than equal to n. So that step is again another step or globalization this is what we meant okay. So let us go back to what we are doing today. So as the inventor, inventor had termed it the success of the concept of this theory of dimension why I am calling this theory because there are other theories also hinges upon successfully strengthening the passage from local to global. This is the topic of this last section wherein we shall be able to reach our goal the final goal of proving that the topological dimension of the pyridian space rn is actually equal to n. So that I call it as a success of the theory. Here is the next step in the passage from local to global okay. So we have to prove all these things now that x be any separable matrix space a is a subset of x of dimension 0. Given any two disjoint closed subsets c1 and c2 there exist a closed subset b of x separating c1 and c2 such that a intersection b is empty. The c1 and c2 are disjoint closed subsets by normality you can separate them by open set that is a different aspect. Of course that will be the starting point in the proof of this. What we are going to do is there is a separation by a closed subset b it does not intersect a at all okay. So this you can call it as really a crunching of of course we have to improve on this also of you know the separation properties being you know globalized globalization problem here. So let us start the proof of this. We have to produce a closed subset b of a b of x contained in the complement of a such that when you throw away this b x minus b can be written as disjoint union of closed one sets v1 and v2 c1 contained inside v1 c2 contained inside v2 that is the meaning of separation of x minus b. Choose open sets ui i equal to 1 and 2 such that u1 bar intersection u2 bar is empty and cis are inside ui so this step is just merely the metric space property here normal as normal property of a metric space okay. So that is all we are using here. Once you have this u1 and u2 look at their intersection with a in fact take ui bar intersection a these will be disjoint subsets of a and a is of dimension 0 so apply the s2 property there okay. We have a equal to c1 prime separation c2 prime here ui bar intersection a is inside c i prime okay so that is the definition for that is the property for the 0 dimensional state. Now you enlarge the c1 prime c2 prime along with c i say put f i equal to c i union c i prime i equal to 1 and 2 okay. Look at these c1s are subsets of a c1 c i prime c1 c2 are subsets of the largest space x okay cis are closed c is are closed inside a but a is that is not closed inside x so there is some problem here otherwise f pi 2 would have been easily you know you can take it as closed subset you are take the a as a closed subset no here this dimension 0 that is the point here so we shall show that f1 and f2 are mutually separated sets which is stronger than saying just disjoint closed subsets okay so I mean they are not closed subsets but f1 bar separates it does not intersect f2 and f2 bar does not intersect f1 so mutually separated subsets so which is slightly weaker hypothesis than having disjoint closed subset disjoint closed subsets are easy to separate so this is into more so for here we will need more than normality namely complete normality wait a minute so let us have this one now how to prove that this one namely f1 bar intersection f2 is empty indeed once you prove this one the other one is symmetrical all these things are conditions are symmetrical in the other one so that will also prove exactly same okay so once you have that use complete normality of the metric space it follows that we get open subset w in x such that f1 is inside w and closure of w does not intersect f2 this way it is easier to state f1 contained inside w closure of w does not intersect f2 you can then take b as the boundary of w automatically boundary of w is closed so this b is a closed subset which will separate f1 and f2 because f1 will be contained inside w does not intersect b f2 does not intersect b because it does not intersect w bar at all so f1 and f2 are inside that one if you take w union complement of w bar that will be precisely equal to the whole x minus boundary of everything x minus b okay so moreover if you look yeah that is what I have told b intersection a okay is empty because why I just why this is this I have not yet shown what I have shown f1 and f2 are different moreover you want to ensure that b intersection a is empty but b intersection a b is boundary of w boundary of w intersection f1 union boundary of w intersection f2 okay because this entire a is union of c1 prime and c2 prime and f1 and f2 contains even prime c2 prime so a is contained in the f1 and f2 therefore this intersection is intersection with f1 union intersection with f2 but both of them are empty okay just now we show them so this b will serve the purpose all right so we have yet to find out that one so here is the schematic picture of what is happening started with c1 and c2 which are disjoint closed subsets these ellipses then enclose them in open subsets u1 and u2 that closure of u1 and closure of u2 are disjoint intersected u1 with a this is my say a is of dimension 0 that is why I have shown like dot here okay intersect that with a1 so there you get some subset here like this okay up to u1 here similarly from here to here all right so they are disjoint inside a and they are disjoint closed subsets so you can separate them by c1 prime and c2 prime so c1 prime and c2 prime may go out of u1 they may be out of this one so here it might come out of u2 also here to here and so on so of course this dot dot dot for dot dot dot ellipses indicate that after all an open subset of subset of a is nothing but some open subset inside x intersection with a so that is what I have shown here then I put f1 equal to c1 union this c1 prime up to here c2 in c2 prime up to here I have to show that they are mutually separated in the picture it is obvious you can't use the picture to prove a theorem okay you can take the help but finally everything should be purely logical and finally what we want is this green thing w such that it is closure okay does not intersect this f2 c1 prime union c2 prime and this boundary if you take boundary if you throw away that you get a separation all right so let us prove first of all that these two namely c1 union c1 prime c2 in c2 prime they are disjoint and mutually separated is what I have shown of injection so let us do that so it remains to prove 31 31 is that this f1 bar intersection f2 is empty okay the other one is similar so by symmetry it is enough to prove that f1 bar intersection f2 is empty first of all f1 bar f1 itself is c1 union c1 prime therefore f1 bar bar denotes now everything denotes happening inside x okay closures are inside the whole space x f1 bar is c1 bar union c c1 prime bar because it is just a union of you know finite union but c1 is already closed so it is c1 union c1 prime bar therefore it is enough to check that c1 intersection f2 okay this is c1 intersection f2 and c1 prime intersection f2 they are empty then f1 bar intersection f2 will be empty so this is the first step I have to show these two things now just to show one of them I have to show these two things okay so let us see now c1 is inside u1 by the very choice u1 u1 is an open subset containing c1 hence c1 intersection a is you contained in u1 intersection a contained in u1 bar intersection a but u1 bar intersection a c1 prime okay and hence c1 intersection c2 prime is contained inside c1 prime intersection c2 prime okay just now c1 intersection a is already in c1 prime c1 intersecting c2 c2 prime is inside the subset of a so I can take intersection with a itself so that is contained in c1 prime intersection c2 prime and that is empty to begin with okay so instead of c1 it is in f2 I have looked at c1 intersection c2 f2 has two two parts right what is c2 and c2 prime the c2 prime okay so c1 intersection what we do here c1 intersection f2 will be c1 intersection c2 which is empty c1 intersection c2 prime okay so this intersection is now c1 intersection c2 prime now what you have what is happening to this one so c1 intersection c2 prime is again a intersection c1 intersection c2 prime everything happening is c2 prime a intersection u1 bar intersection c2 prime because c1 is contained inside u1 bar okay so but that is contained in c1 prime intersection u prime so that is empty okay so I am more or less repeating this one here all right next to show that c1 prime intersection f2 is empty okay so how do you show there are two parts f1 bar has two parts right one we one we showed the second part is this one this one I have to show for each x belonging to f2 we shall produce a neighborhood of x which does not meet c1 prime then it follows that that point is not in the closure of c1 prime because a whole neighborhood does not interact that if x is inside c2 there are two parts to f2 one is c2 and another is c2 prime if c2 you can just take v equal to whole of this u2 for all the points inside c2 take just v because v which is equal to u2 here is contains the whole of c2 okay and then v u2 is not a intersection intersection c1 prime is empty right so u2 intersection a which is contained inside u2 bar intersection a that is contained inside c2 prime and then u2 intersection c1 prime is u2 intersection a intersection c1 prime which is contained in c2 prime intersection c1 prime that is empty so method is the same but argument is different that part takes one part if x is in c2 prime then what do I do but c2 prime is an open subset of a it is actually closed subset right because separation okay c2 prime is also open inside a so we get an open set v this time it is a different v I have to do okay so this v is inside x open such that v intersection a is this c2 prime open subset of a then what happen v intersection c1 prime is v intersection a intersection c1 prime that is c2 prime intersection c1 prime okay that is contained inside so that is also okay so separately we have shown that c1 prime intersection f2 as well as c1 intersection f2 are empty it just means that f1 bar intersection f2 is empty similarly f1 intersection f2 bar is also empty and that completes the proof now we can state a more pliable statement and easy to remember statement start with any separable matrix space take a subset of dimension less than or equal to n where n itself is finite of course I assume bigger than or equal to 0 because if a is empty there is no statement that those things we have seen already so given any two disjoint closed subsets c1 and c2 inside a there exists a closed subset v of x separating c1 and c2 such that the subset a intersection v okay has dimension less than or equal to n minus 1 okay so from 0 dimension we have come to arbitrary dimension here now okay so how do you do that of course x is a matrix space there exist open subsets u and u2 such that c i's are inside ui i equal to 1 and 2 and intersection u and u2 is empty this is normality because c1 and c2 are disjoint closed subsets now suppose n is 0 this n is between 0 and infinity right so n is 0 there are two cases to be handled if a is empty in which case we can take b equal to u and u and u2 complement that's a closed subset and x minus b is just disjoint union of u and u2 over okay that's what when this is 0 otherwise dimension of a is 0 it cannot minus 1 okay in which case the first earlier lemma which we just now that we give you the required result so we have the inductive hypothesis here now suppose n is bigger than 0 using a previous corollary we can write a as union of two subsets d and e where dimension of d is less than or equal to n minus 1 and dimension of e is less than or equal to 0 this is one of the some theorem that we have derived last time right now you use the above lemma to give a closed subset b which separates c1 and c2 such that b intersection e is empty I don't know what is happening to d we'll come to that later but e part is empty that is the the starting point with n equal to 0 but then this implies you take a intersection b that will be now contained inside d but d is of dimension n minus 1 less than or n minus 1 so intersection of a a intersection b is also of dimension less than or n minus 1 okay so this after hard work of lemma this comes quite easily by of course I have to use this crucial thing here namely anything which is of dimension n can be written as union of two subsets one is dimension minus 1 another one is 0 now in the above theorem take a equal to well this was not a theorem it is a proposition it does not matter take a equal to x then what do I get that x be of dimension less than equal to n then any two disjoint closed subsets can be separated by a closed subset of dimension less than or equal to n minus 1 okay so no question of intersecting with a because a is the whole space now we want to improve upon that one let x be of dimension less than equal to n and c1 c2 c3 and similarly this even prime c2 prime etc pairwise disjoint closed subsets c i intersection c i prime is empty that is the meaning of that how many are there n plus 1 this dimension less than equal to n there are n plus 1 pairs of disjoint closed subsets then there exists a closed subset bi there exists closed subsets bi how many n plus 1 of them such that each bi separates ci and ci prime and intersection of bi i rank 1 to n plus 1 is empty all right so how do you get this one it is also easy from previous theorem apply to c1 and c1 prime you get a closed set b1 which separates c1 and c1 prime such that dimension of b1 is less than into n minus 1 now use the proposition or the theorem we get a closed subset of of x a subset b2 separating c2 and c2 prime such that when you intersect it with b1 it is dimension less than to n minus 2 because you already n minus 1 now you keep on doing this you know repeat this step get a b3 such that it separates c3 and c3 prime with dimension of b1 intersection b2 to n minus 3 how far you can go till you get minus 1 and that is empty so you have to have dimension less than equal to n here and there must be n plus 1 of them then only you will actually have all right now we are very close to the end here okay the rectangle or rectangular box minus 1 to plus 1 interval raised to n contained inside rn okay suppose you have c i plus and c i minus denote its faces defined by equation x i equal to plus minus 1 if you just n equal to 1 this is nothing but c i minus 1 and c i minus c i plus is plus 1 that is all if n equal to 2 there will be four faces pair of two of two pairs of opposite faces right so that is the way you have to take these faces defined by the equation the coordinate x i equal to plus minus 1 okay okay now for 1 less than or i less than n there are n pairs here suppose b i is a close subset of jn bar which separates the opposite faces c i plus and c i minus it just means that when you throw away b i from j n bar you get two open subset each of them containing c i plus or c i minus one of them contains the i okay that is separation suppose you have got these b i like this then we want to show that intersection of b i is non-empty the crucial thing here is what there are only n of them the previous theorem says if there are n plus one of such things then it is empty so together they are going to imply a big theorem for us however the proof of this is now based on something different that we did last time namely Brauer 6 point theorem comes here okay let us see how pay attention to the method of proof because that can be used in several other places okay let did you know the Euclidean distance function in R n for each x in u i plus I have overly told you what are u i plus here u i plus is what the open subset containing c i plus and this one containing c i minus u i minus c i minus okay for x belong to u i plus let p i be the root to be the foot of the perpendicular from x to c i minus so this is a one of the faces so go all the way to c i minus from u i plus don't change the ith coordinate that's all the p i and x have same ith coordinate it is the foot of the perpendicular from x to this plane then this is now elementary observation distance between x and b i b i is a subset which separates the two things right it's less than equal to distance between x and p i and this distance is actually put distance between x and c i minus why because p i is the foot of the perpendicular and what is this distance is just one plus the ith coordinate of x okay similarly so what we have proved by this one is distance between x and b i is one plus x i distance between you know x and b i is less than equal to one minus x i for every x inside u i minus this is for u i plus this one u i minus very easy to remember let me justify this one with a small picture here n equal to 2 okay so this is minus 1 to plus 1 minus 1 to plus 1 the product square okay this is square and this is c 1 plus x 1 coordinate this is c 1 minus this is c 2 minus and this is c 2 plus this is my b 2 separate c 1 minus and c 2 minus take a point in u 2 plus this x take its projection on to this axis on to this plane so here is just an axis what is it it is x 1 coordinate will not change what is its y coordinate y coordinate of this point or x 2 coordinate of this point is nothing but this x 2 coordinate here plus this one the distance sorry the distance is this distance is 1 this distance is x 2 right so distance between x and this one is same thing as distance between x and this part which is bigger than distance between x and b 2 and that is precisely what we have to do this x b i less than 2 x b i is a dx c i minus now i define a function here for one less than or less than equal to n define a phi from j and bar to r as follows if x is in u i plus just take it as distance between x and b i that is a continuous function right we know that what is this this is the minimum of distance between x and little b i where little b i runs over b i if x is in u i minus you put a minus sign there so x c is in b i put a 0 look at this one if x is in u i bar this this u i plus it is this one if it tends to a point you know if you take limit tends to a point inside the boundary that this distance becomes 0 right similarly this one also this will become 0 so by these are continuous this f i will be continuous because of that okay now you combine these two in equations here inequality what you conclude is that minus one is always less than x i minus f i of x is always less than good one there are two different cases you may have to work you have to work out is very easy okay you have to just use this and this one according to this because there are different definitions are different the function f i if different you define that's why okay so x i minus f i is between minus and plus one therefore if i define okay f x equal to f 1, f 2, f n and g x equal to x minus f x then what happens this is a function taking the values inside r n but g x will be taking function inside j n bar it will be always between minus and plus one all the coordinates so obviously both of them are continuous okay therefore you have got a function from the closed rectangle j n bar to j n bar you can apply braver six point theorem so we get a point x is that g x equal to x what of that mean f x equal to 0 what of that mean what of that mean f i of x equal to 0 for all i what of that mean x must be inside each of this d i which means x is in the intersection of here okay so that is the statement here okay this theorem is proved now as i told you now you can combine this with the previous theorem you get a wonderful result now tie it thing namely dimension of j n bar has to be equal to n okay let me go through this one it is not so clear if not dimension is always less than equal to n because because of what we know that part we have already proved okay dimension far in less than equal to n and this is a subspace so that part we have already proved so dimension must be less than equal to n minus one as soon as it led minus one by our theorem 9.38 whatever we have proved today it means that so for each i 1 less than 1 less than 1 less than 1 there exists n closed subset b i which separates c i plus and c i minus such that the intersection is empty but braver six point theorem applied now just now the previous theorem says this is unempty so that the contradiction to this theorem right 9.39 therefore dimension is right it has to be equal to n r n being a larger subspace of the you know containing j n or j n bar already so this dimension is also n alright so we have proved that not only this one can you can take now any open subset any take any open subset which is homeomorphic to some d n j n bar and so on contains something etc right any open subset will contain some some copy of this one therefore all open subsets will be of dimension n inside r n all non-empty open subsets of course so here we remark topological dimension whatever you have defined okay is invariant for the class of separable matrizable spaces we have not defined it for arbitrary spaces that is one point you have to remember thus we may also conclude that you know the browser invariance of domain which is a weaker form of this one weaker form of b i d so I am going to give that namely r n and r m if n is not equal to m cannot be homeomorphic because we have just shown that dimension of r n is n dimension of m but dimension is a homeomorphism invariant okay this is weaker form of browser invariance domain of course the the real browser's invariance of domain is the following which is which is a which is a stronger form of this one namely take any two non-empty open subsets non-empty subsets of r n suppose they are homeomorphic then if one of them is open the other one is also open and that is why the name invariance of domain the word domain was used more often than just open subsets in the older days so being open is the same thing as being a domain and that is an invariance so that is why it is called invariance of domain okay unfortunately we are not able to touch this one we have come very close but there is still a big gap here okay so the proof of this will take us much deeper into dimension theory which is beyond the scope of this course original proof due to browser uses another topological dimension theory namely Lebesgue covering dimension in modern times it is fashionable to prove this using homology theory so there are many proofs of this work this great theorem for a proof using only simple shell approximation which is actually a part of it is there in for a visual implicitly you may see my book so here is a easy exercise for you deduce 9.42 from 9.43 namely this I said is a this stronger form I said why so you assume this and prove this one okay we have proved it using the whole thing using the all our you know first two different chapters that we have studied so carefully but assuming this you prove this one that is your exercise okay all right so next time we will start a new topic thank you