 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 topological 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, 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 largeness 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 Lindelof 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 b 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 is D is empty set. This is precisely the condition S2. So similarly 9.15 you can also similar things I just wanted to show you one of the 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 p inside x now I am putting conditions on points of x all together you see I am getting condition like that. Condition is dimension of x prime is n equal to n. What is the relation between points inside x and point the subspace being dimension less than equal to n. So that step is again another step or globalization this is what we meant. So let us go back to what we are doing today. So as the inventor had termed it the success of the concept of this theory of dimension why I am going to 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 R n 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. So we have to prove all these things now that x be any separable matrix of space A is a subset of x of dimension 0. Given any two disjoint closed subsets C 1 and C 2 there exist a closed subset B of x separating C 1 and C 2 such that A intersection B is empty. So C 1 and C 2 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 separation by closed subset B it does not intersect A at all. So this you can call it as really a crunching of of course we have to improve on this also of the separation properties being globalized. Globalization problem here. So let us start the proof of this. You 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 V 1 and V 2 C 1 contained inside V 1 C 2 contained inside V 2 that is the meaning of separation of x minus B. Choose open sets U i i equal to 1 and 2 such that U 1 bar intersection U 2 bar is empty and C i's are inside U i. So this step is just merely the metric space property here normal property of a metric space. So that is all we are using here. Once we have this U 1 and U 2 look at their intersection with A in fact take U i bar intersection A. These will be disjoint subsets of A and A is of dimension 0. So apply the S 2 property there. We have A equal to C 1 prime separation C 2 prime here U i bar intersection A is inside C i i prime. So that is the definition for that is the property for the zero dimension set. Now you enlarge the C 1 prime C 2 prime along with C i. Put F i equal to C i union C i i i equal to 1 and 2. Look at these C 1's are subsets of A C 1 C i prime C 1 C 2's are subsets of the largest space x. C i's are closed C i's are closed inside A but A is not closed inside x. So there is some problem here otherwise F i 2 would have been easily you know you can take it as closed subset. You have to take it as a closed subset. Now here this dimension 0 that is the point here. So we shall show that F 1 and F 2 are mutually separated sets which is stronger than saying just disjoint closed subsets. So I mean they are not closed subsets but F 1 bar separates it does not intersect F 2 and F 2 bar does not intersect F 1. So mutually separated subsets which is slightly weaker hypothesis than having disjoint closed subset. Disjoint closed subsets are easy to separate. So this is 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 F 1 bar intersection F 2 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 way. So once you have that use complete normality of the metric space it follows that we get open subset W in x such that F 1 is inside W and closure of W does not intersect F 2. The same thing as having disjoint closed subsets and so on. This way it is easier to state. F 1 contained inside W closure of W does not intersect F 2. We 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 F 1 and F 2 because F 1 will contain inside W does not intersect B. F 2 does not intersect B because it does not intersect W bar at all. So F 1 and F 2 are inside that one. If you take W union compliment of W bar that will be precisely equal to the whole x minus boundary of W union will be x minus B. So moreover if you look that is what I have shown B intersection A is empty because why this I have not shown. What I have shown F 1 and F 2 are disjoint one by one. Moreover you want to ensure that B intersection A is empty but B intersection A B is boundary of W, boundary of W intersection F 1 union boundary of W intersection F 2. Because this entire A is union of C 1 prime and C 2 prime and F 1 and F 2 contain C 1 prime and C 2 prime. So A is contained in F 1 and F 2. Therefore this intersection is intersection with F 1 union intersection with F 2. Both of them are empty. Just now we show them. So this B will serve the purpose. So we have yet to find out that one. So here is the schematic picture of what is happening. Started with C 1 and C 2 which are disjoint close subsets. These ellipses then enclose them in open subsets C 1 and U 2 that closure of U 1 and closure of U 2 are disjoint. Intersected U 1 with A this is my say A is of dimension 0 that is why I have shown like dot dot dot dot dot here. Intersect that with A 1. So there you get some subset here like this up to U 1 here similarly from here to here. So they are disjoint inside A and they are disjoint close subsets. So you can separate them by C 1 prime and C 2 prime. So C 1 prime and C 2 prime may go out of U 1. They may be out of this one. So here it might come even out of U 2 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 F 1 equal to C 1 union this C 1 prime up to here C 2 in C 2 prime up to here. I have to show that they are mutually separated. In the picture it is obvious. You cannot use the picture to prove a theorem. 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 does not interfact this F 2 C 1 prime union C 2 prime. And this boundary if you take boundary if you throw away that you get a separation. So let us prove first of all that these two namely C 1 union C 1 prime C 2 in C 2 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 F 1 bar intersection F 2 is empty. The other one is similar. So by symmetry it is enough to prove that F 1 bar intersection F 2 is empty. First of all F 1 bar F 1 itself is C 1 union C 1 prime. Therefore F 1 bar bar denotes now everything denotes happening inside X. Closures are inside the whole space X. F 1 bar is C 1 bar union C C 1 prime bar because it is just a union of finite union. But C 1 is already closed so it is C 1 union C 1 prime bar. Therefore it is enough to check that C 1 intersection F 2 this is C 1 intersection F 2 and C 1 prime intersection F 2 they are empty. Then F 1 bar intersection F 2 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. So let us see. Now C 1 is inside U 1 by the very choice U 1. U 1 is an open subset containing C 1. Hence C 1 intersection A is contained in U 1 intersection A contained in U 1 bar intersection A but U 1 bar intersection A is C 1 prime. And hence C 1 intersection C 2 prime is contained inside C 1 prime intersection C 2 prime. Just now C 1 intersection A is already in C 1 prime. C 1 intersection C 2 prime is inside subset of A so I can take intersection with A itself. So that is contained in C 1 prime intersection C 2 prime and that is empty to begin with. So instead of C 1 it is in F 2 I have looked at C 1 intersection C 2 F 2 has two parts. What is C 2 and C 2 prime? So C 1 intersection F 2 will be C 1 intersection C 2 which is empty C 1 intersection C 2 prime. So this intersection is now C 1 intersection C 2 prime. Now what is happening to this one? So C 1 intersection C 2 prime is again A intersection C 1 intersection C 2 prime everything is happening in C 2 prime. A intersection U 1 bar intersection C 2 prime because C 1 is contained inside U 1 bar but that is contained in C 1 prime intersection U prime so that is empty. So I am more or less repeating this one here. Next to show that C 1 prime intersection F 2 is empty. So how do you see? There are two parts F 1 bar has two parts right one we showed. The second part is this one now this one I have to show. For each x belonging to F 2 we shall produce a neighborhood of x which does not meet C 1 prime then it will follows that that point is not in the closure of C 1 prime because the whole neighborhood does not intersect that. If inside C 2 there are two parts to F 2 one is C 2 and one is C 2 prime. If C 2 you can just take V equal to whole of this U 2 for all the points inside C 2 take just V because V which is equal to U 2 here is contains the whole of C 2 okay and then V U 2 is not intersection C 1 prime is empty right. So U 2 intersection A which is contained inside U 2 bar intersection A that is contained inside C 2 prime and then U 2 intersection C 1 prime is U 2 intersection A intersection C 1 prime which is contained in C 2 prime intersection C 1 prime that is empty. So method is the same but argument are completely different. That part takes one part if X is in C 2 prime then what do I do but C 2 prime is an open subset of A it is actually close open subset right because separation okay C 2 prime is also open inside A. So we get an open set V this time it is a different V I have to choose okay so this V is inside X open such that V intersection A is C 2 prime open subset of A. Then what happens V intersection C 1 prime is V intersection A intersection C 1 prime that is C 2 prime intersection C 1 prime okay that is contained in C 1 prime. So that is also okay so separately we have shown that C 1 prime intersection F 2 as well as C 1 intersection F 2 are empty it just means that F 1 bar intersection F 2 is empty. Similarly F 1 intersection F 2 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 equal to 0 because if A is empty there is no statement those things we have seen already. Given any two disjoint closed subsets C 1 and C 2 inside X there exists a closed subset B of X separating C 1 and C 2 such that the subset A intersection B okay has dimension less than or equal to n minus 1 okay. So from zero dimension we have come to arbitrary dimension here now okay so how do you do that? Of course X is a metric space there exists open subsets U and U 2 such that C i's are inside U i i equal to 1 and 2 and intersection U and U 2 is empty this is normality because U and U 2 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 U 2 complement that is a closed subset and X minus B is just disjoint union of U and U 2 over okay that is 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 give 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 even and C2 such that B intersection E is empty I do not know what is happening to D we will come to that later but E part is empty that is 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 equal to n minus 1 so intersection of A intersection B is also of dimension less than or equal to 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 now 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 preparation does not matter take A equal to X then what do I get let X be of dimension less than or 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 X now now we want to improve upon that one let X be of dimension less than or equal to n and C1, C2, C3 and similarly this C1 prime, C2 prime, C3 prime etc pair wise disjoint closed subset 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 or equal to n where n plus 1 pairs of disjoint closed subset then there exists a closed subset B i there exists closed subsets B i how many n plus 1 of them such that each B i separate C i and C i prime and intersection of B i i rank 1 to n plus 1 is empty alright. So how do you get this one this is also easy from previous theorem applied 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 or equal to n minus 1. Now use the proposition or the theorem we get a closed subset of XA subset B2 separating C2 and C2 prime such that when you intersect it with B1 is a dimension less than or equal to n minus 2 because you already n minus 2 dimension. Now you keep on doing this you know repeat this step get a B3 such that it separates C3 and C3 prime the dimension of B1 intersection B2 intersection B3 to n minus 3 how far you can go till you have minus 1 and that is empty. So you have to have dimension less than or equal to n here and there must be n plus 1 of them then only you will actually have alright. Now you are very close to the end here okay 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 4 faces pair of 2 pairs of opposite faces right. So that is the way you have to take these faces defined by the equation coordinate X i equal to plus minus 1 okay. Now for 1 less than or 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 Jn bar you get 2 open subset each of them containing C i plus or C i minus 1 of them containing the i okay that is separation. Suppose you have got these B i's like this then you 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 there are n plus 1 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 D denote the Euclidean distance function in Rn 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 is containing C i minus U i minus U i minus U i minus okay for X belong to U i plus let P i be the root to be the root of the perpendicular from X to C i minus. So this is a one of the phases. So go all the way to C i minus from U i plus do not change the ith coordinate that is all. The P i and X have same ith coordinate it is the root 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 is less than or equal to distance between X and P i and this distance is actually equal distance between X and C i minus why because P i is the root 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 or equal to 1 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 its 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 X B i less than so X B i is a D X C i minus it is M 1 plus X now I define a function here for 1 less than or less than equal to n define F i from J n 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 X is in B i put at 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 then 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 equation 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 use this and this one according because there are different definitions are different the function F i different to define that is why okay so X i minus F i is between minus one 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 a 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 Brauer 6 point theorem so we get a point X such 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 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 type 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 of what we know that part we have already proved okay dimension of R n 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 n minus one by our theorem 9.38 whatever we have proved today it means that for each i 1 less than equal to n there exist n closed subset B i which separates C i plus and C i minus such that the intersection is empty but Brouwer's point theorem applied now just now the previous theorem says this is unempty so that is the contradiction to this theorem right 9.39 therefore dimension has to be equal to n R n being a larger subspace of 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, G n bar and so on contains something etcetera right any open subset will contain some copy of this one therefore all open subsets will be of dimension n inside R n all non-empty open subset so here is a 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 Brouwer's 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 R m but dimension is a homeomorphism invariant okay this is weaker form of Brouwer's invariance domain of course the real Brouwer's invariance of domain is the following which is 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 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 Brouwer uses another topological dimension theory namely Lebesgue covering dimension in the modern times it is fashionable to prove this using homology theory so there are many proofs of this word it is great theorem but a proof using only simplisical approximation which is actually part of it is there in Horowitz-Jewelman implicitly you may see my book so here is a easy exercise for you deduce 9.42 from 9.43 namely this I said 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 if you are okay alright so next time we will start a new topic thank you