 Welcome to NPTEL NOC, an introductory course on points at topology part 2, module 41, dimension of subspaces and unions today. Take a subspace of X, say X prime. Let us have this notation, boundary, I read it as a boundary prime, inside boundary inside X prime, denote the boundary of a subset with respect to X and X prime respectively. Then for any A inside X, we have the boundary of A intersection X prime taken inside X prime, that is a subset of boundary of A. Without any decoration, this boundary just denotes the boundary inside the larger space X. So this is an elementary result which we have seen in the first part itself, but now it becomes very crucial. So let us go through it little carefully. Take a point which is the boundary of A intersection X prime in the subspace X prime. Let us start with that. A can open set U of X, now such that X is inside U. We know that U intersection X prime is a neighborhood of X in X prime and hence U intersection X prime intersects both A intersection X prime and its complement inside X prime because it is a boundary point of intersection X prime. Therefore this U intersection A which is the union of U intersection X prime and A intersection X prime, this has to be non-empty. So we have proved that starting with any open neighborhood of X, U intersection A is non-empty. What else we have to prove? We have to prove that its complement also intersects U. Then it would be a boundary point of A intersection X prime, A itself inside X. So what is the intersection of U with X minus A that contains U intersection X prime as well as complement of A inside X prime. So those which are inside X prime, those which are not inside X prime I have taken, entire U intersection X prime. But the first part is U intersection X prime. The second part is X prime minus A intersection X prime. So one of them has to be non-empty as we have seen. Therefore it follows that X must be inside boundary of A. So one way is true. In fact equality hardly occurs unless A itself is a close subset of A intersection X prime is a close subset of X prime or X prime itself is close and so on. So all that we need is one way in equilibrium here. So we will use this one heavily now. Every subspace of a space of dimension less than or equal to n is of dimension less than or equal to n. When you take subspaces dimension does not increase. So this is the theorem here now. So start with a base for X such that dimension of the boundary of each member is less than or equal to n minus 1. So that is the definition of dimension being less than or equal to n. Then we know that if you take B prime namely B intersection X prime as B ways over B, that family is a base for X prime because subspace of origin. By the lemma we have boundary of B intersection X prime inside X prime is contained inside boundary of B. Now we induct on n. If n is 0 this implies the boundary of B is empty therefore for each B inside B therefore this is also empty. So this proves a statement for theorem of the theorem for n equal to 0. Inductively assume that we have proved a statement for n minus 1. Now this implies the dimension of boundary of B intersection X prime is less than or equal to n minus 1 because it is subspace of the corresponding here. Hence dimension of X prime is less than or equal to n. So all the time we are using this one. If the dimension is say 5 here that will be dimension will be less than or equal to 5. Once we have proved that for 6 dimension we will prove and so on. Starting with 0 we can build up the induction. Next theorem is that X be a subspace of a separable metric space any space for that matter. And separable metric space you start with. Then X as dimension less than or equal to n if and only if given any closed subspace C of X and a point P inside P not inside C outside C. 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 a disjoint U and B with P inside A and C inside B. Remember if remember what is S2. S2 said that points can be separated our closed set and a point can be separated by closed sets. Closed sets where those which such that dimension of the boundary sorry the boundary itself is empty. Now here we have something else. This is extended S2. It is n minus n dimensional version of S2. So dimension of X less than or equal to n you cannot expect closed sets and a point outside it could be separated by closed sets. Closed sets means boundary is empty. Now on that boundary we are going to put condition. So this is elaborately stated in a different way here. You throw away closed subset D of dimension n minus 1. Then you have a separation okay. When you have a closed set the boundary played that role. That is why we did not have to bother about this way stating. Now here we throw away some closed subset of course away from C and P. P and C you have to retain okay. So X minus D is A union B okay. A and B are both closed and both open inside X minus D and P is inside A and C is inside B or the other way around. So this is the generalized version of S2 now you see. So this is equivalent to having dimension n just like S2 is equivalent to having dimension 0. Let us prove this one. Suppose X has dimension less than or equal to n. With C and P as stated means closed subset and a point outside. Take the neighborhood U which is complement of C. So that is the neighborhood of X right. By regularity of X because X is several matrix based after all. We get an open set V in X such that P is inside V. Closure of V is inside complement of C because P is inside complement of C and complement of C is open okay. In between we have what this V and V bar. Since dimension is less than or equal to n we get an open set W inside this V. P belonging to W is contained inside V such that now you take D what is D? D I have to choose namely boundary of W is of dimension less than equal to n minus 1 okay. So this is the statement for dimension I mean condition for dimension being less than equal to n okay. Now look at X minus D. D is a boundary of something so it is a closed subset. So X minus D is obviously union of W and X minus W bar okay. Both of them are open union is this one so both of them are closed in X minus D okay. P is inside W alright and C is inside X minus W bar because of what? The boundary of W bar does not intersect in P okay. V bar is contained inside a complement of C and W is contained inside V. So W bar is also contained inside V bar. So C is contained inside X minus W okay. Conversely starting with any point in X and an open subset U such that P is inside U put C equal to complement of C this time. So C is closed subset and P is outside that. From the given condition we get a closed subset D of dimension less than equal to n minus 1 such that X minus D is separated A separation B A bar B with P is inside A C is inside B. This implies A is an open subset of X okay because D is a closed subset. So A is open inside X minus D and D is closed X minus D is open inside X. So A is open subset of X. P is inside A. A is contained inside U okay because it is disjoint from this C that is all. Also boundary of A will be contained inside D and hence of dimension less than equal to n minus 1. So what we have done starting with any point and an open set we have produced a neighborhood A such that its boundary is of dimension less than equal to n minus 1. So this just means that for every point we have done. So we have a base for X consisting of elements A such that their boundaries are less than equal dimension n minus 1. So that means that dimension of X is less than equal to n alright. Now it is convenient and useful to express the condition for a subspace X prime of a space to have dimension less than equal to n purely in terms of you know condition on larger space X okay. We know what is the condition of subspace. There is a base for subspace with water blah blah blah satisfying okay. So we can say that generalized S2 condition alright. So let us convert that purely in terms of X that will be useful for us. So this lemma says start with a metric space general lemma okay. Let X prime be a subspace suppose A and B are subsets of X prime which are mutually separated in X prime remember mutually separated means A intersection closure of B is empty similarly B intersection closure of A is empty. This time I am taking closures prime inside X prime because there are two spaces in all here you do not know where you are taking if you just take closure or B bar and so on. I will use bar to denote the closures inside X the standard one okay. For subspaces I will use this notation closure of prime okay. Then there exists open subsets W in X such that A is contained inside W, W bar intersection B is empty. So this is again a general result okay while proving a studying metric spaces we have seen such proof but let me recall it because now it is become crucial okay. All that I have to do is use the metric property. By the way this is general thing just about any metric spaces okay. So put W A equal to union of all open bars B R A where A range is inside A and one condition on R namely 3 times R B 3 times R A ball of radius 3 R around A does not intersection B. So points of this you know A must be far away from this one 3 times R A must be empty of course R should be positive alright. This is just the definition W A is union of all such balls. Similarly W B is union of B R B, B range over A same condition now A and B interchange B 3 R B intersection A is empty okay. Then clearly both W and W B are open subsets of X because they are unions of open balls. It suffices to show that A is inside W A intersection W B is empty because B is inside W B and that is open automatically it will imply W A bar intersection B is empty. So after that you can take W equal to this W A. We want W to be an open set containing A such that its closure in X does not intersect B okay. A set does not intersect an open neighborhood around a set will not intersect its closure will not intersect that set okay that is all. So let us prove that okay A is inside W A and this one inside this one provided see we have not yet stated. So what is the hypothesis? Hypothesis A and B are separated mutually separated. We have not used that one yet right. The basic observation we make here is that if D prime is the metric D restricted to X prime the same metric but points are taken inside X prime that is the meaning okay. Then for all X prime inside X prime B prime T X prime and this is a definition T is the radius X prime is the center. So all Y prime in X prime such that D prime of Y prime X prime is less than T nothing but the standard ball in X around X prime and points inside X prime I am just repeating the meaning of what is the restricted metric here. Now A intersection closure of B inside X prime is empty implies see these two are A and closure there exists some point R positive such that the ball of radius 3A of course this prime I am taking only everything working inside X prime now intersection B is empty okay but then B 3 I am talking about one point here for each A B 3A intersection B is B 3R A intersection X prime B because see these things are A and B are inside X after all X prime after all okay this ball is outside inside X if you instead B B is X prime intersection B I can replace once you have X prime this is nothing B B prime of 3R A B prime 3R A intersection B is empty okay. So for each A if you choose R like this then what you have what is that point and this neighbor is inside this so this proves that when you take the union A is contained inside W A argument is similar because A and B have been interchanged in defining W B so B is also contained inside W B alright now you have to prove that W A intersection W B is empty and that is why why we are taking only balls of radius R but condition is on on the ball of you know 3 times larger that B 3R so that the triangle inequality will help you that is all here the standard triangle inequality you have to show suppose your point which is in both W A and W B okay that implies this X this X is inside some B R A as well as in B S B for R and S some positive number right and A and B are inside A and B respectively but then distance between A and B A to X and then X to B will be less than R plus S okay triangle inequality here B 3R A intersection B is empty this implies R plus S must be bigger than 3R right because the ball of 3R A does not intersect this so R plus S must be bigger than this one okay that is what is the meaning of this one consulate one S is bigger than 2R similarly B 3S B intersection A is empty that is the choice of this S and R after all that will imply R process is bigger than 3S okay but now S is bigger than 2R so R is bigger than 2S right so S is bigger than 2R R is bigger than 2S and there are different ways of getting contradiction once you know that the triangle inequality has to be used here okay now let us go to the 0 dimensional dimension theory let X prime contain inside X now I am working with a separable matrix basis okay then X prime has dimension less than equal to n if and only if X prime okay for every point P belonging to X there exist arbitrary small neighbourhoods W of P in X I want everything inside X now okay such that dimension of this W boundary of W is also taken inside X intersected with X prime is less than or n minus okay that is as far as you can go everything trying to do only in terms of X it is not possible somewhere you have to you have to involve X prime and that is involved only at the last moment you have a neighbourhood of point neighbourhood of that point for every point okay inside arbitrary small neighbourhood of that point such that you look at the boundary of that inside X now you intersect with X prime dimension of that must be less than or n minus okay everything if you do not worry about going to X at all inside X prime that is true from that I have to get a neighbourhood W of this point inside X with this property so that is the gist of the theorem okay and the converse converse is obvious here anyway assume that the given condition is satisfied that is part it is converse part here that P belong to X prime and U be a neighbourhood of P in X prime now every neighbourhood can be expanded to a neighbourhood of of the point inside the larger space right by definition of subspace to power X let U be a neighbourhood of P in X such that U prime is U intersection X prime then by the given condition you have a smaller neighbourhood W of P such that W is contained inside U and dimension of W intersection X prime is of dimension less than or n minus 1 this is the stated condition in the theorem now you take V prime equipped with W intersection X prime then U prime is contained inside V prime and we know that the boundary of V prime inside X prime is contained inside boundary of W intersection X prime so this lemma we are using it again here which is given to be of dimension less than or n minus 1 okay so now we can conclude by theorem 9.2 9.2 that the dimension of X prime is less than or n minus 1 okay converse suppose dimension of X prime is less than or n given a point P inside X prime and a neighbourhood opens up set in the larger space X such that P is inside U there exist an opens up set P inside X such that P inside V prime which is V intersection X prime contained inside U prime which is U intersection X prime and dimension of this V prime that boundary of this V prime in X prime has dimension less than or n minus 1 okay this is the meaning of dimension of X prime is less than or equal to n starting with an arbitrary neighbourhood inside that I can find a neighbourhood V prime with this property because dimension of X prime is a boundary of this X prime is taken inside this one you recall that this is nothing but closure inside V prime to closure inside X prime of V prime minus V prime which is same thing as you subtract V does not matter that is the meaning of this one alright now recall that if A is an open set in a topological space Y and B is Y minus A bar then A and B will disjoint open sets they are mutually separated inside Y okay start with an open subset and take B to Y minus A bar automatically boundary of you know A and B are themselves disjoint open subset and the union will be so they are separated boundary V bar that is in whole here applying this with Y could X prime A equal to V prime and B equal to X prime minus the closure of V prime closure taken inside X prime okay this general statement I am applying it for Y could X prime now start with the open subset A which is V prime and B I am taking the complement of its closure closure of A here we conclude that this A and B are mutually separated in X prime therefore by the above lemma previous lemma we get open subset W of X such that V prime is inside W closure of W intersection X prime minus the closure of B is empty so this is B I have taken therefore W bar intersection X prime if you take only this intersection that must be contained inside this subset which you want to make once you throw away this one it is empty so this must be contained inside closure of B prime inside X prime okay we also have B prime is in fact inside contains W now okay they started with such a thing and hence V itself equal to V prime intersection X prime right that is how we started with V being a neighborhood of V being an open subset of X here V prime to say that is contained W intersection X prime so putting these two together what we have is boundary of W intersection X prime will be boundary of W by definition W bar minus W intersection X prime which is W bar intersection X prime minus points of W have to be thrown away W intersection X prime okay but this is nothing but the closure of this is contained is a closure of V prime and that is a large this is a small subset of this one this is a large subset this is subset of this one so this is subset of this one because this is subset of this one this is subset of this one okay so but this is nothing but boundary of V prime taken inside X prime and that is precisely the statement in the corners part they took some set theory set topology here you have to do it carefully now we have a beautiful theorem here not very beautiful because you may expect that dimension of the union is equal dimension of way plus dimension of way however such sweeping things do not work so slight modification is necessary yet it is quite beautiful that is what I want to say A and B are subsets of a separable metric base no other condition then dimension of the union is less than or equal to dimension of A plus dimension of B plus 1 okay only thing is they should be defined in the sense that if one of them is infinite suppose this is infinite then a dimension of A is infinite then this is less than equal to infinite is obvious if this is infinite one of them will be infinite so some such things are there I want to avoid all that I want to take the case where in dimension of A plus B is finite dimension of A is finite dimension of B is finite okay you can examine when what happens when they are infinite and so on no problem you see when you are writing inequality here you have to be careful about something being infinite and if this is infinite and this finally no problem if this is infinite this means that one of them here must be infinite okay I am not covering that case here okay let us do induction induction can work for finite case only right or the sum dimension of A plus dimension of B the least value of dimension of A plus dimension of B both of them empty is minus 2 right and then if both of them are empty is also empty okay minus 2 plus 1 is minus 1 so this is okay you see even at that level empty set dimension of A union B is empty is less than or equal to minus 1 minus 1 would have would not have made sense right see minus 1 is not less than equal to minus 1 plus minus 1 so you have to add one more even at that level this plus 1 is necessary there okay even at the very beginning alright so this case is over the least care alright now assume dimension of A plus B dimension of A plus dimension of B greater than or equal to minus 1 which just means that at least one of A or B must be non-empty that is all okay suppose the statement is true for all pairs of spaces A prime B prime whenever this happens namely for which dimension of A prime plus dimension of B prime is less than dimension of A plus dimension of B this is the induction or induction hypothesis here not just for A and B okay whenever see this is some number which is bigger than equal to minus 1 for numbers which are smaller than this namely sum total the induction is on this one right smaller than that the property should be true namely dimension of A prime even B prime should be less than or equal to 1 plus dimension of A prime plus dimension of B prime that is the induction hypothesis then we want to prove it for one higher namely first of all it follows that A union B is non-empty that is what I have I have told you already take P to be in the union okay B and U be a neighborhood of P by symmetry we may assume P is either in A or in B so you assume A just for writing down the proof by the previous theorem there exist a neighborhood V of P such that V is contained inside U and dimension of V intersection A is less than equal to dimension of A minus 1 the previous theorem enters here you see everything is happening inside X except the last condition then the dimension of W V double V intersection A is of dimension less than or equal to 1 okay so we also have dimension of boundary of V intersection B less than or equal to dimension of B by because subspace is of this dimension here is just subspace okay therefore we can apply the induction to the pair boundary of V intersection A and boundary of V intersection B these by A prime B prime the sum total dimension is smaller than dimension of A plus dimensional B one is smaller strictly A minus 1 the other one is at least that much of sum total will be less than dimension of A plus dimension of B therefore induction hypothesis should be applicable to these two subspaces okay to conclude that the union of these two subspaces is less than equal dimension less than equal to 1 plus dimension of the first one is A minus 1 dimension of A minus 1 here you see the other one is less than equal dimension of B okay so sum total 1 and 1 cancel for dimension of A plus dimension of B okay since this is true for every point P inside union from the previous theorem we conclude that dimension of the space namely A minus B must be 1 plus dimension of A plus dimension of B see this is for the boundary of the neighborhood so 1 plus will be one more with the dimension of the spaces so induction is quite easy here once you have this theorem which gives you you know in terms of the ambient space A in and B instead of I do not have to work with A prime A and B separately the boundary of B is taken inside A and B now we have another easy corollary it is something funny here you can you can say union of N plus 1 subspace is of dimension 0 has dimension less than equal to N you cannot say it is dimension 0 okay dimension when you take union may go up one at a time you have seen that in the previous you know we have seen that that is a statement of previous theorem so this is an easy corollary to the previous thing namely you know take A to 0 dimension of space A the union is less than equal to dimension 1 one take one more less than equal to 1 plus 1 and so on it goes on right so you will get dimension less than equal to N you know here is an example now start with any integer N possibly positive otherwise there will not be anything left here and take M to be smaller than that okay you can take equality also let M upper N lower M curly M upper M lower N denote the space of all points in Rn at most M of whose coordinates are rational here you have curly R Mn earlier where exactly M coordinates were rational and so on now at most so you have to be very careful here okay at most M of whose coordinates are rational and curly L L N M same N M denote the space of all points of in Rn at least M of whose coordinates are rational okay N is the total dimension Rn this M denotes how many coordinates are rational here the first part is for M at most M of them are rational the second part is at most at least M of them are rational okay then dimension of curly M is less than equal to M dimension of curly L less than equal to N minus M okay how do you do that let us look at this space coordinates at most M of them are rational so you can start with Rn 0 which is no coordinate is rational okay it is full all the all the coordinates are rational this Rn 0 we know that is 0 dimensional then Rn 1 exactly one coordinate is at most at most at most M coordinates are rational right exactly one coordinate is rational that is also included here like that 1 2 3 up to M okay exactly now each of them we know we have studied they are 0 dimensional how many are there M plus 1 right so dimension is less than okay exactly similarly we have the other way around here here at least M times M of the coordinates are rational so take all the M of rational okay at least M of whose coordinates are rational Rn M then Rn M plus 1 okay one more one more how many all of them Rn they are all again we have studied they are all 0 dimensional how many are there N minus M plus 1 right therefore the total dimension is less than 1 we will stop here we will continue again next time some theorems for N dimension spaces thank you