 Welcome to NPTEL NOC on point set topology part O, module 42. So, we will now do some theorems for n dimensions. The spirit is the same thing as we did last time, only we are continuing more and more results here. A countable union of closed subspaces of dimensions less than equal to n is of dimension less than equal to n. In other words, countable union of closed subspaces does not increase the dimension, that is what one has to understand. So, I have to take subspaces, where are they, they are all subspaces of one single separable metric space ok. So, consider the following three statements A n, B n, C n. First A n says a countable union of closed subspaces of dimension less than equal to n is of dimension less than equal to n. This is the statement of the previous of the theorem that is stated now, which you want to prove. But we will take some subsidiary statements here, B n says countable union of f sigma subspaces of dimension less than equal to n is of dimension less than equal to n. From closed subspaces, we have improved the statement in a slightly that is not f sigma subspaces, f sigma are countable union of closed subspaces. The third one says any space of dimension less than equal to n is a union of subspace, one subspace of dimension less than equal to n minus 1 and another subspace of dimension 0 ok. So, this is dimension n union of subspace of dimension n minus 1 and a subspace of dimension 0. So, here nothing is said about the subspace being closed that is important here ok. So, what I why I said these two are this looks like an improvement it is not improvement because they are a same because each f sigma is itself a countable unit countable union of closed sets and then you are doing countable union of f sigma. So, that will be countable union of closed sets. So, a n and b n are easily seen to be equivalent, but c n is a somewhat strange thing here let us see what is the role of this c n here. The plan is to prove a n that is what we want to prove by induction, but how do we are going to do this one is the following. Once you have proved a n minus 1, we will prove c n then I will use a n minus 1 and h n to prove a n ok. The statement b n is only subsidiary here it will play a subsidiary role ok. So, this is a plan. So, to begin with a minus 1 is completely obvious a minus 1 means what they are all of dimension minus 1 less than equal to minus 1. So, they are all empty countable union of empty sets is empty set that is nothing to prove that. So, our induction starts at n equal to minus 1. Also we have earlier proved a naught itself remember that I told you today we are not going to prove any new phenomena or anything same phenomena we are proving for higher dimension and so on ok. Total union of closed subsets is of dimension 0 is of dimension 0 less than equal to dimension 0 less than equal to dimension 0. So, this is what we have proved earlier. So, a naught is proved alright. So, we shall prove I mean this is just as I do not have to actually prove this one because my induction starts at a minus 1 itself. So, we shall now prove that a n minus 1 implies c n ok. Here we are going to use separability of the matrix space that we are working explicit plane ok right like x b of dimension less than equal to n. That means there is a basis b for the topology of x consisting of open sets u such that dimension of the boundary of u is less than equal to n minus 1 for all u inside b right. Since x is separable we may assume that b is countable. So, this is the role of separability here. So, I mean just enumerating b as u i i belong to n. Now, put b i equal to boundary of u i ok all of them have dimension less than equal to n minus 1 by the construction here. Now, n minus 1 implies if you take the union of all these b i that is of dimension less than equal to n minus 1 ok. We claim that the complement of b in x namely dimension of x minus b is less than equal to 0 ok. So, how would I prove that? Take x prime equal to x minus b this is just a temporary notation and b prime is a family of all u i intersection x prime. Clearly because b b x prime is a subspace of x b prime will be a basis for x prime the subspace topology. And we have if you take u i intersection x prime and its boundary in x prime then boundary of its containing inside the boundary of u i intersection x prime. And these things are nothing but our notation is just b i intersection x prime and b i intersection x prime are all empty. Why? Because x prime is x minus b all the union of all the b i we have been thrown out ok. So, it follows that the condition in earlier theorem is satisfied namely for n equal to 0 ok. Therefore, dimension of x prime is less than equal to 0 ok. So, this is where we have used a n minus 1 n. So, since x is a x b union x prime we get c n ok. So, remember once again I recall this c n is a string thing c n is any subspace of dimension less than equal to n is the union of subspace of dimension n minus 1 and another subspace of dimension 0. So, c n is proved ok. Now it remains to prove a n minus 1 and c n together implies a n that will complete the proof of the theorem. So, start with x as union of some closed subspaces countably many closed subspaces where each f i is closed dimension of each f i is less than equal to n. So, inductively let us let me define this k i k 1 is f 1 k 2 is f 2 minus f 1 k 3 is f 3 minus f 1 union f 2 and so on. Throw away all the earlier sets because you are using some indexing here ok. So, this is the way you have defined k 1 k n etcetera. Then x is union of k i is also this kind of thing we have used several times. Now k i intersection k j is empty for i not equal to k this is the extra hypothesis we get where that hypothesis is not there on f i ok by construction k i and k j's will all be mutually disjoint. Each k i is f sigma now I cannot say that k i is closed ok each k i is f i minus something. So, it is a open subset inside a closed subset closed subsets are all metric spaces here. So, any open subset of a metric space is f sigma. So, each k i is f sigma ok this is where our b n will play the role that is all. Dimension of k i is less than equal to n because they are subspecies of f i ok. So, I am stating 1 and 2 are obvious set theoretically for 3 if I notice that x is x minus i ring to 1 to f j this is an open subset in x which is a separable metric space and each and see it sigma is x these are x sigma. But, then k i is f i intersection x minus this one. So, that is also f sigma ok directly you can use f i's or metric spaces and apply this one. So, 4 is true because k i is a subset of f i because f i is a dimension less than equal to n ok. Now, we can apply c n to each k i because they are f sigma right k i equal to m i union n i ok that is the meaning of c n where dimension of m i is less than equal to n minus 1 and dimension of n i is less than equal to 0 ok. Put m equal to union of this m i and n equal to union of this n i. Since k i intersection k j these are all mutually disjoint k i intersection k j is empty. Each for each m i which is m intersection k i k i is m i union m i m is union m i. So, if I look at just m i it is just the whole thing intersection k i because they are all disjoint subsets right. So, each of them is f sigma inside m. Therefore, we can apply a n minus 1 is the same thing as b n minus 1 right. So, if it is f sigma it is same thing as they are closed subsets and conclude that dimension of m is less than equal to n minus 1. Whether you apply n minus 1 directly or b n minus 1 you can directly apply b n minus 1 because they are f sigma. So, dimension of m is less than equal to n minus 1. Similarly, each of these n i's are dimension 0. Therefore, dimension of n which is a countable union is less than equal to 0 ok. From 1 it follows that x is m union n what is m for what is 1 x is union of k i ok. Each k i is written like this union of k i will be m i union n i which is m union n that is all. From 9.16 whatever theorem we conclude that dimension of x is less than over to n. So, the inductive proof is completed from countable union from you know union of two things we have proved the same thing for countable union alright. So, this theorem is about now I will repeat this theorem countable union of closed subsets of dimension less than equal to n does not increase the dimension ok. Let us go ahead now if x is union of two subsets each of them dimension less than or equal to n and b is closed then dimension of x is less than over to n. Same kind of result instead of 0 we have now general n here is similar to what we have proved in 9.16. Convert the non-closed set into a countable union ok. We do not convert that itself we take the complement of b inside x and that convert that and take intersection with that that is all. So, it is proof is similar to that one. Now special case here if x is x prime union single term then dimension of x prime is less than equal to n and sorry and dimension of x prime less than equal to n dimension of x less than equal to n because here you know the dimension of x that is 0 ok this is important only if x prime is non-empty after all. So, here is a remark now of our famous and popular example Cranster-Curatovsky space k it is subspace of r to remember and it has proved that it is connected space. In any case it is being substitute of r to it is of dimension less than equal to 2. Now note that for each pin side k there exist an arbitrary small open rectangular with sides parallel to the coordinate axis such that the boundary of r intersection k is a 0 dimensional space ok. Remember how the this Cranster-Curatovsky is constructed ok this point half comma 1 ok from there you are taking lines joining to the points inside the scantor set, but those lines are all perforated either they consist only rational number or they consist only irrational number depending upon what point you are adding to what point in the cantor set you are taking. So, use that property to see that the boundary of any rectangle intersection k is 0 dimensional it will have line segments, but that line segments will be perforated all through right. Therefore k is of dimension less than equal to 1 by our definition because it has a base which is property. Now what are the possibilities minus 1 is not possible because non-empty ok a 0 dimensional space cannot be connected. So it is not 0 dimensional hence dimension of k is equal to 1 ok less than equal to 1 and minus 1 and 0 are not possible. Now you apply this corollary ok that k prime x prime you take it as is k 0 namely obtained by subtracting the apex point 1 by 2 comma 1 from k ok. This also dimension is what I want to claim here ok. So why because if it is of if it is of dimension 0 or minus minus 1 is not possible right because it is non-empty. If it is of dimension 0 then after adding this point this will be also dimension 0 all right therefore this must be already of dimension 1. However notice that it is totally disconnected space ok. So that is some surprise quite often people understand totally disconnected space of dimension 0. So in our definition it does not you know behave like that. So totally disconnected space is can be of higher dimension indeed examples of totally disconnected spaces of dimension n for any finite n these are also known. We will not go into those examples anyway. Another corollary is take a subspace of dimension less than equal to n then every point p inside x not inside x prime. Now I am making a statement about global statement about x itself as arbitrary small neighbourhoods in x whose boundaries have intersection with x prime of dimension less than equal to n. The statement about p belonging to x prime follows by the definition of dimension of x prime less than equal to n. But this is now for all points inside x itself they also have such a neighbourhood only thing is the boundaries are not less than equal to n minus 1 but the boundary intersection x prime is of dimension less than equal to n. So for this you have to apply the above corollary and theorem 9.5 to x prime union p that is all. Here it is in dimension dimension is less than equal to n. So apply there and then then take you know then you can remove the point p that is all. I mean actually you want neighbors of p so you know need to remove the point you can take the boundary. The boundary will be obviously intersection x prime will be all inside x prime. Just above corollary is a direct generalization of 9.50. Okay so let me just recall this 9.50 because several times I am referring to this one. See this is what we had the global characterization of a subspace in terms of what is happening inside x itself. Dimension of x prime is less than equal to n if and only if for every point in x there is a neighbourhood arbitrary small neighbourhood w of p such that dimension of boundary of w intersected with x prime is less than equal to n minus 1. So this is the theorem that we have used there. So thus what we are ready now for ready reference we state the condition cn as a corollary because we have proved that now right n implies a n minus 1 implies cn actually. So let us have that one because it is not just a subsidiary statement in inside the proof. So let us state it separately what is it every space x of dimension less than equal to n is the union of a space of dimension n minus 1 and a space of dimension less than equal to 0. Once again if you take empty set that is this is empty set this itself will be dimension less than equal to n minus 1 that is that will. So in this statement the second part will be automatically non-zero and non-empty okay. The first application of this is interesting namely if you have a you know separable matrix space of dimension less than equal to n but this n must be finite that is important okay then and then only it is the union of n plus 1 subspace is each of which is less than dimension 0 okay. Here I can put equality also provided that is an equality here alright. So we do not have to worry about getting equality so you know so much sometimes it is not possible that is why we have to be careful that is all. Let p and q be any integer bigger than or equal to minus 1 just put p plus q plus 1 equal to n okay this is a definition of n that is all. Given any separable matrix space x of dimension less than equal to n that is p plus q plus 1 that is all separable matrix space of dimension less than equal to n there exists subspace is p q such that x is union of p and q dimension of this capital p is less than equal to p dimension of capital q is less than equal to q. So this is the generalization of now you know writing just arbitrary subsets n and 0 you can break the n into any way you like partition p plus q plus 1 and then you will have this theorem. So how do you do that that is not very difficult you have to use this one cleverly inductively okay theorem 9.26 you can use that okay. Similarly as an easy consequence of the above results we shall prove the following now we have come to product x and y are separable matrix spaces at least one of them non-empty okay if you assume both of them are non-empty well there will be some problem you will see one of them is non-empty other one is empty what happens this will be empty space okay. So what is the statement non-empty and finite dimension okay they are both finite dimension then the dimension of the product is less than 2 dimension of x per dimension of y okay. So if both of them are empty what happens this will be minus 2 but this is always minus 1 you see so if both of them are empty this statement is now not correct only for that reason we have to assume that at least one of them is non-empty so that I have got here 0 and minus 1 and minus 1 that is fine this is empty also alright. So empty one of them is empty it is already taken care so you can assume that both of them are non-empty also by symmetry we may assume that y is non-empty we shall index on dimension of x plus dimension of y as usual the statement is obvious if dimension of x plus dimension of y is minus 1 okay this plus this minus 1 means this must be 0 and this must be 1 okay or anything you like but you cannot have anything else here because dimension cannot be less than minus 1 okay so that statement that part is only this is minus 1 by chance what is it this must be 0 because we are assuming this is non-zero so this must be 0 and this must be empty so assume that all spaces a and b with this property dimension of a plus dimension of b is less than or equal to less than dimension of x plus dimension so this is the induction hypothesis the statement is true okay having verified it for minus 1 now we are assuming this could be any value bigger than or equal to 0 then this one then you take the statement to be true for all the values namely dimension of a plus dimension of b all spaces dimension of a plus dimension of b less than this dimension plus dimension dimension of x plus dimension okay so that is the induction hypothesis for any point x y inside x cross y we have arbitrary small neighbourhoods by the definition of product pair u cross v where u cross v are neighbourhoods are extend by respectively such that dimensions of the boundaries add up to some number less than equal dimension of x equal dimension of y minus 1 each time this minus 1 that minus 1 okay so minus 2 okay. Now what is boundary of u cross v it is by definition u cross boundary of v union boundary of u cross v right boundary of u cross v is that one you can take the closures here boundary of u cross v same thing as boundary of u bar cross v bar u bar cross v bar okay so that you can take the closures here then you get u bar cross boundary of v union boundary of u cross v bar okay by induction hypothesis each of them is of dimension less than equal to dimension of x plus dimension of y minus 1 okay also both of them are closed see here u bar is a full thing that only dimension of this one goes down that is why it is the dimension of x plus dimension of y minus 1 here this one is dimension full dimension of y but this one is dimension was down so this is like 1 to minus 1 because it is a union also both of them are closed thereby corollary 9.20 will apply dimension of u cross boundary of u cross v is less than equal to dimension of this plus dimension of this minus 1 okay so the theorem is proved special case if y is 0 dimension of then dimension of x plus y x plus y is dimension of x plus dimension of y which is just dimension of x so taking product with a 0 dimensional space does not increase the dimension okay so what is y this special case of this one all that I have to do is y is non-empty because it is a 0 dimensional we have x cross y sitting inside x cross capital y for some y inside y okay this is coordinate inclusion therefore dimension of x is same thing as dimension of x cross singleton y which is subspace of this one so this less than dimension of x cross y which is of course dimension of x so equality occurs otherwise we had only less than equal but here equality occurs because it is sitting inside between these two so dimension of x cross y must be equal dimension of x one may anticipate that above statement is true in general just like dimension of rn cross rm is equal to rn plus n right so it would be a nice thing so one may anticipate above statement is true so you can ask this question is dimension of x cross y equal to dimension of x cross dimension of y in general this is not the case as soon as you have huge spaces the huge necessary the meaning of huge is not with respect to dimension here unfortunately okay you do not have to go to infinite dimension if infinite dimension equality will automatically that is not the case so what is the example is our favorite example namely all the points with rational coordinates inside the L2 space QL okay of all points in the Hilbert space L2n coordinates are all rational so this is of dimension one right this is what we have proved earlier but now you can take QL cross QL again isomorphic to QL okay by obvious kind of shifting right x1 x2 xn y1 y2 yn interlace them x1 x1 y1 x2 y2 and so on okay interlace them what you get is again QL cross QL is isomorphic to QL in fact you can you can this this trick you can use QL cross QL cross QL any number finite number of times again isomorphic to QL okay so this is dimension one this dimension one but it is also dimension one so dimension hasn't added up here okay so let us stop here next time we will launch a program to prove that dimension of the real you know for the the Euclidean space Rn is actually equal to n thank you