 Welcome to module 21, completion of a metric space. We have defined the concept of what is a complete metric space. So we want to now define what is the meaning of completion of a given metric space. So we shall describe this construction, a process which assigns to each metric space xb, another metric space x hat d hat that is the notation we have said such that the first condition is this x hat d hat is a complete metric space. It has to do something with the given metric space after all. So all those conditions are explained that in the 1, 2, 3 rest of them. So there is an injective mapping eta from xd to x hat d hat which is distance preserving. In fact any distance preserving map is automatically injective that we know. So injectivity is emphasized here but this is a consequence of distance preserving. The image eta x is dense in x hat. Any isometry f from x1 d1 to x2 d2, x tends to a unique isometry f hat from x1 hat d1 hat to x2 hat d2 hat. So this is what I meant by instead of just you know describing x hat d hat it is a process ok. So I cannot explain it more than that but I will use just to throw it over this is like what one calls canonical or categorical you know it is like a functor and so on ok. So that is the third part here. Any isometry f1 f from x1 to x2 extends uniquely from x1 hat to x2 hat. The fourth one is if xd is already complete then eta itself is an isometry. In other words up to isometry we are not doing any new thing here at all. It is the old xd itself. Something is already complete then there is no completion. Completion itself you know xd itself is its completion. So in what sense isometrically because the construction has to be there for more general space general thing then you have to in the special case it will not be just xd but some copy of xd copy means what again isometrically. So this eta itself is there the canonical embedding of that that will be surjective that is all ok it is injective it is isometric it is surjective the inverse will be also isometric. So these 1, 2, 3, 4 explain the relation between x hat d hat and xd the one which we have started with ok. So let us start with the construction. So this x hat d hat will be called completion of xd that is all. One point I want to remind you if you have done the metric completion of R in your analysis then the process that I am going to do is half of that work only 50% of that work. The metric completion of R starting with Q is exactly like this in the first half and then there is more work to do there. We have less work to do on the other hand if you have not done it at least half of that part you will be learning today. I will not do the other later half but I will later on if time permits I will just explain how that is also done in the case of real numbers. So what I mean to say is here you start with the rational numbers with the standard metric namely distance metric that we know that is not complete right. When you complete it in this process what you get is real numbers. This part we will not prove now ok we will only show the process of constructing the real numbers is similar to this one that is what I want to emphasize that is all. So the motivating fact is the following suppose we have two sequences A and B and in X such that both of them convert to the same point X then of course the first thing is both are Cauchy sequences moreover both the sequences are coming nearer to that point after certain stage that is the meaning of that right. So in particular they will be nearer to each other so how do you express that there exists epsilon for each epsilon there exist n such that after that so m is bigger than equal to a the distance between a m and b m the same m ok this will be less than epsilon ok. This all that you have to do is you have to use triangle inequality here so I will leave it to you to verify this one ok two Cauchy sequences converging then they will be nearer to each other by this definition ok. So that is the starting point which just means that you could take instead of X you could take a neatly chosen Cauchy sequence since there is no way to choose a particular Cauchy sequence what you would like to do is take all sequences Cauchy sequences converging to that point to represent that point. Now when that point is missing since that point is there for every Cauchy sequence then it is complete already when that point is missing the Cauchy sequence will be there still look at all the Cauchy sequences how to pick all of them they must be nearer to each other so that is all we come to this one take X X twiddle denote the set of all Cauchy sequences a n in X now I define a function X at X twiddle cross X twiddle to 0 infinity by the formula d twiddle of a b this a and b are now Cauchy sequences ok take the limit of d of a n b n as n tends to infinity so this is motivated by that if a n and b n were converging to same point then this distance would have limit of distance would have been 0 so that is our motive in general what we do we take this distance to measure their measure the distance between them so this is the measure that is what the d hat of a b so d twiddle of a b the limit of d of a n b n as n tends to infinity now here I want to caution you why this limit exists both a n and b n are Cauchy sequences therefore whether they are convergent or not they are bounded ok when you have both of them are bounded this limit will exist all right so that is our idea ok otherwise you can directly verify this d of a n b n minus d of a m b m ok by use use triangle inequality correctly this difference modulus is always less than equal to d of a n a m plus d of b n b m now if n and m are very large we know that this is less than epsilon these are less than epsilon or both of them are less than epsilon by 2 you can remember ok so here a n b n a m b m this will convert to 0 or some number because you do not know what this number is ok so if they are if they are Cauchy sequences and so on you can you can at least whatever I want to say is this is Cauchy sequence of real number it converges to something both of them this inequality is there so this will be convergent sequence that is all now by the very definition if you interchange b and a here b and a you will interchange the number is the same so limit will be the same so it is symmetric for each a b and c 3 sequences distance between a and c is limit of this one but a n to c n you can write it as a n to b n plus b n to c n so each term can be replaced by some of two terms when you take the limit it is the limit of the the sums right which is some of the limits so therefore what you get is Cauchy inequality d of a n c n is less than d of a n b n plus d of b n c n when you pass to the limit you triangle inequality for d it will triangle inequality is true symmetry is true it is a non-negative function the only thing is why d of a b 0 will imply a equal to b that is not true that is obviously not right you can take just you know just one of the elements you can change same sequence it will have same limit the first element you change the limit will be the same thing so this is not true at all ok however the only thing that is not true is that the positive definiteness positivity is of non-negativity still true the definiteness part is missing here the d 0 is missing here ok d 2 d 3 axiom for d 2 d 3 is true ok so the only thing which means to be d 1 this is one of the exercise given to you earlier if you have spent some time on it even if you have not solved completely these things should be easy for you to understand now in any case whatever is needed here in the proof of this theorem I am going to explain it ok so what we are going to do we make a relation on x hat on x vertical by saying that a is related to b if and only if d hat of a b is 0 just because d hat of a b is 0 does not imply it could be so we are making this relation otherwise there is no need using triangle inequality again you can check that this is a equivalence relation symmetry is obvious by the definition because d hat is symmetric function d d twiddled is symmetric function ok reflexivity is obvious d hat of a a is always 0 alright if d hat of d twiddled of a b is 0 and d twiddled of b c is 0 then you should see that d twiddled of ac is also 0 by triangle inequality that is all so this relation is an equivalence relation let us denote this by square bracket a the class of a the equivalence class of a ok so all Pochini sequences represented by this a they are all what equivalent to in this sense d hat of a comma b is 0 for all b in this class ok let us denote all these equivalence classes by a hat x hat denotes the set of all equivalence classes now I want to modify this definition d hat d hat on two equivalence classes is nothing but d twiddled of a b where a and b are representatives let us put this one to say that this is well defined what I have to do I have to show that this right hand side is independent of what representative I choose so they are all same then this left hand side can be equal to right hand side makes sense otherwise if d hat of a comma b keeps changing as you change the representative then this will not be well defined right well defined as means that so whenever you have choice in the definition you have to you have to verify that the whatever you have taken the right hand side is independent of the choice so suppose a is equivalent to a prime b is equivalent to b prime we must show that d hat of a b same thing d hat of a prime b prime d hat of a a prime is 0 that is the meaning of a is equivalent to a prime similarly d hat of b b d twiddled of b b prime is 0 right because b is equal to b prime again by triangle inequality d hat of a b is d hat of a prime b plus what plus d hat of a a prime so a to a prime I change so a a prime plus this one but a prime is 0 so it is just this one now I am from a a prime a prime b 2 I go to b prime by taking d hat of a prime b plus so a prime b prime plus d hat of b b prime but d hat of b b prime is 0 so this is less than this one so one is less than we could this but this symmetric argument I can start from here and that is less right so they are equal so that is what we wanted to show so you see all once the idea is there all verification is very canonical very easy therefore d hat is well defined and takes only non-negative real values why because d twiddled has that property now d twiddled symmetric also it satisfy triangle inequality also therefore d hat will also satisfy triangle inequality in that one the only thing is why if d hat d hat of you know a a is also 0 that is also clear but d hat of a b is 0 why does it imply that d hat of d sorry a and b are equal that is just a definition that is just a definition equal means what here they are equivalent you see equivalence class is same so this is trick you are using to get equality okay so this here equality is just equivalence classes alright so d hat becomes a metric on x hat alright so this first part is over as you have constructed this one now take a n to be a Cauchy sequence in x hat each a n is a is a Cauchy sequence right the equivalence class is one single element now you take a Cauchy sequence of that okay the Cauchy sequence in x hat let us denote the sequence a n a n is a sequence now okay he is a n m so multi index your views for each fixed n take a n m from 0 to infinity some people write it as a top but I do not want to do that because that may confuse if they are real numbers then the power has something some other meaning okay so for each n n choose a kn such that d of a n kn comma a n m is less than 1 by n so this is where I am using the fact that each a n is a Cauchy sequence okay so it will be less than 1 by n when for some choice of kn large enough for all m bigger than this one this should be true so this is where I have used that each a n is a Cauchy sequence okay this is the first step now look at the sequence a n equal to a n kn this is one element of a n of this sequence which one I have done what is the kn choice of kn it should satisfy this one okay for n 1 for n equal to 1 a 1 k 1 comma a 1 m where m is bigger than k 1 will be less than 1 for 2 it will be less than 1 by 2 similar like that is the meaning of this okay successively you have to make it more and more accurate all right claim is this sequence now this sequence is a sequence in x okay capital a n is a sequence in x these points are in the x right this sequence is a Cauchy in x and its equivalence classes okay sorry this a n is a sequence in here this converges to the equivalence class of this sequence inside x hat okay x hat is what equivalence classes of that this square brackets okay it started with a Cauchy sequence in x hat I have produced I am producing a Cauchy sequence whatever this is an element of x hat anyway the class of Cauchy sequence so that will be the limit of this one so that is the meaning of that is x hat di hat is complete every Cauchy sequence is convergent so this is a claim okay I have to show I have to do this one so the most of the work goes here that showing that it is complete here all right so let us carefully see that how it is coming here is a picture this is your a1 I have chosen this red point here somewhere this is a1 k1 this a2 this is a2 k2 a3 k3 and so on a n kn is this sequence okay so these are all sequence a1 a2 a n what is claimed is this sequence okay is a cluster point of all this namely all these things after certain space in this after certain stage shown here some epsilon ball here actually this is of epsilon by 2 ball so this radius is epsilon all of them are within this the tail end of this one these points may be somewhere the tail end for all of them will be here so something may be here omitted many things are here or omitted but the tail end so the tail ends are here that is the meaning of this one given any epsilon ball so this is what we have to prove this is a picture you can keep it in mind that's all all right so given epsilon that is this is the ball given epsilon tail end I have to find such that this fits all the sequences after certain stage both both ways that's what I have to choose the first choice is given epsilon choose the n1 itself n1 it's the 1 by n1 less than epsilon by 3 you can choose epsilon by 4 also if you want to be careful okay so here there is no nothing this is just real numbers property since a n okay is Cauchy in x hat okay there exists an n2 such that d twiddle of ar as is less than epsilon by 3 for every ar as less than n2 remember this d hat was nothing but d twiddle of the representatives okay these are classes ar you should put a bracket here d hat of ar d hat of d hat of ar and as here brackets that is less than epsilon by 3 is the statement for r and s bigger than equal to n2 but I don't have to put hat here I can directly put d twiddle of ar and ar remember ar and ar themselves Cauchy sequences and what is the definition of d twiddle d twiddle was the limit of as n tends to be arn as n of the original distance between ars and asn okay so finally everything is happening in the metric space x now you take the maximum of this n1 and n2 okay that is n3 I want to show that ar ar as is a Cauchy sequence first of all I haven't yet shown that without that it will not be it's not been x twiddle so it will not be inside x hat the first thing is to show that a itself is a Cauchy sequence that means what for any epsilon I must find a r and s such that the distance between ar and as is less than epsilon right or epsilon by 3 whatever you want to say so that is what I would show to fix ar and s where bigger than n3 for all of them I must show that d ar as this is ordinary distance between the the metric this is the points of x now right is less than epsilon I want to show okay so how do I show that so starting with these choices you see r and s are bigger than n3 so they will be bigger than n1 and n2 so both both these hypotheses will be applicable to d twiddle of ar as is already less than epsilon by 3 because of n they are bigger than n2 okay so this implies that there exists an n4 now I am using the definition of d twiddle because it was taking limits right the limit is less than limit is less than epsilon by 3 that is what I have to show that is what it is so d of ar m as m for large enough m is less than epsilon by 3 r and s are fixed here it is the variable here is the index is m okay for m sufficiently large namely n4 there is some n4 okay so this is one equation okay now you choose m to be bigger than m is bigger than all the time if some it has to be bigger than something I can choose it even bigger than that m to be bigger than k r k s and n4 okay then you can apply both this what is this 18 here see for this one m should be bigger than k r k n right so I am applying it for n equal to r so it will be bigger than k r applying it for n equal to s should be bigger than k s so I am taking it bigger than k r k s as well as to satisfy this bigger than n4 also okay so both 18 and 19 are available to me so I just use triangle inequality now d of ar s what are ar s by definition ar k r as k s so this is the ordinary distance of the metric metric space x right so triangle inequality right it is three different similarly ar k r to ar m okay where m is chosen to be larger than all these plus ar m to asm jump to sm from asm come back to asks so this is triangle inequality okay this is less than or this one from 18 these two things will be less than epsilon by 3 from 19 this should be less than epsilon by 3 so totally less than epsilon so this is a Cauchy sequence is what we have seen the next task is that the original sequence of classes a n converges to the class a that is what I have to show this what okay look at d of a n m and a m what is a m remember this one it is little a m k m right so that is whatever it is this is a number this this is just an element of x they are all elements of x so d of a n m makes sense it is less than d of a n m to a n and then a n to a m so this just triangle inequality okay now the first term on the right hand side this one is less than 1 by n for m bigger than k n right this k n was chosen such that look at this very first choice a n k n a n m is less than 1 by n so that I am applying it here okay so this term is less is as soon as m is bigger than k n this term is less than 1 by n all right this this term the second term can be made any more smaller than any anything once you choose n and m large enough that is right large enough this one this can be made less than m and n because just now we have shown that this is a Cauchy sequence therefore d hat of a n a a capital a here sorry the class a which is by definition 1 d twiddle of a n and a which is again by definition limit as m tends to infinity of d of this sequence a n m and the other sequence a m now here m tends to infinity okay each term is less than epsilon the limit will be less than epsilon after certain states all right so limit will be also less than epsilon okay so for this will apply when n is large enough because I have I have an n here okay so this is epsilon but I have to make a this n less than epsilon by 2 to make it epsilon epsilon so n must be large enough how large 1 by n should be less than epsilon by 2 that is all therefore if this is true for every epsilon look at here this is one sink one one single number here this less than epsilon for every epsilon therefore the limit as n tends to infinity this part d hat of a n a okay is 0 for for large n this is less than epsilon so it should take the limit of this one this should be 0 that is precisely the statement that this this sequence converges to this one okay when does a sequence x n converges x in a matrix space distance between x n and x limit is 0 okay so what we have done we have just completed the the construction and property 0 that x hat is a complete matrix space now we have to verify 1 2 3 4 they are all easy all right hard work is already done so it remains to prove that 4 properties 1 2 3 4 okay so first thing I have defined eta eta of x to x hat that is what I have to define given any x belong to x consider the constant sequence x x x x the nth term is x look at that sequence that is obviously a course is sequence take its class that is an element of x hat okay so this way I get the function eta x is the class of cx obviously eta is injective okay y suppose eta x is equal to eta y cx is in the same class of cy what is the meaning of that that distance the limit of that you see you have to take the limit is equal to 0 but limit is here x y x y d of x y x y so limit itself is d x y that is 0 means x equal to y x is a matrix space okay so this is a this is clearly a injective mapping okay actually we shall prove that we can check that its distance preserving what is the meaning of this take two points x and y cx and cy distance is also d x y so that is the meaning of this distance d hat of cx cy is equal to d of x y because it is both of them are constant sequences okay so it is automatically distance preserving and hence injective mapping what is the second thing here I forgot what is like eta is dense inside x hat the image is dense in x hat take any open ball in x hat you must find a point eta x inside that which might intersect eta capital X so that is the meaning of that right density of eta x okay so here it is we shall prove that every open ball in x hat intersect eta x okay given xn a sequence cosy sequence and its class okay take a r positive and take an open ball around that of radius r okay I want to get an element of the metric space so that the constant function constant sequence the its class of the constant sequence is in that ball that means its distance from xn to that must be for less than r so for that choose a k belonging to n such that this is a cosy sequence d xk xk plus m is less than r for all m okay once you have chosen k for all numbers bigger than k also is easily true so this is definitely true for k k and x k plus m less than r for all m now it follows that once you have chosen this one look at the constant sequence xk that sequence okay has distance less than r from d hat of xn okay so that is why this eta of xk now it belongs to this open ball okay finally take any two metric spaces and an isometry f that isometry itself extends to the whole of x hat to x1 hat to x2 hat so this is stronger than saying that you know x1 hat is unique up to isometry that is the meaning of this that that that is what we want to prove but we are proving say x is strong the actual isometry each isometry extends okay so how do we prove that isometry means distance preserving look at the construction of x1 hat and x2 hat it just depends upon the metric there that is our so in isometry it should metric equivalence right for everything it will be cosy sequence preserved everything will be preserved equivalence and preserved that is why so that is what you have take a sequence cosy sequence in x1 f hat of a which is f of an that sequence that will be cosy inside x2 even even just similarity would have given you this one right here we have an isometry okay since isometry preserved cosy sequence is first of all f hat is well defined from x1 to x2 to x2 okay here take a cosy sequence f of that will be a cosy sequence here okay now you see that this f hat is an isometry of the corresponding d1 twiddle d2 twiddle the pseudo metric spaces the d2 twiddle distance is preserved by this f hat f twiddle d2 twiddle of f of an f of bn by definition the limit as n tends to infinity d2 of f of an f of bn each f of an f of bn is the same thing as d1 distance between these d1 of a and bn right because f itself is an isometry its this limit is d twiddle of a b the given sequence a and bn so that in the meaning of that f is you know distance preserving from d1 to d2 d1 twiddle to d2 twiddle okay clearly f hat also preserves equivalence relations and hence defines an isometry f hat from x1 to x2 why this preserves equivalence what is equivalence an is equivalent to bn if the limit of distance between a and bn is 0 right distance between f of a and b also 0 limit so so everything is preserved that is what you have to verify so you get a map from f twiddle to f hat here from x1 hat to x2 the class here goes to the class there okay and the distances are distances in di at the same thing as distance we have defined for representative so they are preserved so it is preserved so it is an isometry or not so the net conclusion here is the construction depends only on the isometry class of x1 or x whatever starting with if x1 and x2 are isometry x1 hat and x2 hat are isometry okay the last one is suppose x itself was complete then I have to prove that just I have to prove that this eta is surjective then eta itself is an isometry okay so why it is surjective take any xn in x hat okay the class represented by a Cauchy sequence xn in x that is the meaning of this x hat right since x is complete it has a limit this xn has a limit this limit is independent of the Cauchy of the equivalence class all the sequences whatever sequence you take here okay representing this one they will all be convergent they will be convergent to the same x because the distance limit of distance between xn and yn that itself will be 0 so if this is x and that is y x will be equal to y so there is a unique limit here inside x okay now very easy to verify that the constant sequence Cx itself the equivalence class itself will be in the class of xn so you have to show that these two are required what is the meaning of that distance between xn and this Cx that is tends to 0 but that is the definition of that xn converges to x okay so that is the meaning of this one so this just shows that eta axis equal to the class xn so eta is circuit okay so that is more or less completes several things that we have in mind about metric spaces that does not mean that the study of metric space is completed there are many things but now onwards we will concentrate more and more on general topological spaces bringing in metric spaces only to strengthen the results that the topological results will give you about metric spaces okay any questions here are some exercises all straightforward exercises nothing to do with completion of course okay just ordinary exercises this is about the product finite product of metric spaces all right this is about again finite many things then you have to want to take it as the maximum metric just like l infinity norm you can do l infinity the d infinity metric here the d infinity corresponding to maximum that infinity is involved okay so that is the next chapter now so if there are no questions we will stop here thank you