 Hello, welcome to module 23 of NPTEL NOC course on point set topology part 2. So, we started with Stone Weistra's Theorem yesterday we have proved Weistra's approximation theorem and prepared partly some ground for Stone Weistra's theorem. So, today we will continue we will need one more concept which so far we have not introduced. We will introduce this concept only for directly apply it to here in this context. So, we are not going to treat this on its own in general context. So, this is the concept of a lattice. A sub algebra A of Cxr is called a lattice if it satisfies the following condition. Given any two elements inside A the maximum of f and g must be also inside A. Remember, maximum of any two continuous functions is again a continuous function first of all. So, it is an element of Cxr what we want is starting with f and g inside A the maximum must be also inside A. Similarly, minimum must be also inside A. So, then we call it a lattice. In particular what happens if you take instead of just two elements suppose you take finitely many of them then also maximum and minimum are defined and they will be also there by induction. You can f1, f2 first you take the maximum f1, f2, f3 maximum it is same thing as maximum of f1, f2 and f3 and so on. So, you can apply that. So, finitely many elements of A then the maximum and minimum will be also inside A. So, that is the meaning of a lattice. So, let us continue this study. Let A be a closed subalgebra. This concept you have already introduced earlier a closed subalgebra of Cxr then it is a lattice. Therefore, this Cx comes to us free of charge. We want to close subalgebra after all. So, it is automatically a lattice. So, it is only in the property of closed subalgebra in addition property of closed subalgebra that is what we are going to exploit here. That is the content of this lemma. So, first let us say that a closed subalgebra is a lattice. So, that is one line proof here because maximum of f and g can be written as because they are real wide functions after all then only maximum of f and g. So, f plus g plus f minus g modulus divided by 2. Minimum of fg is the same thing f plus g minus f minus g divided by 2. So, you take modulus of minus f minus g of course. So, this you must have seen in many other places like in major theory and so on. So, if we have not seen it spend some time and verify that these two are correct formulas. So, if f and g are inside a, a is a algebra. Therefore, f plus g is there. So, f minus g is also there. The modulus is there while that is what we have proved earlier. There we have used to closed subalgebra. Using Weistras theorem we have proved that if some function is there then its closure is also its modulus is also there provided the algebra is a closed subalgebra. Therefore, f minus g is there modulus of f minus g is there the sum total is there divided by 2 is also there. Similarly, minimum is also there. So, a closed subalgebra is a that is is a easy consequence of what Weistras theorem has been used here namely once f is there mod f is there okay. Let us go ahead now. Now, we want to last time we have introduced this algebra r cross r and we studied all possible algebra of r cross r. So, now we are going to exploit that for each x comma y inside x cross x. Let us make this notation A x y is all ordered pairs H x, H y of real numbers okay. This subset of r 2 now where H range is over this curly A where curly A is the closure subalgebra that we have studied okay. So, this is subspace of r 2 indeed it is a vector subspace is easy to see okay. Indeed it is a subalgebra if A is an algebra right follow that A x y subalgebra of r 2 okay. Why? Because if you multiply this how do you multiply you have to multiply H x 1 H 1 x H 1 y H 2 x H 2 y right how do you multiply H 1 H 2 of x and H 1 H 2 y. So, that is the multiplication of 2 functions inside this algebra A also. So, it will be H 1 H 2 of x and H 1 H 2 of y. So, similar way addition scalar multiplication all these things essentially. So, this A x y for each point x y is dropped down to just a subalgebra of r 2. So, what is happening here will tell you the entire over A complete control over A that is the beauty of this approach here okay. According to this lemma that we have studied last time A x y being a subalgebra of A is equal to 1 of these 5 0 r cross 0 or 0 cross r or the diagonal or the whole of r 2 alright. So, let us go ahead now. Let us put A prime equal to all those f inside A x r such that when you take f x it is a condition f okay f x f y this ordered pair is inside A x y. So, this should be true for all x y inside a subalgebra. This condition on f okay suppose f is already inside A then this is automatic right because the definition of A x y itself is that one okay. So, A prime automatically contains A. So, this A prime looks like a fattening of A something larger than A right. So, we have taken this A prime. So, automatically it contains A okay this curly A. The following lemma is just an alternative description of this A prime okay. So, here we are bringing the point separation or separation of separate points that property in a strange way you have to just watch it out okay. So, first of all I will give you an alternate description of this one. An element f inside C x r is inside A prime if and only for every x comma y there exists a g x y inside A such that f x f y is equal to g x y x g x y y. This g is inside A okay given any two point there is a function g okay g x y of that is equal to f x and g x y of y is equal to f y. This is a strong way of separation point okay x and y are two different things for example two different points especially then you could have chosen this function f x f y some function which has the properties that f x and f y are different then g x y of x and g x y y will be different right. So, in some way this is trying to encode the separation property of A. So, this g x y must be inside A okay but just do not bother about that one just check that this definition this description fits this definition okay. Take an element f it is inside A prime means or f x f y is inside A x y what is A x y A x y are nothing but some g x g y where g is an element of A and that g I am calling it as g x y because it depends upon x and y for different x and y it could be different g that is why g x y. So, this description is obvious okay. Now, here is a picture of the same property that I have told that function f is inside A prime means what this function is f here okay some continuous function take any two points A 1 B 1 A 2 B 2 A 3 B 3 whatever any three any two points pair of points look at f of A 1 and f of B 1 it can be captured by a function g A 1 B 1 this is an element of A now f is an element I do not know f is just a element in the larger algebra c x c x r right this is an element of A and it has a property at B 1 it is f B 1 at A 1 it is f A 1. So, that is the meaning of it similarly A 2 B 2 that would be another g but both g I these g's are inside A so that is the meaning of capture it. So, somehow you know if you choose a lot of points you will get lot of these things they will keep you know some combination of that will give you very you know will give you the function and all the given function okay in approximate way that is the whole idea okay like all approximations like Newton's approximation and so on you choose lot of points and then you choose something going through them that is the approximation for the function okay so that is bringing that is coming here now in an all score way we in a very simple way in a you know without our noticing at all all that you have to do is study the sub algebra of R group okay so you will see that how how to do this one okay now clearly I told you A is containing a prime the crucial topological result which we reverse the inclusion namely A be a closed space of c x r and a lattice in particular if it is closed sub algebra right then these two conditions are satisfied then this A prime is contained inside A in other words these two are equal okay so this is the crucial thing here and here so you have to do a little little more common total topology okay let us see how you show that every element of A prime is inside A let F be inside A prime given epsilon positive we shall show that there exist a g belonging to A such that norm of F minus g is less than epsilon what does the mean what does it mean this means that F is a programator your functions from A that means F is inside A bar but A bar is A because we start with closed subspace okay so this is our idea how to construct the g such that g is inside A so for each x y inside F fix a g x y as in lemma 5.38 okay this is what we have seen for each x y there is a g x y because F is an element of A prime alright so fix g x y as in this lemma now put u x y equal to all z inside A such that F z is less than g x y z plus epsilon okay so this is the left ray b x y is a right ray z belonging to x such that F z is bigger than g x y minus epsilon okay these two rays are around g x y z one is g x y z plus and for minus in mu plus and this one is from g x y z minus epsilon to plus infinity so these two rays are taken of course I have taken the inverse image all z inside z inside I such that F z is inside this ray which means it is the inverse image of the ray under F so this is u x y v x y these are open subsets obviously note that x, y is both inside u x y as well as v x y okay in particular why we are why we are saying F x F x here is g x y x so F x is less than g x y plus epsilon similarly F y also g x y plus epsilon so x and y have this property so they are both inside u x y as well as v x y in particular if you fix y okay and vary x because I have taken all the x u x y will be a cover for x similarly I can fix x and vary y also okay so I will get another cover but x is what compact host surface compactness allows you to take finite cover so you get u x 1 u x 2 u x n which we have x k I have put here okay all suffix further which y is there x 1 y x 2 y 3 y and so on okay so this is a finite sub cover of this one okay now you have finitely many g x y y here correspondingly right you can take the maximum of g x y z as I range for a 1 up to k okay that is my g y the maximum of g x y z is g y so that is inside that all these g x y y they are elements of a therefore the maximum is inside a because a is a lattice okay so what happens to g y it follows g y side a and f is less than g y plus epsilon on the whole of x because f is less than each g x y y and this is maximum okay so it is true for all x for all of them so f is less than g y plus epsilon on the whole of x on the other hand if you go for the other part f is bigger than g y minus epsilon only on the intersection if all of them if if it is if it is bigger than bigger than all of them then only f is bigger than the maximum right therefore it should be true for all the i i could 1 2 3 up to k which means that I have to take the intersection so I take intersection of v y v x i y i here correspondingly I am calling it as v y on this intersection in a small fact f is bigger than g y minus epsilon okay is that clear once this is the case I can do the same thing for now over y now we are fixing x okay once you have what is these v y for each y v y is a neighborhood of y so they cover the whole of x as y rings over x right therefore I can get a finite sub cover of this one now v y 1 v y 2 v y l same okay now for each v y 1 v y 2 I have g v g y i here right so I take the minimum of g y now so I am first maximum and then minimum okay minimum of g z is g z which is g y 1 g y 2 g y l minimum of them okay so this minimum is again a point of way because g y 1 g y 2 g y l are inside a and a is a l at a now what happens f is bigger than g minus epsilon the whole of x now because f is bigger than each of them it should be bigger than minimum of course so it will be valid for the entire effect bigger than x it will be in one of these so it will be bigger than one of them so it is bigger than minimum similar to similar argument at this point okay so it follows that combine this one now f is less than g y on the whole of x g y plus epsilon here g y minus epsilon okay therefore norm of f minus g is strictly less than epsilon now on the whole of it combining these two things so the proof is over okay so it is like a minimax principle here okay compactness is heavily used compactness of the domain space okay so now let us complete the proof of real version of stone wise class like x be a compact of space a be a closer sub-algebra of Cxr which separates points and contains a non-zero constant then a is the entire Cxr contains a non-zero constant is the same thing as contains a sub-algebra r itself you know the real number they as a constant function that is sub-algebra of a always okay so that is the assumption here so a must be there it is possible that a sub-algebra may not have any other non zero will be there any other non-zero constant may not be there even if one non-zero constant is there the entire r will be there okay so this is the standard version of stone wise class theorem what we are going to do is a slight modification of this one a slightly stronger and this is due to Gaddy it is not very old 2016 okay without any extra effort we are going to prove this one so I have put this one so let x be a compact of space a be a closer sub-algebra of Cxr which separates points I am not assuming that it contains non-zero constant okay that is not I am not so I have two different conclusions here then a is Cxr or or what are equal to the maximal ideal mx0 set of all f belongs to Cxr which vanish at x0 for a unique point x0 in x okay so indeed this part is much more you know a lot much more revealing what is happening in the algebra then just studying that it is Cxr under the assumption that there is non-zero constant let us take this one for granted then the standard version follows automatically because as soon as there are non-zero constant it cannot be this one that is all right there is a non-zero constant that will not be equal to 0 so right so mx0 is much smaller after all those only where which vanish at a point the constant function non-zero constant function will not vanish at any point so this part will not occur so it will be always Cxr under the assumption this one so what should have happened if assume that there are no non-zero constant then we must be able to conclude that this a is nothing but mx0 namely you must find out some x0 exactly one point will be there it is like a contraction mapping there is only one point such that fx0 all the functions will vanish at that point okay and a will be precisely equal to that so this is an ideal because if f is such thing then if you take any g inside Cxr g into f will also vanish that is the meaning of an ideal it is a maximal ideal because the moment you take anything which is not 0 at x0 you can produce a non-zero constant once you do that it will be the identity element will be there so it will be the whole of the Cxr okay so let us go towards this one the proof of proof gives a complete picture of Stone Weistras adoption of Weistras theorem of course this use is Weistras theorem just to show that mod f is there as soon as f is there okay if x is a singleton there is nothing to prove what is Cxr it is just r okay so any sub-adjiff of r is either 0 or the whole of r so there is nothing to prove so there is no problem so we assume that x has more than one point okay by Lemma 5.31 we know that a is a lattice so in order to exploit this Lemma 5.33 namely the classification of sub-adjiff of r cross r into 5 classes we have to pay attention to the hypothesis A separate points okay namely we have to come down to this Axy as I explained that is the only thing left out we have done all other preparation so let us do that let us first make some observation all these okay inside r2 now given xy belonging to x cross x we know that Axy is a bulge of r2 and hence we can apply this Lemma and we know what Axy is either 0 or r cross 0 or 0 cross r or the diagonal or the whole of further Axy if it is sub of r cross 0 what does it mean what do you remember what is this fx fy that means fy is all the time 0 inside A as soon as A is inside f fy is 0 what does it mean that A must be inside my remember the definition of my if you have here itself I have taken this definition my is all f such that fy is 0 that is another ideal okay all all mx where for x is a point of it that is an maximal ideal all all of them so this contains my similarly if Axy is 0 chroma that means fx fy fx is 0 for all x that means again A is inside mx okay now the second observation is for all x means for x Axx that means what fx fx so that is already diagonal so it is the entire diagonal or 0 okay so all these are elementary observations but finally you put all of them together you have wonderful results if finally A separates point then there is at most one x such that A is contained inside mx A separates points okay suppose there are two points such that A is contained in mx A is contained in my what does it mean all the elements of A vanish at both x and y so x and y cannot be separated that is all so the uniqueness part comes only by A separates points automatically okay so now consider a case when Ax0x0 is 0 as I have observed Axx is always 0 or delta r right suppose for some x0 I do not know whether it is there or not suppose that that case occurs Ax0x0 is 0 okay automatically it implies that fx0 for all f inside A is 0 right in this case mx0 is contained inside A is the claim A is contained in mx0 is obvious right if you show mx0 is contained inside A that would mean they are equal and we have done we have finished uniqueness part we have already shown here so we would have put part 2 here conclusion in the part 2 of the theorem right so let us prove that in this case mx0 is inside A okay so this is achieved by showing that mx0 is inside A prime okay from A above we get A is inside mx0 this we are saying since A separates point from C we get Axx for any other x if for the delta r it cannot be 0 it must be the delta r as soon as x is 0 equal to x0 also from B B says that for all x Ax is either 0 or delta r right so also from B we get Axx0 now is r cross 0 because if this is 0 that would have means Axx is equal to 0 so Axx for x0 equal to r delta r means Ax0 must be r cross 0 okay and Ax0 x means 0 cross r there is no other way right for all x0 equal to x0 finally from C again we also see that Axy whenever x is not equal to x0 y is not equal to y0 y these are different from x0 that is all then it must be the whole of r2 right right because there are distinct points so fx fy there will be some f for fx fy will be different fx fx fx fy are equal for all of them then only 2 would have been delta r so delta r is not possible this is not possible so it must be the whole of r2 for all xy for x is not and y is not one single point has this property rest of them should have satisfied this one this is what we have concluded okay now let us look at f belonging to x0 I want to show that it is inside A right if xy Axy is r2 then clearly fx fy will be inside Axy because everything in r2 is taken care right so it is the full thing so fx fy is inside Axy if x is x0 or y is y0 one of them since fx0 is 0 then also we see that fx fx0 is inside Ax0 so fx0 fx0 Axx0 the other way around if we interchange them then it will be between Ax0 x0 otherwise it is in that one if Axx is delta r the diagonal then also we see that fx fx c is inside delta r so it is Axx and finally if fx0 fx0 if you look at that is 0 0 so there is also fx0 so what we have concluded is this f is inside A prime this was the criteria for each x it must be inside fxy this is the criteria this is the description of A prime so it is in A prime thus we have shown that mx0 is inside A prime okay now consider the case where Axx is not equal to 0 for any x we just finish the proof for when Ax there is one x0 for which Ax0 x0 is 0 okay that is one first assumption now we are in the assumption that Axx is never 0 okay for any x then what is the choice it must be delta r okay by the way this condition is satisfied if A contains non-zero constant you see then Axx will never be 0 right there will constant term you know will not be 0 non-zero constant so this is just in passing we are telling when this will not be satisfied the case one does not occur if this A contains a non-zero constant that is all so here we claim that this Cxr itself is contained inside A prime okay thereby proving that this A must be the whole of Cxr okay so how do you show that Cxr all elements are inside A prime now so once we have done that one so now look at A condition A again I will just recall this property Axy is r cross 0 inside A is in My similarly xy is in 0 cross A is in Mx that is all I am using here okay so from here it follows that for every xy pair xy and x was x Axy is either r2 or delta r okay 0 cross r r plus 0 is not possible okay since A separates points from B it follows that Axy delta r if it only if x is y see point alpha beta are not equal to or not equal to each other then they are not in the delta right so once you have some other element what is the other choice it must be the whole of r2 so this will delta r only if x is equal to y so now we given f belongs Cxr for x not equal to y what we have fx fy this is Axy this is whole of r2 right and if x is y if x is equal to y then fx fy is fx fx that is in delta r obviously this is Axx so therefore all the points of f all the functions in f verify our condition of the lemma so they are inside A okay so that completes the proof because we have already shown that so I will just recall the important steps here okay where was the definition of A prime this A prime which was obviously larger okay so we have proved that A prime is equal to A the description of A prime is this one we are using that for this is for all f okay fv since here Cxr for all pairs xy fx fy must be Axy then f is inside A prime this is the simple description okay so this has been exploited so stone wise truss theorem when you are taking real value functions this is proved now okay a slightly better version of that one has been proved here okay so next time we shall do the complex case as well as some other extensions wherein we do not assume x is compact but only locally compact thank you.