Advanced mathematical formalisms (operator theory) for quantum field theory (contd.)

0:00 Predictive versus explanatory aspects of scientific theorizing. Generation of incoherence. Subjective probability assignment to predictions, but pre-predictive level. Qua-explanatory power of theory in terms of unification. The sense in which one explanation is better than another. The dynamics of scientific theorizing. The dynamics of... Increase in scientific explanation qua info-theoretic content, not info-theoretic content against correctness of observations, against single observational level. Truth type, truth predicate, cached information, theoretic terms, but rather global, the global character of scientific explanation, concentrate on global character of scientific theorizing, it exists, planetary character is only to be understood in terms of its global characteristics, global features of inter-theory relations structure or larger scale features of inter-theory relations structure rather than in predictive relation to. Single observation level language. Info-theoretic content from that. Of course, there must be that important info-theoretic constraint, a global truth predicate, cached in terms of info-theoretic... Operational imperative, Quay. Relation of partial functions to... Operationalisation of theories, look at operationalisation in some detail, query micro-reductions. I'd argue one from global character scientific theorising, work by Hilpin and others on refinement and exactification of notion of explanation as concentrating on predictive aspects, in fact theoretic sense, modular, global explanatory features.

2:30 I don't know if you've got that written down. Let's do the history of this thing. I don't know if you've got that written down. Let's do the history. Signals is a signal algebra and an axiom, and U is a measure, a probability measure, and X, T are a family of random variables, i.e. a bunch of different kinds of variables with their values in the appropriate range, say, U, T, A, and finite dimensional joint probability measures. In other words, if you take any finite section, say x to the t one after x to the n, for example, and look at the growth probability distribution for example, what we get is that the site is going to use the energy start up. But this theorem is implicitly used in all discussions of probability theory. It could be the most heuristic one in the scope of physics. I don't really mention it. It's always right, but I think it's not making any sense. All right. Now, the actual construction goes as follows. First of all, we construct over here.

5:00 So there we go, equal to what I'll call v, t, by definition, which is the Cartesian product of all of these t's. So omega itself is the Cartesian product of these t's. Let's define x, t, x, t of x is equal by definition to f of t. I remember that the definition of the Cartesian part is simply the set of all functions of the index set in the field. So a simple element of that is a function of the index set in the field. And we're saying that Xt, evaluated on the function, is precisely a function, and there's some evaluation in that. And that's what the Xt is going to be. It is going to be the random derivative. And we've got to show that these things are going to have precisely the same function. So we've already constructed the algorithm. Now, the signal reality we're going to take, as we've defined with some care, you might be tempted to think that the obvious thing to do, as these topological spaces are becoming fit with their braille structure, is to simply take the product topology on the next stage, which is not product topology, and then take the braille sets generated by that topology. Unfortunately, you can't do that. No one would like to be able to do that, but we track the whole of it, so in terms of the next term, we can do that. But in general this is not possible, and it's certainly common knowledge that you don't do this.

7:30 You haven't constructed six values which is something that is much smaller than that, and there are too many of them today. So, a set of the form B cross, B is P minus Pm, which is of course the same thing you know we go, where B is the parallel step. I'll stop going back to the cross of the board now because it seems we've gone out of the field a bit, so we'll go back to the starting point a bit later. So that's, in other words... Let me make a couple of concepts. First of all, I'm really slightly sloppy in notation. Strictly speaking, it shouldn't like be crossed. So a similar set, then, is one which is of this form where you take any Morale set in some finite product state, and you take the product of that with the rest of the B. So it is called a similar set, I think you can understand, because that's what it looks like. For example, if B is an R2, something like that, then B crossed towards left, the R will be a set up here. But of course, the number of dimensions up here is in general infinitely many, and infinitely many too. This is always a finite class collection, so this in a sense is a very meagre piece of a small class of sets, and it's still not a very vast collection of sets, but that's what a cylinder set is called. And they're the sets that we're going to use completely in our sigma algebra. Let's just note incidentally that a class of these The set of all f in omega such that t1 of f up to x tn of f belongs to B. Now that's quite a nice way of writing it if you think about it in terms of probability theory. So a cylinder set is simply the set of all these x.

10:00 Such that this particular random variable, I mean, this collection of random variables belongs to B. So a set of these sets are actually going to be geared to finite collections of random variables. And because the theorem starts off with finite collections of finite random variables, finite dimensional random variables, that's really why certain sets come in, so we can use that way as well. Now, we let curly C know the algebra and comprise of the collection of all certain sets. Index sets TA, I mean all the finite subsets of the index set of BTA cross being T minus TAM, you take all possible B's and all possible finite dimensional subsets and now we just have all the similar sets for all the different sets. Would this be an algebra? Would it contain the whole algebra? Sorry? Would it contain the whole space, I think? Yes, I think for the null set, for example, it's an element of B. That was the empty thing. It wouldn't really matter very much because what we're going to do anyway, which is not a sigma algebra, that's for sure, so we let sigma equal coli s of c, and the sigma algebra 0.85 coli c.

12:30 We explain what x is, x is going to be the infinite product of all of these range spaces, and the sigma algebra is going to be the sigma algebra generated by several new symmetry sets. And in general, that is not the same thing as the Braille sets of the product topology. It is the same thing, and I'll prove that right at the very end, if you use a counter-invector set. If you use a counter-invector set, it turns out that this is exactly the same thing as the product topology. If it's an uncountable collection, then it's not. Excuse me, do you mean that the sphere is equal to the set of things like that? Yeah. Yeah, I haven't done that very well. Is the union of all possible sets of this form, I mean, first of all, four possible Tm, you take all possible Bs in there. I think we can use those for union. Isn't that true? Is the union over all, over all Tm of the set of all possible Bs? In the, in the, how to draw up that Tm, that's right. I've written one underneath the other. I could really like, um, I could really like to use it as a union. This is very straightforward, each one of these is contained in the products, they're subject to the products, so I'm just taking a sketch of all of them, and all of this theory. Now, the first observation is that each x, t is clearly a measurable function from omega, if we do the same thing as b. Now, that's where the reconstruction is, the very definition of the Sillings set guarantees that, particularly if you can see that, right here. But the set of all X, such that X and T belong to anywhere else except B, is certainly, now that we've said the number of all B, it's the very construction of it that is really why it's done. So that, we say it once, these X and T certainly could be random roles because they're multiple functions. So that's, that's one thing.

15:00 Now what we've got to do is define and measure mu on curly C and then extend it, if we can, as a sigma in such a way that we're scripted during one of these sets that really introduces the mu Tn. So we do this by defining mu on curly C by mu of 3 cross T, T minus T, is equal by definition to mu Tn phi. In other words, we define the measure of a cylinder set, a set looking like this, to be simply the measure of the base. We never do that with a vague measure, of course, so there's no chance there's nothing ridiculous there. Now, of course, it's not obvious that that's well-defined, because, for example, there's more ways than one of writing a set like this. This belongs to B of T of A, but there's nothing to stop me writing this as sort of B top another B, if you like, and then locking one off the right-hand side here. So that then, this would really belong not to Tn, but to Tn plus one, so the other question is whether or not mu of Tn plus one on that is equal to mu Tn of this, but in fact the statement of those two are the same, it's exactly the statement of the consistency condition if you started off with it. So this is the role played by the consistency. So I'll just say that this is well defined, Tn are consistent. And again, when I come next week to generalise the capital processes, I shall set things up in terms of projections and Twitterly. And if you don't mind the sort of slightly hazy way of arguing it, we'll be done with the detail there, but I think that's it there. And in fact, in other words, the consistency conditions which you perform are precisely those conditions that make sense. Whatever you decide you need for that to work is what you demand in the original measures of society. Now, clearly, mu is a non-negative additive of U as equal to 1 as the required

17:30 All this part of the proof is really rather straightforward, and it has to be stated in a fair amount of things to happen in this construction. So the mu has effectively all of the topologies we want, except that we have to show, and of course this is the higher part of the theorem, it suffices to show that mu can be extended to this signal. Now of course, we're going to do that using the code of DeLoy. But what we have to show is that the conditions which we have to satisfy in a category extension, which are mainly some sort of continuity, can be lower than or not satisfied, and that is the hard part of this theorem. An interesting observation is that it seems to be essential that we use topological spaces for this. Now, there don't appear to be any forms of conglomerates to it in which B is simply an arbitrary mathematical space, which you might expect when you write your book, actually, when you go to a very mathematical space. But in fact, there doesn't seem to be any version at all which does not use the topological property in a very, very, very intimate sense. I mean, the measure theory and topological theory are completely brought together. And you'll see that the fruit I'm going to do is really primarily topological. It uses the topological property in its extensive measure theory science. Right, so now we start the real part. From, in fact, 1.4.1,

20:00 Let me, remember when I label things as an integer I simply mean a countable collection. A Cn is any countable collection of cylinder sets, which is monotonically decreasing, that is, the decreasing collection of sets, and such that mu of Cn is greater than the sum of sine, which is greater than 0. If that's true, then the intersection from n is equal to 1 up to infinity of the Cn does not equal the empty set. Now I know the original statement about the character of a theorem was that the thing has to be continuous from above or below, but I did show in this corollary that in fact it's sufficient to show that if there are any collection of sets in the ring or algebra concern, which are decreasing, the measures are always bounding below and under the sum of sine, which is strictly greater than zero, plus the new of sine is always greater than zero. And if this is true, the intersection of these sets is not empty. If that is the case, then you have an extension of the measure to the whole of the sigma algebra. So that's what we've got to prove simply by condition. Now the essential reason why the topology comes in, basically, is the fact that one way of showing that a sequence of sets is non-empty is to use the property of compact spaces, if you have any compact space, and you have a countable collection of sets which are non-inclosed, which are non-inclosed sections, and some finite collection of them. And there's that topic which one uses to prove this, and that's what I like to see and really why the quality comes in. And there are no proofs, if not in one sense or another, to use this topic of compactness. Right, let me see how that goes now. As I say, this theorem is absolutely essential. It really is the central theorem, of course, from the point of view of applications. Sets in C are only on a finite number, so what I mean is that they're always of the form of beta plus, where the beta only depends on finitely numbers of... And this means that without loss of generality, we can consider the increasing sequence such that there exists a countable set, which I'm going to call T-infinity, plus a Tm. It's not entirely obvious what that's about. I mean, how can any natural set in the sequence be true?

22:30 Typically, a set surrounding an absolute crop is what you've got already. It's a state we've been following. If you have a Borel set, as I said already, V cross V times T minus Tn, I can write this as V cross V, plus the bit which is left over, and that inches up by one, the dimension of the index set which you're in. So, if you want to, you can always fill in gaps in this sequence of Cn's. In fact, if you don't want that to be obliged and you belong to such a sequence, you can put in a set which all is happening the same, topologically. So this is just my combinatorial curriculum, I don't know what it is, it's just convenient. I don't know, it's strange to think about that, it's not quite so obvious as it looks. Now, UTM I said was regular, and that means it's both in a regular and out a regular, in particular it's in a regular so it can be approximated by compact steps.

25:00 Key terms are called K-Ns, belonging to K-Limits, B-T-M, and are a compact program that allows the K-Ns containing the B-Ns to contain the B-Ns. Minus k is less than epsilon over 2 to the n plus 1. Remember that the condition of an irregularity is that the mu of any set is the supremum of mu over all k of compact sets in the same unit. Therefore, for any finite number greater than zero, we can always find k is less than or equal to greater than zero. This is where the compact sets start to come in. n prime, equal by definition, carries the 1 up to the n of kk. We've got to fiddle around now with these various selections. We construct various types of compactors. The K-imprime is certainly the intersection. Well, it's not obvious actually that that's true. But nevertheless, if you think about it, it's very easy to see. But that's K's subscript N-match, so I want to show that that is in fact less than epsilon.

27:30 So these K-imprimes are just as good as the previous ones, the one with the approximation. Well, this is just a question of straightforward identities. There's a statement that the set C, sometimes getting smaller and smaller, means that the B ends are as well, because the B ends line different spaces. See, that's the whole trouble with this. The reason why we can't instantly prove the theorem we want is that the set and the whole of the form B line in a constant space, and the space itself is changing all the time, and so you've got to sort of get control over this shift in there.

30:00 Let me call this equation sharp again, right hand side is sharp, so the sum from k equal to 1 up to n will appear, that's simply, since this is contained in that, the measure of that is certainly less than or equal to the measure of that, and so I can certainly put that on the right hand side there, so that's ok, but now the whole point about this is that we've got these in the first and the same space, these are now broken onto tk, and so we can instantly... I'll say that this is equal to the sum from k to the 1 up to n of mu tk of vk minus k. And now that is from the consistency condition on the measures mu. But the measure of something off all the rest of the v's is the same thing as the other measure on the something. But we know that we started off with these things from the method of sine over 2 to the k. So this is less than the sine from k to the 1 up to the n, so that's sine over 2 to the k plus 1, which of course is less than the sine over 2 to the n. What we've shown them by this is that these sets k, n, pi, which as I said have certain desirable features, oscillate with b, n's, they're still compact, they oscillate with b, n's, in the usual sets that are less than the sine of k, n, plus 1. The purpose of these will emerge shortly. What we want to show is that these Km primes are actually always non-empty. It's not entirely obvious, but they could actually be empty. It's not important that they're on the reason line. We have to note that nu Tn of Km prime...

32:30 Now the easiest way of showing that a step is non-empty is one way of showing these measures greater than zero. So nu Tn of Km prime is equal to nu Tn of Bm minus nu Tn of... All these measures, of course, are finite, so we can do this without any trouble, because they're finite as a proper sum. Now mu is precisely mu by definition of mu. And this is always greater than epsilon. Remember that condition we have on the mu so they're all bounded below by an epsilon. So this, in fact, is greater than epsilon from here. This thing is always less than epsilon over 2, we've just proven that. Therefore, 50 to the round, the whole thing is greater than this minus epsilon over 2. Therefore, Km' is not empty, and the result of that Vm, which we define to be Km' cross V2 infinity minus Km, so Km' and Vm belongs to the space V2 infinity, so it fills out Km'1 of the beam to the left, there's a decrease, there's a decrease in V2 infinity, it's obvious why it's... The reason why it's decreasing is precisely because of the construction of the K-interfinite. And the reason we constructed the K-interfinite in the way we did is precisely because it would be an increase in the sequence. Since this K-interfinite is contained in this K, which is contained in B,

35:00 an intersection containing the 1 after infinity of D has not been called an empty set. It contains the 1 after infinity of one of these definitions. And therefore, it's sufficient. This isn't what we're trying to prove. It's sufficient to show that this is true. That's the first step in reducing the problem to something which is really compact. It's a standard result of topology, and I'll give you a reference to this written down. Now, exists a sequence called U-N of a relatively compact, in other words, set-through disclosure, relatively compact open subsets of V, which cover V, in other words, the union of them. And we shall have the property that the closure of a rock is in the next one up, or in. In other words, there's an increase in sequence of open sets, each one of which closes compact. And furthermore, the property of each closure is contained in the next set up. This whole rock goes up versus the whole space, and so on and so forth. Furthermore, any compact stuff-setting grid, in some UN, of course, is a factor of this closure.

37:30 So we have this increase in sequence of compact sets, then, which are properly linked to any compact set that's contained in my own sequence. And as a reference for that, you can look at the Novartis general topology, section 9.9, Proposition 15, and the Colony 1 to that proposition. It's a standard, an absolutely standard piece of convoluting, and that's a very excellent reference. This Km' is contained in Vtn. Now I want to produce projection maps of Km' onto the various subspaces V. I'm going to call them Pti, which should be maps Vtn onto Vti. So this is the projection map, so they're continuous anyway. And this is the continuous image in the compact set, it's compact. And hence, this up here, that this is what I'll call such that p of t i a m

40:00 is contained in the closure of this v comma n i, v t i. I hope the notation's going to get further down at this point. Okay, this is a compact set. And therefore, by what I've said up here, there must be one of these u's, which has to be positive, this is contained in the closure of it. The t-i simply indicates that I'm looking at that u as it appears in the i slot, all the slots are the same. And I'm calling that particular number up here, that's the t-i, b, to take the symbol n,i, because it depends on n, also it depends on i. And I'm going to get rid of the i instantly and say this is also contained in the closure of b, n, i. So in other words, all I'm really saying is that from the projection of this compact set onto the various coordinates, there's a common compact set which has the property that it contains each one of the projections. A set like that is obvious. It's obviously true because it's a finite set in a five dimensional subset. And therefore we have that Km' is certainly contained in a Cartesian product. So in other words, all one is really saying, and this is the purpose of this... What we want to do is to have this sequence of sets increasing with n.

42:30 But of course we don't know that's going to be true as it's written. There's no reason why these three M's should be minus-minus sequences of M. They may go down. So we want them to increase with M, and so the way to do that is to take a union over all of them which are less than or equal to a given number. Why don't I do that? I've got all that to do with increasing sequence of sets, as I think I've used numerous times in the past. And of course, plane and prime is similar to the plane, with a g of t1 of n, plus g of t2 of n, plus g of tn of n, and g of n of tj, which is up to the n, for each fixed... Now, the essential statement is that in fact you can shrink this set. This is a rather large set, rather large set, this is a set of mirrors. Well, let's then say that I can in fact shrink these sets down. I can shrink this one right down to g1. I can shrink this one right down to g2. And still have the property that this is contained in there. And this is a, this is the good part of this set, really. And the statement is that by induction, Km prime is in fact containing gt1. G2, G2, G2, G, N. At the end of the day, of course, we can take the limit as N goes to infinity, and then this will be a Cartesian product, and then we can see that this is an increasing compact set, and therefore it will be compact. But we're going to prove that by induction. It's clearly true that N is equal to 1. We suppose it is true that KM plus 1 times,

45:00 it's certainly contained in G1, N plus 1, plus N plus 1, N plus 1, because I've already proved that this is in general true. Well, the object is supposed to shrink down. And also, from the kth m plus one prime is contained in kth m prime cross v e to the m plus one. And that follows from the definition. Remember, I said that the k's have been constructed to be a decreasing sequence of sets, in some appropriate sense. This is precisely the appropriate sense, that this is always true. This is just like many different words. That's contained in that. And it's also contained in G211. Is that it there, please? No, that'd be N. Oh, sorry, yes. And by the inductive step, so it's contained in two sets. It's certainly contained in the intersection. KM plus 1 is contained in GM plus 1, T1, cross GM plus 1, T, M plus 1, GT1, GTM, M, cross... These G's are increasing sequence of sets, and therefore this is contained in that, and so intersection of that with that is precisely that, and so on right up to there. This is in fact therefore y-hand size equal to gt1 cross gt2 cross gtn plus 1 n plus 1.

47:30 So it's the gm, t, j's go upwards and down. Excuse me, sorry, the projection graph, the thing in the brackets, is that tm plus 1? It's tm plus 1. Yes, it's the m plus 1. It's the m plus 1. Okay, so, now that you're saying that there's an increasing 1, so it's simply equal to that, and so on up to there, what happens at the end of the term? The n plus 1 term is that the intersection of gm plus 1 would be equal to gm plus 1. And that always shows that there is truth for n plus 1. And that would be the adaptive proof. It'll take me about three minutes to finish. Is that okay if I do that rather than break through the stage? Then we'll come back to it. That'll be the end of the talk. I'm assuming I've been like this. I couldn't help but look at the proof. Now, what we do now is we refine G between the carcinogen products, mainly the one I'm conspiring with, Now, for me, especially in our film, this is a compact, non-empty compact subset of GTN. Going back to the dn stuff, where the dn wanted to show that interception is 1, 0, we find these to be dn interception genes, where g is this 90-comp-x set, and this is equal to Km-prime-cross-d, d-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime-prime.

52:30 We've just seen that there can be no such matter, and therefore intersection cannot be empty. Intersection can end with one after infinity at the end of intersection g, if you like, and that's precisely what the intersection of the air field is, and that's certainly intersection can end with one after infinity at the end of one after infinity at the end. This is quite, as I say, it's quite a delicate term in a sense, and it's typical. About the existence of such measures, the Hoppe-Lotz theorem, which is the next improvement on this, really does the same sort of thing in a sense, but in terms of defining the measures on the sigma-algebra of a similar set, it defines the measures on the sigma-algebra of the part of topology, which in general is much more clear, and that theorem takes up five, six, seven times as much space and proof as this. But nevertheless, it has as its central feature the construction of certain contacts of the set, marked as we've done here with this G. So the construction of the contacts, which we'll pick up again in four parts later.