Le 27-août-07, à 13:27, David Nyman a écrit :

> > On 16/08/07, Bruno Marchal <[EMAIL PROTECTED]> wrote: > >> If you drop a pen, to >> compute EXACTLY what will happen in principle, you have to consider >> all >> comp histories in UD* (the complete development of the UD) going >> through your actual state (the higher level description of it, which >> exists by comp, but which is actually not knowable by you. Of course >> this cannot be used in practice, but has to be used to derive the more >> usable laws of physics. > > I hope this will become clearer as we proceed. I hope too. Perhaps it would help us if you could tell me which step of the UDA you find unclear. cf the paper: http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL.htm and the single summary slide PDF: http://iridia.ulb.ac.be/~marchal/publications/SANE2004Slide.pdf Normally the first seven steps should not be too much difficult. Remember that we *assume* the comp hyp. Sometimes some people does not understand because they (more or less unconsciously believe that I am arguing for comp, but that is something I am never doing since a very long time: I really take it as a working hypothesis with no (conscious) prejudice about where this can lead us. > >> Empirically we can expect that the 'substitution level" is more >> related >> to a notion of "isolation" than of scaling. Nevertheless, we cannot >> really use this here, given that we have to extract quantum physics >> from the existence of that "level". > > I don't understand this yet. I guess you have some understanding of the first sentence, given that this is something happening in any version of QM without collapse. Sometime ago, some people argued that QM is confined to the microscopic, and they believed that that was the reason why a macroscopic quantum superposition (like Schroedinger's Cat) could not exist. Today we have plenty of evidences that this is not correct, and that it is even quite easy to generate a cat in a superposition eating + drinking (say). Indeed it is enough to *isolate* sufficiently well the cat, and then to force him/her/it to choose between drinking and eating according to the result of a measurement of a quantum superposition state state of some local photon. By linearity the cat will be in the superposition state. What prevent us of seeing the cat in that superposition state is not that the cat is macroscopic but comes from the fact that we cannot isolate the cat sufficiently well from us, so that, very quickly we will find ourselves in a superposition seeing the cat in such state + seeing the cat in the orthe state. The "quickly" here is not due to some magical quick wave collapse, but is due to the rapidity of the decoherence process, which mainly describes the (linear) way superposition are contagiopus to their neighborhood. Now with comp, it is the same. You cannot known, by the first person indeterminacy, which computations support your conscious state among all computations that you cannot discerned. To make this clearer, I will wait you telling me where exactly you have some trouble in the UDA. OK? > >> The >> 3-brain is just not a physical device for producing consciousness, it >> is a local and relative "description" of a state making greater the >> probability that you will be able to manifest your first person >> experience relatively to some "dream", itself being an infinite set of >> histories. > > Do you mean here that: there exists a 'state that [increases] the > probability that you will be able to manifest....etc.' and that the > 3-brain 'is a local and relative "description"' of such a state? A state by itself cannot change the probabilities. It is the relative number of possible continuations of a state, relative to the "number" of comp histories going through that state which counts, up to some (extraordinarily complex) equivalence relation. > >> Can you explain why the set of all binary sequences *is* closed >> for diagonalization > > Because any additional members generated by diagonalisation must also > be binary sequences? OK. > >> and why any *enumerable* set of binary sequences >> is *not* close for diagonalization? > > Because new members can always be generated by diagonalisation that go > outside the original enumerable set (as distinct from the larger set > of *all* sequences)? OK. > >> A bit more difficult: can you show that for any set A, the set of >> functions from A to {0,1} is bigger than A? > > Could you please elucidate "functions from A to {0,1}" ? I recall that *a* function (without "s") from a set A to a set B, is just any association to each member of A of a member of B, in such a way that no element of A is associate to more than one element of B. It is usual to describe a function by either a table of associations, or by a graph, etc. I will represent them by the set of associations. For example, the function FACTORIAL from N to N is represented by the infinite set: { (0,1) (1,1) (2, 2) (3, 6) (4, 24) (5, 120) (6, 720), ...}. I will write sometimes FACTORIAL = { (0,1) (1, 1) (2, 2) (3, 6) (4, 24) (5, 120) (6, 720), ...} instead of the more cumbersome FACTORIAL can be represented by the set { (0,1) (1, 1) (2, 2) (3, 6) (4, 24) (5, 120) (6, 720), ...}. Here we were asking for ALL functions from one set to another. Example: let B be {0,1} like above. Let us choose A to be: 1) the set {a} which contains just one element a. There are two functions: f_1 = {(a, 0)} and f_2 = {(a, 1)} So the number of functions from A to {0,1}, when A = {a} is 2. 2) the set {David, Bruno} which contains the two elements David, Bruno. There are four functions: The two constant functions f_1 and f_2: f_1 = {(David, 0) (Bruno, 0)} f_2 = {(David, 1) (Bruno, 1)} The two "variable" functions f_3 and f_4: f_3 = {(David, 0) (Bruno, 1)} f_4 = {(David, 1) (Bruno, 0)} and no more. Can you compute how many functions from A to B there are, in case A has n elements and B has m elements? Answer: m^n. Can you see that? If not, find how many functions there are from A = {a, b, c} to {0,1}, and then some other simple example, and then try to make a conjecture, and then try to find some argument which makes you think the negation of the conjecture has to be inconsistent (that is: find a proof of the conjecture). By "proof" here, I mean an argument which convinces you, or better, an argument which you have the feeling that it can be used to convince your "little sister" (which I suppose not to be a mathematician). Bruno http://iridia.ulb.ac.be/~marchal/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---