Re: Theory of Everything based on E8 by Garrett Lisi
Bruno Marchal skrev: Le 29-nov.-07, à 17:22, Torgny Tholerus a écrit : There is a difference between unlimited and infinite. Unlimited just says that it has no limit, but everything is still finite. If you add something to a finite set, then the new set will always be finite. It is not possible to create an infinite set. Come on! Now you talk like a finitist, who accepts the idea of potential infinity (like Kronecker, Brouwer and the intuitionnist) and who rejects only the so called actual infinities, like ordinal and cardinal numbers (or sets). Yes, I am more like a finitist than an ultrafinitist in this respect. I accept that something can be without limit. But I don't want to use the word potential infinity, because infinity is a meaningless word for me. At the ontic level, (or ontological, I mean the minimum we have to bet on at the third person pov), comp is mainly finitist. Judson Webb put comp (he calls it mechanism) in Finitism. But that is no more ultrafinitism. With finitism: every object of the universe is finite, but the universe itself is infinite (potentially or actually). With ultrafinitism, every object is finite AND the universe itself is finite too. Here I am an ultrafinitist. I believe that the universe is strictly finite. The space and time are discrete. And the space today have a limit. But the time might be without limit, that I don't know. Jesse wrote: My instinct would be to say that a well-defined criterion is one that, given any mathematical object, will give you a clear answer as to whether the object fits the criterion or not. And obviously this one doesn't, because it's impossible to decide where R fits it or not! But I'm not sure if this is the right answer, since my notion of well-defined criteria is just supposed to be an alternate way of conceptualizing the notion of a set, and I don't actually know why the set of all sets that are not members of themselves is not considered to be a valid set in ZFC set theory. Frege and Cantor did indeed define or identify sets with their defining properties. This leads to the Russell's contradiction. (I think Frege has abandoned his work in despair after that). One solution (among many other one) to save Cantor's work from that paradox consists in formalizing set theory, which means using belongness as an undefined symbol obeying some axioms. Just two examples of an axiom of ZF (or its brother ZFC = ZF + axiom of choice): is the extensionality axiom: AxAyAz ((x b z - y b z) - x = y) b is for belongs. It says that two sets are equal if they have the same elements. AxEy(z included-in x - z b y) with z included-in x is a macro for Ar(r b z - r b x). This is the power set axiom, saying that the set of all subsets of some set is also a set). For me belongness is not a problem, because everything is finite. For me the axiom of choice always is true, because you can always do a chioce in a finite world. Paradoxes a-la Russell are evacuated by restricting Jesse's well-defined criteria by 1) first order formula (in the set language, that is with b as unique relational symbols (+ equality) ... like the axioms just above. 2) but such first order formula have to be applied only to an already defined set. This 2) rule is a very important restriction, and it is just this that my type theory is about. When you construct new things, those things can only be constructed from things that are already defined. So when you construct the set of all sets, then that new set will not be included in the new set. For example, you can defined the set of x such that x is in y and has such property P(x). With P defined by a set formula, and y an already defined set. Also, ZFC has the foundation axiom which forbids a set to belong to itself. This is a natural consequence of my type theory. When you construct a set, that set can never belong to itself, because that set is not defined before it is constructed. In particular the informal collection of all sets which does not belongs to themselves is the universe itself, which cannot be a set (its power set would be bigger than the universe!). Yes, the set of all sets which does not belongs to themselves is the universe itself. But this is not a problem for me, because you can always extend the universe by creating new objects. So you can create the power set, and the power set will then be bigger than the universe. But this power set will not be part of the universe. -- Torgny --~--~-~--~~~---~--~~ 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
Re: Theory of Everything based on E8 by Garrett Lisi
[EMAIL PROTECTED] skrev: On Nov 28, 9:56 pm, Torgny Tholerus [EMAIL PROTECTED] wrote: You only need models of cellular automata. If you have a model and rules for that model, then one event will follow after another event, according to the rules. And after that event will follow another more event, and so on unlimited. The events will follow after eachother even if you will not have any implementation of this model. Any physics is not needed. You don't need any geometric properties. In this model you may have a person called Torgny writing a message on a google group, and that event may be followed by a person called Marc writing a reply to this message. And you don't need any implementation of that model. A whole lot of unproven assumptions in there. For starters, we don't even know that the physical world can be modelled solely in terms of cellular automata at all. Why can't our universe be modelled by a cellular automata? Our universe is very complicated, but why can't it be modelled by a very complicated automata? An automata where you have models for protons and electrons and photons and all other elementary particles, that obey the same laws as the particles in our universe? Digital physics just seems to be the latest 'trendy' thing, but actual evidence is thin on the ground. Mathematics is much richer than just discrete math. Discrete math deals only with finite collections, and as such is just a special case of algebra. Isn't it enough with this special case? You can do a lot with finite collections. There is not any need for anything more. Algebraic relations extend beyond computational models. Finally, the introduction of complex analysis, infinite sets and category theory extends mathematics even further, beyond even algebraic relations. So you see that cellular automata are only a small part of mathematics as a whole. There is no reason for thinking for that space is discrete and in fact physics as it stands deals in continuous differential equations, not cellular automata. The reason why physics deals in continuous differential equations is that they are a very good approximation to a world where the distance between the space points and the time points are very, very small. And if you read a book in Quantum Field Theory, they often start from a discrete model, and then take the limit when the distances go to zero. Further, the essential point I was making is that an informational model of something is not neccesserily the same as the thing itself. An informational model of a person called Marc would capture only my mind, not my body. The information has to be super-imposed upon the physical, or embodied in the physical world. If the model models every atom in your body, then that model will describe your body. That model will describe how the atoms in your body react with eachother, and they will describe all your actions. -- Torgny --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
N^N is not enumerable (post before key post bis)
Hi,
I recall the proof that N^N is not enumerable.
I recall that N^N is the set of functions from N to N. Such functions
associate a natural number to each natural number. Example:
factorial = {(0, 1) (1, 1) (2, 2) (3, 6) (4, 24) (5 120) ...}
of course the same information is provided by the sequence:
1, 1, 2, 6, 24, 120, which is an element of the infinite
cartesian product NXNXNXNXNX
So N^N is in bijection with NXNXNXNXNX , and giving that we are
interested in cardinality, for all practical purpose we can identify
both sets.
Theorem; N^N is not enumerable
Proof 1 (by absurdo):
If a bijection exists between N and N^N, then it will look like (cf the
identification above):
0 --4 1 2 6 24 57 ...
1 --0 0 5 0 457 ...
2 --0 9 7 010 ...
...
Well, at least in the mind of a God capable of seeing the whole
infinite table.
Again, for such a God, the diagonal is entirely well defined:
4 1 8
So we can consider the sequence of the diagonal elements, each added
to one:
4+1 1+1 8+1 ..., that is
5 2 9 ...
By construction this sequence is not in the list above, because it
differs:
from the 0th sequence, at the 0th decimal let us say,
from the1st sequence, at the 1st decimal,
from the 2nd sequence, at the 2nd decimal,
from the 3rd sequence, at the 3nd decimal
from the 4th sequence, at the 4th decimal
from the 5th sequence, at the 5th decimal
from the 6th sequence, at the 6th decimal
from the 7th sequence, at the 7th decimal
from the 8th sequence, at the 8th decimal
...
from the n-th sequence, at the n-th decimal
...
All right? This works for any tentative bijection proposed by any Gods.
Now I give you the same proof, but with the traditional functional
notation. This is for helping those who could have some notation
problems. Please make you sure that you see I am giving the same proof,
but with better notations:
proof 2 (by absurdo)
If a bijection exists between N and N^N, then it means there is an
enumeration of all functions from N to N, it looks like:
f_0 f_1 f_2 f_3 f_4 f_5 f_6 ...
All those f_i are well-defined (in Platonia) functions from N to N. So,
the following function g is also well defined. By definition:
g(n) = f_n(n) + 1 (this is the diagonal +1 described above).
Now g cannot be in the list f_0 f_1 f_2 ...
Why?
Because if g is in the list, it means there is a number k such that g =
f_k (and thus for all n, g(n) = f_k(n)).
But g(n) = f_k(n) + 1 by definition of g. Now we have, by applying g on
its code k (cf g = f_k):
g(k) = f_k(k) (by the assumption that g is in the list), and
g(k) = f_k(k) + 1 (by definition of g)
Thus (by Leibniz rule)
f_k(k) = f_k(k)+1.
But those are well defined numbers (cf: the f_k are functions from N
to N), thus we can subtract f_k(k) on both sides, and this gives
0 = 1
Contradiction. Thus g cannot be in the list, and appears as a sheep
without a rope. N^N has more element than N, and is thus non
enumerable. QED (OK?).
NEXT: the key post. It should be even more simple, because, although we
will be obliged to stay in Platonia, we will not invoke any God
anymore. Instead we will invoke its quasi antipode, not the devil (!),
but the humble finite earthly creature. The idea is to concentrate
ourself on computable functions from N to N, instead of *all* functions
from N to N.
Computable?
By who? Computability is an epistemic notion, it seems to involve a
subject.
Well, as you can guess, computable will mean computable by that humble
finite earthly creature ... Now, what does that mean ...
Soon on a screen near you ;) ..., asap ...
Good week-end,
Bruno
http://iridia.ulb.ac.be/~marchal/
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RE: Theory of Everything based on E8 by Garrett Lisi
Date: Fri, 30 Nov 2007 09:00:17 +0100 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: Theory of Everything based on E8 by Garrett Lisi Jesse Mazer skrev: Date: Thu, 29 Nov 2007 19:55:20 +0100 From: [EMAIL PROTECTED] As soon as you say the set of ALL numbers, then you are forced to define the word ALL here. And for every definition, you are forced to introduce a limit. It is not possible to define the word ALL without introducing a limit. (Or making an illegal circular definition...) Why can't you say If it can be generated by the production rule/fits the criterion, then it's a member of the set? I haven't used the word all there, and I don't see any circularity either. What do you mean by a well-defined criterion? Is this a well-defined criterion? : The set R is defined by: (x belongs to R) if and only if (x does not belong to x). If it fits the criterion (x does not belong to x), then it's a member of the set R. Then we ask the question: Is R a member of the set R?. How shall we use the criterion to answer that question? If we substitute R for x in the criterion, we will get: (R belongs to R) if and only if (R does not belong to R)... What is wrong with this? My instinct would be to say that a well-defined criterion is one that, given any mathematical object, will give you a clear answer as to whether the object fits the criterion or not. And obviously this one doesn't, because it's impossible to decide where R fits it or not! But I'm not sure if this is the right answer, since my notion of well-defined criteria is just supposed to be an alternate way of conceptualizing the notion of a set, and I don't actually know why the set of all sets that are not members of themselves is not considered to be a valid set in ZFC set theory. Jesse --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
Re: Theory of Everything based on E8 by Garrett Lisi
Jesse Mazer skrev: Date: Thu, 29 Nov 2007 19:55:20 +0100 From: [EMAIL PROTECTED] As soon as you say the set of ALL numbers, then you are forced to define the word ALL here. And for every definition, you are forced to introduce a limit. It is not possible to define the word ALL without introducing a limit. (Or making an illegal circular definition...) Why can't you say If it can be generated by the production rule/fits the criterion, then it's a member of the set? I haven't used the word all there, and I don't see any circularity either. What do you mean by a well-defined criterion? Is this a well-defined criterion? : The set R is defined by: (x belongs to R) if and only if (x does not belong to x). If it fits the criterion (x does not belong to x), then it's a member of the set R. Then we ask the question: Is R a member of the set R?. How shall we use the criterion to answer that question? If we substitute R for x in the criterion, we will get: (R belongs to R) if and only if (R does not belong to R)... What is wrong with this? -- Torgny --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
