Re: Bijections (was OM = SIGMA1)
Le 30-janv.-08, à 13:43, Mirek Dobsicek wrote (in different posts): 2\ Bruno, you recently wrote that you do not agree with Wolfram's Principle of Computational Equivalence. As I understand to that principle, Wolfram says that universe is a big cellular automata. What is the evidence that it is unlikely this way? Wolfram is very vague about his Principle of Comp Equiv (PCE): http://www.wolframscience.com/nksonline/page-6-text The problem is that Wolfram seems to take a natural world for granted, and does not seem to be aware that each of us (us = Lobian machine or even just Lobian entity) cannot discern about 2^ALEPH_0 locally equivalent computations. So Wolfram is either not aware of the consequence of the computationalist hypothesis, or not aware about what we can expect nature to be after knowing that Bell's inequality are violated, quantum measurement problem and its MW solution, etc. Wolfram's idea is inconsistent for the same reason that , as a TOE Schmidhuber's constructive approach is inadequate or at least incomplete. Such theories entails COMP. But COMP entails the physical world cannot be entirely a constructive structure. Physics or physicalness emerges from an infinite sum (the nature of which is still under scrutiny) of computational histories, observed in relative perspectives (points of view). There is no reason to believe this leads to a constructive universe. On the contrary, its geographical and local aspects could have verifiable non computational feature (like when we repeat spin measurements for example). The problem is that, like many, not only Wolfram and Schmidhuber, seem to take a physical universe for granted, but they take also a sort of identity thesis (brain/mind) for granted. The Universal Dovetailer sequence of thought experiences is supposed to explain in all details why such an identity thesis just can't go through ... Now, Mirek, I don't know if you really want to dig on the UDA and the philosophy-of-mind/theology issue, because you can enjoy the math per se. Many people dislike or get stuck in front of the idea that comp makes us duplicable, or by the same token that QM makes us 100% duplicable too, although not 100% clonable. But the UDA does explain why we have to take into account the different points of view (first singular, first plural, third, zeroth, ... ). The UDA, in english, can be found here: The Origin of Physical Laws and Sensations, (Invited Talk SANE 2004). Click on that title, or copy the following in your browser: http://iridia.ulb.ac.be/~marchal/publications/ SANE2004MARCHALAbstract.html (if you study it I would suggest you print the slider too, so that you could perhaps tell me which step you would find hard to go through ). Uu, reading about cardinals and ordinals on Wikipeadia did not helped me at this point. Could you please elaborate more on this? Of course, only relatively to its importance towards CT ... I will. Slowly because I will be more and more busy, and also I should write papers, I should extend a bit the Plotinus' one, correct the typo error, submit, etc. I still don't know if it is the physicists, the logicians, or the theologians who will grasp the UDA/AUDA highly interdisciplinary or trans-disciplinary reasoning ... Got not enough feed-back ... I will also solve the combinator crashing problem: it does illustrate another form of use of the diagonalization idea ... (this is a hint) ... Best, 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 -~--~~~~--~~--~--~---
RE: Bijections (was OM = SIGMA1)
Date: Tue, 20 Nov 2007 19:01:38 +0100 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: Bijections (was OM = SIGMA1) Bruno Marchal skrev: But infinite ordinals can be different, and still have the same cardinality. I have given examples: You can put an infinity of linear well founded order on the set N = {0, 1, 2, 3, ...}. The usual order give the ordinal omega = {0, 1, 2, 3, ...}. Now omega+1 is the set of all ordinal strictly lesser than omega+1, with the convention above. This gives {0, 1, 2, 3, ... omega} = {0, 1, 2, 3, 4, {0, 1, 2, 3, 4, }}. As an order, and thus as an ordinal, it is different than omega or N. But as a cardinal omega and omega+1 are identical, that means (by definition of cardinal) there is a bijection between omega and omega+1. Indeed, between {0, 1, 2, 3, ... omega} and {0, 1, 2, 3, ...}, you can build the bijection: 0omega 10 21 32 ... n --- n-1 ... All right?- represents a rope. An ultrafinitist comment: In the last line of this sequence you will have: ? - omega-1 But what will the ? be? It can not be omega, because omega is not included in N... -- Torgny There is no such ordinal as omega-1 in conventional mathematics. Keep in mind that ordinals are always defined as sets of previous ordinals, with 0 usually defined as the empty set {}...So, 0 = {} 1 = {0} = {{}} 2 = {0, 1} = {{}, {{}}} 3 = {0, 1, 2} = {{}, {{}}, {{}, {{ ...and so forth. In thes terms, the ordinal omega is the set of finite ordinals, or: omega = {0, 1, 2, 3, 4, ... } = too much trouble for me to write out in brackets How would the set omega-1 be defined? It doesn't make sense unless you believe in a last finite ordinal, which of course a non-ultrafinitist will not believe in. 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: Bijections (was OM = SIGMA1)
Le 20-nov.-07, à 17:59, meekerdb a écrit : Bruno Marchal wrote: . But infinite ordinals can be different, and still have the same cardinality. I have given examples: You can put an infinity of linear well founded order on the set N = {0, 1, 2, 3, ...}. What is the definition of linear well founded order? I'm familiar with well ordered, but how is linear applied to sets? Just curious. By linear, I was just meaning a non branching order. A tree can be well founded too, meaning all its branches have a length given by an ordinal. 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Le 19-nov.-07, à 17:00, Torgny Tholerus a écrit : Torgny Tholerus skrev: If you define the set of all natural numbers N, then you can pull out the biggest number m from that set. But this number m has a different type than the ordinary numbers. (You see that I have some sort of type theory for the numbers.) The ordinary deduction rules do not hold for numbers of this new type. For all ordinary numbers you can draw the conclusion that the successor of the number is included in N. But for numbers of this new type, you can not draw this conclusion. You can say that all ordinary natural numbers are of type 0. And the biggest natural number m, and all numbers you construct from that number, such that m+1, 2*m, m/2, and so on, are of type 1. And you can construct a set N1 consisting of all numbers of type 1. In this set there exists a biggest number. You can call it m1. But this new number is a number of type 2. It may look like a contradiction to say that m is included in N, and to say that all numbers in N have a successor in N, and to say that m have no successor in N. But it is not a constrdiction because the rule all numbers in N have a successor in N can be expanded to all numbers of type 0 in N have a successor in N. And because m is a number of type 1, then that rule is not applicable to m. You can comapre this with the Russell's paradox. This paradox says: Construct the set R of all sets that does not contain itself. For this set R there will be the rule: For all x, if x does not contain itself, then R contains x. If we here substitute R for x, then we get: If R does not contain itself, then R contains R. This is a contradiction. The contradiction is caused by an illegal conclusion, it is illegal to substitute R for x in the For all x-quantifier above. This paradox is solved by type theory. If you say that all ordinary sets are of type 0, then the set R will be of type 1. And every all-quantifiers are restricted to objects of a special type. So the rule above should read: For all x of type 0, if x does not contain itself, then R contains x. In this case you will not get any contradiction, because you can not substitute R for x in that rule. This points on one among many ways to handle Russell's paradox. Type Theories (TT) are nice, but many logicians prefer some untyped set theory, like ZF, or a two types theory like von Neuman Bernays Godel (VBG). Or Cartesian closed categories, toposes, etc. But set theory is a bit out of the scope of this thread. All such theories (ZF, VBG, TT) are example of Lobian Machine, and my goal is to study all such machine without choosing one in particular, and using traditional math, instead of working really in some particular theories. Another solution for many paradoxes consists in working with constructive objects. Soon, this is what we will do, by focusing on the set of computable functions instead of the set of all functions. The reason is not to escape paradoxes though. The reason is to learn something about machines (which are finite or constructive object). Just wait a bit. I will first explain Cantor's diagonal, which is simple but rather transcendental. Compare this with the case of the biggest natural number: Construct the set N of all natural numbers. For this set N there will be the rule: For all x, if N contains x, then N contains x+1. Suppose that there exists a biggest natural number m in N. If we substitute m for x, then we get: If N contains m, then N contains m+1. This is a contradiction, because m+1 is bigger than m, so m can not be the biggest number then. But the contradiction is caused by an illegal conclusion, it is illegal to substitute m for x in the For all x-quantifier above. This paradox is solved by type theory. If you say that all ordinary natural numbers are of type 0, then the natural number m will be of type 1. And every all-quantifiers are restricted to objects of a special type. So the rule above should read: For all x of type 0, if N contains x, then N contains x+1. In this case you will not get any contradiction, because you can not substitute m for x in that rule. === Do you see the similarities in both these cases? Except that naive usual number theory does not lead to any paradox, unlike naive set theory. *you* got a paradox because of *your* ultrafinistic constraint. So you are proposing a medication which could be worst that the disease I'm afraid. Very few people have any trouble with the potential infinite N = omega = {0, 1, 2, 3, 4, 5, ...}. With comp it can be shown that you don't need more at the ontological third person level. What will happen is that infinities come back in the first person point of views, and are very useful and lawful. Now, sound Lobian Machines can disagree on the real status of some of those infinities, but this does
Re: Bijections (was OM = SIGMA1)
Bruno Marchal skrev: To sum up; finite ordinal and finite cardinal coincide. Concerning infinite number there are much ordinals than cardinals. In between two different infinite cardinal, there will be an infinity of ordinal. We have already seen that omega, omega+1, ... omega+omega, omega+omega+1, 3.omega, ... 4.omega omega.omega . omega.omega.omega, .omega^omega . are all different ordinals, but all have the same cardinality. Was it not an error there? 2^omega is just the number of all subsets of omega, and the number of all subsets always have bigger cardinality than the set. So omega^omega can not have the same cardinality as omega. -- 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Le 20-nov.-07, à 12:14, Torgny Tholerus a écrit : Bruno Marchal skrev: To sum up; finite ordinal and finite cardinal coincide. Concerning infinite number there are much ordinals than cardinals. In between two different infinite cardinal, there will be an infinity of ordinal. We have already seen that omega, omega+1, ... omega+omega, omega+omega+1, 3.omega, ... 4.omega omega.omega . omega.omega.omega, .omega^omega . are all different ordinals, but all have the same cardinality. Was it not an error there? 2^omega is just the number of all subsets of omega, and the number of all subsets always have bigger cardinality than the set. Yes, that is true. So omega^omega can not have the same cardinality as omega. But addition, multiplication, and thus exponentiation are not the same operation for ordinals and cardinals. I should have written omega^omega, or something like that. That is why I have written 3.omega instead of 3*omega. We can come back on ordinal later, but now I will focus the attention on the cardinals, and prove indeed that 2^omega, or 2^N, or equivalently the infinite cartesian product (of sets) 2X2X2X2X2X2X2X2X... , is NOT enumerable (and indeed vastly bigger that the ordinal omega^omega. You can look at the thread on the growing functions for a little more on the ordinals. Actually my point was to remind people of the difference between ordinal and cardinal, and, yes, they have different addition, multiplication, etc. 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Bruno Marchal wrote: . But infinite ordinals can be different, and still have the same cardinality. I have given examples: You can put an infinity of linear well founded order on the set N = {0, 1, 2, 3, ...}. What is the definition of linear well founded order? I'm familiar with well ordered, but how is linear applied to sets? Just curious. Brent Meeker --~--~-~--~~~---~--~~ 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: Bijections (was OM = SIGMA1)
Bruno Marchal skrev: But infinite ordinals can be different, and still have the same cardinality. I have given examples: You can put an infinity of linear well founded order on the set N = {0, 1, 2, 3, ...}. The usual order give the ordinal omega = {0, 1, 2, 3, ...}. Now omega+1 is the set of all ordinal strictly lesser than omega+1, with the convention above. This gives {0, 1, 2, 3, ... omega} = {0, 1, 2, 3, 4, {0, 1, 2, 3, 4, }}. As an order, and thus as an ordinal, it is different than omega or N. But as a cardinal omega and omega+1 are identical, that means (by definition of cardinal) there is a bijection between omega and omega+1. Indeed, between {0, 1, 2, 3, ... omega} and {0, 1, 2, 3, ...}, you can build the bijection: 0omega 10 21 32 ... n --- n-1 ... All right?- represents a rope. An ultrafinitist comment: In the last line of this sequence you will have: ? - omega-1 But what will the ? be? It can not be omega, because omega is not included in N... -- 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Quentin Anciaux skrev: Hi, Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit : What do you mean by "each" in the sentence "for each natural number"? How do you define ALL natural numbers? There is a natural number 0. Every natural number a has a natural number successor, denoted by S(a). What do you mean by "Every" here? Can you give a *non-circular* definition of this word? Such that: "By every natural number I mean {1,2,3}" or "By every naturla number I mean every number between 1 and 100". (This last definition is non-circular because here you can replace "every number" by explicit counting.) How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, By definition there exists no biggest number unless you add an axiom saying there is one but the newly defined set is not N. I can prove by induction that there exists a biggest number: A) In the set {m} with one element, there exists a biggest number, this is the number m. B) If you have a set M of numbers, and that set have a biggest number m, and you add a number m2 to this set, then this new set M2 will have a biggest number, either m if m is bigger than m2, or m2 if m2 is bigger than m. C) The induction axiom then says that every set of numbers have a biggest number. Q.E.D. -- Torgny Tholerus --~--~-~--~~~---~--~~ 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: Bijections (was OM = SIGMA1)
Le Friday 16 November 2007 09:33:38 Torgny Tholerus, vous avez écrit : Quentin Anciaux skrev: Hi, Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit : What do you mean by each in the sentence for each natural number? How do you define ALL natural numbers? There is a natural number 0. Every natural number a has a natural number successor, denoted by S(a). What do you mean by Every here? Can you give a *non-circular* definition of this word? Such that: By every natural number I mean {1,2,3} or By every naturla number I mean every number between 1 and 100. (This last definition is non-circular because here you can replace every number by explicit counting.) I do not see circularity here... every means every, it means all natural numbers possess this properties ie (having a successor), that means by induction that N does contains an infinite number of elements, if it wasn't the case that would mean that there exists a natural number which doesn't have a successor... well as we have put explicitly the successor rule to defined N I can't see how to change that without changing the axioms. How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, By definition there exists no biggest number unless you add an axiom saying there is one but the newly defined set is not N. I can prove by induction that there exists a biggest number: A) In the set {m} with one element, there exists a biggest number, this is the number m. B) If you have a set M of numbers, and that set have a biggest number m, and you add a number m2 to this set, then this new set M2 will have a biggest number, either m if m is bigger than m2, or m2 if m2 is bigger than m. C) The induction axiom then says that every set of numbers have a biggest number. Q.E.D. -- Torgny Tholerus Hmm I don't understand... This could only work on finite set of elements. I don't see this as a proof that N is finite (because it *can't* be by *definition*). Quentin Anciaux -- All those moments will be lost in time, like tears in the rain. --~--~-~--~~~---~--~~ 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: Bijections (was OM = SIGMA1)
Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit : Bruno Marchal skrev:Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit : What do you mean by ...? Are you asking this as a student who does not understand the math, or as a philospher who, like an ultrafinist, does not believe in the potential infinite (accepted by mechanist, finistist, intuitionist, etc.). I am asking as an ultrafinitist. Fair enough. I am not sure there are many ultrafinitists on the list, but just to let John Mikes and Norman to digest the bijection post, I will say a bit more. A preliminary remark is that I am not sure an ultrafinitist can really assert he is ultrafinitist without acknowledging that he does have a way to give some meaning on But I have a more serious question below. I have already explained that the meaning of ...' in {I, II, III, , I, II, III, , I, ...} is *the* mystery. Do you have the big-black-cloud interpretation of ...? By that I mean that there is a big black cloud at the end of the visible part of universe, Concerning what I am trying to convey, this is problematic. The word universe is problematic. The word visible is also problematic. and the sequence of numbers is disappearing into the cloud, so that you can only see the numbers before the cloud, but you can not see what happens at the end of the sequence, because it is hidden by the cloud. I don't think that math is about seeing. I have never seen a number. It is a category mistake. I can interpret sometimes some symbol as refering to number, but that's all. For example, the function which sends x on 2*x, for each x in N is such a bijection. What do you mean by each x here? I mean for each natural number. What do you mean by each in the sentence for each natural number? How do you define ALL natural numbers? By relying on your intuition of finiteness. I take 0 as denoting a natural number which is not a successor. I take s(0) to denote the successor of 0. I accept that any number obtained by a *finite* application of the successor operation is a number. I accept that s is a bijection from N to N \ {0}, and things like that. How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, There is no biggest number in N. By definition of N we accept that if x is in N, then x+1 is also in N, and is different from x. How do you know that m+1 is also in N? By definition. You say that for ALL x then x+1 is included in N, but how do you prove that m is included in ALL x? I say for all x means for all x in N. If you say that m is included in ALL x, then you are doing an illegal deduction, and when you do an illegal deduction, then you can prove anything. (This is the same illegal deduction that is made in the Russell paradox.) ? (if you believe this then you have to accept that Peano Arithmetic, or even Robinson arithmetic) is inconsistent. Show me the precise proof. then there will be no corresponding number 2*m in E, because 2*m is not a number. Of course, but you are not using the usual notion of numbers. If you believe that the usual notion of numbers is wrong, I am sorry I cannot help you. I am using the usual notion of numbers. You are not. By definition of the usual natural numbers, all have a successor. But m+1 is not a number. This means that you believe there is a finite sequence of s of the type A = s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s( s(s(s(s( s(0)...) where ... here represents a finite sequence, and which is such that s(A) is not a number. But you can define a new concept: number-2, such that m+1 is included in that new concept. And you can define a new set N2, that contains all natural numbers-2. This new set N2 is bigger than the old set N, that only contains all natural numbers. Torgny, have you followed my fairy tale which I have explain to Tom Caylor. There I have used transfinite sequence of growing functions to name a big but finite natural number, which I wrote F_superomega(999), or OMEGA+[OMEGA]+OMEGA. My serious question is the following: is your biggest number less, equal or bigger than a well defined finite number like F_superomega(999). If yes, then a big part of the OM = SIGMA_1 thread will be accessible to you, except for the final conclusion. Indeed, you will end up with a unique finite bigger universal machine (which I doubt). If not, let us just say that your ultrafinitist hypothesis is too strong to make it coherent with the computationalist hypo. It means that you have a theory which is just different from what I propose. And then I will ask you to be ultra-patient, for I prefer to continue my explanation, and to come back on the discussion on
Re: Bijections (was OM = SIGMA1)
Bruno Marchal wrote: ... If not, let us just say that your ultrafinitist hypothesis is too strong to make it coherent with the computationalist hypo. It means that you have a theory which is just different from what I propose. And then I will ask you to be ultra-patient, for I prefer to continue my explanation, and to come back on the discussion on hypotheses after. OK. Actually, my conversation with Tom was interrupted by Norman who fears people leaving the list when matter get too much technical; Pay no attention to Norman. :-) I attend to this list because I learn things from it and I learn a lot from your technical presentations. I'm also doubtful of infinities, but they make things simpler; so my attitude is, let's see where the theory takes us. Brent Meeker --~--~-~--~~~---~--~~ 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: Bijections (was OM = SIGMA1)
Le 16-nov.-07, à 09:33, Torgny Tholerus a écrit : There is a natural number 0. Every natural number a has a natural number successor, denoted by S(a). What do you mean by Every here? Can you give a *non-circular* definition of this word? Such that: By every natural number I mean {1,2,3} or By every naturla number I mean every number between 1 and 100. (This last definition is non-circular because here you can replace every number by explicit counting.) How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, By definition there exists no biggest number unless you add an axiom saying there is one but the newly defined set is not N. I can prove by induction that there exists a biggest number: A) In the set {m} with one element, there exists a biggest number, this is the number m. B) If you have a set M of numbers, and that set have a biggest number m, and you add a number m2 to this set, then this new set M2 will have a biggest number, either m if m is bigger than m2, or m2 if m2 is bigger than m. C) The induction axiom then says that every set of numbers have a biggest number. What do you mean by every here? You just give us a non ultrafinitistic proof that all numbers are finite, not that the set of all finite number is finite. Bruno Q.E.D. -- Torgny Tholerus 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Bruno Marchal skrev: Le 15-nov.-07, 14:45, Torgny Tholerus a crit : But m+1 is not a number. This means that you believe there is a finite sequence of "s" of the type A = s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s( s(0)...) where "..." here represents a finite sequence, and which is such that s(A) is not a number. Yes, exactly. When you construct the set of ALL natural numbers N, you have to define ALL these numbers. And you can only define a finite number of numbers. See more explanations below. BTW, do you agree that 100^(100^(100^(100^(100^(100^(100^(100^100)], and 100^(100^(100^(100^(100^(100^(100^(100^100)] +1 are numbers? I am just curious, Yes, I agree. All explicitly given numbers are numbers. The biggest number is bigger than all by human beeings explicitly given numbers. If you define the set of all natural numbers N, then you can pull out the biggest number m from that set. But this number m has a different "type" than the ordinary numbers. (You see that I have some sort of "type theory" for the numbers.) The ordinary deduction rules do not hold for numbers of this new type. For all ordinary numbers you can draw the conclusion that the successor of the number is included in N. But for numbers of this new type, you can not draw this conclusion. You can say that all ordinary natural numbers are of type 0. And the biggest natural number m, and all numbers you construct from that number, such that m+1, 2*m, m/2, and so on, are of type 1. And you can construct a set N1 consisting of all numbers of type 1. In this set there exists a biggest number. You can call it m1. But this new number is a number of type 2. There is some sort of "temporal" distinction between the numbers of different type. You have to "first" have all numbers of type 0, "before" you can construct the numbers of type 1. And you must have all numbers of type 1 "before" you can construct any number of type 2, and so on. The construction of numbers of type 1 presupposes that the set of all numbers of type 0 is fixed. When the set N of all numbers of type 0 is fixed, then you can construct new numbers of type 1. It may look like a contradiction to say that m is included in N, and to say that all numbers in N have a successor in N, and to say that m have no successor in N. But it is not a constrdiction because the rule "all numbers in N have a successor in N" can be expanded to "all numbers of type 0 in N have a successor in N". And because m is a number of type 1, then that rule is not applicable to m. -- 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit : Bruno Marchal skrev: 0) Bijections Definition: A and B have same cardinality (size, number of elements) when there is a bijection from A to B. Now, at first sight, we could think that all *infinite* sets have the same cardinality, indeed the cardinality of the infinite set N. By N, I mean of course the set {0, 1, 2, 3, 4, ...} What do you mean by ...? Are you asking this as a student who does not understand the math, or as a philospher who, like an ultrafinist, does not believe in the potential infinite (accepted by mechanist, finistist, intuitionist, etc.). I have already explained that the meaning of ...' in {I, II, III, , I, II, III, , I, ...} is *the* mystery. A beautiful thing, which is premature at this stage of the thread, is that accepting the usual meaning of ... , then we can mathematically explained why the meaning of ... has to be a mystery. By E, I mean the set of even number {0, 2, 4, 6, 8, ...} Galileo is the first, to my knowledge to realize that N and E have the same number of elements, in Cantor's sense. By this I mean that Galileo realized that there is a bijection between N and E. For example, the function which sends x on 2*x, for each x in N is such a bijection. What do you mean by each x here? I mean for each natural number. How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, There is no biggest number in N. By definition of N we accept that if x is in N, then x+1 is also in N, and is different from x. then there will be no corresponding number 2*m in E, because 2*m is not a number. Of course, but you are not using the usual notion of numbers. If you believe that the usual notion of numbers is wrong, I am sorry I cannot help you. Bruno Now, instead of taking this at face value like Cantor, Galileo will instead take this as a warning against the use of the infinite in math or calculus. -- Torgny Tholerus 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 -~--~~~~--~~--~--~---
Re: Bijections (was OM = SIGMA1)
Bruno Marchal skrev: Le 14-nov.-07, 17:23, Torgny Tholerus a crit : What do you mean by "..."? Are you asking this as a student who does not understand the math, or as a philospher who, like an ultrafinist, does not believe in the potential infinite (accepted by mechanist, finistist, intuitionist, etc.). I am asking as an ultrafinitist. I have already explained that the meaning of "...'" in {I, II, III, , I, II, III, , I, ...} is *the* mystery. Do you have the big-black-cloud interpretation of "..."? By that I mean that there is a big black cloud at the end of the visible part of universe, and the sequence of numbers is disappearing into the cloud, so that you can only see the numbers before the cloud, but you can not see what happens at the end of the sequence, because it is hidden by the cloud. For example, the function which sends x on 2*x, for each x in N is such a bijection. What do you mean by "each x" here? I mean "for each natural number". What do you mean by "each" in the sentence "for each natural number"? How do you define ALL natural numbers? How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, There is no biggest number in N. By definition of N we accept that if x is in N, then x+1 is also in N, and is different from x. How do you know that m+1 is also in N? You say that for ALL x then x+1 is included in N, but how do you prove that m is included in "ALL x"? If you say that m is included in "ALL x", then you are doing an illegal deduction, and when you do an illegal deduction, then you can prove anything. (This is the same illegal deduction that is made in the Russell paradox.) then there will be no corresponding number 2*m in E, because 2*m is not a number. Of course, but you are not using the usual notion of numbers. If you believe that the usual notion of numbers is wrong, I am sorry I cannot help you. I am using the usual notion of numbers. But m+1 is not a number. But you can define a new concept: "number-2", such that m+1 is included in that new concept. And you can define a new set N2, that contains all natural numbers-2. This new set N2 is bigger than the old set N, that only contains all natural numbers. -- Torgny Tholerus --~--~-~--~~~---~--~~ 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: Bijections (was OM = SIGMA1)
Hi, Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit : Bruno Marchal skrev: Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit : What do you mean by each x here? I mean for each natural number. What do you mean by each in the sentence for each natural number? How do you define ALL natural numbers? There is a natural number 0. Every natural number a has a natural number successor, denoted by S(a). There is no natural number whose successor is 0. Distinct natural numbers have distinct successors: if a ≠ b, then S(a) ≠ S(b). You need at least the successor axiom. N = {0 ,1 ,2 ,3 ,... ,N ,N+1, ..} All natural numbers are defined by the above. How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, By definition there exists no biggest number unless you add an axiom saying there is one but the newly defined set is not N. Quentin Anciaux -- All those moments will be lost in time, like tears in the rain. --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
Bijections (was OM = SIGMA1)
Hi David, Hi John, OK, here is a first try. Let me know if this is too easy, too difficult, or something in between. The path is not so long, so it is useful to take time on the very beginning. I end up with some exercice. I will give the solutions, but please try to be aware if you can or cannot do them, so as not missing the train. John, if you have never done what has been called modern math, you could have slight notation problem, please ask any question. I guess for some other the first posts in this thread could look too much simple. Be careful when we will go from this to the computability matter. I recall the plan, where I have added the bijection thread: Plan 0) Bijections 1) Cantor's diagonal 2) Does the universal digital machine exist? And for much later, if people are interested or ask question: 3) Lobian machines, who and/or what are they? 4) The 1-person and the 3- machine. 5) Lobian machines' theology 6) Lobian machines' physics 7) Lobian machines' ethics But my main goal first is to explain that Church thesis is a very strong postulate. I need first to be sure you have no trouble with the notion of bijection. 0) Bijections Suppose you have a flock of sheep. Your neighbor too. You want to know if you have more, less or the same number of sheep, but the trouble is that neither you nor your neighbor can count (nor anyone around). Amazingly enough, perhaps, it is still possible to answer that question, at least if you have enough pieces of rope. The idea consists in attaching one extremity of a rope to one of your sheep and the other extremity to one of the neighbor's sheep, and then to continue. You are not allow to attach two ropes to one sheep, never. In the case you and your neighbor have a different number of sheep, some sheep will lack a corresponding sheep at the extremity of their rope, so that their ropes will not be attached to some other sheep. Exemple (your flock of sheep = {s, r, t, u}, and the sheep of the neighborgh = {a, b, c, d, e, f}. s - a r -- d u - c t --- f e b and we see that the neighbor has more sheep than you, because b and e have their ropes unable to be attached to any remaining sheep you have, and this because there are no more sheep left in your flock. You have definitely less sheep. In case all ropes attached in that way have a sheep well attached at both extremities, we can say that your flock and your neighbor's flock have the same number of elements, or same cardinality. In that case, the ropes represent a so called one-one function, or bijection, between the two flocks. If you have less sheep than your neighbor, then there is a bijection between your flock and a subset of your neighbor's flock. If those flocks constitute a finite set, the existence of a bijection between the two flocks means that both flocks have the same number of sheep, and this is the idea that Cantor will generalize to get a notion of same number of element or same cardinality for couples of infinite sets. Given that it is not clear, indeed, if we can count the number of element of an infinite set, Cantor will have the idea of generalizing the notion of same number of elements, or same cardinality by the existence of such one-one function. The term bijection denotes one-one function. Definition: A and B have same cardinality (size, number of elements) when there is a bijection from A to B. Now, at first sight, we could think that all *infinite* sets have the same cardinality, indeed the cardinality of the infinite set N. By N, I mean of course the set {0, 1, 2, 3, 4, ...} By E, I mean the set of even number {0, 2, 4, 6, 8, ...} Galileo is the first, to my knowledge to realize that N and E have the same number of elements, in Cantor's sense. By this I mean that Galileo realized that there is a bijection between N and E. For example, the function which sends x on 2*x, for each x in N is such a bijection. Now, instead of taking this at face value like Cantor, Galileo will instead take this as a warning against the use of the infinite in math or calculus. Confronted to more complex analytical problems of convergence of Fourier series, Cantor knew that throwing away infinite sets was too pricy, and on the contrary, will consider such problems as a motivation for its set theory. Dedekind will even define an infinite set by a set which is such that there is a bijection between itself and some proper subset of himself. By Z, I mean the set of integers {..., -3, -2, -1, 0, 1, 2, 3, ...} Again there is a bijection between N and Z. For example, 0 1 2 3 4 56 7 8 9 10 0 -1 1 -2 2 -3 3 -4 4 -5 5 ... or perhaps more clearly (especially if the mail does not respect the blank uniformely; the bijection, like all function, is better represented as set of couples: bijection from N to Z = {(0,0) (1, -1) (2, 1) (3 -2)
Re: Bijections (was OM = SIGMA1)
Bruno Marchal skrev: 0) Bijections Definition: A and B have same cardinality (size, number of elements) when there is a bijection from A to B. Now, at first sight, we could think that all *infinite* sets have the same cardinality, indeed the cardinality of the infinite set N. By N, I mean of course the set {0, 1, 2, 3, 4, ...} What do you mean by ...? By E, I mean the set of even number {0, 2, 4, 6, 8, ...} Galileo is the first, to my knowledge to realize that N and E have the same number of elements, in Cantor's sense. By this I mean that Galileo realized that there is a bijection between N and E. For example, the function which sends x on 2*x, for each x in N is such a bijection. What do you mean by each x here? How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, then there will be no corresponding number 2*m in E, because 2*m is not a number. Now, instead of taking this at face value like Cantor, Galileo will instead take this as a warning against the use of the infinite in math or calculus. -- Torgny Tholerus --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---