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)
Bruno Marchal wrote: > > 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. 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 ... Cheers, Mirek --~--~-~--~~~---~--~~ 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
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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/
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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
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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
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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 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)
Hi Mirek,
Le 19-nov.-07, à 20:14, Mirek Dobsicek a écrit :
>
> Hi Bruno,
>
> thank you for posting the solutions. Of course, I solved it by myself
> and it was a fine relaxing time to do the paper work trying to be
> rigorous, however, your solutions gave me additional insights, nice.
>
> I am on the board for the sequel.
Thanks.
I will explain soon (this "afternoon") how Cantor managed to show that
some infinite set "have more elements" than the infinite set N, in the
sense that there will be no bijection from N to such set, despite
obvious bijection between N and subset of such set.
I am sure most of you know that proof by diagonal. However, the goal
will be to show later how a similar reasoning can put a serious doubt
on the existence of universal machine, and on serious constraints such
machine have to live with in case we continue to believe in their
existence.
Before doing that, I want to explain briefly the difference between
ordinal and cardinal. This explanation is not necessary for the sequel,
but it could help.
I will also use the set representation of numbers and ordinals. So I
will represent the number 0 by the empty set, and the number n by the
set of numbers strictly little than n.
So 0 = { }, 1 = {0} = {{}}, 2 = {0, 1} = {{} {{}}}, 3 = {0, 1, 2},
4 = {0, 1, 2, 3}, ...
then omega = N = {0, 1, 2, 3 ...} is the least infinite ordinal. The
advantage of such a representation is that "belongness" modelizes the
strictly-lesser-than relation, and subsetness modelizes the
lesser-than-or equal. I recall that A is a subset of B, if for all x, x
belongs to A entails that x belongs to B. In particular for all set A x
belongs to A entails x belongs to A, so all sets are subset of
themselves.
An ordinal is defined by being a linear well founded order.
Well-foundness means that all subsets have a least element.
The finite ordinal are thus the natural numbers. They all have
different cardinals. That is, two different natural numbers (=
different finite ordinal) have different cardinality (different
"number" of elements). Take 7 and 5, there is no bijection between
them, for example.
So in the finite realm, ordinal and cardinal coincide.
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.
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.
Don't worry, we will not use that.
Question: are there really two different infinite cardinals? That is,
are there two infinite sets with different cardinality? That is, are
there two different infinite sets A and B without any bijection in
between ? The answer is yes, and that is what cantor has discovered by
its diagonal construction, and that is the object of the next post. All
what I did want to say here, is that automatically, in between A and B,
there will be an infinite amount of different ordinals.
Bruno
http://iridia.ulb.ac.be/~marchal/
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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 lawf
Re: Bijections (was OM = SIGMA1)
Hi Bruno, thank you for posting the solutions. Of course, I solved it by myself and it was a fine relaxing time to do the paper work trying to be rigorous, however, your solutions gave me additional insights, nice. I am on the board for the sequel. Best, Mirek > > I give the solution of the little exercises on the notion of bijection. > > > So the sequel is: > > 1) Cantor's diagonal > 2) are there universal computing machine? (Kleene's diagonal, and > Church thesis) > > 3) A fundamental theorem about universal computing machines. (All such > machine are imperfect, or insecure) > > Please ask questions. To miss math due to notation problem is like to > miss travels due to mishandling of the use of maps, or to miss love by > mishandling of the use of clothes ... It is missing a lot, for > mishandling a few I wanna say. > --~--~-~--~~~---~--~~ 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)
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. == 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? -- 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 16-nov.-07, à 18:41, meekerdb (Brent Meeker) a écrit :
>
> 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.
Fair enough, thanks.
I give the solution of the little exercises on the notion of bijection.
1) is there a bijection between N and N?
Of course! The identity function is a bijection from N to N, and
actually any identity function defined on any set is a bijection from
that set with itself. Here is a drawing of a bijection from N to N (I
don't represent the "ropes" ...)
0 1 2 3 4 5 6 7 8 ...
0 1 2 3 4 5 6 7 8 ...
So all sets "have the same number of element" than itself.
2) Q is the set of rational numbers, that is length of segment which
can be measured by the ratio of natural numbers (or integers).
There is a bijection between N and Q. We have already seen a
bijection between N and ZXZ, which I called the spiraller (I put only
the image of the bijection, imagine the rope 0(0,0), 1-(0,1),
2-(-1,1), ... :
(0,0) (0,1) (-1,1) (-1,0) (-1,-1) (0,-1) (1,-1) (1,0) (1,1) (1,2) (0,2)
(-1,2) (-2,2) (-2,-1)
We can transform that bijection between N and ZXZ into a bijection
between N and Q in the following way. We start from the bijection above
between N and ZXZ, but we interpret (x,y) as the fraction x/y, throwing
it out in case we already met the corresponding rational number, or
when we met an indeterminate fraction (like 0/0) or spurious one (like
1/0, 2/0, ...), this gives first
0/0 0/1 -1/1 -1/0 -1/-1 0/-1 1/-1 1/0 1/1 1/2 0/2 -1/2
-2/2 -2/1 , and after the throwing out of the repeated rationals:
0 -1 1 1/2 -1/2 -2 ...
OK?
3) We have seen a bijection between N and NXN. We can use it to provide
a bijection between N and NX(NXN). Indeed, you can zigzag on NX(NXN)
like we have zigzag on NXN, starting from:
...
5
4
3
2
1
0
(0,0) (0,1) (1,0) (2,0) (1,1) (0,2) (0,3) (1,2) (2,1) (3,0) (4,0)
(3,1) ...
All right?
Obviously there is a bijection between NXNXN and NX(NXN)); just send
(x,y,z) on (x,(y,z)).
In the same manner, you can show the existence of a bijection and any
of NXNXNXN, NXNXNXNXN, NXNXNXNXNXN,
4) (The one important for the sequel). Take any finite alphabet, like
{0,1}, {a, b}, {a, b, c, ... z} or all keyboard keys. Then the set of
all finite words build on any such alphabet is in bijection with N.
Indeed, to be sure of enumerating all the words, on {a,b,c}, say, just
enumerate the words having length 1, then length 2, etc. And order just
alphabetically the words having the same length. On the alphabet
{a,b,c}, this gives
a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc, aaa, aab, aac, aba, abb,
abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc, ...
I hope you see that this gives a bijection between N and A° (the set of
words on A, which in this example is {a, b, c}. Purist would add the
empty word in front.
From this you can give another proof that Q is enumerable (= in
bijection with N). Indeed all rational numbers, being a (reduced)
fraction, can be written univocally as a word in the finite alphabet
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, /}. Example 456765678 / 9898989 (I have
added blank for reason of readibility, but those are not symbols taken
from the alphabet).
Another important example, the set of all programs in any programming
language. For some language, like Python for example, you have to put
explicitly the "enter", or "carriage return" key symbol in the
alphabet, of course.
5) Let N be the set {0, 1, 2, 3, 4, 5, 6, ...}, as usual. Let N' be a
sort of copy or relabeling of N, that is N' = {0', 1', 2', 3', 4', 5',
6', ...}.
What are really object like 5' is not relevant except that we consider
that any x is different from any x'. (usually mathematician formalize
N' by NX{0}, or NX{1}, but that are details).
Is N U N' in bijection with N ?
Sure, NUN' = {0, 0', 1, 1', 2, 2', 3, 3', 4, 4', 5, 5', 6, 6', 7, 7',
8, 8', ...}, and
0, 0', 1, 1', 2, 2', 3, 3', 4, 4', 5, 5', 6, 6', 7, 7', 8, 8', ... is
indeed an enumeration of NUN' (bijection from N to the set NUN'). You
can see it as the result of a zigzagging between
0, 1, 2, 3, 4, 5, 6, ...
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/
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Re: Bijections (was OM = SIGMA1)
Bruno Marchal skrev: Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit : 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 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
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
--
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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
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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
--
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Re: Bijections (was OM = SIGMA1)
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.
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
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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/
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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
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