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
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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)
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
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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)
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)
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)
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/
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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/
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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
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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
--
All those moments will be lost in time, like tears in the rain.
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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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