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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The big-black-cloud-interpretation.
Bruno Marchal skrev:
Le 15-nov.-07, 14:45, Torgny Tholerus a crit :
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.
A way to prove the consistency of a theory is to make a "visualization"
of the theory. If you can visualize all that happens in the theory,
then you know the theory is consistent.
To visualize the natural numbers, you can think of them as a long
sequence {0,1,2,3,4,5,...}, and this sequence is going far, far, away.
But you can only visualize finite sequences. So you can think that you
have a finite sequence of numbers, and you have a big black cloud far,
far, away. You see the first part of the sequence {0,1,2,...,m} before
the cloud. But inside the cloud you can imagine that you have the
finite sequence {m+1,m+2,...,4*m-1,4*m}. This whole sequence
{0,1,2,...,m,m+1,...4*m} is what you call the set N of all natural
numbers.
>From that set N you construct the true subset
{0,2,4,6,...,2*m,2*m+2,...,4*m}, which you call the set E of all even
numbers. The visible part of the set E is then {0,2,4,...,2*m}, and
the hidden part of that sequence is {2*m+2,...,4*m}.
Now you define a new concept INNFINITE, that is defined by:
If you have a bijection from all visible numbers of a set S, to all
visible numbers of a true subset of S, then you say that the set S in
INNFINITE.
Then you can use this concept INNFINITE, and you will get a consistent
theory with no contradictions, because you have a finite visualization
of this theory.
--
Torgny
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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 -~--~~~~--~~--~--~---
Introduction
Hello, I came upon this group entirely by chance and concluded that I had somehow found evidence of Intelligent Life on the Internet. I had been doing an Internet search on memory in single-celled organisms, such as the amoeba. Amoebas seem to move purposefully, which would suggest that they have at least one rudimentary sensory organ and some kind of memory. My search was not fruitful, except that it found a 'hit' in one of the discussions in this group. Fortunately, I do not possess any specialized knowledge that you would not understand, so this introduction is really unnecessary. I am retired from a university laboratory and have done a little reading in logic after retirement. It is possible that I might contribute here occasionally when I notice some point that seems susceptible to simple logical analysis. Gene Ledbetter --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
