Re: The seven step-Mathematical preliminaries
Brian Tenneson skrev:
On Thu, Jun 4, 2009 at 8:27 AM, Torgny Tholerus tor...@dsv.su.se
mailto:tor...@dsv.su.se wrote:
Brian Tenneson skrev:
Torgny Tholerus wrote:
It is impossible to create a set where the successor of every
element is
inside the set, there must always be an element where the
successor of
that element is outside the set.
I disagree. Can you prove this?
Once again, I think the debate ultimately is about whether or not to
adopt the axiom of infinity.
I think everyone can agree without that axiom, you cannot build or
construct an infinite set.
There's nothing right or wrong with adopting any axioms. What
results
is either interesting or not, relevant or not.
How do you handle the Russell paradox with the set of all sets
that does
not contain itself? Does that set contain itself or not?
If we're talking about ZFC set theory, then the axiom of foundation
prohibits sets from being elements of themselves.
I think we agree that in ZFC, there is no set of all sets.
But there is a set of all sets. You can construct it by taking all
sets, and from them doing a new set, the set of all sets. But note,
this set will not contain itself, because that set did not exist before.
My answer is that that set does not contain itself, because no set can
contain itself. So the set of all sets that does not contain
itself, is
the same as the set of all sets. And that set does not contain
itself.
This set is a set, but it does not contain itself. It is exactly the
same with the natural numbers, *BIGGEST+1 is a natural number, but it
does not belong to the set of all natural numbers. *The set of
all sets
is a set, but it does not belong to the set of all sets.
How can BIGGEST+1 be a natural number but not belong to the set of all
natural numbers?
One way to represent natural number as sets is:
0 = {}
1 = {0} = {{}}
2 = {0, 1} = 1 union {1} = {{}, {{}}}
3 = {0, 1, 2} = 2 union {2} = ...
. . .
n+1 = {0, 1, 2, ..., n} = n union {n}
. . .
Here you can then define that a is less then b if and only if a belongs
to b.
With this notation you get the set N of all natural numbers as {0, 1, 2,
...}. But the remarkable thing is that N is exactly the same as
BIGGEST+1. BIGGEST+1 is a set with the same structure as all the other
natural numbers, so it is then a natural number. But BIGGEST+1 is not a
member of N, the set of all natural numbers. BIGGEST+1 is bigger than
all natural numbers, because all natural numbers belongs to BIGGEST+1.
What the largest number is depends on how you define natural
number.
One possible definition is that N contains all explicit numbers
expressed by a human being, or will be expressed by a human
being in the
future. Amongst all those explicit numbers there will be one
that is
the largest. But this largest number is not an explicit number.
This raises a deeper question which is this: is mathematics
dependent
on humanity or is mathematics independent of humanity?
I wonder what would happen to that human being who finally expresses
the largest number in the future. What happens to him when he wakes
up the next day and considers adding one to yesterday's number?
This is no problem. If he adds one to the explicit number he
expressed
yesterday, then this new number is an explicit number, and the number
expressed yesterday was not the largest number. Both 17 and 17+1 are
explicit numbers.
This goes back to my earlier comment that it's hard for me to believe
that the following statement is false:
every natural number has a natural number successor
We -must- be talking about different things, then, when we use the
phrase natural number.
I can't say your definition of natural numbers is right and mine is
wrong, or vice versa. I do wonder what advantages there are to the
ultrafinitist approach compared to the math I'm familiar with.
The biggest advantage is that everything is finite, and you can then
really know that the mathematical theory you get is consistent, it does
not contain any contradictions.
--
Torgny Tholerus
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Re: The seven step-Mathematical preliminaries
Kory Heath skrev: On Jun 4, 2009, at 8:27 AM, Torgny Tholerus wrote: How do you handle the Russell paradox with the set of all sets that does not contain itself? Does that set contain itself or not? My answer is that that set does not contain itself, because no set can contain itself. So the set of all sets that does not contain itself, is the same as the set of all sets. And that set does not contain itself. This set is a set, but it does not contain itself. It is exactly the same with the natural numbers, BIGGEST+1 is a natural number, but it does not belong to the set of all natural numbers. The set of all sets is a set, but it does not belong to the set of all sets. So you're saying that the set of all sets doesn't contain all sets. How is that any less paradoxical than the Russell paradox you're trying to avoid? The secret is the little word all. To be able to use that word, you have to define it. You can define it by saying: By 'all sets' I mean that set and that set and that set and When you have made that definition, you are then able to create a new set, the set of all sets. But you must be carefull with what you do with that set. That set does not contain itself, because it was not included in your definition of all sets. If you call the set of all sets for A, then you have: For all x such that x is a set, then x belongs to A. A is a set. But it is illegal to substitute A for x, so you can not deduce: A is a set, then A belongs to A. This deductuion is illegal, because A is not included in the definition of all x. -- Torgny Tholerus --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: The seven step-Mathematical preliminaries
2009/6/5 Torgny Tholerus tor...@dsv.su.se: Kory Heath skrev: On Jun 4, 2009, at 8:27 AM, Torgny Tholerus wrote: How do you handle the Russell paradox with the set of all sets that does not contain itself? Does that set contain itself or not? My answer is that that set does not contain itself, because no set can contain itself. So the set of all sets that does not contain itself, is the same as the set of all sets. And that set does not contain itself. This set is a set, but it does not contain itself. It is exactly the same with the natural numbers, BIGGEST+1 is a natural number, but it does not belong to the set of all natural numbers. The set of all sets is a set, but it does not belong to the set of all sets. So you're saying that the set of all sets doesn't contain all sets. How is that any less paradoxical than the Russell paradox you're trying to avoid? The secret is the little word all. To be able to use that word, you have to define it. I call that secret bullshit, and to understand that word (bullshit), you have to define it. Sorry but I think we're talking in english here, all means all not what you decide it means. Quentin. You can define it by saying: By 'all sets' I mean that set and that set and that set and When you have made that definition, you are then able to create a new set, the set of all sets. But you must be carefull with what you do with that set. That set does not contain itself, because it was not included in your definition of all sets. If you call the set of all sets for A, then you have: For all x such that x is a set, then x belongs to A. A is a set. But it is illegal to substitute A for x, so you can not deduce: A is a set, then A belongs to A. This deductuion is illegal, because A is not included in the definition of all x. -- Torgny Tholerus -- All those moments will be lost in time, like tears in rain. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Cognitive Theoretic Model of the Universe
Bruno: I understand a little better. is there a citition for a version of Church Thesis that all algorithm can be written in FORTRAN? Ronald On Jun 4, 10:49 am, Bruno Marchal marc...@ulb.ac.be wrote: Hi Ronald, On 02 Jun 2009, at 16:45, ronaldheld wrote: Bruno: Since I program in Fortran, I am uncertain how to interpret things. I was alluding to old, and less old, disputes again programmers, about which programming language to prefer. It is a version of Church Thesis that all algorithm can be written in FORTRAN. But this does not mean that it is relevant to define an algorithm by a fortran program. I thought this was obvious, and I was using that known confusion to point on a similar confusion in Set Theory, like Langan can be said to perform. In Set Theorist, we still find often the error consisting in defining a mathematical object by a set. I have done that error in my youth. What you can do, indeed, is to *represent* (almost all) mathematical objects by sets. Langan seems to make that mistake. The point is just that we have to distinguish a mathematical object and the representation of that object in some mathematical theory. I will have the opportunity to give a precise example in the 7th thread later. In usual mathematical practice, this mistake is really not important, yet, in logic it is more important to take into account that distinction, and then in cognitive science it is *very* important. Crucial, I would say. The error consisting in identifying consciousness and brain state belongs to that family, for example. To confuse a person and its body belongs to that family of error too. All such error are of the form of the confusion between the Moon and the finger which point to the moon, or the confusion between a map and the territory. I have nothing against the use of FORTRAN. On the contrary I have a big respect for that old venerable high level programming language :) 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 everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Cognitive Theoretic Model of the Universe
Well as FORTRAN is a turing complete language, then you can. As long as the programming language is universal/turing complete you can. http://en.wikipedia.org/wiki/Turing_completeness Regards, Quentin 2009/6/5 ronaldheld ronaldh...@gmail.com: Bruno: I understand a little better. is there a citition for a version of Church Thesis that all algorithm can be written in FORTRAN? Ronald On Jun 4, 10:49 am, Bruno Marchal marc...@ulb.ac.be wrote: Hi Ronald, On 02 Jun 2009, at 16:45, ronaldheld wrote: Bruno: Since I program in Fortran, I am uncertain how to interpret things. I was alluding to old, and less old, disputes again programmers, about which programming language to prefer. It is a version of Church Thesis that all algorithm can be written in FORTRAN. But this does not mean that it is relevant to define an algorithm by a fortran program. I thought this was obvious, and I was using that known confusion to point on a similar confusion in Set Theory, like Langan can be said to perform. In Set Theorist, we still find often the error consisting in defining a mathematical object by a set. I have done that error in my youth. What you can do, indeed, is to *represent* (almost all) mathematical objects by sets. Langan seems to make that mistake. The point is just that we have to distinguish a mathematical object and the representation of that object in some mathematical theory. I will have the opportunity to give a precise example in the 7th thread later. In usual mathematical practice, this mistake is really not important, yet, in logic it is more important to take into account that distinction, and then in cognitive science it is *very* important. Crucial, I would say. The error consisting in identifying consciousness and brain state belongs to that family, for example. To confuse a person and its body belongs to that family of error too. All such error are of the form of the confusion between the Moon and the finger which point to the moon, or the confusion between a map and the territory. I have nothing against the use of FORTRAN. On the contrary I have a big respect for that old venerable high level programming language :) Bruno http://iridia.ulb.ac.be/~marchal/ -- All those moments will be lost in time, like tears in rain. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
RE: The seven step-Mathematical preliminaries
Date: Fri, 5 Jun 2009 08:33:47 +0200
From: tor...@dsv.su.se
To: everything-list@googlegroups.com
Subject: Re: The seven step-Mathematical preliminaries
Brian Tenneson skrev:
On Thu, Jun 4, 2009 at 8:27 AM, Torgny Tholerus tor...@dsv.su.se
mailto:tor...@dsv.su.se wrote:
Brian Tenneson skrev:
Torgny Tholerus wrote:
It is impossible to create a set where the successor of every
element is
inside the set, there must always be an element where the
successor of
that element is outside the set.
I disagree. Can you prove this?
Once again, I think the debate ultimately is about whether or not to
adopt the axiom of infinity.
I think everyone can agree without that axiom, you cannot build or
construct an infinite set.
There's nothing right or wrong with adopting any axioms. What
results
is either interesting or not, relevant or not.
How do you handle the Russell paradox with the set of all sets
that does
not contain itself? Does that set contain itself or not?
If we're talking about ZFC set theory, then the axiom of foundation
prohibits sets from being elements of themselves.
I think we agree that in ZFC, there is no set of all sets.
But there is a set of all sets. You can construct it by taking all
sets, and from them doing a new set, the set of all sets. But note,
this set will not contain itself, because that set did not exist before.
My answer is that that set does not contain itself, because no set can
contain itself. So the set of all sets that does not contain
itself, is
the same as the set of all sets. And that set does not contain
itself.
This set is a set, but it does not contain itself. It is exactly the
same with the natural numbers, *BIGGEST+1 is a natural number, but it
does not belong to the set of all natural numbers. *The set of
all sets
is a set, but it does not belong to the set of all sets.
How can BIGGEST+1 be a natural number but not belong to the set of all
natural numbers?
One way to represent natural number as sets is:
0 = {}
1 = {0} = {{}}
2 = {0, 1} = 1 union {1} = {{}, {{}}}
3 = {0, 1, 2} = 2 union {2} = ...
. . .
n+1 = {0, 1, 2, ..., n} = n union {n}
. . .
Here you can then define that a is less then b if and only if a belongs
to b.
With this notation you get the set N of all natural numbers as {0, 1, 2,
...}. But the remarkable thing is that N is exactly the same as
BIGGEST+1. BIGGEST+1 is a set with the same structure as all the other
natural numbers, so it is then a natural number. But BIGGEST+1 is not a
member of N, the set of all natural numbers.
Here you're just contradicting yourself. If you say BIGGEST+1 is then a
natural number, that just proves that the set N was not in fact the set of
all natural numbers. The alternative would be to say BIGGEST+1 is *not* a
natural number, but then you need to provide a definition of natural number
that would explain why this is the case.
The biggest advantage is that everything is finite, and you can then
really know that the mathematical theory you get is consistent, it does
not contain any contradictions.
Even if you define natural number in such a way that there are only a finite
number of them (which you haven't actually done, you've just asserted it
without providing any specific definition), you still could have an infinite
number of *propositions* about them if you allow each proposition to contain an
unlimited number of AND and OR operators. For example, even if I say that the
only natural numbers are 1,2,3, I can still make arbitrarily long propositions
like ((31) AND (21)) OR (31)) AND ((23) OR (31)) AND ((23) OR ((13) OR
((21) OR ((13) OR (31). Of course a non-finitist would be able to prove
that these infinite number of propositions are consistent, but I don't know if
an ultrafinitist would (likewise a non-finitist can accept a proof that
something like the Peano axioms are consistent based on an understanding of
their application to a model dealing with rows of dots, even if the Peano
axioms cannot be used to formally prove their own consistency).
Jesse
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Re: Cognitive Theoretic Model of the Universe
On 04 Jun 2009, at 21:23, Brent Meeker wrote: Bruno Marchal wrote: ... Bruno Marchal wrote: The whole point of logic is to consider the Peano's axioms as a mathematical object itself, which is studied mathematically in the usual informal (yet rigorous and typically mathematica) way. PA, and PA+GOLDBACH are different mathematical objects. They are different theories, or different machines. Now if GOLDBACH is provable by PA, then PA and PA+GOLDBACH shed the same light on the same arithmetical truth. In that case I will identify PA and PA+GOLDBACH, in many contexts, because most of the time I identify a theory with its set of theorems. Like I identify a person with its set of (possible) beliefs. If GOLDBACH is *true, but not provable* by PA, then PA and PA +GOLDBACH still talk on the same reality, but PA+GOLDBACH will shed more light on it, by proving more theorems on the numbers and numbers relations than PA. I do no more identify them, and they have different set of theorems. If GOLDBACH is false. Well GOLBACH is PI_1, that is, its negation is SIGMA_1, that is, it has the shape it exist a number such that it verify this decidable property. Indeed the negation of Goldbach conjecture is it exists a number bigger than 2 which is not the sum of two primes. This, if true, is verifiable already by the much weaker RA (Robinson arithmetic). So, if GOLDBACH is false PA + GOLDBACH is inconsistent. That is a mathematical object quite different from PA! So what then is the status of the natural numbers? Are there many different objects in Platonia which we loosely refer to as the natural numbers or is there only one such object and the Goldbach conjecture is either true of false of this object? Nobody can answer this question in your place. But if you believe that the principle of excluded middle can be applied to closed arithmetical sentences, like 99,999% of the mathematician, then you have to believe that the Goldbach conjecture is either true or false. Even intuitionist will admit that Goldabch conjecture is true or false, given its Sigma_1 character. This means that, about the (true- or-false) nature of GOLDBACH is doubtable only for an ultrafinitist. BTW, Goldbach conjecture asserts that all female (even) numbers can be written as a sum of two primes, except the number two. (I forget the word even in my enunciation above!). Here, you would have taken the twin primes conjecture, and things would have been different, and more complex. Because, even if it is false, it cannot be proven false by exhibiting an example? Yes. And this entails that both PA+TPC and PA + (~TPC) could be consistent, yet one of those theory has to be unsound, or if you prefer has to enunciate false arithmetical statements (yet consistent with PA). Sound is relative to the usual understanding of the natural numbers which is presupposed in any work in mathematical logic or computer science, like it is presupposed in any part of any physical theory. That usual meaning is taught in primary school without any trouble. In model theory, this notion of soundness can be made more precise, through the notion of standard model of PA for example, but this presupposes, in the meta-theory, an understanding of that usual notion of numbers. Nobody doubts the consistency and soundness of the theories like RA and PA. (Even Torgny, who fakes that he doubts them for a philosophical purpose unrelated to our discussion, like he fakes to be a faking zombie, etc. This is clear from older post by Torgny). Note that a theory of set like ZF shed even much more large light on arithmetical truth, (and is still incomplete on arithmetic, by Gödel ...). Incidentally it can be shown that ZF and ZFC, although they shed different light on the mathematical truth in general, does shed exactly the same light on arithmetical truth. They prove the same arithmetical theorems. On the numbers, the axiom of choice add nothing. This is quite unlike the ladder of infinity axioms. I would say it is and will be particularly important to distinguish chatting beings like RA, PA, ZF, ZFC, etc... and what those beings are talking about. Bruno Do you mean PA talks about the natural numbers but PA+theorems is a different mathematical object than N? I am not sure I understand what you mean. PA is an (immaterial) machine, or a program if you want. I guess that, by PA+theorems, you mean the set of theorems of PA. In some context we can identify PA and PA+theorems, because the context makes things unambiguous. But strictly speaking those are different mathematical object: PA is finite (well, as I defined it usually), But PA+theorems is infinite. Both talk about N, and both are different of N. Indeed PA is a finite (or infinite in the usual first order presentation) set of axioms and rules, PA+theorems is an infinite set of formula, and N is an infinite
Re: The seven step-Mathematical preliminaries
Torgny Tholerus wrote:
Brian Tenneson skrev:
On Thu, Jun 4, 2009 at 8:27 AM, Torgny Tholerus tor...@dsv.su.se
mailto:tor...@dsv.su.se wrote:
Brian Tenneson skrev:
Torgny Tholerus wrote:
It is impossible to create a set where the successor of every
element is
inside the set, there must always be an element where the
successor of
that element is outside the set.
I disagree. Can you prove this?
Once again, I think the debate ultimately is about whether or not to
adopt the axiom of infinity.
I think everyone can agree without that axiom, you cannot build or
construct an infinite set.
There's nothing right or wrong with adopting any axioms. What
results
is either interesting or not, relevant or not.
How do you handle the Russell paradox with the set of all sets
that does
not contain itself? Does that set contain itself or not?
If we're talking about ZFC set theory, then the axiom of foundation
prohibits sets from being elements of themselves.
I think we agree that in ZFC, there is no set of all sets.
But there is a set of all sets. You can construct it by taking all
sets, and from them doing a new set, the set of all sets. But note,
this set will not contain itself, because that set did not exist before.
If that set does not contain itself then it is not a set of all sets.
My answer is that that set does not contain itself, because no set can
contain itself. So the set of all sets that does not contain
itself, is
the same as the set of all sets. And that set does not contain
itself.
This set is a set, but it does not contain itself. It is exactly the
same with the natural numbers, *BIGGEST+1 is a natural number, but it
does not belong to the set of all natural numbers. *The set of
all sets
is a set, but it does not belong to the set of all sets.
How can BIGGEST+1 be a natural number but not belong to the set of all
natural numbers?
One way to represent natural number as sets is:
0 = {}
1 = {0} = {{}}
2 = {0, 1} = 1 union {1} = {{}, {{}}}
3 = {0, 1, 2} = 2 union {2} = ...
. . .
n+1 = {0, 1, 2, ..., n} = n union {n}
. . .
Here you can then define that a is less then b if and only if a belongs
to b.
With this notation you get the set N of all natural numbers as {0, 1, 2,
...}. But the remarkable thing is that N is exactly the same as
BIGGEST+1. BIGGEST+1 is a set with the same structure as all the other
natural numbers, so it is then a natural number. But BIGGEST+1 is not a
member of N, the set of all natural numbers. BIGGEST+1 is bigger than
all natural numbers, because all natural numbers belongs to BIGGEST+1.
Right, so n+1 is a natural number whenever n is.
What the largest number is depends on how you define natural
number.
One possible definition is that N contains all explicit numbers
expressed by a human being, or will be expressed by a human
being in the
future. Amongst all those explicit numbers there will be one
that is
the largest. But this largest number is not an explicit number.
This raises a deeper question which is this: is mathematics
dependent
on humanity or is mathematics independent of humanity?
I wonder what would happen to that human being who finally expresses
the largest number in the future. What happens to him when he wakes
up the next day and considers adding one to yesterday's number?
This is no problem. If he adds one to the explicit number he
expressed
yesterday, then this new number is an explicit number, and the number
expressed yesterday was not the largest number. Both 17 and 17+1 are
explicit numbers.
This goes back to my earlier comment that it's hard for me to believe
that the following statement is false:
every natural number has a natural number successor
We -must- be talking about different things, then, when we use the
phrase natural number.
I can't say your definition of natural numbers is right and mine is
wrong, or vice versa. I do wonder what advantages there are to the
ultrafinitist approach compared to the math I'm familiar with.
The biggest advantage is that everything is finite, and you can then
really know that the mathematical theory you get is consistent, it does
not contain any contradictions.
From what you said earlier, BIGGEST={0,1,...,BIGGEST-1}. Then
BIGGEST+1={0,1,...,BIGGEST-1} union {BIGGEST} = {0,1,...,BIGGEST}.
Why would {0,1,...BIGGEST} not be a natural number while
{0,1,...,BIGGEST-1} is?
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Re: Cognitive Theoretic Model of the Universe
On 05 Jun 2009, at 14:23, ronaldheld wrote: Bruno: I understand a little better. is there a citition for a version of Church Thesis that all algorithm can be written in FORTRAN? The original Church Thesis, (also due to Post, Turing, Markov, Kleene, and others independently) is this: A function is computable if and only if it is programmable in LAMDA CALCULUS. Then it is an easy but tedious exercise of programing to show that you can simulate LAMDA CALCULUS with FORTRAN, and that you can simulate FORTRAN with LAMBDA CALCULUS. So they compute the same functions. And the same is true with LISP, or JAVA, or ALGOL, or C++, etc... in the place of FORTRAN. A thorough introduction to Church thesis, and I would say one far deeper than usual, is integrally part of the seventh step of UDA. So we will come back on this soon or later. Church thesis is really the key and the motor of both UDA and AUDA. I have discovered that it is rarely well understood, even by many experts. Like Gödel's theorem, Church's thesis is often deformed or misused. 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 everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: The seven step-Mathematical preliminaries
From what you said earlier, BIGGEST={0,1,...,BIGGEST-1}. Then
BIGGEST+1={0,1,...,BIGGEST-1} union {BIGGEST} = {0,1,...,BIGGEST}.
Why would {0,1,...BIGGEST} not be a natural number while
{0,1,...,BIGGEST-1} is?
If {0, 1, ... , BIGGEST-1} is a natural number, then {0,1,...,BIGGEST} is
too, and then so is {0, 1, ... , BIGGEST+1}, etc. There's no such thing as
a largest natural number: that's the whole point of the construction. The
set of all natural numbers is an infinite set, unbounded above. The set N
has no largest element within it: it is the set of all finite ordinals. N
(usually called omega when treated as an ordinal) has no predecessor,
because it is formed by taking the limit of all the ordinals below it, *not*
by applying the successor function x+ = x U {x}. This is the way
well-ordering works...it's not symmetric. So any set described {a, b, ... ,
z} in the standard way is not N.
N is not the successor of any natural number; rather, it contains them all.
This allows us to talk about (and prove things about) all natural numbers.
This isn't an arbitrary mathematical choice. Without infinite sets, we
would be unable to rigorously prove things by induction, which is necessary
for a wide array of basic arithmetical proofs. This is because a finite set
of natural numbers cannot be closed under successor (or addition or
multiplication, for that matter). If you relied on only finitely many
numbers, your functions could take natural numbers and hand you back
something that isn't a number at all. This makes even basic math untenable.
Taking the closure of {} under successor is the solution.
(There are non-standard models of the natural numbers that contain numbers
other than the elements of N, but these are not well-ordered.)
Anna
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Re: The seven step-Mathematical preliminaries 2
Bruno, Thanks for the encouragement. I intend to follow your instructions and it's a relief to know that some of my answers were correct. However, I will be away for two weeks and unable to work on the lessons. I'll try to make up for it when I return. Best, marty a. - Original Message - From: Bruno Marchal To: everything-list@googlegroups.com Sent: Friday, June 05, 2009 10:03 AM Subject: Re: The seven step-Mathematical preliminaries 2 Hi Marty, On 05 Jun 2009, at 00:30, m.a. wrote: Bruno, I don't have dyslexia Good news. but my keyboard doesn't contain either the UNION symbol or the INTERSECTION symbol Nor do mine! (unless I want to go into an INSERT pull down menu every time I use those symbols). Like I have to do too. I don't need you to switch to English symbols, but I would like to see the English equivalents of the symbols you use (so that I can use them). I gave them. I would also like a reference table defining each term in both your symbols and their English equivalents which I could look back to when I get confused. I suggest you do this by yourself. It is a good exercise and it will help you not only in the understanding, but in the memorizing. Then you submit it to the list, and I can verify the understanding. Please include examples. Up to now, I did it for any notions introduced. Just ask me one or two or (name your number) examples more in case you have a doubt. If I send too much posts, and if there are too long, people will dismiss them. try to ask explicit question, like you did, actually. I tend to be somewhat careless when dealing with very fine distinctions This means that a lot of work is awaiting for you. It is normal. Everyone can understand what I explain, but some have more work to do. and may type the wrong symbol while intending to type the correct one. That is unimportant. I am used to do typo errors too. One of my favorite book on self-reference (the one by Smorynski) contains an average of two or three typo error per page. Of course, once a typo error is found, it is better to correct it. Also, I must admit that the lessons are going too fast for me and are moving ahead before I've mastered the previous material. We have all the time, and up to now I did not proceed without having the answer of all exercises. You make no faults in the first set of seven exercise, and that is why I have quickly proceed to the second round. For that one, you make just one error, + the dismiss of a paragraph on UNION. To slow me down it is enough to tell me things like I don't understand what you mean by this or that and you quote the unclear passage. If you can't do an exercise, just wait for some other (Kim?) to propose a solution. Or try to guess one and submit, or just ask. I will not proceed to new matters before I am sure you grasp all what has been already presented. What is possible is that you understand, but fail ti memorize. This will lead to problems later. So you have to make your own summary and be sure you can easily revise the definition. If I'm requesting too much simplification, please let me know because I'm quite well adjusted to my math disabilities and won't take offence at all. Thanks, marty a. I think that there is no problem at all. I am just waiting for explicit question from the second round. You can ask any question, and slow me down as much as you want so that we proceed at your own rhythm. Don't ask me to slow down in any abstract way. You are the one who have to slow me down by pointing on what you don't understand in a post. take it easy, and take all your time. Don't try to understand the more advanced replies I give to people who have a bigger baggage. You did show me that you have understood the notion of set, and the notion of intersection of sets. Have you a problem with the notion of union of sets? If that is the case, just quote the passage of my post that you don't understand, or the example that I gave, and I will explain. Try to keep those post in some well ranged place so as to re-access them easily. I ask this to Kim too, and any one interested: just let me know what you don't understand, so that I can explain, give other examples, etc. Take it easy, you seem quite good, you suffer just of a problem of familiarity with notations. You read the post too quickly, I suspect also. Are you OK? I can understand you could be afraid of the amount of work, but given that we have all the time, there is no exams, nor deadline, I am not sure there is any problem.
