tommy1729 set axioms
amy666
2) axiom of the empty set
3) axiom of pairing
4) axiom of union
5) non-standard elements within the set of real numbers always have properties that correspond to the properties of infinitesimal and unlimited elements.
6) this set theory is game theory in disguise , whatever holds in game theory can be restated in this set theory.
this set theory is complete and consistant apart from solving equations.
this set theory is the analogue of a computer with oo resourses (as should be since math is about computation)
7) axiom of logic : x = [x] ( also if x = empty )
regards
tommy1729
ps : it has been requested that i explicitly gave my axioms , now i did.
|R|^2 = |R| is consistant with it.
Lumpy Hernandez
imbecile, crawling in here and pooping all over
the floor, then crawling away for a few months.
You have nothing to say, and you are saying it.
"amy666" <tomm...@hotmail.com> wrote in message
news:6617782.12159631717...@nitrogen.mathforum.org...
> 1) Fountain of idiocy
>
> 2) Geyser of lies
>
Jesse F. Hughes
> 1) axiom of extensionality
>
> 2) axiom of the empty set
>
> 3) axiom of pairing
>
> 4) axiom of union
>
> 5) non-standard elements within the set of real numbers always have
> properties that correspond to the properties of infinitesimal and
> unlimited elements.
>
> 6) this set theory is game theory in disguise , whatever holds in
> game theory can be restated in this set theory.
(5) and (6) are mighty cute.
> this set theory is complete and consistant apart from solving
> equations.
I don't know what you mean by "consistant apart from solving
equations", but the theory is *not* consistent. See below.
> this set theory is the analogue of a computer with oo resourses (as
> should be since math is about computation)
>
> 7) axiom of logic : x = [x] ( also if x = empty )
Funny axiom, this. You haven't defined what [x] means, but if it is
supposed to mean the singleton containing x (usually written "{x}"),
then the theory is provably inconsistent.
Let's write out the definition for [x] explicitly and you can tell me
if this is what you mean. Here's what I think you mean.
(A x)(A y)(y in [x] <-> y = x).
Let's write out the axiom of empty set, too.
(A y)( NOT y in [] ).
Here's the proof that it is inconsistent.
By the definition of [[]], we have:
(A y)(y in [[]] <-> y = [])
and hence
[] in [[]] <-> [] = [].
Now, since [] = [], we see that [] in [[]]. But by the axiom of empty
set, NOT [] in []. By axiom (7), [] = [[]] and hence NOT [] in [[]].
Thus, we have proved
[] in [[]] AND NOT [] in [[]]
Same contradiction we have discussed repeatedly.
--
Jesse F. Hughes
"How come there's still apes running around loose and there are
humans? Why did some of them decide to evolve and some did not? Did
they choose to stay as a monkey or what?" -Kans. Board of Ed member
David C. Ullrich
wrote:
>1) axiom of extensionality
>
>2) axiom of the empty set
>
>3) axiom of pairing
>
>4) axiom of union
>
>5) non-standard elements within the set of real numbers always have properties that correspond to the properties of infinitesimal and unlimited elements.
>
>6) this set theory is game theory in disguise , whatever holds in game theory can be restated in this set theory.
The idea that you think (5) and (6) are precisely enough stated to
count as "axioms" is remarkable.
>this set theory is complete and consistant apart from solving equations.
Well, Jesse has already pointed out that it's not consistent. And I'm
afraid to ask what you mean by "apart from solving equations".
So I'll just ask this: Exactly what do you mean when you say that
this set theory is "complete"?
Please answer the question, instead of making some crack about
how I don't know what "complete" means. It means many different
things... what do you mean by the word _here_, exactly?
>this set theory is the analogue of a computer with oo resourses (as should be since math is about computation)
>
>7) axiom of logic : x = [x] ( also if x = empty )
>
>regards
>
>tommy1729
>
>ps : it has been requested that i explicitly gave my axioms , now i did.
>
>|R|^2 = |R| is consistant with it.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
hagman
> 1) axiom of extensionality
Presumably meaning
Ax Ay (x=y <-> Az (zex <-> zey))
>
> 2) axiom of the empty set
Presumably meaning
Ex Ay ~(yex)
>
> 3) axiom of pairing
Presumably meaning
Ax Ay Ez At (tez <-> t=x v t=y)
>
> 4) axiom of union
Presumably meaning
Ax Ey Az (zey <-> Et (zet & tex))
>
> 5) non-standard elements within the set of real numbers always have properties that correspond to the properties of infinitesimal and unlimited elements.
>
I didn't know that the set of real numbers contains non-standard
elements.
Do you have any idea of the meaning of any of the terms used
especially in 5)?
> 6) this set theory is game theory in disguise , whatever holds in game theory can be restated in this set theory.
Is this an *attempt* to restate Conway numbers?
>
> this set theory is complete and consistant apart from solving equations.
>
> this set theory is the analogue of a computer with oo resourses (as should be since math is about computation)
>
> 7) axiom of logic : x = [x] ( also if x = empty )
>
> regards
>
> tommy1729
>
> ps : it has been requested that i explicitly gave my axioms , now i did.
>
> |R|^2 = |R| is consistant with it.
... in the sense that ~(|R|^2 = |R|) cannot be proved from the above
axioms?
I just note that
|R|^2 = |R|
is also consistant with the axiom
0) tommy1729 is stupid
and of course it is also consistant with ZFC.
hagman
amy666
> wrote:
> > 1) axiom of extensionality
>
> Presumably meaning
> Ax Ay (x=y <-> Az (zex <-> zey))
> >
> > 2) axiom of the empty set
>
> Presumably meaning
> Ex Ay ~(yex)
>
> >
> > 3) axiom of pairing
>
> Presumably meaning
> Ax Ay Ez At (tez <-> t=x v t=y)
>
> >
> > 4) axiom of union
>
> Presumably meaning
> Ax Ey Az (zey <-> Et (zet & tex))
>
> >
> > 5) non-standard elements within the set of real
> numbers always have properties that correspond to the
> properties of infinitesimal and unlimited elements.
> >
>
> I didn't know that the set of real numbers contains
> non-standard
> elements.
> Do you have any idea of the meaning of any of the
> terms used
> especially in 5)?
yes.
>
>
> > 6) this set theory is game theory in disguise ,
> whatever holds in game theory can be restated in this
> set theory.
>
> Is this an *attempt* to restate Conway numbers?
no.
>
> >
> > this set theory is complete and consistant apart
> from solving equations.
> >
> > this set theory is the analogue of a computer with
> oo resourses (as should be since math is about
> computation)
> >
> > 7) axiom of logic : x = [x] ( also if x = empty )
> >
> > regards
> >
> > tommy1729
> >
> > ps : it has been requested that i explicitly gave
> my axioms , now i did.
> >
> > |R|^2 = |R| is consistant with it.
>
> ... in the sense that ~(|R|^2 = |R|) cannot be proved
> from the above
> axioms?
> I just note that
> |R|^2 = |R|
> is also consistant with the axiom
> 0) tommy1729 is stupid
are we getting personal again hmm ??
i already proved |R|^2 = R many times by now !!
it seems you didnt pay attention in the past :
x axis : 0.10101101010101010101...
y axis : 0.11101111010010101001...
now take the first digit in x and the first in y
gives (00) do the same with second (11) and third (01) etc
then you get (00).(11)(01)(11)...
which is a base 4 number where each digit is expressed in binary.
take away the brackets .. that preserves cardinality ofcourse -> ONE binary number.
= proof |R|^2 = |R|
and i did not use ZF nor AC and not even CH.
just binary representation and coordinate system.
thus -> axiom 0) hagman is stupid.
or should i say hagman is stupid QED ?
( sorry you asked for it dude !! )
> and of course it is also consistant with ZFC.
yes i know.
a set theory that disagrees with all of ZFC would be silly of course.
>
> hagman
>
regards
tommy1729
amy666
> <tomm...@hotmail.com>
> wrote:
>
> >1) axiom of extensionality
> >
> >2) axiom of the empty set
> >
> >3) axiom of pairing
> >
> >4) axiom of union
> >
> >5) non-standard elements within the set of real
> numbers always have properties that correspond to the
> properties of infinitesimal and unlimited elements.
> >
> >6) this set theory is game theory in disguise ,
> whatever holds in game theory can be restated in this
> set theory.
>
> The idea that you think (5) and (6) are precisely
> enough stated to
> count as "axioms" is remarkable.
yes it is , isnt it.
its quite uncommen i admit.
however its a self-reference which only has one unique interpretation together with the other axioms.
so i dont think 5) and 6) could be replaced without loosing its meaning.
i cant blame you for having doubts about it , i once had it myself.
but im sure now.
>
> >this set theory is complete and consistant apart
> from solving equations.
>
> Well, Jesse has already pointed out that it's not
> consistent.
you mean not consistant with ZFC.
it is perfectly self-consistant.
although it may be hard for a non-constructivist to see this.
i think computer scientist will understand better ...
And I'm
> afraid to ask what you mean by "apart from solving
> equations".
> So I'll just ask this: Exactly what do you mean when
> you say that
> this set theory is "complete"?
any statement in set theory consistant with the axioms can be proven or disproven ( not both -> consistant )
with the axioms , apart from questions equivalent to CH, halting, existance and expressibility.
>
> Please answer the question, instead of making some
> crack about
> how I don't know what "complete" means. It means many
> different
> things... what do you mean by the word _here_,
> exactly?
i hope the answer was clear.
btw i assume everyone understands what complete means even JSH :)
it does indeed mean different things in different contexts yes.
so it is hard not to understand it as one of these many interpretations.
>
> >this set theory is the analogue of a computer with
> oo resourses (as should be since math is about
> computation)
> >
> >7) axiom of logic : x = [x] ( also if x = empty )
> >
> >regards
> >
> >tommy1729
> >
> >ps : it has been requested that i explicitly gave
> my axioms , now i did.
> >
> >|R|^2 = |R| is consistant with it.
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
amy666
>
> > 1) axiom of extensionality
> >
> > 2) axiom of the empty set
> >
> > 3) axiom of pairing
> >
> > 4) axiom of union
> >
> > 5) non-standard elements within the set of real
> numbers always have
> > properties that correspond to the properties of
> infinitesimal and
> > unlimited elements.
> >
> > 6) this set theory is game theory in disguise ,
> whatever holds in
> > game theory can be restated in this set theory.
>
> (5) and (6) are mighty cute.
yep.
>
> > this set theory is complete and consistant apart
> from solving
> > equations.
>
> I don't know what you mean by "consistant apart from
> solving
> equations", but the theory is *not* consistent. See
> below.
>
> > this set theory is the analogue of a computer with
> oo resourses (as
> > should be since math is about computation)
> >
> > 7) axiom of logic : x = [x] ( also if x = empty )
>
> Funny axiom, this.
yes very funny.
only a few people on sci.math understand it.
You haven't defined what [x]
> means, but if it is
> supposed to mean the singleton containing x (usually
> written "{x}"),
> then the theory is provably inconsistent.
[x] = is the set that contains x ( only )
however x may be a set itself.
>
> Let's write out the definition for [x] explicitly and
> you can tell me
> if this is what you mean. Here's what I think you
> mean.
>
> (A x)(A y)(y in [x] <-> y = x).
if x is not a set containing other sets or elements than y and if y is not a set containing other sets or elements than x.
>
> Let's write out the axiom of empty set, too.
>
> (A y)( NOT y in [] ).
>
> Here's the proof that it is inconsistent.
>
> By the definition of [[]], we have:
>
> (A y)(y in [[]] <-> y = [])
>
> and hence
>
> [] in [[]] <-> [] = [].
>
> Now, since [] = [], we see that [] in [[]]. But by
> the axiom of empty
> set, NOT [] in []. By axiom (7), [] = [[]] and hence
> NOT [] in [[]].
> Thus, we have proved
>
> [] in [[]] AND NOT [] in [[]]
ok.
NOT [] = []
just like -1 * 0 = +1 * 0
[] does not contain anything thus NOT 5 NOT 4 etc
NOT's denote no element rather than to fill up a set with elements.
9 in binary : 1 NOT 1 NOT 1 1
the name " NOT " should be " NOT yet a member" to avoid problems when taking unions of NOT 5 and 5.
perhaps i should drop the axiom the empty set or clarify :
empty set exists.
i dont support these NOT's , at least not the way you used them ( see below )
but im not sure i have too !! since i can defend much better !!! :
(quote)
> Thus, we have proved
>
> [] in [[]] AND NOT [] in [[]]
what is the union of [] and NOT [] ?
only [] can be the answer.
thus [[]] = [[] U NOT []] => [[],NOT[]] ( 4 U 5 = 4,5 )
both [] and NOT[] in [[]].
which invalidates your disproof !!!
>
> Same contradiction we have discussed repeatedly.
and you still dont understand ... repeatedly ...
> --
> Jesse F. Hughes
> "How come there's still apes running around loose and
> there are
> humans? Why did some of them decide to evolve and
> some did not? Did
> they choose to stay as a monkey or what?" -Kans.
> Board of Ed member
regards
tommy1729
amy666
impressive hmm.
tommy1729
MoeBlee
> i was able to defend my axioms against david c ullrich , jesse f hughes and hagman.
>
> impressive hmm.
No, it's sad that you think you've given clear axioms, let alone that
you think you've defended anything.
Why are you so averse to informing yourself as to some basic
mathematical logic by which you'd come to actually understand
something about the axiomatic method?
MoeBlee
Jesse F. Hughes
>> [] in [[]] AND NOT [] in [[]]
>
> what is the union of [] and NOT [] ?
You're confused about what NOT means here. NOT is the negation that
applies to formulas, i.e., I mean:
[] in [[]] and NOT ([] in [[]]).
NOT [] is meaningless.
--
Jesse F. Hughes
"If the world weren't rather strange, by now I should at least be with
some research group talking about my number theory research."
-- James S. Harris learns the world is a funny place
Jesse F. Hughes
>> Now, since [] = [], we see that [] in [[]]. But by
>> the axiom of empty
>> set, NOT [] in []. By axiom (7), [] = [[]] and hence
>> NOT [] in [[]].
>> Thus, we have proved
>>
>> [] in [[]] AND NOT [] in [[]]
>
> ok.
>
> NOT [] = []
Golly.
--
Jesse F. Hughes
"Philosophy, Socrates, if pursued in moderation and at the proper age,
is an elegant accomplishment, but too much philosophy is the ruin of
human life." -- Callicles, in "Gorgias"
Jesse F. Hughes
> i was able to defend my axioms against david c ullrich , jesse f hughes and hagman.
>
> impressive hmm.
Impressive definition of "defend", yes.
--
Jesse F. Hughes
"But you probably aren't a person with the ability to make any kind of
checks for yourself. But you do talk a lot in posts on Usenet where
you probably live out some fantasy." --James S. Harris is funning, no?
Jesse F. Hughes
> You haven't defined what [x]
>> means, but if it is
>> supposed to mean the singleton containing x (usually
>> written "{x}"),
>> then the theory is provably inconsistent.
>
> [x] = is the set that contains x ( only )
> however x may be a set itself.
>
>
>
>
>>
>> Let's write out the definition for [x] explicitly and
>> you can tell me
>> if this is what you mean. Here's what I think you
>> mean.
>>
>> (A x)(A y)(y in [x] <-> y = x).
>
> if x is not a set containing other sets or elements than y and if y
> is not a set containing other sets or elements than x.
>
You seem to be equivocating on the word "contains". Do you mean "x
contains y" if y is an element of x or "x contains y" if y is a subset
of x?
--
"If I were only dafter
I might be making hymns
To the liquor of your laughter
And the lacquer of your limbs." -- Emanuel Morgan, Opus 6
lwa...@lausd.net
> 1) axiom of extensionality
> 2) axiom of the empty set
> 3) axiom of pairing
> 4) axiom of union
OK, so tommy1729 is trying to restate TST again. Well, I
assume that no one in this thread has any problem with
Axioms 1-4, since these are just axioms of Z.
> 5) non-standard elements within the set of real numbers always have properties that correspond to the properties of infinitesimal and unlimited elements.
It hasn't escaped my notice that there is no Axiom of
Infinity, so I'm wondering how there could be any _standard_
real numbers, much less nonstandard reals.
> 6) this set theory is game theory in disguise , whatever holds in game theory can be restated in this set theory.
Now hagman has already suggested that this refers to
Conway's surreal numbers. I'm wondering how tommy1729
hopes to accomplish this.
We know that Conway constructed his numbers by taking
the single primitive of ZFC, e, and replace it with two new
primitives, L (is to the left of) and R (is to the right of). I'm
wondering how tommy1729 can construct surreal numbers
without the primitives L and R.
And even if tommy1729 does intend on including both
primitives L and R, then how would one rewrite the Z
Axioms 1-4 so that they refer to these primitives?
Axiom 2, Empty Set, most likely would become:
Ex (Ay (~yLx & ~yRx))
and this corresponds to the surreal number zero, {|}.
Axiom 3, Pairing, could be cast in terms of left and right:
Ax (Ay (Ez (At ((tLz <-> t=x) & (tRz <-> t=y)))
But Conway has forbidden such combinations as {0|0},
which is not a surreal number -- though still a game. Then
again, tommy1729 mentioned games, not surreal numbers.
But what about Axiom 1, Extensionality? In Conway, two
surreals may be equal even if they have distinct left- or
right-elements, such as {|} = {-1|1} = 0. So technically
speaking, Conway doesn't obey Extensionality, although
tommy1729 is claiming it as part of his theory.
And without an Axiom of Infinity, we still can't prove the
existence of any infinitesimals or unlimited surreals.
> 7) axiom of logic : x = [x] ( also if x = empty )
Not even Conway's numbers can follow anything like
Axiom 7. One can't put 0 into either to the left or to
the right and expect 0 as a result.
Therefore, taking hagman's suggestion and trying to
use Conway to intepret tommy1729 still leads to problems.
David C. Ullrich
wrote:
>> On Sun, 13 Jul 2008 11:31:55 EDT, amy666
>> <tomm...@hotmail.com>
>> wrote:
>>
>> >1) axiom of extensionality
>> >
>> >2) axiom of the empty set
>> >
>> >3) axiom of pairing
>> >
>> >4) axiom of union
>> >
>> >5) non-standard elements within the set of real
>> numbers always have properties that correspond to the
>> properties of infinitesimal and unlimited elements.
>> >
>> >6) this set theory is game theory in disguise ,
>> whatever holds in game theory can be restated in this
>> set theory.
>>
>> The idea that you think (5) and (6) are precisely
>> enough stated to
>> count as "axioms" is remarkable.
>
>yes it is , isnt it.
>
>its quite uncommen i admit.
>
>however its a self-reference which only has one unique interpretation together with the other axioms.
>
>so i dont think 5) and 6) could be replaced without loosing its meaning.
>
>i cant blame you for having doubts about it , i once had it myself.
>
>but im sure now.
Whether you're sure or not doesn't matter - the "axioms" are
vague enough that they simply cannot count as actual
axioms for a theory. Because nobody else can tell for
certain whether this or that is in fact a correct application
of the axioms.
>>
>> >this set theory is complete and consistant apart
>> from solving equations.
>>
>> Well, Jesse has already pointed out that it's not
>> consistent.
>
>you mean not consistant with ZFC.
No, I mean not consistent.
>it is perfectly self-consistant.
And where's the error in Jesse's proof of inconsistency?
(Also where's your proof that it _is_ consistent?)
>although it may be hard for a non-constructivist to see this.
>
>i think computer scientist will understand better ...
>
> And I'm
>> afraid to ask what you mean by "apart from solving
>> equations".
>> So I'll just ask this: Exactly what do you mean when
>> you say that
>> this set theory is "complete"?
>
>any statement in set theory consistant with the axioms can be proven or disproven ( not both -> consistant )
>with the axioms ,
Ah. And how do you _prove_ that it's complete in this sense?
>apart from questions equivalent to CH, halting, existance and expressibility.
Huh? So you're not saying it's complete after all.
And again, what you're claiming is so vague that it's actually
meaningless.
David C. Ullrich
At least this time you're not trying to explain what he
really means - that's progress. But the stuff about trying
to figure out how he's going to justify various things
shows you haven't been paying attention: he thinks
he can prove things just by wishing they're so.
David C. Ullrich
wrote:
>i was able to defend my axioms against david c ullrich , jesse f hughes and hagman.
That's very funny. You "defense" consisted of simply _stating_ that
there's no problem.
>impressive hmm.
>
>tommy1729
amy666
>
> > amy666 <tomm...@hotmail.com> writes:
> >
> > >> [] in [[]] AND NOT [] in [[]]
> > >
> > > what is the union of [] and NOT [] ?
> >
> > You're confused about what NOT means here. NOT is
> > the negation that
> > applies to formulas, i.e., I mean:
> >
> > [] in [[]] and NOT ([] in [[]]).
> >
> > NOT [] is meaningless.
>
>
> then i will have a second look on what you wrote.
>
it still holds.
[] and NOT [] both in [[]].
no problem.
>
> >
> > --
> > Jesse F. Hughes
> > "If the world weren't rather strange, by now I
> should
> > at least be with
> > some research group talking about my number theory
> > research."
> > -- James S. Harris learns the world
> > learns the world is a funny place
>
> regards
>
> tommy1729
regards
tommy1729
amy666
> wrote:
> > i was able to defend my axioms against david c
> ullrich , jesse f hughes and hagman.
> >
> > impressive hmm.
>
> No, it's sad that you think you've given clear
> axioms, let alone that
> you think you've defended anything.
i dont have the illusion that they are clear at first sight.
in fact the axioms only become a bit clear all together , individually they are vague.
but they are somewhat lets say descriptive , you really need to think about them.
see them as a whole.
>
> Why are you so averse to informing yourself as to
> some basic
> mathematical logic by which you'd come to actually
> understand
> something about the axiomatic method?
because things are not so well defined as commenly believed.
e.g. what is a "set" ?
my set axioms are not intented to work with a specific type of logic , they should work with most types of logic.
there is freedom within my set axioms towards both logic and set theory.
dont confuse that freedom with poorly defined.
just like ZF is independant of AC does not mean ZF is poorly defined.
a similar thing occurs here.
>
> MoeBlee
>
regards
tommy1729
amy666
lol.
But the stuff about
> trying
> to figure out how he's going to justify various
> things
> shows you haven't been paying attention: he thinks
> he can prove things just by wishing they're so.
but they are or at least represent axioms.
you assume provability ?
perhaps you mean self-consistant ...
amy666
> On Jul 13, 8:31 am, amy666 <tommy1...@hotmail.com>
> wrote:
> > 1) axiom of extensionality
> > 2) axiom of the empty set
> > 3) axiom of pairing
> > 4) axiom of union
>
> OK, so tommy1729 is trying to restate TST again.
yes.
> Well, I
> assume that no one in this thread has any problem
> with
> Axioms 1-4, since these are just axioms of Z.
Jesse assumes a problem combining axiom 4 with axiom 7.
>
> > 5) non-standard elements within the set of real
> numbers always have properties that correspond to the
> properties of infinitesimal and unlimited elements.
>
> It hasn't escaped my notice that there is no Axiom of
> Infinity, so I'm wondering how there could be any
> _standard_
> real numbers, much less nonstandard reals.
even ZFC without the axiom of infinity is consistant.
a said in the OP we end up with infinitesimals and limits or similar.
combining this with digits we can make all real numbers.
a real number is just an infinite amount of digits.
TST is in harmony with basic calculus.
>
> > 6) this set theory is game theory in disguise ,
> whatever holds in game theory can be restated in this
> set theory.
>
> Now hagman has already suggested that this refers to
> Conway's surreal numbers. I'm wondering how tommy1729
> hopes to accomplish this.
when hagman asked if i try to accomplish conway's surreals i answered : no.
>
> We know that Conway constructed his numbers by taking
> the single primitive of ZFC, e, and replace it with
> two new
> primitives, L (is to the left of) and R (is to the
> right of). I'm
> wondering how tommy1729 can construct surreal numbers
> without the primitives L and R.
as said : i dont try to construct them.
singularity theory might however answer your question.
indeed BUT that is not the goal of my axioms.
i dont desire proof for them , they are an optional model.
>
> > 7) axiom of logic : x = [x] ( also if x = empty )
>
> Not even Conway's numbers can follow anything like
> Axiom 7. One can't put 0 into either to the left or
> to
> the right and expect 0 as a result.
but im not working with conway numbers.
>
> Therefore, taking hagman's suggestion and trying to
> use Conway to intepret tommy1729 still leads to
> problems.
simple : no conway numbers intended.
regards
tommy1729
hagman
Then please enlighten me:
What is the set of real number in your theory?
Just some complete ordered field?
Do you have an axiom that states the existence of such a set?
How do you define "nonstandard element of R"?
Etc. etc.
But you cannot really do that from the axioms you gave without
a proper defintion of the terms occurring here.
I'm getting problems if I simply use my definitions of terms,
e.g.
|A|=|B| :<=> there exists a bijective map f:A->B
<=> there exists a set F such that all elements of F are
(Kuratowsky) ordered pairs (a,b) with a in A and b in B and
whenever (a,b) in F and (a,b') in F then b=b' and
whenever (a,b) in F and (a',b) in F then b=b' and
for all a in A exists b in B such that (a,b) in F and
for all b in B exists a in A such that (a,b) in F.
I doubt that it can be shown from your axioms that
|R| = |R| holds (let alone |R|^2 = |R|).
I wonder how could you exhibit a bijective map F:R->R?
I don't see that the usual suspect, the identit map,
can be shown to exist from your axioms.
>
> it seems you didnt pay attention in the past :
Probably not.
>
> x axis : 0.10101101010101010101...
> y axis : 0.11101111010010101001...
>
> now take the first digit in x and the first in y
>
> gives (00) do the same with second (11) and third (01) etc
>
> then you get (00).(11)(01)(11)...
>
> which is a base 4 number where each digit is expressed in binary.
>
> take away the brackets .. that preserves cardinality ofcourse -> ONE binary number.
>
> = proof |R|^2 = |R|
This is not based on your axiom system.
This does not exactly show |R|^2 = |R| but rather |2^N|^2 = |2^N|
since you
failed to deal with the 0.11111... = 1.0000... problem.
The missing step is no big deal but should be mentioned.
>
> and i did not use ZF nor AC and not even CH.
You used Continuous Handwaving, but of course the Continuum
Hypothesis
is not needed. You did not use ZF only in the sense that you did not
make your proof very formal.
>
> just binary representation and coordinate system.
>
> thus -> axiom 0) hagman is stupid.
>
> or should i say hagman is stupid QED ?
>
> ( sorry you asked for it dude !! )
Maybe, but the statement "hagman is stupid" seems impossible to
translate into the language of your theory and therefore
cannot be proved from it.
Even worse: Guess what happens if you can prove hagman is stupid
from it and it turns out that in one model (the Real World)
hagman fails to be stupid?
>
> > and of course it is also consistant with ZFC.
>
> yes i know.
>
> a set theory that disagrees with all of ZFC would be silly of course.
Do you want to imply that a set theory that disagress with some part
of ZFC might not be silly?
Then you are right (e.g. think of ZF+~AC or of non-founded theories)
hagman
hagman
> > Jessy wrote :
>
> > > amy666 <tommy1...@hotmail.com> writes:
>
> > > >> [] in [[]] AND NOT [] in [[]]
>
> > > > what is the union of [] and NOT [] ?
>
> > > You're confused about what NOT means here. NOT is
> > > the negation that
> > > applies to formulas, i.e., I mean:
>
> > > [] in [[]] and NOT ([] in [[]]).
>
> > > NOT [] is meaningless.
>
> > oh i see ...
>
> > then i will have a second look on what you wrote.
>
> it still holds.
>
> [] and NOT [] both in [[]].
>
> no problem.
>
?? You still misinterprete NOT.
Your theory gives two contradictory answers to the
question "Is [] an element of [[]]?", namely "yes" and "no" at the
same time.
(Provided our interpretation of the brackets you introduced is right)
"Yes" because in general x is an elment of [x] by definition
"No" because no x can be an element of the empty set [] (=[[]] by
Axiom)
by definition of empty set, especially if x=[].
amy666
sure both yes and no since [] = NOT [].
amy666
read the other replies , i answer this one to lwalke i believe.
evidently Q = Q for all Q !!!
are you saying Q can =/= Q ???
do i need to add as axiom Q = Q for all Q ??
that axiom is not in ZFC either !
tommy1729
Jesse F. Hughes
> lwalke3 wrote :
>
>> On Jul 13, 8:31 am, amy666 <tommy1...@hotmail.com>
>> wrote:
>> > 1) axiom of extensionality
>> > 2) axiom of the empty set
>> > 3) axiom of pairing
>> > 4) axiom of union
>>
>> OK, so tommy1729 is trying to restate TST again.
>
> yes.
>
>
>> Well, I
>> assume that no one in this thread has any problem
>> with
>> Axioms 1-4, since these are just axioms of Z.
>
> Jesse assumes a problem combining axiom 4 with axiom 7.
^^^^^^^^^^^^^^^^^
I think you meant "demonstrated the obvious contradiction". But I did
it with (2) and (7).
Really, I don't need credit for this. The fact that (2) and (7) are
inconsistent is plain to see. Also, (4) and (7) are problematic.
Using (4) and (7), we can prove (Ax)(Ay)(x = y). Probably, that's not
what you want with a set theory.
--
Conservative, n:
A statesman who is enamored of existing evils, as distinguished
from the Liberal who wishes to replace them with others.
-- Ambrose Bierce
MoeBlee
> > On Jul 14, 1:46 pm, amy666 <tommy1...@hotmail.com>
> > wrote:
> > > i was able to defend my axioms against david c
> > ullrich , jesse f hughes and hagman.
>
> > > impressive hmm.
>
> > No, it's sad that you think you've given clear
> > axioms, let alone that
> > you think you've defended anything.
>
> i dont have the illusion that they are clear at first sight.
Except for the ones that restate known axioms of set theory, they're
far from clear in any sight.
> in fact the axioms only become a bit clear all together , individually they are vague.
Please read up just a little to find out how modern axiomatic systems
work.
> but they are somewhat lets say descriptive , you really need to think about them.
>
> see them as a whole.
You need to learn how modern axiomatic systems work. I don't need to
think about your slop; I already have a reading list of a lifetime of
mathematics to study, including alternatives to standard mathematics.
Such mathematics is informed, intellectually organized, and well
written. That's what you need to learn.
> > Why are you so averse to informing yourself as to
> > some basic
> > mathematical logic by which you'd come to actually
> > understand
> > something about the axiomatic method?
>
> because things are not so well defined as commenly believed.
However well defined things are or are not, you only deprive yourself
by not studying the basics of the subject.
> e.g. what is a "set" ?
(1) If we had to, we can do all of set theory without ever mentioning
the word 'set'. The primitive of set theory is 'e' (read "element
of"). That (and '=', read "is equal to") are not defined. All other
predicate and function symbols are defined. (2) Anyway, in a theory
such as Z set theory, a technical definition of 'is a set' can be
provided:
First, we prove that there exists exactly one object that has no
members:
E!xAy ~yex
Then we define:
0=x <-> Ay ~yex
Then we define:
x is an element <-> Ey xey
C is a class <-> (Ey yec v c=0)
S is a set <-> (S is a class & Ey Sey)
C is a proper class <-> (C is a class & ~C is a set)
u is an urelement <-> ~u is a class
> my set axioms are not intented to work with a specific type of logic , they should work with most types of logic.
You don't know what you're talking about. And even if you leave the
logic unspecified, you still have to have a language for the theory.
> there is freedom within my set axioms towards both logic and
> set theory.
>
> dont confuse that freedom with poorly defined.
I don't. And to call yours "poorly defined" would be to give it more
credit than it deserves.
> just like ZF is independant of AC does not mean ZF is poorly
> defined.
>
> a similar thing occurs here.
No, I don't conflate independence with definition. Your slop is just
not defined. Has nothing to do with independence, not even by analogy.
Grow up. Start by reading a good textbook on predicate logic and then
one on set theory and then one on mathematical logic.
MoeBlee
galathaea
i'm surprised that conway was the first thought
re: tommy's mention of game theoretical semantics
i just expected hintikka would get the props here...
nobody here love jaakko? :(
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
MoeBlee
> Except for the ones that restate known axioms of set theory,
> they're
> far from clear in any sight.
Correction: The known axioms 1-4 are clear, and 7 is clear when I find
out that you use '[]' for '{}'.
And it's trivial that 7 and 2 together are inconsistent with basic
identity theory.
MoeBlee
David C. Ullrich
wrote:
Actually they don't - you wish they did, but
they're much too vague.
In any case, I didn't anything about your proving
the axioms. Your proofs of consistency and
completeness are just wishes.
David C. Ullrich
wrote:
No. If there were such a thing as "NOT []"
then there'd be no problem with both []
and "NOT []" being an element of the same
set. At least no purely logical problem.
But nobody but you has said anything about
this silly "NOT []" thing. People have pointed
out that your theory proves that "[] is an elemeent
of [[]]" is true and it _also_ proves that the very
same statement is false.
How you choose to define "NOT []" is
irrelevant to that. Your theory proves that
a certain statement is true and it also proves
the same statement is false. That means the
theory is inconsistent, by definition.
amy666
you have a point there...
but in this case it seems false and true are actually the same.
its like 5 + 0 contains 0 although it equals 5 + 0 - 0.
but by formal definition you and jesse have a point.
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
amy666
i very much doubt that !!
axiom 4 is basicly: a union of a set is a set.
[y] U [x] = [y,x]
y U x = y,x
then we still get [y,x] = y,x
this reminds me of the original post that started all this TST , which wasnt posted by me ! , but someone who challenged ZFC and the posters here on sci.math to resolve his paradox ...
only i could give a consistant answer.
but with x = [x] instead of ZF(C).
i forgot the title though.
about a year ago that was.
>
> --
> Conservative, n:
> A statesman who is enamored of existing evils, as
> s distinguished
> from the Liberal who wishes to replace them with
> h others.
> -- Ambrose Bierce
tommy1729
amy666
thus a change that will look better i assume :
1) axiom of extensionality
2) axiom of the empty set
3) axiom of pairing
4) axiom of union
5) non-standard elements within the set of real
numbers always have properties that correspond to the
properties of infinitesimal and unlimited elements.
6) axiom of regularity
7) axiom of infinity
8) axiom of logic : x = [x] ( also if x = empty )
regards
tommy1729
Jesse F. Hughes
Too bad. Here is the trivial proof.
Axiom (4) says the following, right?
(Ax)(Ay)(Ez)(Aw)(w in z <-> (w = x or w = y))
And axiom (7) says the following, right?
(Ax)(Ay)(y in x <-> y = x)
Let x and y be given and let z be the pair {x,y}, i.e.,
(Aw)(w in z <-> w = x or w = y).
In particular, notice that x in z and y in z.
Now, by axiom z, we see that
(Aw)(w in z <-> w = z).
In particular, x in z -> x = z and y in z -> y = z. Thus, since x in
z and y in z, we see that x = z and y = z.
Consequently, x = y.
As I said, axioms (4) and (7) prove (Ax)(Ay)(x = y).
--
"Even if a man has good parts, still, if he carries philosophy into
later life, he is necessarily ignorant of all those things which a
gentleman and a person of honor ought to know."
--Callicles, in "Gorgias"
Jesse F. Hughes
> you have a point there...
>
> but in this case it seems false and true are actually the same.
>
> its like 5 + 0 contains 0 although it equals 5 + 0 - 0.
>
> but by formal definition you and jesse have a point.
Oh, well that's all right then. After all, if false and true happen
to be the same, everything's just fine.
--
"I'd step through arguments in such detail that it was like I was
teaching basic arithmetic and some poster would come back and act like
I hadn't said anything that made sense. For a while I almost started
to doubt myself." -- James S. Harris, so close and yet....
MoeBlee
> 1) axiom of extensionality
>
> 2) axiom of the empty set
>
> 3) axiom of pairing
>
> 4) axiom of union
>
> 5) non-standard elements within the set of real
> numbers always have properties that correspond to the
> properties of infinitesimal and unlimited elements.
(a) Without some form of comprehenion and a power set axiom, it's not
known how you define 'set of real numbers'. (b) A first order
axiomatization does not refer to 'properties' except by a schema
quantifying not over properties but rather formulas. (c) 'non-
standard', 'correspond to', 'infinitesimal' and 'unlimited' are
undefined, especially in context where there is no form of
comprehension and no power set axiom to use to define 'the set of real
numbers'.
But even stronger, your axiom 8 entails that every set has exactly one
member, so there is no 'set of real numbers' or anything like that.
> 6) axiom of regularity
Without any form of comprehension axiom to prove the existence of
intersections, I suppose 6 would be stated:
~x=0 -> EmexAy~(yem & yex)
> 7) axiom of infinity
>
> 8) axiom of logic : x = [x] ( also if x = empty )
I don't know why you call that 'axiom of logic'.
Your system is inconsistent using just axioms 2, 3, 8, and basic
identity theory:
From axiom 2 we have a set S that has no members.
From axiom 3, there is the set {S} such that S is a member of {S}.
From axiom 8, S={S}.
So S is a member of S, which contradicts that S has no members.
And axiom 8 is inconsistent with and axioms 3 and 6:
Let x={x}. So Az(Ay~yez -> ~x=z) (i.e., even without extensionaliy, we
know that x is not an empty set since it has a member, viz. x). So for
all m in x, there is no set that is in both m and in x. But the only
member of x is x, so m=x. So x in m and in x, contradicting that there
is no set that is in m and in x.
Also, axiom 8 entails that that every set has exactly one member. So
your axiom of infinity fizzles and there is no 'set of real numbers'
or any of those kinds of things.
Clearly, you have no idea what you're doing. I really don't understand
why you don't get some good textbooks that would teach you how to set
up your own axiom system.
MoeBlee
MoeBlee
P.S. Jesse's proof is correct that there is exactly one object in your
theory. Though, since your theory is inconsistent we can also prove
ANYHING in the language of your theory anyway.
So you have an inconsistent theory that also says - just on the face
of it, before we've even looked for inconsistency - that every set has
exactly one member and there is only one set that exists, and there is
a unique empty set, so there is only one set, which is the unique
empty set that has no members but has exactly one member, which is
itself, which is a contradiction...and yet you think the set of real
numbers and all kinds of other UNDEFINED stuff happens in your theory.
Well, since your theory is inconsistent, yeah, you do get that there
exists a set of real numbers, and also that there does NOT exist a set
of real numbers.
You're pathetic. Grow up. Get some good books.
MoeBlee
galathaea
considering tommy has regularly mentioned mereology
(after i pointed out to him that [some of] his ideas were consistent
with mereologies used in some computer science data structures)
i find it strange that you think "basic identity theory"
is even applicable here
the whole point seems to be to introduce a different-but-consistent
notion
of the relationship between parts and their whole
basically unnested collection
it is certainly possible to do that
i'm not arguing any of tommy's other "points"
or against some of the objections on cardinalities represented
(except to note that axioms not present doesn't mean a model can't
obey them)
but i find it strange that this particular issue is the one several
have focused on
(excuse if double-posted due to lame gg-errors in posting)
MoeBlee
> considering tommy has regularly mentioned mereology
> (after i pointed out to him that [some of] his ideas were consistent
> with mereologies used in some computer science data structures)
> i find it strange that you think "basic identity theory"
> is even applicable here
Oh come on, if he's got some other axioms or definiition for '=', then
he needs to state them, rather than let the reader guess that the
ordinary idenity axioms are eschewed and some other merelogical axioms
(that might or might have been mentioned in some thread somewhere or
another) apply instead.
If he wants to give an entirely new slant on anything at all, that's
fine with me; but it is unreasonable not to expect that unless
otherwise stated, the reader will regard '=' as being handled in the
ordinary manner.
Moreover, I greatly doubt that the poster of this thread actually DOES
eschew (let alone systematically eschews) ordinary identity principles
in his own mathematical reasonings (such as they are).
MoeBlee
lwa...@lausd.net
Actually, hagman was the first to mention
Conway for game theory. I decided to
add on to what hagman mentioned, since
tommy1729 mentioned infinitesimals and
unlimited numbers, which are included in
Conway's surreal numbers.
lwa...@lausd.net
> On Jul 16, 7:46 am, amy666 <tommy1...@hotmail.com> wrote:
> > 5) non-standard elements within the set of real
> > numbers always have properties that correspond to the
> > properties of infinitesimal and unlimited elements.
> (a) Without some form of comprehenion and a power set axiom, it's not known how you define 'set of real numbers'.
Actually, it _is_ possible to define the real
numbers without Powerset. Indeed, Randall
Holmes came up with a set theory, called
Pocket Set Theory, that does exactly that.
lwa...@lausd.net
> > On Tue, 15 Jul 2008 18:41:19 EDT, amy666
> > irrelevant to that. Your theory proves that
> > a certain statement is true and it also proves
> > the same statement is false. That means the
> > theory is inconsistent, by definition.
> you have a point there...
> but in this case it seems false and true are actually the same.
> its like 5 + 0 contains 0 although it equals 5 + 0 - 0.
It appears that there is some confusion
regarding the use of the word "NOT."
Here tommy1729 seems to feel that the
"NOT" is a unary operator similar to the
negation operator in Z (where Z denotes
the ring of integers). This is evident
from his example of how 0 = -0.
But the others are using "NOT" as a
logical operator, as in NOT P (or ~P),
the negation of a property P.
I suppose that there is nothing wrong
with defining a NOT operator on sets,
so that if x is a set, then NOT(x) is
also a set. Indeed, Han de Bruijn does
exactly that in his own set theory
(introduced on sci.math months ago),
where NOT(x) is actually the set
complement of x (relative to some
universal set U). Indeed, perhaps
in tommy1729's set theory, we could
have NOT([]) = [].
The confusion appears to be that even
though there can exist an integer n
such that n = -n, there _can't_ exist
a property P such that P <-> ~P -- at
least, not in a consistent theory. So,
as much as I like to defend tommy1729
in his pursuit of a new set theory,
his line:
> but in this case it seems false and true are actually the same.
is indefensible. "False" and "true"
can only be the same in an inconsistent
theory, and there is no comparison
between "0 = -0" and "P <-> ~P."
Of course, the reason for this subthread
is that tommy1729 wants the axiom
"Ax (x = [x])," which is incompatible
with set theory but, as galathaea points
out, could make sense in a mereology
theory similar to the ones zuhair posted
at sci.logic earlier this year.
MoeBlee
> (except to note that axioms not present doesn't mean a model can't
> obey them)
If I correctly understand what you mean, then I agree. I don't know
why you would think I wouldn't agree on point.
MoeBlee
MoeBlee
Interesting, but Tommy's system lacks both any form of comprehension
and power set, so I don't know how he thinks he can get a set of real
numbers out that.
MoeBlee
Aatu Koskensilta
In Holmes's theory there is no set of real numbers. Individual reals
-- e.g. in the form of Dedekind cuts -- are sets, and we can, with
some care, reformulate most of analysis in the theory, the same way we
do in e.g. ACA_0. In effect, the collection of the reals is treated as
a proper class.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
David C. Ullrich
<14159848.1216215907...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
That's correct.
> but in this case it seems false and true are actually the same.
Jesus. You're making even less sense than usual today.
> its like 5 + 0 contains 0 although it equals 5 + 0 - 0.
5+0 does not "contain" 0.
> but by formal definition you and jesse have a point.
In particular, your assertion that the system is
consistent is simply wrong. Which means it's not right,
by the way - right and wrong are not the same.
>
> >
> > David C. Ullrich
> >
> > "Understanding Godel isn't about following his formal
> > proof.
> > That would make a mockery of everything Godel was up
> > to."
> > (John Jones, "My talk about Godel to the post-grads."
> > in sci.logic.)
>
> regards
>
> tommy1729
--
David C. Ullrich
David C. Ullrich
<9a4e7c12-e9dd-4ec1...@x35g2000hsb.googlegroups.com>,
lwa...@lausd.net wrote:
Erm, that's somewhat indefensible as well.
False and true are not the same in an inconsistent
theory. (Actually saying they're the same or not
the same "in a certain theory" makes vert little
sense. Anyway, let T be the collection of all
sentences in some (standard) formal language.
Then T is an inconsistent theory - false and true
are not the same, "in T" or otherwise.)
>and there is no comparison
> between "0 = -0" and "P <-> ~P."
>
> Of course, the reason for this subthread
> is that tommy1729 wants the axiom
> "Ax (x = [x])," which is incompatible
> with set theory but, as galathaea points
> out, could make sense in a mereology
> theory similar to the ones zuhair posted
> at sci.logic earlier this year.
--
David C. Ullrich
MoeBlee
wrote:
> MoeBlee <jazzm...@hotmail.com> writes:
> > On Jul 16, 4:35 pm, lwal...@lausd.net wrote:
>
> > > Actually, it _is_ possible to define the real
> > > numbers without Powerset. Indeed, Randall
> > > Holmes came up with a set theory, called
> > > Pocket Set Theory, that does exactly that.
>
> > >http://math.boisestate.edu/~holmes/holmes/pocket.txt
>
> > Interesting, but Tommy's system lacks both any form of comprehension
> > and power set, so I don't know how he thinks he can get a set of real
> > numbers out that.
>
> In Holmes's theory there is no set of real numbers. Individual reals
> -- e.g. in the form of Dedekind cuts -- are sets, and we can, with
> some care, reformulate most of analysis in the theory, the same way we
> do in e.g. ACA_0. In effect, the collection of the reals is treated as
> a proper class.
Thanks for that (though, I think the point is even better addressed to
lwal). Moreover, I noticed that that theory has a quite powerful
impredicative-over-proper-classes comprehension schema; so it's hardly
akin to Tommy's fantasy in which he has no comprehension (not even
separation or replacement, etc.) axioms at all, as well as no power
set axiom.
MoeBlee
amy666
> On Jul 16, 7:46 am, amy666 <tommy1...@hotmail.com>
> wrote:
>
> > 1) axiom of extensionality
> >
> > 2) axiom of the empty set
> >
> > 3) axiom of pairing
> >
> > 4) axiom of union
> >
> > 5) non-standard elements within the set of real
> > numbers always have properties that correspond to
>
(snip)
>
> But even stronger, your axiom 8 entails that every
> set has exactly one
> member, so there is no 'set of real numbers' or
> anything like that.
NO !!
you got it all wrong :
its not true that a set can have only element.
and not only Jack Markan but most here got it wrong and deduce other wrong things based on that misconception.
lets repeat axiom 8 : x = [x]
x may also be empty or denote a set itself.
thus we have
= [] = [[]]
and
x = [x] = [[x]]
now lets consider [x] where this set does contain two elements.
x = a,b
then we get [x] = x = a,b = [a,b]
and [a,b] clearly has 2 elements : a and b.
you see ? perfectly consistant !!
here you repeat yourself again !
you still dont understand that axiom 8 still allows sets with more than one member.
and thus everything based on that misconception is a misconception.
since you repeat your arguments in a single post , so will i :
you got it all wrong :
its not true that a set can have only element.
and not only Jack Markan but most here got it wrong and deduce other wrong things based on that misconception.
lets repeat axiom 8 : x = [x]
x may also be empty or denote a set itself.
thus we have
= [] = [[]]
and
x = [x] = [[x]]
now lets consider [x] where this set does contain two elements.
x = a,b
then we get [x] = x = a,b = [a,b]
and [a,b] clearly has 2 elements : a and b.
you see ? perfectly consistant !!
So
> your axiom of infinity fizzles and there is no 'set
> of real numbers'
> or any of those kinds of things.
>
> Clearly, you have no idea what you're doing. I really
> don't understand
> why you don't get some good textbooks that would
> teach you how to set
> up your own axiom system.
clearly , you have no idea what axiom 8 is about.
why i dont get a textbook that teaches me to set up my own axiom system ?
well , simply because there no such textbook , its all ZFC or " worse ".
there is no textbook containing axiom 8 , nor any real decent NEW axiom system.
and in fact we could do set theory from those books without even defining what a set is.
however this is considered quite important by me !!
axiom 8 explains what a set and what an element REALLY is.
so in a sence my set axioms are more fundamental then any book.
axiom 5 and axiom 7 extends to infinity limits reals and calculus.
axiom 2 includes the existance of the empty set.
axiom 4 allows a logical union operator to exist.
and with the remaining axioms we make a consistant whole.
( maybe axiom 9 ~[] = [] for clarity )
>
> MoeBlee
>
>
>
regards
tommy1729
amy666
a similar thing applies to my set axioms
regards
tommy1729
MoeBlee
Are you DEAF? I already belabored that your set axioms have no
comprehension schema AT ALL in contrast to Holmes's theory that has
just about as strong a comprehension schema as you could have short of
inconsistency.
What, you just look at someone elses theory and declare, "And my
theory has that too!", when your theory does not having anything even
like it.
A New Yorker magazine style cartoon about you would be something like
this:
Two down and out, ragged bums are slouched against a wall. A Rolls
Royce driven by a chauffer pulls up to the curb and a rich, handsome,
debonair guy in tuxedo with a beautiful blonde on his elbow gets out
of the backseat of the car. The one bum says, "See, that guy has all
that, so I have it too!"
REALLY, why don't you just get good book so that you could learn how
axiom systems actually work?
MoeBlee
MoeBlee
> Jack Markan wrote :
Who is Jack Markan? Maybe the name is a pseudonym that I once used in
some signup somewhere? In any case, I don't know why you're getting it
in your messages by me. I don't recall anyone ever responding to my
messages as if my message were written under that name. For sake of
keeping straight who wrote what, my messages should display as written
by 'MoeBlee' or by my email address or truncation of my email
address.
> > But even stronger, your axiom 8 entails that every
> > set has exactly one
> > member, so there is no 'set of real numbers' or
> > anything like that.
>
> NO !!
>
> you got it all wrong :
>
> its not true that a set can have only element.
Please state the EXACT point in my proof that you THINK is wrong.
> lets repeat axiom 8 : x = [x]
>
> x may also be empty or denote a set itself.
>
> thus we have
>
> = [] = [[]]
When you say "x may be empty", that will be taken as "if x=0", not as
'x' reprented by a blank character.
You've not provided for any notation that is a BLANK character. Please
define such notation from your primitives or take it as primitive and
give axioms for it.
> and
> x = [x] = [[x]]
That's fine, except that if it holds for ALL x, you get an
inconsistent theory with the rest of your axioms, as I PROVED and as
you IGNORED even commenting on the actual content of my proof.
> now lets consider [x] where this set does contain two elements.
>
> x = a,b
The notation 'a, b' is undefined. Please eaither define it from the
primitives in your theory, or add it (the comma) as a new primitive
operation symbol with axioms for it. Otherwise, no one can say what
just 'a,b' is supposed to mean as a term onto itself.
> > Your system is inconsistent using just axioms 2, 3,
> > 8, and basic
> > identity theory:
>
> > From axiom 2 we have a set S that has no members.
> > From axiom 3, there is the set {S} such that S is a
> > member of {S}.
> > From axiom 8, S={S}.
> > So S is a member of S, which contradicts that S has
> > no members.
>
> > And axiom 8 is inconsistent with and axioms 3 and 6:
>
> > Let x={x}. So Az(Ay~yez -> ~x=z) (i.e., even without
> > extensionaliy, we
> > know that x is not an empty set since it has a
> > member, viz. x). So for
> > all m in x, there is no set that is in both m and in
> > x. But the only
> > member of x is x, so m=x. So x in m and in x,
> > contradicting that there
> > is no set that is in m and in x.
>
> > Also, axiom 8 entails that that every set has exactly
> > one member.
>
> here you repeat yourself again !
What did I repeat that you had already made clear you understand?
And it is clear that you didn't even read those proofs.
> you still dont understand that axiom 8 still allows sets with
> more than one member.
No, YOU don't understand what a mathematical argument is.
> and thus everything based on that misconception is a
> misconception.
No, please state the EXACT step in my proofs that you THINK is
incorrect.
> since you repeat your arguments in a single post , so will i :
Childish.
And you didn't even read my proofs, let alone find anything repeated
that you had already recognized as understood by you.
> why i dont get a textbook that teaches me to set up my own
> axiom system ?
>
> well , simply because there no such textbook , its all ZFC or "
> worse ".
There are textbooks in mathematical logic that would inform you as to
how axiom systems work.
> there is no textbook containing axiom 8 , nor any real decent
> NEW axiom system.
A textbook doesn't have to contain a SPECIFIC axiom to teach you about
the axiomatic method! My god, you're silly.
MoeBlee
David C. Ullrich
wrote:
>update : i discovered axiom 6 could be restated in 2 others without violation of the other ones.
>
>thus a change that will look better i assume :
>
>
>
>1) axiom of extensionality
>
>2) axiom of the empty set
>
>3) axiom of pairing
>
>4) axiom of union
>
>5) non-standard elements within the set of real
>numbers always have properties that correspond to the
>properties of infinitesimal and unlimited elements.
This was much too vague to count as making sense the first time,
and it still is the second time.
>6) axiom of regularity
>
>7) axiom of infinity
>
>8) axiom of logic : x = [x] ( also if x = empty )
>
>regards
>
>tommy1729
>
>|R|^2 = |R| is consistant with it.
Right. This time are you going to show us your
proof that TST is consistent?
Giggle.
David C. Ullrich
wrote:
I don't know if you ever noticed, but when you
use words that you don't understand you almost
always make a fool of yourself.
Showing that your theory proves that there exists
a set with two elements does not show the theory
is consistent. In fact, since there is also a proof in
you theory that every set has exactly one element
what you have proved here is that your theory
is _inconsistent_.
You seem to think that the definition of "consistent"
is "I can use it to prove the things I want to be true".
That's not the definition.
David C. Ullrich
MoeBlee
Here's a P.S. to my previous post about this:
> lets repeat axiom 8 : x = [x]
>
> x may also be empty or denote a set itself.
>
> thus we have
>
> = [] = [[]]
> and
> x = [x] = [[x]]
>
> now lets consider [x] where this set does contain two elements.
>
> x = a,b
>
> then we get [x] = x = a,b = [a,b]
>
> and [a,b] clearly has 2 elements : a and b.
As I mentioned, 'a,b' is undefined. And also aside from the fact that
your theory is inconsistent, in your argument above you assumed a is
not equal to b. But you have no prior proof that there do exist at
least two different objects. So your argument commits the fallacy of
question begging. You REALLY do need to learn some basic logic.
MoeBlee
galathaea
eventually tommy will discover ur elements and names
and they will probably start to be mentioned more in TST
again
i stress
just because a theory does not have an axiom
does not mean a model will not obey them
there is a huge historical record of just this type of math theorising
like cantor's original work
or zermelo's first stabs
or ...
where theories don't have everything at first
not explicitly
and i watch tommy here
and you call him pathetic
and maybe he does deserve some pathos
most people do at some point
but i watch him
and i have every expectation that he will continue posting
and continue getting ideas about what works and what doesn't
and he's laid out his plan of what he wants in TST
so eventually
after enough rounds
he will have his TST
whether it will be new or not will be seen
but i see nothing that would prevent him from getting there
and even though many will have helped him along the way
it will be his vision that guides the whole thing
and it will be uniquely his result
and then i look at you
moeblee
and i wonder when you will get an MST
because i don't see you trying yet
despite your superior literature cites
maybe that's not your desire
maybe like a concert pianist
you are satisfied playing back the masters
but if i were looking for a model
of how to become an original mathematician
i would be looking at tommy right now over you
despite the uncomfortable errors
and hideous expectations of impending doom
because it is clear that tommy's vocabulary is growing
and he does apparently "listen" intently
for indications of where he is right
and generally
despite the bluster and bravado
he doesn't seem to hold on to the things that are inherently wrong
i'm just telling you this
moeblee
so you understand that your taunts of patheticness
appear to you as they appear to me
that they come from someone still too socially constrained
to be making the public mistakes they taunt others for
sure
bravado and bluster need their own comeuppance
but you have not been showing much competence in your attacks
and you have been far more aggressive
than your own lack of results would seem to warrant
so maybe both you and tommy can read up
on holmes arguments for ur elements being more natural foundations
and when the next update to TST comes around
maybe the objections to it can come from someone
who has their own collection of MSTs as exemplars
amy666
> On Jul 17, 11:33 am, amy666 <tommy1...@hotmail.com>
> wrote:
> > Jack Markan wrote :
>
> Who is Jack Markan? Maybe the name is a pseudonym
> that I once used in
> some signup somewhere? In any case, I don't know why
> you're getting it
> in your messages by me. I don't recall anyone ever
> responding to my
> messages as if my message were written under that
> name. For sake of
> keeping straight who wrote what, my messages should
> display as written
> by 'MoeBlee' or by my email address or truncation of
> my email
> address.
>
well jack markan ,
jack markan is your own name , genius !!! :p
at least here on sci.math.
and it is listed with every post you make.
tomorrow tommy1729 will ask who amy666 is :)
> > > But even stronger, your axiom 8 entails that
> every
> > > set has exactly one
> > > member, so there is no 'set of real numbers' or
> > > anything like that.
> >
> > NO !!
> >
> > you got it all wrong :
> >
> > its not true that a set can have only element.
>
> Please state the EXACT point in my proof that you
> THINK is wrong.
>
> > lets repeat axiom 8 : x = [x]
> >
> > x may also be empty or denote a set itself.
> >
> > thus we have
> >
> > = [] = [[]]
>
> When you say "x may be empty", that will be taken as
> "if x=0", not as
> 'x' reprented by a blank character.
thats strange ? since when does empty mean 0 ?
the empty set is not the set [0] ?
so whats with this empty means x = 0 ?
maybe its this way of representing integers by sets , a thing i dont like and dont support.
( and is not possible since x = [x] )
>
> You've not provided for any notation that is a BLANK
> character. Please
> define such notation from your primitives or take it
> as primitive and
> give axioms for it.
isnt the axiom of the empty set sufficient for that ??
>
> > and
> > x = [x] = [[x]]
>
> That's fine, except that if it holds for ALL x, you
> get an
> inconsistent theory with the rest of your axioms, as
> I PROVED and as
> you IGNORED even commenting on the actual content of
> my proof.
give a specific example x where this gives inconsistancy with the other axioms.
>
> > now lets consider [x] where this set does contain
> two elements.
> >
> > x = a,b
>
> The notation 'a, b' is undefined. Please eaither
> define it from the
> primitives in your theory, or add it (the comma) as a
> new primitive
> operation symbol with axioms for it. Otherwise, no
> one can say what
> just 'a,b' is supposed to mean as a term onto itself.
so a,b is undefined ?
a,b are just two elements.
like the solution of a polynomial (x-5)(x-3)=0 has solution 5,3. or [5,3]
ZFC has only 9 axioms , i can imagine i a bit need more , but either does a,b really need an axiom ?
and if so , what axiom in ZFC explains "a,b" ???
an axiom that is not in my set axiom ?
regards
tommy1729
MoeBlee
> again
> i stress
> just because a theory does not have an axiom
> does not mean a model will not obey them
So what? Who said otherwise?
> there is a huge historical record of just this type of math theorising
> like cantor's original work
> or zermelo's first stabs
> or ...
> where theories don't have everything at first
>
> not explicitly
Oh please. Yes, there is value in informal and imperfect attempts. But
that doesn't entail that just any pile of clueless slop, such as
Tommy's, is the seed of value.
> and then i look at you
> moeblee
> and i wonder when you will get an MST
>
> because i don't see you trying
Exactly. It's not my goal.
> yet despite your superior literature cites
I'm not interested in making an impression for myself by citing
literature. I cite it for the sake of its value as a rescource.
> maybe that's not your desire
> maybe like a concert pianist
> you are satisfied playing back the masters
A mere hobbyist such as me, who has never been in his life what you
would call "a math person", would be THRILLED even to attain that
analogous level of competence.
> but if i were looking for a model
> of how to become an original mathematician
> i would be looking at tommy right now over you
You'd be foolish to look at EITHER of us. He lacks the motivation to
properly learn the subject though he may fancy being an innovator, and
I lack the motivation to be an innovator though I do have the
motivation to properly learn the subject.
> i'm just telling you this
> moeblee
> so you understand that your taunts of patheticness
> appear to you as they appear to me
>
> that they come from someone still too socially constrained
> to be making the public mistakes they taunt others for
>
> sure
What are you talking about? I don't shy from posting things that may
expose that I've made mistakes or even have basic misunderstandings.
On the contrary, I post always with the hope that any substantive
error of mine be pointed out to me. Moreover, sometimes I do post
philosophical (or "meta-explanatory", for lack of a better word I can
think of right now) notions I have, some of which may be at least a
little bit novel or at least take some extra measure from a
conventional idea; and I welcome reasoned criticism when I do go out
on a bit of limb like that.
> bravado and bluster need their own comeuppance
> but you have not been showing much competence in your attacks
You've shown nothing incompetent in my remarks. By the way, here's
another one: Since his system has no comprehension axiom, if we set
aside the source of contradiction in his system, his axiom of infinity
not only does not ensure a set of natural numbers, since there is only
one object, but even if there WERE more than one object, without a
comprehension axiom, he has no way of cutting the inductive set down
to one that excludes members that are not natural numbers.
> so maybe both you and tommy can read up
> on holmes arguments for ur elements being more natural foundations
I wish to read up on TONS of stuff, ranging from the most straight
ahead standard Z set theory to dialetheism, from non-monotonic logic
to intuitonistic type theory, all kinds of stuff. The paper by Holmes
is just one of hundreds and hundreds I wish to study eventually.
> and when the next update to TST comes around
> maybe the objections to it can come from someone
> who has their own collection of MSTs as exemplars
That's silly. One doesn't have to be an innovator oneself in this
context. For examplars of alternative mathematics there is all kinds
of stuff published. But, oops, Tommy is in no position to appreciate
it since he lacks the SINCERITY of motivation to properly learn the
material one needs to know to read those innovations.
MoeBlee
MoeBlee
> jack markan is your own name , genius !!! :p
No, it's not, idiot. I TOLD YOU, it's a name I registered a long time
ago. And I used it only very briefly. It is not the name for the
account under which I am posting now. Of course, I can't control that
some system may still be putting that name on my posts; but I don't
recall anyone other than you posting back to that name since I
abandoned it long ago, so I wonder whether you're posting from a news
interface that is errantly putting that name on my posts.
I think I might have used the name at Math Forum. Are you posting
through that interface? It might be that Math Forum is attaching my
old name to posts made from my current account made away from Math
Forum.
> and it is listed with every post you make.
It should not be. As I mentioned, are you posting through Math Forum?
Or do you only sometimes use Math Forum and other times some other
news interface? Because I notice now that sometimes you respond to my
posts as they are identified as by 'MoeBlee' and sometimes as they are
identified as 'Jack Markan'.
> thats strange ? since when does empty mean 0 ?
For a long, long time, '0' has been used for the empty set.
> the empty set is not the set [0] ?
NO! 0 is the empty set. {0} is NOT the empty set. {0} is the set that
has 0 as the only member.
Except in your system, we DO have 0={0}, but that is INCONSISTENT with
the empty set axiom.
> so whats with this empty means x = 0 ?
I don't use that expression. What I use is standard terminology:
the empty set = 0.
Actually, more formally:
0=x <-> Ay~yex
I don't refer to just 'empty' as if it is an object. The object is the
empty SET.
> maybe its this way of representing integers by sets , a thing i dont like and dont support.
It doesn't matter. We're talking about the axioms you set up and the
language behind those axioms.
> > You've not provided for any notation that is a BLANK
> > character. Please
> > define such notation from your primitives or take it
> > as primitive and
> > give axioms for it.
>
> isnt the axiom of the empty set sufficient for that ??
No, of course not! The empty set axiom is this:
ExAy ~yex.
That doesn't say ANYTHING about the blank character being either a
primitive or defined character of any language.
> > > and
> > > x = [x] = [[x]]
>
> > That's fine, except that if it holds for ALL x, you
> > get an
> > inconsistent theory with the rest of your axioms, as
> > I PROVED and as
> > you IGNORED even commenting on the actual content of
> > my proof.
>
> give a specific example x where this gives inconsistancy with the other axioms.
I DID! You didn't read my proof! The EMPTY SET is such an x. From YOUR
axioms we (I, anybody who wishes to) may prove that the empty set has
no members and that the empty set has a member, which is a
contradiction.
PLEASE go back and understand my proofs (and Jesse's too).
> > > now lets consider [x] where this set does contain
> > > two elements.
>
> > > x = a,b
>
> > The notation 'a, b' is undefined. Please eaither
> > define it from the
> > primitives in your theory, or add it (the comma) as a
> > new primitive
> > operation symbol with axioms for it. Otherwise, no
> > one can say what
> > just 'a,b' is supposed to mean as a term onto itself.
>
> so a,b is undefined ?
>
> a,b are just two elements.
Then you've not specified any syntax that would make:
x = a,b
well formed.
> like the solution of a polynomial (x-5)(x-3)=0 has solution 5,3. or [5,3]
The solution SET is {5 3}. And 5 is a solution and 3 is a solution.
So you can have
x=5
x=3
x={5 3}
x=t53 (where 't' is any 2-place function symbol of the language), but
x=5,3
is not even well formed, unless, possibly you mean it to be informal
for:
x=5 v x=3
(i.e. "x=3 or x=5")
But the context of your argument suggest what you really mean is one I
mentioned above:
x={5 3}.
So in your argument
x=a,b
probably means
x={a b}.
But the mistake in your argument is that you ASSUME that a and b are
not equal. But that is the very thing you're trying to prove: that
there are at least two sets not equal to one another. That's called
'the fallcy of begging the question', or 'the fallacy of question
begging', or, more fancy 'petitio principii'.
> ZFC has only 9 axioms ,
Actually, ZFC has INFINITELY many axioms. The 'F' in ZFC stands for
the replacement schema, which is a definition of a certain denumerable
(countably infinite) set of axioms.
Moreover, ZFC is (in one treatement) and extension of identity theory,
so ZFC has all the axioms (if we use axioms rather than purely rules)
of predicate logic and of identity, plus the set theoretic axioms:
extensionality
union
power set
replacement schema (denumerably many axioms)
infinity
regularity
choice
Note: Here we use a "strong form" of replacement that can derive the
schema of separation and empty set.
And there are other axiomatizations of ZFC too. I can list some of
those axiomatizations if you would like me to.
> i can imagine i a bit need more , but either does a,b really
> need an axiom ?
Either it needs a DEFINITION, or it needs to be taken as a primitive
and then you need axioms for it to have any effect. Or, you need to
tell me how you're using it as informal notation.
> and if so , what axiom in ZFC explains "a,b" ???
An axiom sets "conditions", though "explains" might be another way of
looking at it. I could say a lot more about that, more technically,
but you really need first to learn some stuff even more basic.
> > No, please state the EXACT step in my proofs that you
> > THINK is
> > incorrect.
No reply from you on that.
MoeBlee
lwa...@lausd.net
> On Jul 17, 6:56 am, Aatu Koskensilta <aa...@alatheia.dsl.inet.fi>
> > -- e.g. in the form of Dedekind cuts -- are sets, and we can, with
> > some care, reformulate most of analysis in the theory, the same way we
> > do in e.g. ACA_0. In effect, the collection of the reals is treated as
> > a proper class.
> Thanks for that (though, I think the point is even better addressed to
> lwal). Moreover, I noticed that that theory has a quite powerful
> impredicative-over-proper-classes comprehension schema; so it's hardly
> akin to Tommy's fantasy in which he has no comprehension (not even
> separation or replacement, etc.) axioms at all, as well as no power
> set axiom.
You got me there.
PST has a comprehension schema, whereas TST -- at least the version
of TST in this thread -- has no schemata at all.
lwa...@lausd.net
> On Jul 18, 1:05 pm, amy666 <tommy1...@hotmail.com> wrote:
> > jack markan is your own name , genius !!! :p
> through that interface? It might be that Math Forum is attaching my
> old name to posts made from my current account made away from Math
> Forum.
> > and it is listed with every post you make.
> It should not be. As I mentioned, are you posting through Math Forum?
> Or do you only sometimes use Math Forum and other times some other
> news interface? Because I notice now that sometimes you respond to my
> posts as they are identified as by 'MoeBlee' and sometimes as they are
> identified as 'Jack Markan'.
You're right. "Jack Markan" appears at Math Forum. Indeed,
several posters have different names at Math Forum from
through other newsreaders, including Bob Kolker and Lester Zick.
And this of course reignites the long standing debate between
free newsreaders (Google Groups, Math Forum, Outlook Express,
etc.) and the full-priced news servers. In particular, this will give
proponents of the full-priced news servers another reaon to reject
free newsreaders (and possibly killfile those who use them) --
namely, that they give incorrect user names.
Jesse F. Hughes
> And this of course reignites the long standing debate between
> free newsreaders (Google Groups, Math Forum, Outlook Express,
> etc.) and the full-priced news servers. In particular, this will give
> proponents of the full-priced news servers another reaon to reject
> free newsreaders (and possibly killfile those who use them) --
> namely, that they give incorrect user names.
It hasn't a damn thing to do with free news readers. It has to do
with one strange news server.
--
Jesse F. Hughes
"The future is a fascinating thing, and so is history. And you people
are a fascinating part of history, for those in the future."
-- James S. Harris is fascinating, too
Aatu Koskensilta
> I think I might have used the name at Math Forum. Are you posting
> through that interface? It might be that Math Forum is attaching my
> old name to posts made from my current account made away from Math
> Forum.
"Jack Markan" is given as your name in the profile at Math Forum. It
doesn't appear anywhere in the headers of your posts posted through
Google.
Aatu Koskensilta
> And this of course reignites the long standing debate between
> free newsreaders (Google Groups, Math Forum, Outlook Express,
> etc.) and the full-priced news servers.
Google Groups and Math Forum are not newsreaders. Outlook Express is,
and is used with a news server of the users choice. Price is
irrelevant; there are many free news servers.
MoeBlee
> You're right. "Jack Markan" appears at Math Forum.
Thanks for checking it.
MoeBlee
MoeBlee
wrote:
> "Jack Markan" is given as your name in the profile at Math Forum.
Thanks for that info. I don't recall my log-in info there, so I
wonder, if I wrote to Math Forum, whether they would take the time to
fix it so that that name is not on future posts.
MoeBlee
hagman
> On Jul 15, 12:57 pm, amy666 <tommy1...@hotmail.com> wrote:
>
> > > On Jul 14, 1:46 pm, amy666 <tommy1...@hotmail.com>
> > > wrote:
> > > > i was able to defend my axioms against david c
> > > ullrich , jesse f hughes and hagman.
>
> > > > impressive hmm.
>
> > > No, it's sad that you think you've given clear
> > > axioms, let alone that
> > > you think you've defended anything.
>
> > i dont have the illusion that they are clear at first sight.
>
> Except for the ones that restate known axioms of set theory, they're
> far from clear in any sight.
>
> > in fact the axioms only become a bit clear all together , individually they are vague.
>
> Please read up just a little to find out how modern axiomatic systems
> work.
>
> > but they are somewhat lets say descriptive , you really need to think about them.
>
> > see them as a whole.
>
> You need to learn how modern axiomatic systems work. I don't need to
> think about your slop; I already have a reading list of a lifetime of
> mathematics to study, including alternatives to standard mathematics.
> Such mathematics is informed, intellectually organized, and well
> written. That's what you need to learn.
>
> > > Why are you so averse to informing yourself as to
> > > some basic
> > > mathematical logic by which you'd come to actually
> > > understand
> > > something about the axiomatic method?
>
> > because things are not so well defined as commenly believed.
>
> However well defined things are or are not, you only deprive yourself
> by not studying the basics of the subject.
>
> > e.g. what is a "set" ?
>
> (1) If we had to, we can do all of set theory without ever mentioning
> the word 'set'. The primitive of set theory is 'e' (read "element
> of"). That (and '=', read "is equal to") are not defined. All other
> predicate and function symbols are defined. (2) Anyway, in a theory
> such as Z set theory, a technical definition of 'is a set' can be
> provided:
>
> First, we prove that there exists exactly one object that has no
> members:
>
> E!xAy ~yex
>
> Then we define:
>
> 0=x <-> Ay ~yex
>
> Then we define:
>
> x is an element <-> Ey xey
>
> C is a class <-> (Ey yec v c=0)
>
> S is a set <-> (S is a class & Ey Sey)
>
> C is a proper class <-> (C is a class & ~C is a set)
>
> u is an urelement <-> ~u is a class
>
Erm, doesn't this deny the existence of urelements?
Then
u is an urelement
<-> ~ u is a class
<-> ~(Ey yeu v u=0)
<-> ~(Ey yeu v Ay ~yeu)
<-> ~(Ey yeu v ~Ey yeu)
<-> ~ true
hagman
hagman
> hagman wrote :
>
>
>
> > On 14 Jul., 21:40, amy666 <tommy1...@hotmail.com>
> > wrote:
> > > > On 13 Jul., 17:31, amy666 <tommy1...@hotmail.com>
> > > > wrote:
> > > > > 1) axiom of extensionality
>
> > > > Ax Ay (x=y <-> Az (zex <-> zey))
>
> > > > > 2) axiom of the empty set
>
> > > > Presumably meaning
> > > > Ex Ay ~(yex)
>
> > > > > 3) axiom of pairing
>
> > > > Presumably meaning
> > > > Ax Ay Ez At (tez <-> t=x v t=y)
>
> > > > > 4) axiom of union
>
> > > > Presumably meaning
> > > > Ax Ey Az (zey <-> Et (zet & tex))
>
> > > > > 5) non-standard elements within the set of real
> > > > numbers always have properties that correspond to
> > > > properties of infinitesimal and unlimited
> > elements.
>
> > contains
> > > > non-standard
> > > > elements.
> > > > Do you have any idea of the meaning of any of the
> > > > terms used
> > > > especially in 5)?
>
> > > yes.
>
> > Then please enlighten me:
> > What is the set of real number in your theory?
> > Just some complete ordered field?
> > Do you have an axiom that states the existence of
> > such a set?
> > How do you define "nonstandard element of R"?
> > Etc. etc.
>
> read the other replies , i answer this one to lwalke i believe.
>
>
>
>
>
> > > > > 6) this set theory is game theory in disguise ,
> > > > whatever holds in game theory can be restated in
> > this
> > > > set theory.
>
> > > > Is this an *attempt* to restate Conway numbers?
>
> > > no.
>
> > > > > this set theory is complete and consistant
> > apart
> > > > from solving equations.
>
> > > > > this set theory is the analogue of a computer
> > with
> > > > oo resourses (as should be since math is about
> > > > computation)
>
> > > > > 7) axiom of logic : x = [x] ( also if x =
> > empty )
>
> > > > > regards
>
> > > > > tommy1729
>
> > > > > ps : it has been requested that i explicitly
> > gave
> > > > my axioms , now i did.
>
> > > > > |R|^2 = |R| is consistant with it.
>
> > proved
> > > > from the above
> > > > axioms?
> > > > I just note that
> > > > |R|^2 = |R|
> > > > is also consistant with the axiom
> > > > 0) tommy1729 is stupid
>
> > > are we getting personal again hmm ??
>
> > > i already proved |R|^2 = R many times by now !!
>
> > But you cannot really do that from the axioms you
> > gave without
> > a proper defintion of the terms occurring here.
> > I'm getting problems if I simply use my definitions
> > of terms,
> > e.g.
> > |A|=|B| :<=> there exists a bijective map f:A->B
> > <=> there exists a set F such that all elements of F
> > are
> > (Kuratowsky) ordered pairs (a,b) with a in A and b in
> > B and
> > whenever (a,b) in F and (a,b') in F then b=b' and
> > whenever (a,b) in F and (a',b) in F then b=b' and
> > for all a in A exists b in B such that (a,b) in F and
> > for all b in B exists a in A such that (a,b) in F.
>
> > I doubt that it can be shown from your axioms that
> > |R| = |R| holds (let alone |R|^2 = |R|).
>
> evidently Q = Q for all Q !!!
>
> are you saying Q can =/= Q ???
It depends.
It is not necessarily the case that the "=" sign
is "the" equal sign, but some prefer to view this as a
matter of writing "|<foo>|=|<bar>|" for the clumsier
"There exists a bijection between <foo> and <bar>".
(Compare with the "=" in "cos(x) = 1 + O(x^2)", which
is a similar (and "even wronger") abuse of notation).
Since your theoretical framework hardly allows one to
talk about the smallest ordinal a set can be bijected to,
I don't see how you want to define a "free-standing"
cardinality |R|; in fact I don't even see how you
can define "There exists a bijection..." on
your grounds.
>
> do i need to add as axiom Q = Q for all Q ??
No, you need to define |.|
hagman
hagman
Your argument is that [a,b] "clearly" has 2 elements.
You never bothered to define most of your notation, but apparently
the "number of elements" of something enclosed in square barackets
is essentially simply the number of commas plus 1.
Thus [a,b] has two elements.
(What fun we can have in the special case a=b, I do not
dare to imagine)
And [x] has one element.
Then your statement [x] = x = a,b = [a,b]
claims that the same object has one element and two elements
in the same time.
hagman
MoeBlee
Yes, in certain theories, it is a theorem that there are no
urelements. In other theories, it is undetermined whether there are
urelements, and in other theories there are urelements.
> u is an urelement
> <-> ~ u is a class
> <-> ~(Ey yeu v u=0)
> <-> ~(Ey yeu v Ay ~yeu)
> <-> ~(Ey yeu v ~Ey yeu)
> <-> ~ true
Right.
To make it is undetermined whether there are urelements, we could have
a primitive predicate 'is a set', then modify the axiom of
extensionality:
(x is a set & y is a set) -> (x = y <-> Az(zex <-> zey))
(Other axioms may require modification also.)
Then to ensure that there are urelements, we could adopt an axiom:
ExAy(~yex & ~x is a class)
Or (with that aforementioned modified axiom of extensionality) to
ensure that there are no urelements, we could adopt as an axiom (the
axiom of purity) the negation of the above formula:
~ExAy(yex & ~x is a class)
which is equivalent to
Ax x is a class
MoeBlee
amy666
> Your argument is that [a,b] "clearly" has 2 elements.
yes of course.
> the "number of elements" of something enclosed in
> square barackets
> is essentially simply the number of commas plus 1.
> Thus [a,b] has two elements.
right.
> (What fun we can have in the special case a=b, I do
> not
> dare to imagine)
well in traditional set theory how many elements does [3,5,5] have ? 3 and 5 or 3 and 5 and 5.
which zero's does (x-1)^2 * (x-2) = 0 have ?
1 and 2 or 1 and 1 and 2.
what are the primefactors of 3*3*5 ? 3 and 5 or 3 and 3 and 5.
its just a matter of convention and not such a big deal in general.
the set of integers U [2] = the set of integers
or
[1,2,2,3,4,5,6,7,8,...]
your remark is important in a way , to get any detail clear , on the other hand it clearly is just a detail.
> And [x] has one element.
x since [x] = x
>
> Then your statement [x] = x = a,b = [a,b]
> claims that the same object has one element and two
> elements
> in the same time.
indeed. and this is no paradox since x is also a set.
>
> hagman
regards
tommy1729
MoeBlee
> hagman wrote ( in part )
>
> > Your argument is that [a,b] "clearly" has 2 elements.
>
> yes of course.
No, because it might be the case that a=b, in which case {a b} has
only one element.
First, from the way you've described in your initial post, you're
using [a b] just as we would {a b}. So I don't know why you don't just
use the ordinary notation {a b}.
Second, if a=b then {a b} has one element. And, in your arguments
about this, you have not proven that there exists an a and b such that
~a=b. Well, it turns out you can prove that, but only because your
system is inconsistent anyway.
MoeBlee
MoeBlee
> then modify the axiom of
> extensionality:
>
> (x is a set & y is a set) -> (x = y <-> Az(zex <-> zey))
Or, if the existence of proper classes is provable:
(x is a class & y is a class) -> (x = y <-> Az(zex <-> zey))
/
In any case, in a given theory, whether there exist memberless objects
other than the empty set is not determined by defintions, but rather
by axioms.
By defining 'urelement' as 'not a class' is equivalent with, just by
defintion 'a memberless object other than the empty set'. Then whether
there are such objects is up to the axioms to decide or leave
undecided.
MoeBlee
hagman
Okay, if you don't think this is a paradox...
Thus you will eventually define |A|, denoting
"the cardinality of" the set A.
(At least since you seem to wish to use your theory
to at least express |R^2| = |R|)
And we will have of course that A=B implies |A| = |B|
(at least you laughed at me when I tried to be careful
in this place).
And we will have in your theory |[x]| = 1 and |[a,b]|=2
with the example above.
And we have [x]=[a,b] with the same example.
Thus - in your theory - we have 1=2
(at least if your example is valid).
Your opinion is that this is a sound theory?
And since 2=1, you need no 2nd opinion on that...
hagman
amy666
> Thus you will eventually define |A|, denoting
> "the cardinality of" the set A.
maybe not.
> (At least since you seem to wish to use your theory
> to at least express |R^2| = |R|)
i originally used R^2 = R.
people here on sci.math insisted on |R| or C.
thats a big difference !
> And we will have of course that A=B implies |A| = |B|
if | | denotes elements and subelements.
elso , no not at all !
> (at least you laughed at me when I tried to be
> careful
> in this place).
> And we will have in your theory |[x]| = 1 and
> |[a,b]|=2
> with the example above.
as said | | must also include subelements since x = [x].
thus |x| is 2 since |[a,b]| = 2.
> And we have [x]=[a,b] with the same example.
> Thus - in your theory - we have 1=2
no, we have 2 = 2
> (at least if your example is valid).
>
> Your opinion is that this is a sound theory?
> And since 2=1, you need no 2nd opinion on that...
2 = 2 , i am the second opinion on your post :)
>
> hagman
regards
tommy1729
Jesse F. Hughes
complication, I think. Let's just give a simple proof from the axioms
you've mentioned.
I will prove the following theorem:
(Ax)(Ay)(x = y).
I will use the following axioms:
Pairing:
(Ax)(Ay)(Ez)(Aw)(w in z <-> (w = x or w = y))
Tommy's neato axiom:
x = [x] is how you phrase it, but let me make it more explicit:
(Az)(Aw)( w in z <-> w = z )
If that's not an accurate formalization of your axiom, then what is?
Proof:
Let x and y be given and let z be the set guaranteed by pairing, i.e.,
(Aw)(w in z <-> (w = x or w = y)) (*)
We can see that
x in z and y in z (**)
follows immediately from (*). Now, apply the neato axiom to z and we
see
(Aw)( w in z <-> w = z ). (***)
Instantiating (***) with both x and y, it follows that
x in z <-> x = z and y in z <-> y = z. (****)
From (**) and (****), it follows that
x = z and y = z. (OOPS)
From OOPS, it follows that x = y. Since x and y were arbitrary, we
see that (Ax)(Ay)(x = y).
No need to use anything but two axioms you accept.
--
"Kim liked the math I did for her and gave me quite a few
groceries... likely so many groceries that they would have cost Kim
about what she pays for two whole packages of cigarettes. Few people
have ever rewarded me for my work as much as Kim did." -- Usenet nut
amy666
> Hagman's introduction of cardinality really is an
> unnecessary
> complication, I think. Let's just give a simple
> proof from the axioms
> you've mentioned.
>
> I will prove the following theorem:
>
> (Ax)(Ay)(x = y).
another so-called disproof.
it only proofs you guys dont get it :)
>
> I will use the following axioms:
>
> Pairing:
>
> (Ax)(Ay)(Ez)(Aw)(w in z <-> (w = x or w = y))
>
> Tommy's neato axiom:
>
> x = [x] is how you phrase it, but let me make it
> it more explicit:
>
> (Az)(Aw)( w in z <-> w = z )
>
> If that's not an accurate formalization of your
> ur axiom, then what is?
same mistakes over and over again.
i said it many times before, i explained it many times before ;
w in z does not mean w = z.
let z = [w,a]
now [z] still is equal to z.
and w is in z.
but it does not follow that w = z.
z also contains 'a' without any paradox.
>
> Proof:
>
> Let x and y be given and let z be the set guaranteed
> by pairing, i.e.,
>
> (Aw)(w in z <-> (w = x or w = y))
> )) (*)
>
> We can see that
>
> x in z and y in z
> z (**)
>
> follows immediately from (*). Now, apply the neato
> axiom to z and we
> see
>
> (Aw)( w in z <-> w = z ).
> ). (***)
>
> Instantiating (***) with both x and y, it follows
> that
>
> x in z <-> x = z and y in z <-> y = z.
> z. (****)
>
> From (**) and (****), it follows that
>
> x = z and y = z.
> z. (OOPS)
>
> From OOPS, it follows that x = y. Since x and y were
> arbitrary, we
> see that (Ax)(Ay)(x = y).
>
> No need to use anything but two axioms you accept.
from OOPS its follows you still dont get it.
>
> --
> "Kim liked the math I did for her and gave me quite a
> few
> groceries... likely so many groceries that they would
> have cost Kim
> about what she pays for two whole packages of
> cigarettes. Few people
> have ever rewarded me for my work as much as Kim
> did." -- Usenet nut
regards
tommy1729
MoeBlee
> same mistakes over and over again.
No, Jesse ASKED you to give an EXPLICIT formulation of what you intend
by your axiom with the notation '[]'. But you still have not done that
in your remarks below:
> i said it many times before, i explained it many times before ;
>
> w in z does not mean w = z.
>
> let z = [w,a]
>
> now [z] still is equal to z.
>
> and w is in z.
>
> but it does not follow that w = z.
>
> z also contains 'a' without any paradox.
Please unpack into primitive terms of your language, the axiom
x=[x]
and in way that is consistent with your remark:
(T) "[x] = is the set that contains x ( only )
however x may be a set itself."
Given that remark (actually itself ineptly formed with the redundant
'is' after '='), you could not expect Jesse or any reasonable person
NOT to infer:
[x] = the z such that Aw(wez <-> w=x)
which is perfect rendering of your:
"[x] = the set that contains x ( only )" [taking out the redundant
'is']
and
[x] = the z such that Aw(wez <-> w=x)
is equivalent with
[x]=z <-> Aw(wez <-> w=x)
so that '[]' is just ordinary set theoretic '{}'.
So your axiom
x=[x]
which, more pednatically, is understood as quantified (and just
changing bound variables):
Az z=[z].
So we have
Az z=[z] ... axiom
AzAw(wez <-> w=z) ... by (T) (instantiating x to z)
which is exactly the formula Jesse used.
If you mean something else, then please state so by means of a precise
formulation. (Actually, the problem is that you don't know how to do
that.)
MoeBlee
Jesse F. Hughes
> Jesse F Hughes wrote :
>
>> Tommy's neato axiom:
>>
>> x = [x] is how you phrase it, but let me make it
>> it more explicit:
>>
>> (Az)(Aw)( w in z <-> w = z )
>>
>> If that's not an accurate formalization of your
>> ur axiom, then what is?
>
> same mistakes over and over again.
>
> i said it many times before, i explained it many times before ;
>
> w in z does not mean w = z.
Tell me what the heck x = [x] *means* then. The natural
interpretation of [x] is: the singleton containing x, i.e., the set z
satisfying
* x is in z
* nothing but x is in z.
Evidently, you don't mean that, so what the heck do you mean?
> let z = [w,a]
>
> now [z] still is equal to z.
>
> and w is in z.
>
> but it does not follow that w = z.
>
> z also contains 'a' without any paradox.
Why not either define the "[ ... ]" notation or at least give some
axioms? Just saying x=[x] isn't doing it. There is an obvious
interpretation of [x], which I gave above, but you claim it's wrong.
So, what the hell does [x] mean?
--
Jesse F. Hughes
"A gorgeous display of homoerotic lust."
-- Review blurb found on the back of a
Chinese black market "Dawn of the Dead" DVD
Jesse F. Hughes
> If you mean something else, then please state so by means of a precise
> formulation. (Actually, the problem is that you don't know how to do
> that.)
It's possible that when Tommy writes [...], he means the transitive
closure of ... .
It's also possible that he means nothing in particular and he's simply
mightily confused.
Place your bets!
--
"And I wish some of you would grow past thinking that you've discovered
some extraordinary thing [...] as if you found the Holy Grail or
something, when I acknowledge a mistake. After all, I've had to do it
quite a few times. It's not like it's news." --James S. Harris
MoeBlee
> Tell me what the heck x = [x] *means* then. The natural
> interpretation of [x] is: the singleton containing x, i.e., the set z
> satisfying
>
> * x is in z
>
> * nothing but x is in z.
>
> Evidently, you don't mean that,
But he did post that is what he means:
"[x] = is the set that contains x ( only )
however x may be a set itself."
Though maybe he didn't really mean THAT.
MoeBlee
MoeBlee
> MoeBlee <jazzm...@hotmail.com> writes:
> It's possible that when Tommy writes [...], he means the transitive
> closure of ... .
Ha ha.
> It's also possible that he means nothing in particular and he's simply
> mightily confused.
He posted that he meant:
"[x] = is the set that contains x ( only )
however x may be a set itself."
But maybe he didn't mean it.
MoeBlee
Jesse F. Hughes
> On Jul 23, 12:15 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> MoeBlee <jazzm...@hotmail.com> writes:
>
>> It's possible that when Tommy writes [...], he means the transitive
>> closure of ... .
>
> Ha ha.
I hate to say it, but I was being serious. But it still wouldn't make
sense of his claims, I guess.
Let's try again. He seems to want
[x] = [ y | y in x ].
I'm going to use the brackets and the braces hereafter, but I'm
treating the brackets as an operation on sets and the braces as the
usual notion on sets. So, when I write
[ y | y in x ]
I really mean [ {y | y in x} ] and similarly [ a,b ] = [ {a,b} ].
What about [x]? Well, I want it to mean "apply [] to x" instead of
"apply [] to {x}", but it won't matter, since they both amount to the
same thing. So, this explains the sloppy notation, I hope.
Now, suppose for a moment that our set theory has urelements. Define
[x] = { y in trans_clos(x) | y is an urelement }
Not a very difficult notion to understand and it makes sense of *some*
of what Tommy says, but not all.
Bah. Who'm I fooling? There's no guessing what Tommy means because
he's too coherent to mean *anything*. I'll leave these useless
conjectures to lwalker.
>> It's also possible that he means nothing in particular and he's simply
>> mightily confused.
>
> He posted that he meant:
>
> "[x] = is the set that contains x ( only )
> however x may be a set itself."
>
> But maybe he didn't mean it.
Right. And maybe he did. And maybe he did and didn't.
And maybe it's both true and false but that's okay because true =
false in this special case.
--
God made the bees
And the bees make honey.
The miller's man does all the work,
But the miller makes the money. --- Mother Goose
MoeBlee
> Now, suppose for a moment that our set theory has urelements.
Suppose he didn't have the added axioms that make his theory
inconsistent. He still has extensionality (not relativized to a
primitive 'is a set'), so there are no urelements (at most one
memberless set, and that's the empty set, by his axioms).
> Define
>
> [x] = { y in trans_clos(x) | y is an urelement }
I forgot what we need for every set to have its transitive closure.
Don't we need regularity? What about schema of replacement? He has
neither.
> > He posted that he meant:
>
> > "[x] = is the set that contains x ( only )
> > however x may be a set itself."
>
> > But maybe he didn't mean it.
>
> Right. And maybe he did. And maybe he did and didn't.
>
> And maybe it's both true and false but that's okay because true =
> false in this special case.
I think what he really means is that his mommy told him that he's a
very special little boy, so now that he's a grownup, he doesn't need
to read and study math, 'cause he can make up his own math that is
even better and truer than anything in books.
MoeBlee
Jesse F. Hughes
>> Now, suppose for a moment that our set theory has urelements.
>
> Suppose he didn't have the added axioms that make his theory
> inconsistent. He still has extensionality (not relativized to a
> primitive 'is a set'), so there are no urelements (at most one
> memberless set, and that's the empty set, by his axioms).
Sure, but I was just playing a silly game of what-if. I could've
what-if'ed a fix to his axioms allowing urelements, too.
>> Define
>>
>> [x] = { y in trans_clos(x) | y is an urelement }
>
> I forgot what we need for every set to have its transitive closure.
> Don't we need regularity? What about schema of replacement? He has
> neither.
We don't need regularity, but we have to take countable unions so I
suspect replacement is needed.
But I didn't want to introduce the trans_clos operator in its full
glory. The construction I mention above can be had for a small axiom
or two, something like:
y in [x] <-> y in x or (E z)(z in x & y in [z]).
Anyway, I was just puzzling over whether we could say something
sensible about what Tommy's thinking. Just temporarily silly, really.
--
"Sorry, wakeup to the real world. You're on your own dependent on me
as your guide. Luckily for you, I'm self-correcting to a large extent,
so if the proof were wrong, I'd tell you. It's not wrong."
--- James Harris confirms that his proof is correct.
lwa...@lausd.net
> Bah. Who'm I fooling? There's no guessing what Tommy means because
> conjectures to lwalker.
One way to interpret [...] is by U({...}) -- that is, the (unary)
union
of the bracket's contents.
Then Ax (x = [x]) becomes Ax (x = U({x})), which is a theorem of ZFC.
But of course, Jesse's comments in the post I quoted, once
again, echo his belief that I'm being too "charitable" in my
interpretation of tommy1729. Once again, the reason I
do this is because I _really_ want to discuss the pros and
cons of set theories other than ZFC. The so-called
"cranks" are the only sci.math posters who are discussing
alternate theories to ZFC, and, as Jesse points out, they
aren't sufficiently coherent enough to for everyone to
understand what they are saying.
But what else can I do? I can't _force_ tommy1729 to be
any more coherent! All I can do is make an educated guess
at what he's saying so that there can be some sort of
theory that's worth discussing.
I wish that someone could post an alternative to ZFC in a
sufficiently coherent manner, with sufficient rigor so that
no one would ask what the poster _means_ by the symbols
and terminology that he's using, and just focus on whether
the theory's consistent, and compare and contrast it with ZFC.
But until that day, this is all I have to work with! All I know is
that tommy1729 desires a theory in which the concepts of
elementhood and subsethood are unified into a single concept.
Jesse F. Hughes
> On Jul 23, 6:41 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Bah. Who'm I fooling? There's no guessing what Tommy means because
>> he's too [in]coherent to mean *anything*. I'll leave these useless
>> conjectures to lwalker.
>
> One way to interpret [...] is by U({...}) -- that is, the (unary)
> union
> of the bracket's contents.
>
> Then Ax (x = [x]) becomes Ax (x = U({x})), which is a theorem of
> ZFC.
Right, *maybe* that's what Tommy means. But Tommy also said that [x]
is a set whose only element is x, and x is *not* an element of U({x}).
So that suggestion makes no more sense than mine.
[...]
> But until that day, this is all I have to work with! All I know is
> that tommy1729 desires a theory in which the concepts of elementhood
> and subsethood are unified into a single concept.
I think that's right, more or less. Either Tommy desires such a
theory or he's too damned confused to realize that there are two
relations there.
--
"This sucks," said a Pennsylvania State University student [...] " Why
can't the college let me do what I want to do with my computer? The
school computer security guys are being way more annoying than the
spyware was." -- A student pines for his disabled spyware
hagman
> > Thus you will eventually define |A|, denoting
> > "the cardinality of" the set A.
>
> maybe not.
Indeed, I'm afraid you'll never define all the terms you
want to use in your theory.
>
> > (At least since you seem to wish to use your theory
> > to at least express |R^2| = |R|)
>
> i originally used R^2 = R.
>
> people here on sci.math insisted on |R| or C.
>
> thats a big difference !
Well, there *are* sets A such that A^2 = A, even in
standard set theory. Typically, in such cases the equality
is unaffected by whether you Kuratowsky or any other
breed of pairs for A^2 - or view it as the set of
mappings 2->A.
>
> > And we will have of course that A=B implies |A| = |B|
>
> if | | denotes elements and subelements.
>
> elso , no not at all !
If | | denotes anything function-like.
This is a basic property of identity.
(Wasn't it you who claimed that the usual properties
of "=" should not all need to be stated explicitly?
Well, just saying thyt the theory is based on FOL with identity
is enough to ensure the statement above)
>
> > (at least you laughed at me when I tried to be
> > careful
> > in this place).
> > And we will have in your theory |[x]| = 1 and
> > |[a,b]|=2
> > with the example above.
>
> as said | | must also include subelements since x = [x].
Nope, you admitted that the comma-counting method is valid
>
> thus |x| is 2 since |[a,b]| = 2.
What happens if a = [c,d]? Still |[a,b]| = 2?
If you want to define |x| by |[]|=0
and then recursively |x| = |Ux|
(this needs a bit of regularity),
then we might have |x|=0 for all x
(in fact, this would be the result in ZFC without urelements)
MoeBlee
> I forgot what we need for every set to have its transitive closure.
> Don't we need regularity? What about schema of replacement? He has
> neither.
Now that I refreshed my reading, replacement we need, I think, and not
regularity.
MoeBlee
MoeBlee
> We don't need regularity, but we have to take countable unions so I
> suspect replacement is needed.
Right.
MoeBlee
MoeBlee
> The so-called
> "cranks" are the only sci.math posters who are discussing
> alternate theories to ZFC,
(1) There are non-crank discussions about alternatives to ZFC in
sci.logic. (2) Cranks don't offer theories. Usually they offer
polemics, and at best, bungled, vaguely formed facsimiles of theories.
(3) They're not just so-called cranks; they're cranks.
> and, as Jesse points out, they
> aren't sufficiently coherent enough to for everyone to
> understand what they are saying.
Indeed.
> But what else can I do? I can't _force_ tommy1729 to be
> any more coherent!
You can give him the advice to learn how to be coherent.
> All I can do is make an educated guess
> at what he's saying so that there can be some sort of
> theory that's worth discussing.
Best to give it up as a representation of what he's unable to express.
Rather, better to say that you have your OWN idea that was caused by
ruminating about some of his posts.
> I wish that someone could post an alternative to ZFC in a
> sufficiently coherent manner, with sufficient rigor so that
> no one would ask what the poster _means_ by the symbols
> and terminology that he's using, and just focus on whether
> the theory's consistent, and compare and contrast it with ZFC.
In sci.logic you can find discussions that, usually after some
intitial round of clarifications about notation and that kind of
thing, do then get around to examining for consistency.
MoeBlee
amy666
> that tommy1729 desires a theory in which the concepts
> of
> elementhood and subsethood are unified into a single
> concept.
indeed.
in fact they are the same !
x = [x]
gotta run.
David C. Ullrich
>[...]
>
>I wish that someone could post an alternative to ZFC
Why do you wish that? Is there some particular problem
with ZFC that you see?
(Note the question marks above...)
>in a
>sufficiently coherent manner, with sufficient rigor so that
>no one would ask what the poster _means_ by the symbols
>and terminology that he's using, and just focus on whether
>the theory's consistent, and compare and contrast it with ZFC.
>
>But until that day, this is all I have to work with! All I know is
>that tommy1729 desires a theory in which the concepts of
>elementhood and subsethood are unified into a single concept.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Herman Jurjus
> On Thu, 24 Jul 2008 01:06:59 -0700 (PDT), lwa...@lausd.net wrote:
>
>> [...]
>>
>> I wish that someone could post an alternative to ZFC
>
> Why do you wish that?
Every data structure can be represented in one way or another using XML.
But i'm glad that that's not the only format available. Because now that
we have more means than just one, it's more readily understandable that
the XML specs are /not/ the 'foundations of computer science'.
> Is there some particular problem with ZFC that you see?
The fact that it has no alternatives next to it?
--
Cheers,
Herman Jurjus
David C. Ullrich
wrote:
>David C. Ullrich wrote:
>> On Thu, 24 Jul 2008 01:06:59 -0700 (PDT), lwa...@lausd.net wrote:
>>
>>> [...]
>>>
>>> I wish that someone could post an alternative to ZFC
>>
>> Why do you wish that?
>
>Every data structure can be represented in one way or another using XML.
>But i'm glad that that's not the only format available. Because now that
>we have more means than just one, it's more readily understandable that
>the XML specs are /not/ the 'foundations of computer science'.
Who in the world has ever said that it is?
Your "now that we have more than just one" sounds like you
think that computer science began with XML and other
data formats have been developed only recently...
> > Is there some particular problem with ZFC that you see?
>
>The fact that it has no alternatives next to it?
First, that's simply not true. Second, if it were true that
wouldn't be an answer to my question - if there are no
particular problems with ZFC then the supposed fact
that there are no alternatives would not be a problem.
Herman Jurjus
> On Fri, 25 Jul 2008 17:14:21 +0200, Herman Jurjus <hju...@hetnet.nl>
> wrote:
>
>> David C. Ullrich wrote:
>>> On Thu, 24 Jul 2008 01:06:59 -0700 (PDT), lwa...@lausd.net wrote:
>>>
>>>> [...]
>>>>
>>>> I wish that someone could post an alternative to ZFC
>>> Why do you wish that?
>> Every data structure can be represented in one way or another using XML.
>> But i'm glad that that's not the only format available. Because now that
>> we have more means than just one, it's more readily understandable that
>> the XML specs are /not/ the 'foundations of computer science'.
>
> Who in the world has ever said that it is?
>
> Your "now that we have more than just one" sounds like you
> think that computer science began with XML and other
> data formats have been developed only recently...
Thanks for informing me how it sounds. I'll rephrase it on a next occasion.
>>> Is there some particular problem with ZFC that you see?
>> The fact that it has no alternatives next to it?
>
> First, that's simply not true. Second, if it were true that
> wouldn't be an answer to my question - if there are no
> particular problems with ZFC then the supposed fact
> that there are no alternatives would not be a problem.
> David C. Ullrich
FYI, you missed my point.
--
Cheers,
Herman Jurjus
amy666
> On 23 Jul., 15:31, amy666 <tommy1...@hotmail.com>
> wrote:
> > > Thus you will eventually define |A|, denoting
> > > "the cardinality of" the set A.
> >
> > maybe not.
>
> Indeed, I'm afraid you'll never define all the terms
> you
> want to use in your theory.
that might not be the reason.
>
> >
> > > (At least since you seem to wish to use your
> theory
> > > to at least express |R^2| = |R|)
> >
> > i originally used R^2 = R.
> >
> > people here on sci.math insisted on |R| or C.
> >
> > thats a big difference !
>
> Well, there *are* sets A such that A^2 = A, even in
> standard set theory. Typically, in such cases the
> equality
> is unaffected by whether you Kuratowsky or any other
> breed of pairs for A^2 - or view it as the set of
> mappings 2->A.
my argument remains valid.
>
>
> >
> > > And we will have of course that A=B implies |A| =
> |B|
> >
> > if | | denotes elements and subelements.
> >
> > elso , no not at all !
>
> If | | denotes anything function-like.
> This is a basic property of identity.
no it isnt.
> (Wasn't it you who claimed that the usual properties
> of "=" should not all need to be stated explicitly?
> Well, just saying thyt the theory is based on FOL
> with identity
> is enough to ensure the statement above)
wrong again.
anything function like ??
Q(x) = 1 if x contains a 1.
Q(x) = 0 if x does not contain a 1.
Q(A) = 1
Q(B) = 1
THUS Q(A) = Q(B)
BUT
A = 17 , B = 13.
so your generalized (nonsense) of anything functionlike fails BIGTIME.
>
> >
> > > (at least you laughed at me when I tried to be
> > > careful
> > > in this place).
> > > And we will have in your theory |[x]| = 1 and
> > > |[a,b]|=2
> > > with the example above.
> >
> > as said | | must also include subelements since x =
> [x].
as explained above , your " anyfunctionlike " fails.
>
> Nope, you admitted that the comma-counting method is
> valid
i never ever mentioned " comma - counting ".
>
> >
> > thus |x| is 2 since |[a,b]| = 2.
>
> What happens if a = [c,d]? Still |[a,b]| = 2?
>
> If you want to define |x| by |[]|=0
of course i dont.
> and then recursively |x| = |Ux|
> (this needs a bit of regularity),
neither that.
> then we might have |x|=0 for all x
> (in fact, this would be the result in ZFC without
> urelements)
nonsense. ( no offense , but your still in the dark )
>
> >
> > > And we have [x]=[a,b] with the same example.
> > > Thus - in your theory - we have 1=2
> >
> > no, we have 2 = 2
> >
> > > (at least if your example is valid).
> >
> > > Your opinion is that this is a sound theory?
> > > And since 2=1, you need no 2nd opinion on that...
> >
> > 2 = 2 , i am the second opinion on your post :)
> >
> >
> >
> > > hagman
> >
> > regards
> >
> > tommy1729
>
sigh , you still dont understand.
regards
tommy1729
amy666
> <je...@phiwumbda.org> wrote:
>
> > Tell me what the heck x = [x] *means* then. The
> natural
> > interpretation of [x] is: the singleton containing
> x, i.e., the set z
> > satisfying
> >
> > * x is in z
> >
> > * nothing but x is in z.
> >
> > Evidently, you don't mean that,
why not ?
why do you emphase on singleton ?
i said x could be a set too !
>
> But he did post that is what he means:
>
> "[x] = is the set that contains x ( only )
> however x may be a set itself."
>
> Though maybe he didn't really mean THAT.
what ?
why not ?
>
> MoeBlee
regards
tommy1729