For All x
apoorv
all values in the domain under consideration.Does it mean that we know
beforehand what is in the domain and what is not?
As an example, consider a set theory with only two of the usual
axioms :
1) Ez:Ax~x e z ( existence of 0 )
2)Az Ey Ax x e y <-> x e z & 0 e x (specific instance of separation)
This system, at the object level, is agnostic about the existence of
a successor of 0.
If z in axiom 2 can range only over the sets explicitly guaranteed
to
exist by the axiom system without reference to the domain ,then 2)
reduces to
Ey Ax xe y <-> x e 0 & 0 e x , so that y=0.The two axioms , by
themselves can
assert the existence of only the set 0
If the domain is specified beforehand to contain 0, &{S0} , then 2)
can assert the existence of
both 0 (z=0) and {S0} (z={S0})
which of these ways of interpreting 'for all x' , namely x ranging
over sets whose existence
can be proved at the object level, OR x ranging over a pre-specified
domain --is the correct one? shall appreciate a clarification.
-apoorv
William Elliot
> When we use the quantifier 'For All x' the variable x can range over
> all values in the domain under consideration.Does it mean that we know
> beforehand what is in the domain and what is not?
In ZF, there is no domain given nor does any FOL have a domain.
Domains are a part of model theory and will vary from model to
model for the same FOL theory.
> As an example, consider a set theory with only two of the usual
> axioms :
> 1) Ez:Ax~x e z ( existence of 0 )
> 2)Az Ey Ax x e y <-> x e z & 0 e x (specific instance of separation)
That seems wrong. I'll presume you mean
Az Ey Ax (x e y <-> x e z & 0 e x)
Don't you also need, a = b iff for all x, (x in a iff x in b) ?
> This system, at the object level, is agnostic about the existence of a
> successor of 0.
>
> If z in axiom 2 can range only over the sets explicitly guaranteed to
> exist by the axiom system without reference to the domain ,then 2)
> reduces to Ey Ax xe y <-> x e 0 & 0 e x , so that y=0.The two axioms ,
> by themselves can assert the existence of only the set 0
>
Again you repeat the same mistake.
Ey Ax (xe y <-> x e 0 & 0 e x)
It is obvious that y = 0.
> If the domain is specified beforehand to contain 0, &{S0} , then 2)
> can assert the existence of
> both 0 (z=0) and {S0} (z={S0})
>
If you add the axiom, for all x, x = 0;
I don't know what &{S0} means;
the rest is vague beyond rescue.
> which of these ways of interpreting 'for all x' , namely x ranging over
> sets whose existence can be proved at the object level, OR x ranging
> over a pre-specified domain --is the correct one? shall appreciate a
> clarification. -apoorv
>
The two axioms have a model with a domain of one element.
This means that you cannot prove the existence of a set
from the axioms, other than 0.
george
> When we use the quantifier 'For All x' the variable x can range over
> all values in the domain under consideration.Does it mean that we know
> beforehand what is in the domain and what is not?
We don't know and we don't care.
Aatu Koskensilta and some other people whom you may respect
will say otherwise (which is why you need a clarification), but they
are wrong.
apoorv
I was not able to convey my question precisely. One more attempt.
Suppose we have a set theory with just one axiom:
Ez Ax ~x e z [i.e there is a null set 0]
Would every set that contains 0 be a domain for a model of this
theory?
In particular, would the truth value of Ax x=0 depend on the pre-
selected domain?
-apoorv
apoorv
Suppose we work with ZFC minus the axiom of infinity.
How do we interpret ~Ez Ax xez ?
Would it mean that the set of all finite sets does not exist?
-apoorv
Herbert Newman
>
> When we use the quantifier 'For All x' the variable x can range over
> all values in the domain under consideration. Does it mean that we know
> beforehand what is in the domain and what is not?
>
No. We don't know that. But we might have one "in mind", an "intended"
domain (universe of discourse) if you like. (Very often this is the domain
of the so called "standard interpretation".)
>
> As an example, consider a set theory with only two of the usual
> axioms :
>
> 1) ...
> 2) ...
>
> This system, at the object level, is agnostic about the existence of
> a successor of 0.
>
Whatever you mean with _successor_ of 0. But let's CLAIM that it (whatever
it might actually be) is "agnostic" about the existence of, say, {0}.
>
> If z in axiom 2 can range only over ...
>
Actually, it can range over any domain. (Though we might especially be
interested in the interpretations -and hence domains- which make those two
axioms true. Such interpretations are called _models_.)
>
> which of these ways of interpreting 'for all x', namely x ranging
> over sets whose existence can be proved at the object level, OR x
> ranging over a pre-specified domain -- is the correct one?
>
The second one. In any _interpretation_ (considered) the "for all x" is
restricted to the domain of this very interpretation. (Though on the other
hand we are free to consider any nonempty interpretation -and hence domain-
we want.)
Herb
Herbert Newman
>
> Suppose we have a set theory with just one axiom:
>
> Ez Ax ~x e z
"There is an empty set."
>
> Would every set that contains 0 be a domain for a model of this
> theory?
>
Yes - if, say, we interpret "e" as /e/ (at meta level). With other words,
the fact that our domain contains the empty set and that the empty set does
not contain any other set (i.e. any other element in the domain) together
with interpreting "e" as /e/ is what makes the statement/formula
Ez Ax ~x e z
true (in this interpretation).
>
> In particular, would the truth value of Ax x=0 depend on the pre-
> selected domain?
>
Well, you did not define "0" so far. So let me help you out here. Consider
"0" to be an _undefined_ notion of your system. And let's just consider the
set theory consisting of the single axiom:
Ax x !e 0.
"0 is empty."
Of course the truth value of
Ax(x = 0) (*)
Will depend on the interpretation (and hence domain) considered.
If the domain of a certain interpretation contains more that just one
element then (*) will be false (in this interpretation).
Herb
MoeBlee
> Suppose we have a set theory with just one axiom:
> Ez Ax ~x e z [i.e there is a null set 0]
> Would every set that contains 0 be a domain for a model of this
> theory?
A set alone is not a model of a theory whose language has a 2-place
predicate symbol. Rather, a set with a binary relation on that set is
needed to be a model for the language, and then that will be a model
of the theory iff all members of the theory are true in that model.
For example:
map 'A' to {0}.
map 'e' to {<0 0>}.
The universe of that model has 0 as a member, but EzAx ~xez is not
true in that model.
> In particular, would the truth value of Ax x=0 depend on the pre-
> selected domain?
With the standard interpretation of '=', "Ax x=0" is true in all and
only those models whose universe is a singleton.
MoeBlee
MoeBlee
> Suppose we work with ZFC minus the axiom of infinity.
> How do we interpret ~EzAx xez ?
> Would it mean that the set of all finite sets does not exist?
Aside from any matter of interpretation,
~EzAx xez is a theorem of ZFC with or without the axiom of infinity.
It's even a theorem of certain theories weaker than ZFC without the
axiom of infinity.
Also, ~ExAy(y is finite -> yex) is a theorem of ZFC without the axioim
of infinity and of certain theories weaker than ZFC without the axiom
of infinity. It should be clear to you how to prove that.
MoeBlee
apoorv
> On Tue, 3 Feb 2009 07:40:22 -0800 (PST), apoorv wrote:
>
> > Suppose we have a set theory with just one axiom:
>
> > Ez Ax ~x e z
>
> "There is an empty set."
>
> > Would every set that contains 0 be a domain for a model of this
> > theory?
>
> Yes - if, say, we interpret "e" as /e/ (at meta level). With other words,
> the fact that our domain contains the empty set and that the empty set does
> not contain any other set (i.e. any other element in the domain) together
> with interpreting "e" as /e/ is what makes the statement/formula
>
> Ez Ax ~x e z
>
> true (in this interpretation).
>
>
>
> > In particular, would the truth value of Ax x=0 depend on the pre-
> > selected domain?
>
> Well, you did not define "0" so far. So let me help you out here. Consider
> "0" to be an _undefined_ notion of your system. And let's just consider the
> set theory consisting of the single axiom:
>
> Ax x !e 0.
>
> "0 is empty."
>
> Of course the truth value of
>
> Ax(x = 0) (*)
>
> Will depend on the interpretation (and hence domain) considered.
'If the domain 'contains' only 0, then Ax(x=0) is true.'
This appears to be a statement at the meta-level ;
At the object level, is there any way of asserting that
any "'x' is or not in in the domain" except for those whose existence
is explicitly asserted / denied by axioms at the object level?
-apoorv
apoorv
> On Feb 3, 7:45 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > Suppose we work with ZFC minus the axiom of infinity.
> > How do we interpret ~EzAx xez ?
> > Would it mean that the set of all finite sets does not exist?
>
> Aside from any matter of interpretation,
>
> ~EzAx xez is a theorem of ZFC with or without the axiom of infinity.
> It's even a theorem of certain theories weaker than ZFC without the
> axiom of infinity.
infinity,
how do we interpret 'the set of all ordinals'? Does it exist in ZFC
without
the axiom of infinity?
-apoorv
MoeBlee
> What I am trying to understand is if we work without the axiom of
> infinity,
> how do we interpret 'the set of all ordinals'? Does it exist in ZFC
> without
> the axiom of infinity?
Even just Z set theory (even without the axiom of infinity) proves
~ExAy(y is an ordinal -> yex).
A proof is not too difficult you should be able to do it yourself (cf.
Burali-Forti paradox).
MoeBlee
apoorv
an infinite ordinal as its value? In other words, does y in All y
range only over finite ordinals,
whose existence can be explicitly proved at the object level?
-apoorv
Dan Christensen
> all values in the domain under consideration.Does it mean that we know
> beforehand what is in the domain and what is not?
[snip]
As I understand it, the domain under consideration can be arbitrary
(ranging over all objects), as in set theory or logic, or it can be
thought of as specific set of objects, as in geometry (e.g. the points
in a plane).
The domain under consideration is not a formal construct. It is a
convenience in the case of areas of study like the geometry of a
plane. In this case, it saves having to repeatedly write "for all x IN
THE PLANE..." and so on. If the domain under consideration is
understood to be the points in a plane, you can write simply "for all
x..." saving many lines in lengthy formal proofs establishing that the
object in question is indeed is a point in the plane. If we are
restricting the domain under consideration in this way, extra care
must be taken in constructing axioms. In the case of plane geometry,
the domain of consideration cannot be both points AND lines in the
plane.
Dan
Download my DC Proof software at http://www.dcproof.com
MoeBlee
> On Feb 5, 10:32 pm, MoeBlee <jazzm...@hotmail.com> wrote:> On Feb 5, 12:37 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > What I am trying to understand is if we work without the axiom of
> > > infinity,
> > > how do we interpret 'the set of all ordinals'? Does it exist in ZFC
> > > without
> > > the axiom of infinity?
>
> > Even just Z set theory (even without the axiom of infinity) proves
> > ~ExAy(y is an ordinal -> yex).
>
> > A proof is not too difficult you should be able to do it yourself (cf.
> > Burali-Forti paradox).
>
> an infinite ordinal as its value?
Variables range over objects PER a given model. And it is not ruled
out that an infinite ordinal is a member of the universe of a model of
ZF(without axiom of infinity).
> In other words, does y in All y
> range only over finite ordinals,
> whose existence can be explicitly proved at the object level?
It depends on what PARTICULAR model we're dealing with. The objects of
the model don't have to be objects proven to exist "at the object
theory level".
For example, suppose our theory is Z(without the axiom of infinity but
with an added axiom that no infinite sets exist). That theory has the
theorem that all ordinals are finite. But, if I am not mistaken, the
universe of a model of that theory could be the set of ordinals ('w'
here stand for omega) {w, w+1, w+1...}.
I strongly recommend that to understand how theories and models work,
you should get a good textbook on mathematical logic and carefully
follow the definitions and proofs.
MoeBlee
apoorv
> On Feb 6, 3:51 am, apoorv <sudhir...@hotmail.com> wrote:
>
>
>
>
>
> > On Feb 5, 10:32 pm, MoeBlee <jazzm...@hotmail.com> wrote:> On Feb 5, 12:37 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > > What I am trying to understand is if we work without the axiom of
> > > > infinity,
> > > > how do we interpret 'the set of all ordinals'? Does it exist in ZFC
> > > > without
> > > > the axiom of infinity?
>
> > > Even just Z set theory (even without the axiom of infinity) proves
> > > ~ExAy(y is an ordinal -> yex).
>
> > > A proof is not too difficult you should be able to do it yourself (cf.
> > > Burali-Forti paradox).
>
> > Sure. But in the absence of Axiom of Infinity, can the variable y take
> > an infinite ordinal as its value?
>
> Variables range over objects PER a given model. And it is not ruled
> out that an infinite ordinal is a member of the universe of a model of
> ZF(without axiom of infinity).
>
> > In other words, does y in All y
> > range only over finite ordinals,
> > whose existence can be explicitly proved at the object level?
>
> It depends on what PARTICULAR model we're dealing with. The objects of
> the model don't have to be objects proven to exist "at the object
> theory level".
domain
whose objects are known to us at the meta level.Then, to interpret
ZFC, we
need a model with a domain which at the meta level contains an
infinity of ordinals
and an infinite ordinal. Having assumed implicitly at the meta level
an infinite domain and an infinite ordinal, we assert at the object
level, through the
axiom of infinity the existence of an infinite ordinal .
-apoorv
> For example, suppose our theory is Z(without the axiom of infinity but
> with an added axiom that no infinite sets exist). That theory has the
> theorem that all ordinals are finite. But, if I am not mistaken, the
> universe of a model of that theory could be the set of ordinals ('w'
> here stand for omega) {w, w+1, w+1...}.
>
> I strongly recommend that to understand how theories and models work,
> you should get a good textbook on mathematical logic and carefully
> follow the definitions and proofs.
>
>
> - Show quoted text -
apoorv
wrote:
> On Feb 2, 6:41 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > When we use the quantifier 'For All x' the variable x can range over
> > all values in the domain under consideration.Does it mean that we know
> > beforehand what is in the domain and what is not?
>
> [snip]
>
> As I understand it, the domain under consideration can be arbitrary
> (ranging over all objects), as in set theory or logic, or it can be
> thought of as specific set of objects, as in geometry (e.g. the points
> in a plane).
>
> The domain under consideration is not a formal construct. It is a
> convenience in the case of areas of study like the geometry of a
> plane. In this case, it saves having to repeatedly write "for all x IN
> THE PLANE..." and so on. If the domain under consideration is
> understood to be the points in a plane, you can write simply "for all
> x..." saving many lines in lengthy formal proofs establishing that the
> object in question is indeed is a point in the plane. If we are
> restricting the domain under consideration in this way, extra care
> must be taken in constructing axioms. In the case of plane geometry,
> the domain of consideration cannot be both points AND lines in the
> plane.
>
> Dan
> Download my DC Proof software athttp://www.dcproof.com
the meta level;
the 'e' at the object level cannot express any relation between x and
D.
Indeed , D does not even exist at the object level.
-apoorv
>
>
> > As an example, consider a set theory with only two of the usual
> > axioms :
> > 1) Ez:Ax~x e z ( existence of 0 )
> > 2)Az Ey Ax x e y <-> x e z & 0 e x (specific instance of separation)
>
> > This system, at the object level, is agnostic about the existence of
> > a successor of 0.
>
> > If z in axiom 2 can range only over the sets explicitly guaranteed
> > to
> > exist by the axiom system without reference to the domain ,then 2)
> > reduces to
> > Ey Ax xe y <-> x e 0 & 0 e x , so that y=0.The two axioms , by
> > themselves can
> > assert the existence of only the set 0
>
> > If the domain is specified beforehand to contain 0, &{S0} , then 2)
> > can assert the existence of
> > both 0 (z=0) and {S0} (z={S0})
>
> > which of these ways of interpreting 'for all x' , namely x ranging
> > over sets whose existence
> > can be proved at the object level, OR x ranging over a pre-specified
> > domain --is the correct one? shall appreciate a clarification.
Frederick Williams
>
> When we use the quantifier 'For All x' the variable x can range over
> all values in the domain under consideration.Does it mean that we know
> beforehand what is in the domain and what is not?
Has anyone every cooked up a first-order (or other) theory in which
there wasn't an intended interpretation?
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
Jan Burse
> apoorv wrote:
>> When we use the quantifier 'For All x' the variable x can range over
>> all values in the domain under consideration.Does it mean that we know
>> beforehand what is in the domain and what is not?
>
> Has anyone every cooked up a first-order (or other) theory in which
> there wasn't an intended interpretation?
>
Each number n can represents a FOL theory. Namely n
can encode a partial recursive function, and thus
recursively enumerates a FOL theory.
Which of those do you think have an intended
interpretation? Not to speak of those FOL theories
which are true sets and not r.e.?
Anyway, some of them are not consistent, but I
assume, that there are still many consistent
left, and they thus have SOME model.
Bye
MoeBlee
I don't know that I'd put it that way, but I guess that's okay.
> Then, to interpret
> ZFC, we
> need a model with a domain which at the meta level contains an
> infinity of ordinals
> and an infinite ordinal.
I didn't say that. Are you claiming it yourself?
> Having assumed implicitly at the meta level
> an infinite domain and an infinite ordinal, we assert at the object
> level,
I don't see that anything that can't be explicit in this regard.
> through the
> axiom of infinity the existence of an infinite ordinal .
Yes, the axiom of infinity along with certain other axioms does prove
there exists an infinite ordinal.
> > I strongly recommend that to understand how theories and models work,
> > you should get a good textbook on mathematical logic and carefully
> > follow the definitions and proofs.
Have you done this? Would you tell me what books you're working with?
I can better address your questions in context of at least one
systematic treatment of the subject.
MoeBlee
apoorv
far
as the relation 'e' is concerned , 0 and D are the same?
-apoorv
>
>
>
>
> > > As an example, consider a set theory with only two of the usual
> > > axioms :
> > > 1) Ez:Ax~x e z ( existence of 0 )
> > > 2)Az Ey Ax x e y <-> x e z & 0 e x (specific instance of separation)
>
> > > This system, at the object level, is agnostic about the existence of
> > > a successor of 0.
>
> > > If z in axiom 2 can range only over the sets explicitly guaranteed
> > > to
> > > exist by the axiom system without reference to the domain ,then 2)
> > > reduces to
> > > Ey Ax xe y <-> x e 0 & 0 e x , so that y=0.The two axioms , by
> > > themselves can
> > > assert the existence of only the set 0
>
> > > If the domain is specified beforehand to contain 0, &{S0} , then 2)
> > > can assert the existence of
> > > both 0 (z=0) and {S0} (z={S0})
>
> > > which of these ways of interpreting 'for all x' , namely x ranging
> > > over sets whose existence
> > > can be proved at the object level, OR x ranging over a pre-specified
> > > domain --is the correct one? shall appreciate a clarification.
> > > -apoorv- Hide quoted text -
>
>
> - Show quoted text -- Hide quoted text -
apoorv
well, you seem to be in agreement with what i said.
> > > I strongly recommend that to understand how theories and models work,
> > > you should get a good textbook on mathematical logic and carefully
> > > follow the definitions and proofs.
>
> Have you done this? Would you tell me what books you're working with?
> I can better address your questions in context of at least one
> systematic treatment of the subject.
You have too much belief (faith?) in your books.
-apoorv
MoeBlee
That is a quite dishonest answer.
(1) I don't just believe whatever I read in books, not even in logic
and math books, since I scrutinize each proof before committing that
it is correct, and I take as only provisional results that I haven't
yet gone through the proofs, but with the reasonable supposition that
there are not fatal errors in proofs that have been scrutinized by
many other mathematicians, while, still, I maintaind at least a
nominal distinction between what I have personally verified and what I
can only take on good credit. (And at the level of our current
discussion, there is nothing, unless otherwise mentioned, that I
haven't verified myself.)
(2) You completely evaded the purpose of my question. My purpose was
not to select some book that we should believe as a matter of faith or
even take as definitive authority, but rather, as I alluded, my
purpose is to know the basis of your understanding and to give us a
context for our exchanges. Having at least one book to refer to for
definitions and explanations of the various terminology and concepts
facilitates communication on the subject.
So, I don't see why you won't just say what are the main books you are
using for your study of this subject. I'm not interested in twisting
your arm to do that; but I really don't appreciate the dishonesty of
twisting my question into a pretext to make a bogus claim about the
degree of my belief.
MoeBlee
MoeBlee
P.S. If one doesn't learn set theory from books and/or organized
lectures, then just how does one learn the subject? By flitting from
one Wikipedia page to another, with no overall conceptual and
technical context among them, complemented by various ad hoc
conversations with experts, sincere amateurs, and outright
ignoramuses, and then supplemented by watching one's favorite episodes
of "Friends"?
MoeBlee
Herbert Newman
>
> You have too much belief (faith?) in your books.
>
Idiot.
apoorv
Well, I am only saying that we need to be alive to other perspectives.
(after all, for 1500 years, the sun went round the earth) In any model
of ZF set theory, the variables range over a domain D specified
outside the theory. The domain D is itself a set, yet its existence is
postulated outside the theory. Further, when we say, for example,
Ax ~x e x, we are actually saying, Ax in D, ~x e x.
So, we actually are using D as a constant in the theory ( is that
permissible always?), and two different relations
Namely, “in” between objects and domain, and “e” between objects.
If we are looking for a self contained set theory, the existence of
the domain D
must be postulated within the theory.
Consider a theory with these axioms:
1)E z Ax x e z (existence of domain D)
2) E z Ax ~ x e z (existence of null set 0)
3)Ax E z Ay ( ye z <-> y e x or y=x) (existence of successor Sx of
x)
Axioms 2 &3) together imply that D is infinite and a separate Axiom
of infinity is not
needed. Because of 1) D contains itself. We consider a model with the
smallest possible domain D. D would be the
set D=[0,1,2 . . .D}
Having postulated the existence of the domain D, we need to give up
full fledged separation. In particular, we can see that
~E y Ax [x ey <->x e D and ~x =D}, that is, the set D-{D} =
{0,1,2 . . .} does not
exist.
In this model,
a) There is only one infinite set, which is its own successor .
b) Infinity cannot be separated from itself.
If “e” is interpreted as ‘less or equal’ and Sx as any function ,
then every sequence
In D is a model for the theory and D is the limit for that sequence..
So, we have a choice , we either accept full fledged separation, and a
unending escalator
Into the transfinite, or restrict separation and work with a single
infinite domain.
-apoorv
PHPBABY3
> When we use the quantifier 'For All x' the variable x can range over
> all values in the domain under consideration.Does it mean that we know
> beforehand what is in the domain and what is not?
You're referrring to the fact that Predicate Calculus is poorly
designned. A much better way to quantify varfiables is to realize
that it is just
referring to set equality. Let TRUE(a) be the universal set. Then:
(all X)P(X)
is really
P = TRUE
and (exists X)P(X)
is
~( P = ~TRUE).
Then (all X)P(X) and (exists X)P(X) are expressed as functions of P,
and results such as Rice's Theorem are easier to express and
understand.
E.g. an Index Set is simply f(yes(I)) where yes(I) is the set of
inputs accepted by Turing Machine I.
C-B
> -apoorv
MoeBlee
> On Feb 12, 1:00 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > Re Feb 10, 12:02 pm, MoeBlee <jazzm...@hotmail.com>:
>
> > P.S. If one doesn't learn set theory from books and/or organized
> > lectures, then just how does one learn the subject? By flitting from
> > one Wikipedia page to another, with no overall conceptual and
> > technical context among them, complemented by various ad hoc
> > conversations with experts, sincere amateurs, and outright
> > ignoramuses, and then supplemented by watching one's favorite episodes
> > of "Friends"?
> Well, I am only saying that we need to be alive to other perspectives.
> (after all, for 1500 years, the sun went round the earth)
I don't know how that answers the question of how one would learn the
basics of set theory other than by books, classes, or tutoring.
> In any model
> of ZF set theory, the variables range over a domain D specified
> outside the theory. The domain D is itself a set, yet its existence is
> postulated outside the theory.
Or proven to exist in a meta-theory.
> Further, when we say, for example,
> Ax ~x e x, we are actually saying, Ax in D, ~x e x.
Sure, IF we have fixed upon a particular domain D.
> So, we actually are using D as a constant in the theory ( is that
> permissible always?),
Wrong. When we prove theorems in the theory itself, we do NOT
reference any particular domain D. It is when we DISCUSS the theory
and (in a meta-theory) prove things ABOUT the theory and its relation
with a particular domain D is when we reference D, which is defined
not in the theory but in the meta-theory. (The case, if one
countenances it, where the meta-theory may be itself the theory is
more complicated, but we don't need to explain such an instance just
to note that proving theorems in set theory does NOT require
mentioning any particular domain.)
> and two different relations
> Namely, “in” between objects and domain, and “e” between objects.
Okay, fair enough.
> If we are looking for a self contained set theory, the existence of
> the domain D
> must be postulated within the theory.
If you mean a domain D and a relation R such that all the axioms of
the theory are claimed to be true in D with 'e' interpreted as R, then
the theory (such as a sufficiently rich and recursively axiomatized
set theory) will be inconsistent since it will prove its own
consistency.
So, no one promises you a "self contained set theory".
> Consider a theory with these axioms:
> 1)E z Ax x e z (existence of domain D)
Your parenthetical comment is not part of the theory itself.
> 2) E z Ax ~ x e z (existence of null set 0)
Your parenthetical comment is not part of the theory itself.
> 3)Ax E z Ay ( ye z <-> y e x or y=x) (existence of successor Sx of
> x)
Your parenthetical comment is not part of the theory itself.
> Axioms 2 &3) together imply that D is infinite
Do you mean that the conjunction (2)&(3) has no finite models? Are you
sure? What is your proof that (2)&(3) has no finite models? (Remember,
'e' can be interpreted to a relation other than membership on the
domain.)
> and a separate Axiom
> of infinity is not
> needed. Because of 1) D contains itself.
No, you confused meta and object again.
> We consider a model with the
> smallest possible domain D. D would be the
> set D=[0,1,2 . . .D}
You persist to argue based on a fundamental confusion you have.
> Having postulated the existence of the domain D, we need to give up
> full fledged separation. In particular, we can see that
> ~E y Ax [x ey <->x e D and ~x =D}, that is, the set D-{D} =
> {0,1,2 . . .} does not
> exist.
> In this model,
All that based on your incorrect premises above.
I'm not going to bother sorting through the rest of your remarks.
MoeBlee
MoeBlee
> apporv wrote:
> > Consider a theory with these axioms:
> > 1)E z Ax x e z (existence of domain D)
>
> Your parenthetical comment is not part of the theory itself.
>
> > 2) E z Ax ~ x e z (existence of null set 0)
>
> Your parenthetical comment is not part of the theory itself.
>
> > 3)Ax E z Ay ( ye z <-> y e x or y=x) (existence of successor Sx of
> > x)
>
> Your parenthetical comment is not part of the theory itself.
>
> > Axioms 2 &3) together imply that D is infinite
>
> Do you mean that the conjunction (2)&(3) has no finite models? Are you
> sure? What is your proof that (2)&(3) has no finite models? (Remember,
> 'e' can be interpreted to a relation other than membership on the
> domain.)
Hmm, this particular claim of yours might be correct, but still
requires proof. However, even if correct, the rest of your arguments
(such as that the domain must be a member of itself) are still not
supported.
MoeBlee
apoorv
>
> - Show quoted text -
1)E z ax x e z
Ax x e D (existential instantiation )
D e D.
-apoorv
MoeBlee
> 1)E z ax x e z
> Ax x e D (existential instantiation )
That's not a finished proof of anything. Existential instantiation is
not an ENDING line for a proof (unless the proof ends with
undischarged assumptions). If you're going to use terminology of the
predicate calculus, PLEASE, why don't you get a good book and properly
learn the predicate calculus?
And the above is certainly not a proof in the object-theory (1)&(2)&
(3) about any domain of discourse for a model for the language of that
theory.
If you think it is, then you're confusing object and meta-language
again.
'D' used as a symbol of object language is in a different context from
'D' used as a symbol of the meta-language.
It is in the META-theory that we discuss the domain D for a model for
the language of the theory (1)&(2)&(3). In that context, 'D' is symbol
of the META-language. But if you use 'D' also as a symbol of the
object language, then you don't get to assume that 'D' used in that
context refers to the set that you named (in the META-theory) 'D'.
Anway existential instantiation is not an ending line of a proof
(unless the proof ends with undischarged assumptions) and existential
instantiation (toward the aim of discharging itself) does NOT permit
instantiating to a variable or constant, such as 'D', that has already
been mentioned in the proof - so must be 'D' is new to the proof. Thus
we are not to take 'D' as standing for anything in particular such as
a domain of discourse for a model of the language of the theory (1)&(2)
&(3), and we can't do that anyway since 'model', 'domain', etc. are
terminology of a META-theory about the object theory (1)&(2)&(3) and
not terminology you've defined IN that object theory.
MoeBlee
Daryl McCullough
>> > > Consider a theory with these axioms:
>> > > 1)E z Ax x e z (existence of domain D)
>> > > 2) E z Ax ~ x e z (existence of null set 0)
>> > > 3)Ax E z Ay ( ye z <-> y e x or y = x) (existence of successor
[stuff deleted]
>1)E z ax x e z
>Ax x e D (existential instantiation )
>D e D.
There is a subtle point here. The D that is the domain
in the sense of model theory need not be the universal
set whose existence follows from the axiom
E z Ax x e z
A general model for a language with one binary predicate symbol,
e, and no constant or function symbols is a pair <D,R> where
D is some set, R is a set of ordered pairs <x,y> such that
x and y are both elements of D.
Your axiom
E z Ax x e z
implies that for any D,R satisfying your axioms,
E z in D, A x in D, <x,z> in R
It does *not* follow that <D,D> is in R.
For example, here is a model of the above axioms:
Let D = { 0, 1, 2, .... } union { omega }
where 0 = empty set, 1 = { 0 }, 2 = { 0, 1 }, etc.
and where omega = { 0, 1, 2, ... }
Let R = { <x,y> | (x and y are naturals, and x < y), or y = omega }
Then the structure <D,R> satisfies your axioms, I believe.
But D is not equal to the universal set, omega.
--
Daryl McCullough
Ithaca, NY
MoeBlee
wrote:
> apoorv says...
>
> >> > > Consider a theory with these axioms:
> >> > > 1)E z Ax x e z (existence of domain D)
> >> > > 2) E z Ax ~ x e z (existence of null set 0)
> >> > > 3)Ax E z Ay ( ye z <-> y e x or y = x) (existence of successor
>
> [stuff deleted]
>
> >1)E z ax x e z
> >Ax x e D (existential instantiation )
> >D e D.
>
> There is a subtle point here. The D that is the domain
> in the sense of model theory need not be the universal
> set whose existence follows from the axiom
>
> E z Ax x e z
By the way, his axioms don't entail that there is UNIQUE universal set
(or, better, a unique object of which every object is e-related to
it), since he doesn't have the axiom of extensionality.
MoeBlee
Daryl McCullough
>By the way, his axioms don't entail that there is UNIQUE universal set
>(or, better, a unique object of which every object is e-related to
>it), since he doesn't have the axiom of extensionality.
That's true.
ross.fi...@gmail.com
wrote:
Hi Daryl,
Hey, how's it going, Jesse, Moe.
When there is the universal quantifier then a usual observation is
that there isn't the universal truth in axiomatics, everything is
based in terms of definitions of the objects of the theory.
Mathematics is a particularly useful way to forward application of
theory.
So, there is not the universal truth, for example in a regular set
theory, there isn't a predicate that applies to every set, even though
there would be expected to be that thing from the union of all the
others, because otherwise there aren't all the others. That is true
in many theories.
True(X) doesn't return true for every true statement, so of the
objects that embody theorems in models defined in axiomatic theories,
in the model there are only representations of the objects, so, the
objects there can't have all their true properties, and also, there
properties in the model could be invalidated by their properties
outside the model. Skolemization is a shelter from model containment,
that there is computing outside the model.
Thus, it's convenient to use the shortest path in axiomatics to
establish causal relations, with being aware the rules could be wrong.
In universal quantification in mathematical theories, there is
cancellation among product groups with the maintenance of the transfer
principle application, where each of everything has a property, as do
they all. Then, for example, a mathematical container is of
containers, etcetera. While that is so in representation, it is also
exactly and completely, in the completion theorem guarantee, the
generation of all the intermediate structures or those sufficient to
define them, maintaining fantastical units in transfer cancellation.
Thus, maintenance of containers is information-arithmetic.
Regards,
Ross F.
apoorv
wrote:
> apoorv says...
>
> >> > > Consider a theory with these axioms:
> >> > > 1)E z Ax x e z (existence of domain D)
> >> > > 2) E z Ax ~ x e z (existence of null set 0)
> >> > > 3)Ax E z Ay ( ye z <-> y e x or y = x) (existence of successor
>
> [stuff deleted]
>
> >1)E z ax x e z
> >Ax x e D (existential instantiation )
> >D e D.
>
> There is a subtle point here. The D that is the domain
> in the sense of model theory need not be the universal
> set whose existence follows from the axiom
>
> E z Ax x e z
from being
the domain for a model of the theory.
> A general model for a language with one binary predicate symbol,
> e, and no constant or function symbols is a pair <D,R> where
> D is some set, R is a set of ordered pairs <x,y> such that
> x and y are both elements of D.
>
> Your axiom
>
> E z Ax x e z
>
> implies that for any D,R satisfying your axioms,
>
> E z in D, A x in D, <x,z> in R
>
> It does *not* follow that <D,D> is in R.
>
> For example, here is a model of the above axioms:
>
> Let D = { 0, 1, 2, .... } union { omega }
>
> where 0 = empty set, 1 = { 0 }, 2 = { 0, 1 }, etc.
> and where omega = { 0, 1, 2, ... }
>
> Let R = { <x,y> | (x and y are naturals, and x < y), or y = omega }
>
> Then the structure <D,R> satisfies your axioms, I believe.
> But D is not equal to the universal set, omega.
we have w R w , or w e w ,so
w = {0,1,2 . . .w}, the braces representing inclusion as now defined.
-apoorv
Daryl McCullough
>
>On Mar 3, 11:32=A0pm, stevendaryl3...@yahoo.com (Daryl McCullough)
>wrote:
>> There is a subtle point here. The D that is the domain
>> in the sense of model theory need not be the universal
>> set whose existence follows from the axiom
>>
>> E z Ax x e z
>But it could be and there seems to be nothing that prevents that set
>from being the domain for a model of the theory.
Well, it depends on what set theory you are using to do your
model theory. If the model theory is being done in ZFC, then
no set can be an element of itself.
>> A general model for a language with one binary predicate symbol,
>> e, and no constant or function symbols is a pair <D,R> where
>> D is some set, R is a set of ordered pairs <x,y> such that
>> x and y are both elements of D.
>>
>> Your axiom
>>
>> E z Ax x e z
>>
>> implies that for any D,R satisfying your axioms,
>>
>> E z in D, A x in D, <x,z> in R
>>
>> It does *not* follow that <D,D> is in R.
>>
>> For example, here is a model of the above axioms:
>>
>> Let D = { 0, 1, 2, .... } union { omega }
>>
>> where 0 = empty set, 1 =3D { 0 }, 2 =3D { 0, 1 }, etc.
>> and where omega =3D { 0, 1, 2, ... }
>>
>> Let R = { <x,y> | (x and y are naturals, and x < y), or y =3D omega }
>>
>> Then the structure <D,R> satisfies your axioms, I believe.
>> But D is not equal to the universal set, omega.
>
>At the object level ,with the definition of R as proposed,
> we have w R w , or w e w ,so
>w = {0,1,2 . . .w}, the braces representing inclusion as now defined.
No, you don't have that. The symbol "w" does not appear in the
object language, and neither does the notation {0,1,...}. So such
a sentence as w = { 0, 1, 2, ..., w} is not a sentence in object
language.
apoorv
> apoorv says...
>
>
>
> >On Mar 3, 11:32=A0pm, stevendaryl3...@yahoo.com (Daryl McCullough)
> >wrote:
> >> There is a subtle point here. The D that is the domain
> >> in the sense of model theory need not be the universal
> >> set whose existence follows from the axiom
>
> >> E z Ax x e z
> >But it could be and there seems to be nothing that prevents that set
> >from being the domain for a model of the theory.
>
> Well, it depends on what set theory you are using to do your
> model theory. If the model theory is being done in ZFC, then
> no set can be an element of itself.
regularity,
the set D defined as (with w as a shorthand notation for omega as
usually understood) D= w U {D} would exist. This is the domain for a
model of theory
comprising the 3 axioms . Pl also note that in the theory,
D is the universal set defined by axiom 1) and w =D-{D} does not
exist.
>
>
>
>
> >> A general model for a language with one binary predicate symbol,
> >> e, and no constant or function symbols is a pair <D,R> where
> >> D is some set, R is a set of ordered pairs <x,y> such that
> >> x and y are both elements of D.
>
> >> Your axiom
>
> >> E z Ax x e z
>
> >> implies that for any D,R satisfying your axioms,
>
> >> E z in D, A x in D, <x,z> in R
>
> >> It does *not* follow that <D,D> is in R.
>
> >> For example, here is a model of the above axioms:
>
> >> Let D = { 0, 1, 2, .... } union { omega }
>
> >> where 0 = empty set, 1 =3D { 0 }, 2 =3D { 0, 1 }, etc.
> >> and where omega =3D { 0, 1, 2, ... }
>
> >> Let R = { <x,y> | (x and y are naturals, and x < y), or y =3D omega }
>
> >> Then the structure <D,R> satisfies your axioms, I believe.
> >> But D is not equal to the universal set, omega.
>
> >At the object level ,with the definition of R as proposed,
> > we have w R w , or w e w ,so
> >w = {0,1,2 . . .w}, the braces representing inclusion as now defined.
>
> No, you don't have that. The symbol "w" does not appear in the
> object language, and neither does the notation {0,1,...}. So such
> a sentence as w = { 0, 1, 2, ..., w} is not a sentence in object
> language.
I used 'w' as a usual shorthand for 'omega' , which by your
definition is 'in' the domain , and therefore part of the object
language. Also, by your definition of
'e', we have Ax x e w and therefore, w e w , and w ={0,1,2 ...w}
-apoorv
Daryl McCullough
>> Well, it depends on what set theory you are using to do your
>> model theory. If the model theory is being done in ZFC, then
>> no set can be an element of itself.
> Of course, we are not doing model theory in ZFC. But in ZFC-
>regularity,
When did you say that?
I just want to note that you *can* use ZFC (with regularity) to
construct models of non-well-founded theories. In that case,
the relation R will not be the membership relation.
>> No, you don't have that. The symbol "w" does not appear in the
>> object language, and neither does the notation {0,1,...}. So such
>> a sentence as w = { 0, 1, 2, ..., w} is not a sentence in object
>> language.
>
> I used 'w' as a usual shorthand for 'omega,
I know. But omega is not a term in your object language.
>which by your definition is 'in' the domain,
>and therefore part of the object
>language.
No, it isn't. The object language has one binary predicate symbol, e,
no constant symbols and no function symbols. That's it. There is no
symbol "omega".
>Also, by your definition of
>'e', we have Ax x e w and therefore, w e w , and w = {0,1,2 ...w}
Look, we are talking about two different theories, with two different
languages. omega is a term of the *metatheory*. e is a relation of
the *object* theory. So a sentence such as w e w is completely nonsensical.
It doesn't make sense in either theory.
The sentence <w,w> in R makes sense in the metatheory.
The sentence "Exists z, Forall x, x e z" makes sense in the object theory.
The sentence "Exists z, z e z" makes sense in the object theory.
The sentence "w e w" makes no sense in either theory.
If you want, you can certainly have a constant D in your object theory.
Then you can have axioms:
1. Forall x, x e D
2. Exists x, Forall y, not (y e x)
3. Forall x, (x e D -> Exists y, y e D and Forall z, z e y <-> z e x or z=x)
So in this theory, you can certainly prove
D e D.
But the "D" of the object theory is *not* necessarily
the domain in the metatheory.
MoeBlee
> On Mar 3, 11:32 pm, stevendaryl3...@yahoo.com (Daryl McCullough)
> wrote:
>
> > apoorv says...
>
> > >> > > Consider a theory with these axioms:
> > >> > > 1)E z Ax x e z (existence of domain D)
> > >> > > 2) E z Ax ~ x e z (existence of null set 0)
> > >> > > 3)Ax E z Ay ( ye z <-> y e x or y = x) (existence of successor
>
> > [stuff deleted]
>
> > >1)E z ax x e z
> > >Ax x e D (existential instantiation )
> > >D e D.
>
> > There is a subtle point here. The D that is the domain
> > in the sense of model theory need not be the universal
> > set whose existence follows from the axiom
>
> > E z Ax x e z
>
> But it could be and there seems to be nothing that prevents that set
> from being
> the domain for a model of the theory.
Your theory does not even provide a unique universal set. So when you
say "that set", it's quite ambiguous.
A model of your theory is a <D R> with a nonempty set D and relation
R on D such that youre three axioms are true when 'e' maps to R. Your
theory does not provde that there is a unique object z such that Ax
xez. In your theory, unless you prove otherwise, we cannot assume that
there are not MANY objects z such that Ax xez, so (unless proven
otherwise) we cannot assume that there are not many objects b in D
such that for all members m of D we have <m b> in R.
Moreover, if Z set theory is our meta-theory in which we discuss
models for your theory, then the axiom of regularity rules out there
is a D such that D is a member of D. Moreover, even WITHOUT the axiom
of regularity, you'll have no chance to prove that there is such a D,
since the axiom of regularity is independent of the rest of the axioms
of Z.
> > A general model for a language with one binary predicate symbol,
> > e, and no constant or function symbols is a pair <D,R> where
> > D is some set, R is a set of ordered pairs <x,y> such that
> > x and y are both elements of D.
>
> > Your axiom
>
> > E z Ax x e z
>
> > implies that for any D,R satisfying your axioms,
>
> > E z in D, A x in D, <x,z> in R
>
> > It does *not* follow that <D,D> is in R.
>
> > For example, here is a model of the above axioms:
>
> > Let D = { 0, 1, 2, .... } union { omega }
>
> > where 0 = empty set, 1 = { 0 }, 2 = { 0, 1 }, etc.
> > and where omega = { 0, 1, 2, ... }
>
> > Let R = { <x,y> | (x and y are naturals, and x < y), or y = omega }
>
> > Then the structure <D,R> satisfies your axioms, I believe.
> > But D is not equal to the universal set, omega.
>
> At the object level ,with the definition of R as proposed,
> we have w R w , or w e w ,so
What are you talking about? Do you mean 'w' as the set of natural
numbers? You've not defined anything like that in YOUR OBJECT theory.
> w = {0,1,2 . . .w}, the braces representing inclusion as now defined.
That does not follow by any axioms or system of reasoning you've
proposed.
What satisfaction do you get from cluttering yourself with nonsense
when instead you could be acquiring a disciplined understanding of the
subject?
MoeBlee
MoeBlee
> On Mar 4, 4:44 pm, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
>
> > apoorv says...
>
> > >On Mar 3, 11:32=A0pm, stevendaryl3...@yahoo.com (Daryl McCullough)
> > >wrote:
> > >> There is a subtle point here. The D that is the domain
> > >> in the sense of model theory need not be the universal
> > >> set whose existence follows from the axiom
>
> > >> E z Ax x e z
> > >But it could be and there seems to be nothing that prevents that set
> > >from being the domain for a model of the theory.
>
> > Well, it depends on what set theory you are using to do your
> > model theory. If the model theory is being done in ZFC, then
> > no set can be an element of itself.
>
> Of course, we are not doing model theory in ZFC. But in ZFC-
> regularity,
Okay, but I don't see why that is "of course". But anyway, even in ZFC-
regularity, you will not be able to provide a D such that D in D,
since regularity is independent of the axioms of ZFC. Or, more
directly put, from ZFC-regularity, we cannot prove nor disprove:
Es ses.
> the set D defined as (with w as a shorthand notation for omega as
> usually understood) D= w U {D} would exist.
No. WRONG. You cannot prove the existence of such a set in ZFC-
regularity.
> This is the domain for a
> model of theory
> comprising the 3 axioms .
No, it's not. Rather, you merely ASSERT that there exists such a set
while you do not PROVE that there exists such a set.
> l also note that in the theory,
> D is the universal set defined by axiom 1)
You've not proven in your theory that there even IS a set to be called
"THE universal set".
And if 'D' is defined in the META-theory then you cannot use the
letter 'D' in the object theory and cohrently think that all the
properties of the set D in the meta-theory are even EXPRESSIBLE in the
object theory. You are mixing meta and object in a typically ignorant
way.
> and w =D-{D} does not
> exist.
There is a D that satisfies the above equation.
Let D = w. Then w = D\{D}.
What are you talking about? w may or not be in D, depending on what
exact set D is.
>, and therefore part of the object
> language.
No it's not! You haven't defined 'w' in your OBJECT theory.
> Also, by your definition of
> 'e',
There is NO definition of 'e'. Rather, 'e' is PRIMITIVE. Or, if you
want to define 'e', then you must do so - in the object theory - from
primitives. But the only other primitive you have is '='.
> we have Ax x e w and therefore, w e w , and w ={0,1,2 ...w}
What complete nonsense! You are only fooling yourself with this rot.
What a waste of whatever intellect you have.
MoeBlee
apoorv
my understanding is that If the existence of an object with a
particular
attribute can be asserted,then that object could be referred to by a
constant symbol .For example, (assuming extensionality),consider
Ez Ax ~x e z,
Instead of saying, "the z such that Ax ~x e z", we use 0.
Could we not use a constant symbol ,say w or D for "the z such
that Ax x e z"?
> If you want, you can certainly have a constant D in your object theory.
> Then you can have axioms:
>
> 1. Forall x, x e D
> 2. Exists x, Forall y, not (y e x)
> 3. Forall x, (x e D -> Exists y, y e D and Forall z, z e y <-> z e x or z=x)
Maybe ,We could state the three axioms as you suggest.But would we
have to
assert the existence of D before 1) could be stated?On the whole,
the formulation
Ez Ax xe z seems preferable.
> So in this theory, you can certainly prove
> D e D.
>
> But the "D" of the object theory is *not* necessarily
> the domain in the metatheory.
But that D could be ,and is the most appropriate domain in the
metatheory.
We are then not required to postulate the existence of a 'domain' in
the metatheory
that does not exist in the object theory;we are spared the unending
layers of infinities and metatheories.
This D is a closer fit to our intuitive notion of infinity as all
encompassing,
closer to notion of infinity in ordinary arithmetic (infinity +1=
infinity)and more
in consonance with the notion of an iteratively concieved universe
(one does not start off with postulating an infinite domain in the
metatheory)
-apoorv
Daryl McCullough
> my understanding is that If the existence of an object with a
>particular attribute can be asserted,then that object could be
>referred to by a constant symbol.
Yes, you can extend the original language with a new constant
symbol to get a new language that is equivalent to the old
language. But it doesn't make sense to call this new constant
"omega", because omega already has a standard definition,
and that is the set of all natural numbers.
>> If you want, you can certainly have a constant D in your object theory.
>> Then you can have axioms:
>>
>> 1. Forall x, x e D
>> 2. Exists x, Forall y, not (y e x)
>> 3. Forall x, (x e D -> Exists y, y e D and Forall z, z e y <-> z e x or z=
> Maybe ,We could state the three axioms as you suggest. But would we
>have to assert the existence of D before 1) could be stated?
Not if it is an axiom.
>On the whole, the formulation Ez Ax xe z seems preferable.
Why? They are equivalent.
>> So in this theory, you can certainly prove
>> D e D.
>>
>> But the "D" of the object theory is *not* necessarily
>> the domain in the metatheory.
>But that D could be, and is the most appropriate domain in the
>metatheory.
Well, I guess I don't know what your objectives are here.
I don't see any advantage at all for doing that. The point
of using regular ZF as the metatheory is because one might
question whether a theory with D e D is consistent. If you
can interpret the theory in well-founded set theory, then
you know that it is consistent if regular ZF is.
There is a theory, Quine's NF, that has a universal set.
Maybe you should look at that.
apoorv
This theory also offers a possible way out of Zeno's paradox.
Because D={0,1,2. . .D} and D-{D}={0,1,2 . . } does not exist,
infinite sequences contain necessarily the limit point.
If the hare in Zeno's paradox starts at (position) 0 and the dog at 1
, the set of hare's positions is {1,3/2,7/4,. . .2} and that of the
dog is {0,1,3/2,7/4,. . .2}, so that the dog always catches the hare!
It also offers a way of understanding the continuing debate in this
forum
about whether the smallest infinite set contains an infinite member--
in ZFC, the smallest infinite set has no infinite member; in this
theory, the only infinite set has itself as a member.
--apoorv
MoeBlee
> Well, conceptually, it appeals to have just one single infinity.
Your three axioms?
You have not yet proven even the existenced of a unique object of ANY
kind, let alone a unique "infinity".
> This theory also offers a possible way out of Zeno's paradox.
Your three axioms?
You have not yet proven even that there is a number 1, let alone
anything of greater mathematical interest.
MoeBlee
apoorv
> On Mar 6, 2:48 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > Well, conceptually, it appeals to have just one single infinity.
>
> Your three axioms?
>
> You have not yet proven even the existenced of a unique object of ANY
> kind, let alone a unique "infinity".
> > This theory also offers a possible way out of Zeno's paradox.
>
> Your three axioms?
>
> You have not yet proven even that there is a number 1, let alone
> anything of greater mathematical interest.
the existence of any 'object' any less than ZFC or its variant 'prove'
the existence of a zoo of transfinites.is it that you believe that the
domain for any model of any theory should be a set in some variant of
Z -set theory?
-apoorv
MoeBlee
> We add extensionality
That will facilitate things. And you now can define '1'. But it still
is not enough for your claim that DeD.
> > You have not yet proven even that there is a number 1, let alone
> > anything of greater mathematical interest.
>
> I am really curious to know why you believe that this theory 'proves'
> the existence of any 'object' any less than ZFC
I'll believe it proves all the existence theorems of ZFC when you
prove that it does. Meanwhile, since your theory is weaker than ZFC,
it's pretty ease to see that it doesn't prove all of existence
theorems of ZFC.
For example, please show your proof of this existence theorem of ZFC:
AzExAy(yex <-> Aj(jey -> jez))
> or its variant 'prove'
> the existence of a zoo of transfinites.is it that you believe that the
> domain for any model of any theory should be a set in some variant of
> Z -set theory?
I'm not sure I understand your question. But if a declared meta-theory
R for a theory T is such that "Ax x is set" (where, by definition, 'x
is a set <-> (x = the_y(Az ~zey) v Ezy(zexey))') then, perforce, any
domain of a model is a set.
Generally in mathematical logic, the DEFINITION of a model includes
that the domain is a set; and the theory in which mathematical logic
and model theory is formalized is presumed to be some kind of set
theory. However, of course, one could adopt a class theory such as NBG
as the theory in which to formalize, or even some other kind of formal
foundational theory; and one could adopt a different defintion of
'model' so that proper classes could be domains. Yet, this is not
without some difficulty at least. (See, for example, remarks in Chang
& Keisler, where Morse class theory is used as the official meta-
theory.)
So, if you have some alternative meta-theory, you should declare it so
that we can evaluate your claims in that regard.
MoeBlee
is a theorem of R, then
MoeBlee
> your theory is weaker than ZFC
CORRECTION: Since you have the axiom "ExAy yex", of course, we can't
say that your theory is weaker than ZFC. I meant to say that we don't
have a proof that your theory is stronger than ZFC.
MoeBlee
MoeBlee
And, by the way, if your theory is consistent, then it is not stronger
than ZFC (since ZFC has the theorem "~ExAy yex"), so your theory is
not comparable in strength with ZFC. In any case, back to the
substantive question, you've not given any evidence your theory proves
existence of even a fair sampling of even the more "ordinary" objects
that ZFC proves to exist.
MoeBlee
apoorv
> On Mar 8, 11:48 pm, apoorv <sudhir...@hotmail.com> wrote:
>
> > We add extensionality
>
> That will facilitate things. And you now can define '1'. But it still
> is not enough for your claim that DeD.
see the basis for your assertion that the theory does not prove D e
D.
> > > You have not yet proven even that there is a number 1, let alone
> > > anything of greater mathematical interest.
>
> > I am really curious to know why you believe that this theory 'proves'
> > the existence of any 'object' any less than ZFC
>
> I'll believe it proves all the existence theorems of ZFC when you
> prove that it does. Meanwhile, since your theory is weaker than ZFC,
> it's pretty ease to see that it doesn't prove all of existence
> theorems of ZFC.
theorems of ZFC. Indeed, the main idea is to set up a theory which
allows for the
exisrence of just one infinite set and no more.
> For example, please show your proof of this existence theorem of ZFC:
>
> AzExAy(yex <-> Aj(jey -> jez))
>
> > or its variant 'prove'
> > the existence of a zoo of transfinites.is it that you believe that the
> > domain for any model of any theory should be a set in some variant of
> > Z -set theory?
>
> I'm not sure I understand your question. But if a declared meta-theory
> R for a theory T is such that "Ax x is set" (where, by definition, 'x
> is a set <-> (x = the_y(Az ~zey) v Ezy(zexey))') then, perforce, any
> domain of a model is a set.
x e x
for a set.
> Generally in mathematical logic, the DEFINITION of a model includes
> that the domain is a set; and the theory in which mathematical logic
> and model theory is formalized is presumed to be some kind of set
> theory. However, of course, one could adopt a class theory such as NBG
> as the theory in which to formalize, or even some other kind of formal
> foundational theory; and one could adopt a different defintion of
> 'model' so that proper classes could be domains. Yet, this is not
> without some difficulty at least. (See, for example, remarks in Chang
> & Keisler, where Morse class theory is used as the official meta-
> theory.)
>
> So, if you have some alternative meta-theory, you should declare it so
> that we can evaluate your claims in that regard.
theory to supply a domain for a meaningful interpretation of its
variables. Nonetheless,
if the domain has to be a set in a more conventionally formulated
theory, we
could consider ZFc-regularity +~regularity , with the ~regularity
being incorporated in the theory by explicitly asserting the existence
D defined by x e D iff (x is a finite ordinal or x =D)
-apoorv
apoorv
transfinites.
This is a theory which proves the existence of a single infinite set
ehich contains
itself and is the domain for the theory.
-apoorv
MoeBlee
> On Mar 10, 12:55 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 8, 11:48 pm, apoorv <sudhir...@hotmail.com> wrote:
>
> > > We add extensionality
>
> > That will facilitate things. And you now can define '1'. But it still
> > is not enough for your claim that DeD.
>
> since by the axioms ,a (unique) universal set exists, i am unable to
> see the basis for your assertion that the theory does not prove D e
> D.
I've explained it half a dozen times already. 'D' is defined in the
the meta-theory, not in your object theory.
Yes, your theory proves Ex xex. But that says NOTHING about D, which
is something discussed in a DIFFERENT theory.
> > > You have not yet proven even that there is a number 1, let alone
> > > > anything of greater mathematical interest.
>
> > > I am really curious to know why you believe that this theory 'proves'
> > > the existence of any 'object' any less than ZFC
>
> > I'll believe it proves all the existence theorems of ZFC when you
> > prove that it does. Meanwhile, since your theory is weaker than ZFC,
> > it's pretty ease to see that it doesn't prove all of existence
> > theorems of ZFC.
>
> There is no attempt to show that this theory proves all the existence
> theorems of ZFC. Indeed, the main idea is to set up a theory which
> allows for the
> exisrence of just one infinite set and no more.
But previously you referred to all kinds of things as if defined in
your object theory. Please make up your mind.
And now how do you define 'infinite' in your object theory and prove
Ex x is infinite in your object theory?
> > I'm not sure I understand your question. But if a declared meta-theory
> > R for a theory T is such that "Ax x is set" (where, by definition, 'x
> > is a set <-> (x = the_y(Az ~zey) v Ezy(zexey))') then, perforce, any
> > domain of a model is a set.
>
> If I read your definition right, you have not excluded the possibilty
> x e x
> for a set.
It depends on T.
> > Generally in mathematical logic, the DEFINITION of a model includes
> > that the domain is a set; and the theory in which mathematical logic
> > and model theory is formalized is presumed to be some kind of set
> > theory. However, of course, one could adopt a class theory such as NBG
> > as the theory in which to formalize, or even some other kind of formal
> > foundational theory; and one could adopt a different defintion of
> > 'model' so that proper classes could be domains. Yet, this is not
> > without some difficulty at least. (See, for example, remarks in Chang
> > & Keisler, where Morse class theory is used as the official meta-
> > theory.)
>
> > So, if you have some alternative meta-theory, you should declare it so
> > that we can evaluate your claims in that regard.
>
> The theory defines its own domain and need not call upon any other
> theory to supply a domain for a meaningful interpretation of its
> variables.
Bull. Pure unsupported assertion.
> Nonetheless,
> if the domain has to be a set in a more conventionally formulated
> theory, we
> could consider ZFc-regularity +~regularity , with the ~regularity
> being incorporated in the theory
Here, by 'the theory', we're talking about the META-theory.
> by explicitly asserting the existence
> D defined by x e D iff (x is a finite ordinal or x =D)
"ASSERTING existence". I.e., taking as an axiom for the META-theory.
Fine.
Now specify a 2-place relation R on D such that when the universal
quantifier of your object language maps to D, and 'e' (from your
object language) maps to R, we have that every axiom of your OBJECT
theory is true by the just mentioned mapping (i.e., the just mentioned
structure for the object language).
MoeBlee
MoeBlee
> This is a theory which proves the existence of a single infinite set
> ehich contains
> itself and is the domain for the theory.
Define (in your OBJECT THEORY) 'x is infinite', and prove (in your
OBJECT THEORY) the formula "E!x(x is infinite & xex)".
Then, using your specified meta-theory ZFC-R+~R+"EdAy(y in d <-> (y is
a finite ordinal v y=d))", specify a 2-place relation R on D such that
when the universal quantifier of your object language maps to D, and
'e' (from your object language) maps to R, we have that every axiom of
your OBJECT theory is true by the just mentioned mapping (i.e., the
just mentioned structure for the object language).
Then, tell us signifcance you find in this?
MoeBlee
apoorv
> On Mar 12, 3:13 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > On Mar 10, 12:55 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 8, 11:48 pm, apoorv <sudhir...@hotmail.com> wrote:
>
> > > > We add extensionality
>
> > > That will facilitate things. And you now can define '1'. But it still
> > > is not enough for your claim that DeD.
>
> > since by the axioms ,a (unique) universal set exists, i am unable to
> > see the basis for your assertion that the theory does not prove D e
> > D.
>
> I've explained it half a dozen times already. 'D' is defined in the
> the meta-theory, not in your object theory.
>
> Yes, your theory proves Ex xex. But that says NOTHING about D, which
> is something discussed in a DIFFERENT theory.
universal set in the theory and clearly D e D. I am unable to see
where is this different theory coming from.
> > > > You have not yet proven even that there is a number 1, let alone
> > > > > anything of greater mathematical interest.
>
> > > > I am really curious to know why you believe that this theory 'proves'
> > > > the existence of any 'object' any less than ZFC
>
> > > I'll believe it proves all the existence theorems of ZFC when you
> > > prove that it does. Meanwhile, since your theory is weaker than ZFC,
> > > it's pretty ease to see that it doesn't prove all of existence
> > > theorems of ZFC.
>
> > There is no attempt to show that this theory proves all the existence
> > theorems of ZFC. Indeed, the main idea is to set up a theory which
> > allows for the
> > exisrence of just one infinite set and no more.
>
> But previously you referred to all kinds of things as if defined in
> your object theory. Please make up your mind.
you have been in this thread continuously , it is hard to believe that
you missed that the theory
was designed to allow for the existence of one infinite set and no
more.
My own sense is that you are so mesmerized by this garden of the
transfinite and your own (believed) felicity with some of the
symbolism that you are unable to see or would not see any other
perspective. Pleasse wake up!
> And now how do you define 'infinite' in your object theory and prove
> Ex x is infinite in your object theory?
successor.
> > > I'm not sure I understand your question. But if a declared meta-theory
> > > R for a theory T is such that "Ax x is set" (where, by definition, 'x
> > > is a set <-> (x = the_y(Az ~zey) v Ezy(zexey))') then, perforce, any
> > > domain of a model is a set.
>
> > If I read your definition right, you have not excluded the possibilty
> > x e x
> > for a set.
>
> It depends on T.
adequate for the possibilty x ex.
>
>
>
>
> > > Generally in mathematical logic, the DEFINITION of a model includes
> > > that the domain is a set; and the theory in which mathematical logic
> > > and model theory is formalized is presumed to be some kind of set
> > > theory. However, of course, one could adopt a class theory such as NBG
> > > as the theory in which to formalize, or even some other kind of formal
> > > foundational theory; and one could adopt a different defintion of
> > > 'model' so that proper classes could be domains. Yet, this is not
> > > without some difficulty at least. (See, for example, remarks in Chang
> > > & Keisler, where Morse class theory is used as the official meta-
> > > theory.)
>
> > > So, if you have some alternative meta-theory, you should declare it so
> > > that we can evaluate your claims in that regard.
>
> > The theory defines its own domain and need not call upon any other
> > theory to supply a domain for a meaningful interpretation of its
> > variables.
>
> Bull. Pure unsupported assertion.
> > Nonetheless,
> > if the domain has to be a set in a more conventionally formulated
> > theory, we
> > could consider ZFc-regularity +~regularity , with the ~regularity
> > being incorporated in the theory
>
> Here, by 'the theory', we're talking about the META-theory.
>
> > by explicitly asserting the existence
> > D defined by x e D iff (x is a finite ordinal or x =D)
>
> "ASSERTING existence". I.e., taking as an axiom for the META-theory.
> Fine.
>
> Now specify a 2-place relation R on D such that when the universal
> quantifier of your object language maps to D, and 'e' (from your
> object language) maps to R, we have that every axiom of your OBJECT
> theory is true by the just mentioned mapping (i.e., the just mentioned
> structure for the object language).
-apoorv
apoorv
of the transfinite ' does not appear as significant to you, than all i
can say is that you are
well and truly bewitched.
-apoorv
MoeBlee
> On Mar 12, 11:35 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Mar 12, 3:13 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > On Mar 10, 12:55 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 8, 11:48 pm, apoorv <sudhir...@hotmail.com> wrote:
>
> > > > > We add extensionality
>
> > > > That will facilitate things. And you now can define '1'. But it still
> > > > is not enough for your claim that DeD.
>
> > > since by the axioms ,a (unique) universal set exists, i am unable to
> > > see the basis for your assertion that the theory does not prove D e
> > > D.
>
> > I've explained it half a dozen times already. 'D' is defined in the
> > the meta-theory, not in your object theory.
>
> > Yes, your theory proves Ex xex. But that says NOTHING about D, which
> > is something discussed in a DIFFERENT theory.
>
> i think we are covering the same ground again ;D is the unique
> universal set in the theory and clearly D e D. I am unable to see
> where is this different theory coming from.
The meta-theroy comes from your talk about D being a universe of a
model for your object theory. It is in a meta-theory for an object
theory that we define structures (models) for the language of the
object theory and probe which, if any, of those is a model of the
theory.
I have no problem with, in your object theory, defining 'D' as a
constant symbol by D = the unique x such Ay yex. That is justified by
the existence and uniquenss theorem E!xAy yex, since we've added the
axiom of extenstionality to your object theory. But making that
definition doesn't simply allow us to claim that D itself is a
universe for a model of the object theory.
> > > > > You have not yet proven even that there is a number 1, let alone
> > > > > > anything of greater mathematical interest.
>
> > > > > I am really curious to know why you believe that this theory 'proves'
> > > > > the existence of any 'object' any less than ZFC
>
> > > > I'll believe it proves all the existence theorems of ZFC when you
> > > > prove that it does. Meanwhile, since your theory is weaker than ZFC,
> > > > it's pretty ease to see that it doesn't prove all of existence
> > > > theorems of ZFC.
>
> > > There is no attempt to show that this theory proves all the existence
> > > theorems of ZFC. Indeed, the main idea is to set up a theory which
> > > allows for the
> > > exisrence of just one infinite set and no more.
>
> > But previously you referred to all kinds of things as if defined in
> > your object theory. Please make up your mind.
>
> I think you are deliberately obfuscating and misinterpreting. since
> you have been in this thread continuously , it is hard to believe that
> you missed that the theory
> was designed to allow for the existence of one infinite set and no
> more.
You've not shown a single thing I've obfuscated or misunderstood.
Merely saying that a theory was "designed" for providing X is not a
PROOF that the theory actually DOES provide X.
If you claim that either your meta-thoery or your object theory prove
certain things, then you need to show the proofs.
> My own sense is that you are so mesmerized by this garden of the
> transfinite and your own (believed) felicity with some of the
> symbolism that you are unable to see or would not see any other
> perspective.
Your own sense about me is irrelevent. We're talking about your object
therory and your meta-theory. Whether I am or am not mesmerized by
anything is irrelevent to the fact that you've not provided proofs of
your claims. Also, I don't claim any speciality "felicity" with
symbolization beyond, say, the level of an attentive undergraduate.
Moreover, you're wrong anyway about what you think is an attraction I
have for the transfinite. I'd be quite happy to see an axiomatization
that is as easy to use as set theory while still comprehensive enough
for mathematics for the sciences but without infinite sets. Moreover,
as to greater and greater infinities, I do NOT consider that to be,
for me PERSONALLY, a desirable aspect of a theory; however, I
recognize that it does comes as part of the price of set theory as an
easy to use axiomatization for mathematics for the sciences. Moreover,
as to different perspectives, I do strive to appreciate them, but I
don't credit yours since it is confused, uninformed, and dogmatic (you
merely claim, but not prove).
> Pleasse wake up!
Wake up to what? To your false sense about myself? To your UNPROVEN
claims about your theories?
> And now how do you define 'infinite' in your object theory and prove
> > Ex x is infinite in your object theory?
>
> A simple definition is obvious. it is the only set that is its own
> successor.
You're confused. First, 'infinite' is a predicate, so its definiens is
not of the form you just gave. Second, you may wish to define a
constant by setting it to "the only set that is its own successor",
but first you need to prove that there is a unique set that is its own
successor. Third, such a definition, pertaining to the term 'infinite'
would not at all be OBVIOUS, since that is not an ordinary definiens
of 'infinite'.
> > > I'm not sure I understand your question. But if a declared meta-theory
> > > > R for a theory T is such that "Ax x is set" (where, by definition, 'x
> > > > is a set <-> (x = the_y(Az ~zey) v Ezy(zexey))') then, perforce, any
> > > > domain of a model is a set.
>
> > > If I read your definition right, you have not excluded the possibilty
> > > x e x
> > > for a set.
>
> > It depends on T.
>
> you specified T by "Ax x is aset " and defined 'x is a set'. That is
> adequate for the possibilty x ex.
I did not specify ALL of T. I merely mentioned what we have if a
theory T proves a couple of prosaic theorems. I did not exlude that
such a theory T has the axiom of regularity, which does foreclose
having Ex xex, nor did I exclude that such a theory T does not have
the axiom of regularity, so that Ex xex is not foreclosed upon. THAT,
for example, is why I said, in response to your question, "it depends
on T".
Now, aren't you glad I'm willing to take the time to point out such
OBVIOUS considerations to you even though the main point of your
latest post is to spew false characterization of my beliefs and
inclinations.
> > > > Generally in mathematical logic, the DEFINITION of a model includes
> > > > that the domain is a set; and the theory in which mathematical logic
> > > > and model theory is formalized is presumed to be some kind of set
> > > > theory. However, of course, one could adopt a class theory such as NBG
> > > > as the theory in which to formalize, or even some other kind of formal
> > > > foundational theory; and one could adopt a different defintion of
> > > > 'model' so that proper classes could be domains. Yet, this is not
> > > > without some difficulty at least. (See, for example, remarks in Chang
> > > > & Keisler, where Morse class theory is used as the official meta-
> > > > theory.)
>
> > > > So, if you have some alternative meta-theory, you should declare it so
> > > > that we can evaluate your claims in that regard.
>
> > > The theory defines its own domain and need not call upon any other
> > > theory to supply a domain for a meaningful interpretation of its
> > > variables.
>
> > Bull. Pure unsupported assertion.
>
> I reciprocate the sentiments.
You don't need to reciprocate sentiments. You just need to prove your
claims about your theories.
> > > Nonetheless,
> > > if the domain has to be a set in a more conventionally formulated
> > > theory, we
> > > could consider ZFc-regularity +~regularity , with the ~regularity
> > > being incorporated in the theory
>
> > Here, by 'the theory', we're talking about the META-theory.
>
> > > by explicitly asserting the existence
> > > D defined by x e D iff (x is a finite ordinal or x =D)
>
> > "ASSERTING existence". I.e., taking as an axiom for the META-theory.
> > Fine.
>
> > Now specify a 2-place relation R on D such that when the universal
> > quantifier of your object language maps to D, and 'e' (from your
> > object language) maps to R, we have that every axiom of your OBJECT
> > theory is true by the just mentioned mapping (i.e., the just mentioned
> > structure for the object language).
>
> i would think that x R y <--> x e y adequately meets the requirement.
Then interpreting 'R' to stand for the membership relation, all you
have to do now is PROVE that <D membership-relation> is a model of
your object theory. And keep in mind that what 'D' stands for in the
meta-theory cannot be assumed to be as 'D' is defined in the object
theory.
MoeBlee
MoeBlee
Aside from whatever super-natural spells you think I might be under,
and aside from your utterly nebulous claims about freedom form
enchanted gardens, I just would like to know the SPECIFIC MATHEMATICAL
signficance you would point out were you even to accomplish proving
that the particular model you formulated in your particular meta-
theory is a model of your particular object theory. Of course, if you
did that, at least you would show a willingness to prove your claims;
but still, even if you prove, in your particular meta-theory, that <D
membership-relation on D> ('D' not necessarily standing here for the
same as 'D' does in your object theory) is a model of your object
theory, you still have not shown that you've devised anything that
remotely presents as an alternative axiomatization for mathematics for
the sciences. Moreover, your meta-theorem about your object theory is
itself in a theory STRONGER than ZFC-regularity!
MoeBlee
apoorv
anything
that in your view would bar or prevent D from being a domain for a
model of
the theory? is it because D exists in the object theory that it is
barred from being a
domain for a model of the object theory?
theory,
you do come out as defending that with some amount of zeal.You also
overstate the case when you that it comes as a price for an easy to
use axiomatization for mathematics for the sciences. Practically,
sciences work on a 'finite precision arithmetic' . Sciences have used
mathematics but also been pragmatic enough
in interpreting mathematical results. The trajectory of a projectile
is not an open interval and when a mason lay tiles, it is not that the
first tile coincides with the
interval [0,1] and the next with (1,2].
> Moreover, as to different perspectives, I do strive to appreciate them, but I
> don't credit yours since it is confused, uninformed, and dogmatic (you
> merely claim, but not prove).
it is the prism through which you see that colours your view.
You are only repeating what i said -that the possibility Ex x ex is
not
excluded. I did not say that it was or was not the case.
What makes you think that <D, membership relation> does not satisfy
the
axioms of the theory?
-apoorv
apoorv
theory
nescessarily has to be different from d in the meta-theory and why
<D,e> is not a model for the theory;
-apoorv
MoeBlee
> > The meta-theroy comes from your talk about D being a universe of a
> > model for your object theory. It is in a meta-theory for an object
> > theory that we define structures (models) for the language of the
> > object theory and probe which, if any, of those is a model of the
> > theory.
You skipped right past this on the way to your questions below.
> > I have no problem with, in your object theory, defining 'D' as a
> > constant symbol by D = the unique x such Ay yex. That is justified by
> > the existence and uniquenss theorem E!xAy yex, since we've added the
> > axiom of extenstionality to your object theory. But making that
> > definition doesn't simply allow us to claim that D itself is a
> > universe for a model of the object theory.
>
> We could work with different models with different domains. is there
> anything
> that in your view would bar or prevent D from being a domain for a
> model of
> the theory? is it because D exists in the object theory that it is
> barred from being a
> domain for a model of the object theory?
D has to be defined in your META-theory for it to elgible to be a
universe for a model for the language of your object theory. All of
the discussion about models and universes takes place in the META-
theory. And simply using the same letter 'D' as a constant symbol in
the language of your meta-theory and defining D to be a certain set in
your meta-theory does not carry over to have any connection with using
the letter 'D' as a constant symbol in the language of your object
theory. Now, of course, once you defined D to be a certain non-empty
set in your meta-theory, you can define a model for the langauge of
your object theory such that D is the universe for the model. But that
does not in itself entail that the model is a model of your object
THEORY. It requires PROOF to show that a model (i.e. structure) FOR
the LANGUAGE of a theory is also a model OF a particular THEORY in
that language.
These kinds of things would be clear to you if you just read a good
book on the subject. I don't see why you don't have the intellectual
self-reliance just to read a book rather than to remain continually
confused and only intermittently informed by ad hoc explanations by
other posters.
> > Your own sense about me is irrelevent. We're talking about your object
> > therory and your meta-theory. Whether I am or am not mesmerized by
> > anything is irrelevent to the fact that you've not provided proofs of
> > your claims. Also, I don't claim any speciality "felicity" with
> > symbolization beyond, say, the level of an attentive undergraduate.
> > Moreover, you're wrong anyway about what you think is an attraction I
> > have for the transfinite. I'd be quite happy to see an axiomatization
> > that is as easy to use as set theory while still comprehensive enough
> > for mathematics for the sciences but without infinite sets. Moreover,
> > as to greater and greater infinities, I do NOT consider that to be,
> > for me PERSONALLY, a desirable aspect of a theory; however, I
> > recognize that it does comes as part of the price of set theory as an
> > easy to use axiomatization for mathematics for the sciences.
>
> Even if you do not consider transfinites a desirable aspect of a
> theory,
> you do come out as defending that with some amount of zeal.
Please site a PARTICULAR quote in which you think I show "zeal" in
defense of infinite sets.
What I am zealous about is refuting specious arguments that the
existence of infinite sets is inconsistent and things along those
lines. That is quite differerent from a "zealous defense for the claim
that there exist infinite sets".
> You also
> overstate the case when you that it comes as a price for an easy to
> use axiomatization for mathematics for the sciences. Practically,
> sciences work on a 'finite precision arithmetic' . Sciences have used
> mathematics but also been pragmatic enough
> in interpreting mathematical results. The trajectory of a projectile
> is not an open interval and when a mason lay tiles, it is not that the
> first tile coincides with the
> interval [0,1] and the next with (1,2].
You said "practically". That's different from the question of formal
axiomatization. Yes, one could do all kinds of practical calculations
without even having a notion of 'axiom'. But AS TO the specific
question of axiomatization, Z set theory does provide such an
axiomatization.
Moreover, I did not claim that there does NOT exist other
axiomatizations, with only finite sets, that may also be easy. Only
that Z set theory happens to be easy and that any other axiomatization
with only finite sets may be compared, whether favorably or
unfavorably, to Z set theory on criteria such as ease, simplicity,
intuitiveness, faithfulness to reasonable foundational conceptions,
etc.
> > Moreover, as to different perspectives, I do strive to appreciate them, but I
> > don't credit yours since it is confused, uninformed, and dogmatic (you
> > merely claim, but not prove).
>
> it is the prism through which you see that colours your view.
You should seriously think about submitting that for one of those
"Thought ForThe Day" calendars. Meanwhile, it does nothing to refute
anything I've said.
> > > Pleasse wake up!
>
> > Wake up to what? To your false sense about myself? To your UNPROVEN
> > claims about your theories?
>
> > > And now how do you define 'infinite' in your object theory and prove
> > > > Ex x is infinite in your object theory?
>
> > > A simple definition is obvious. it is the only set that is its own
> > > successor.
>
> > You're confused. First, 'infinite' is a predicate, so its definiens is
> > not of the form you just gave. Second, you may wish to define a
> > constant by setting it to "the only set that is its own successor",
> > but first you need to prove that there is a unique set that is its own
> > successor. Third, such a definition, pertaining to the term 'infinite'
> > would not at all be OBVIOUS, since that is not an ordinary definiens
> > of 'infinite'.
No reply by you.
> > > > > I'm not sure I understand your question. But if a declared meta-theory
> > > > > > R for a theory T is such that "Ax x is set" (where, by definition, 'x
> > > > > > is a set <-> (x = the_y(Az ~zey) v Ezy(zexey))') then, perforce, any
> > > > > > domain of a model is a set.
>
> > > > > If I read your definition right, you have not excluded the possibilty
> > > > > x e x
> > > > > for a set.
>
> > > > It depends on T.
>
> > > you specified T by "Ax x is aset " and defined 'x is a set'. That is
> > > adequate for the possibilty x ex.
>
> > I did not specify ALL of T. I merely mentioned what we have if a
> > theory T proves a couple of prosaic theorems. I did not exlude that
> > such a theory T has the axiom of regularity, which does foreclose
> > having Ex xex, nor did I exclude that such a theory T does not have
> > the axiom of regularity, so that Ex xex is not foreclosed upon. THAT,
> > for example, is why I said, in response to your question, "it depends
> > on T".
>
> You are only repeating what i said -that the possibility Ex x ex is
> not
> excluded. I did not say that it was or was not the case.
No, I said it depends on T whether Ex xex is excluded. Then you
commented as if there is anything at all askance in my statement that
it depends on T whether Ex xex is excluded. Then I explained for you
why it depends on T whether Ex xex is excluded.
MoeBlee
MoeBlee
> > Then interpreting 'R' to stand for the membership relation, all you
> > have to do now is PROVE that <D membership-relation> is a model of
> > your object theory. And keep in mind that what 'D' stands for in the
> > meta-theory cannot be assumed to be as 'D' is defined in the object
> > theory.
>
> What makes you think that <D, membership relation> does not satisfy
> the
> axioms of the theory?
I didn't say it doesn't. But if you claim it does, then you need to
prove it. And then if you prove it, what is the signficance? You will
have proven something in your own personal meta-theory. That doesn't
prove anything about such ordinary theories as Z set theory and the
like.
MoeBlee
MoeBlee
You just twisted what I said. I didn't say that they are necessarily
different. I said it is not necessarily the case that they are the
same.
"X is necessarily different from Y" (which is what I did NOT say) is
not equivalent with "X is not necessarily the same as Y" (which is
what I DID say).
> and why
> <D,e> is not a model for the theory;
I take it you mean <D e> being the set D defined in your meta-theory
and the membership relation on D as a model of your object theory.
I didn't say that <D e_on_D> is not a model of your object theory. I'm
simply pointing out that you haven't PROVEN, in your meta-theory, that
<D e_on_D> is a model of your object theory, and I ask what would be
the significance even if you did prove this, since what goes on in
your personal meta-theory doesn't automatically carry over to anything
about Z and other ordinary theories.
MoeBlee
hurb...@live.fr
> On Feb 12, 1:00 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > Re Feb 10, 12:02 pm, MoeBlee <jazzm...@hotmail.com>:
>
> > P.S. If one doesn't learn set theory from books and/or organized
> > lectures, then just how does one learn the subject? By flitting from
> > one Wikipedia page to another, with no overall conceptual and
> > technical context among them, complemented by various ad hoc
> > conversations with experts, sincere amateurs, and outright
> > ignoramuses, and then supplemented by watching one's favorite episodes
> > of "Friends"?
>
> > MoeBlee
>
> Well, I am only saying that we need to be alive to other perspectives.
> (after all, for 1500 years, the sun went round the earth) In any model
> of ZF set theory, the variables range over a domain D specified
> outside the theory. The domain D is itself a set, yet its existence is
> postulated outside the theory. Further, when we say, for example,
> Ax ~x e x, we are actually saying, Ax in D, ~x e x.
> So, we actually are using D as a constant in the theory ( is that
> permissible always?), and two different relations
> Namely, “in” between objects and domain, and “e” between objects.
> If we are looking for a self contained set theory, the existence of
> the domain D
> must be postulated within the theory.
> Consider a theory with these axioms:
> 1)E z Ax x e z (existence of domain D)
> 2) E z Ax ~ x e z (existence of null set 0)
> x)
>
> Axioms 2 &3) together imply that D is infinite and a separate Axiom
> of infinity is not
> needed. Because of 1) D contains itself. We consider a model with the
> smallest possible domain D. D would be the
> set D=[0,1,2 . . .D}
>
> Having postulated the existence of the domain D, we need to give up
> full fledged separation. In particular, we can see that
> ~E y Ax [x ey <->x e D and ~x =D}, that is, the set D-{D} =
> {0,1,2 . . .} does not
> exist.
> In this model,
> a) There is only one infinite set, which is its own successor .
> b) Infinity cannot be separated from itself.
> If “e” is interpreted as ‘less or equal’ and Sx as any function ,
> then every sequence
> In D is a model for the theory and D is the limit for that sequence..
>
> So, we have a choice , we either accept full fledged separation, and a
> unending escalator
> Into the transfinite, or restrict separation and work with a single
> infinite domain.
>
> -apoorv
But it is impossible to "postulate the existence" of any D of a theory
within the theory itself. If it is something to be done, it is to be
done only with inconsistent theories!
apoorv
> On Mar 20, 12:24 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > The meta-theroy comes from your talk about D being a universe of a
> > > model for your object theory. It is in a meta-theory for an object
> > > theory that we define structures (models) for the language of the
> > > object theory and probe which, if any, of those is a model of the
> > > theory.
>
> You skipped right past this on the way to your questions below.
has to score
debating points.Please tell me whether there is a bar on the object
theory being
a meta theory for itself---from one of your own posts in this
thread ,you accept that possibility.
Why do Z set theories need a meta theory different from the object
theory? That is
because there is no constant (set ) in the object theory which
'contains' all the objects in the domain for the variables in any
model .For a model , you need a domain ,and there is no entity in the
object theory [with 'e' of object theory carrying
over to 'in' of the meta theory] which could be a domain.On the other
hand ,if the object theory has a universal set U, <U,e> naturally map
over to <D,in> of the meta theory-the object theory itself is the meta
theory.
theories
as the 'gold standard'. One would have liked you referring to other
possible
perspectives , say on the questions -does Z theory allow for an
iterative
conception of the universe or whether the smallest infinite set
contains an infinite
set.
> > You also
> > overstate the case when you that it comes as a price for an easy to
> > use axiomatization for mathematics for the sciences. Practically,
> > sciences work on a 'finite precision arithmetic' . Sciences have used
> > mathematics but also been pragmatic enough
> > in interpreting mathematical results. The trajectory of a projectile
> > is not an open interval and when a mason lay tiles, it is not that the
> > first tile coincides with the
> > interval [0,1] and the next with (1,2].
>
> You said "practically". That's different from the question of formal
> axiomatization. Yes, one could do all kinds of practical calculations
> without even having a notion of 'axiom'. But AS TO the specific
> question of axiomatization, Z set theory does provide such an
> axiomatization.
>
> Moreover, I did not claim that there does NOT exist other
> axiomatizations, with only finite sets, that may also be easy. Only
> that Z set theory happens to be easy and that any other axiomatization
> with only finite sets may be compared, whether favorably or
> unfavorably, to Z set theory on criteria such as ease, simplicity,
> intuitiveness, faithfulness to reasonable foundational conceptions,
> etc.
you to assume an infinite domain and then another infinite set
contained in the first assumed set, easy and intuitive . If an
unending series of meta theories are required to interpret the theory,
i hope you do not find it 'faithful to reasonable foundational
conceptions'.
-apoorv
> > > Moreover, as to different perspectives, I do strive to appreciate them, but I
> > > don't credit yours since it is confused, uninformed, and dogmatic (you
> > > merely claim, but not prove).
>
> > it is the prism through which you see that colours your view.
>
> You should seriously think about submitting that for one of those
> "Thought ForThe Day" calendars. Meanwhile, it does nothing to refute
> anything I've said.
debate.naturally,
some of the thoughts need more working or even given up as one
progresses through a discussion. if one is exploring, one is more
likely to be tentative,
rather than when one is just moving on the beaten path.And, you can go
through the books, and get all the technique and no insight.
> MoeBlee- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
apoorv
The proof is obvious. If U is the universal set in the object
theory,'e' is the membership relation, D ={0,1,2. . .D} in the meta
theory and 'R' is the membership relation E in the meta theory
restricted to members of D
i.e Axy[x Ry iff xE D&yED &x Ey] , then clearly,
Ax x R d,
Ax ~x R 0
and
Ax Ey [Az (z R y<->z R x or Z=x]
I used Z-regularity +~regularity to construct D just to show that D
can be constructed in a conventional manner. i still believe that if
the object theory has a universal set,
then the object theory itself can be the meta theory.
Before going to the significance of this or that theory, let me put
forth the axioms
of the object theory with one addition.
1)extensionality.
2) Ax x e U [existence of universal set/def of constant U]
3)Ax ~x e 0[existence of null set/def of constant 0]
4)Ax Ey [Az(z ey <->z e x or z=x][existence of successor;y=Sx and x
=Py if x exists,x is the predecessor of y ]
Def : x is finite <->[ x=0 or (~x=0& x has a finite predessor y and
~y=x)]
5) Ax [x e U <-> x is finite or x=U]
Axiom 5) ensures that there is a unique infinite set U that contains
itself
Also, the set of all finite x does not exist.
As to significance, i wonder what happens to all the 'diagonal
arguments'
in such a theory. and it would be nice resolution to the question
'whether
the smallest infinite set contains an infinite set' that it depends on
the axioms you
choose.
-apoorv
apoorv
There could be theories with a Universal set. Thatset could be a
domain.
-apoorv
hurb...@live.fr
No!
Of course there are set theories with a universal set but that
universal set is very different from the set of the domain and their
epsilons are repectively different!
Frederick Williams
> There could be theories with a Universal set.
Plenty: http://math.boisestate.edu/~holmes/holmes/setbiblio.html.
> Thatset could be a
> domain.
No.
--
Science is a differential equation.
Religion is a boundary condition.
--Alan Turing
MoeBlee
> On Mar 21, 12:34 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 20, 12:24 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > > The meta-theroy comes from your talk about D being a universe of a
> > > > model for your object theory. It is in a meta-theory for an object
> > > > theory that we define structures (models) for the language of the
> > > > object theory and probe which, if any, of those is a model of the
> > > > theory.
>
> > You skipped right past this on the way to your questions below.
>
> Well, I see this as a forum for discussion and not debate where one
> has to score
> debating points.
But the discussion is not productive when you skip too much of the
substance. I'm sitting here explaining certain things for your
benefit, yet you're not absorbing much. And when you just skip
replying to some of my more important remarks I have no way of knowing
why it is that they have not gotten through to you.
Also, of course, the notion of an open ended discussion with you is
silly since you continue to refuse to familiarize yourself with the
basics of the subject.
And you belie your own claim not be in a debate mode when you try to
slip in various incorrect cheap shot ad hominems about me, such as
your false claims as to how I regard Z set theories and an insinuation
that I'm not open minded enough to have a wider perspective, which is
very ironic, since I, much much more than you, have spent a fair
amount of time acquainting myself with a wide range of philosophies of
mathematics and with different theories and foundational proposals,
even if I have not mastered all that material yet as I at least
continue to strive to learn more and more about it, in contrast with
you as you won't even read a first introductory textbook.
> Please tell me whether there is a bar on the object
> theory being
> a meta theory for itself---from one of your own posts in this
> thread ,you accept that possibility.
I don't opine that it is impossible, but for you to reasonably assert
that you have an object theory that acts as its own metatheory, you
have to SHOW how that is. So, in your object theory you have to define
all kinds of things, such as 'language', 'formula', 'proof', 'theory',
'model for a language', 'true in a model', etc. - which are the kinds
of things that go into claiming that a certain structure is a model of
a certain theory. But you've not done anything like that with your
object theory.
> Why do Z set theories need a meta theory different from the object
> theory? That is
> because there is no constant (set ) in the object theory which
> 'contains' all the objects in the domain for the variables in any
> model .For a model , you need a domain ,and there is no entity in the
> object theory [with 'e' of object theory carrying
> over to 'in' of the meta theory] which could be a domain.On the other
> hand ,if the object theory has a universal set U, <U,e> naturally map
> over to <D,in> of the meta theory-the object theory itself is the meta
> theory.
See my paragraph above; for your object theory to be a meta-theory of
ANYTHING, let alone of itself, we have to be able to define and prove
in your object theory all kinds of things that you haven't shown how
to define and prove in your object theory.
You skipped this. It's not a matter of "debate" but rather that if you
just skip too much of my core remarks then, of course, there's little
hope that I'll be able to get you to understand this matter.
> > These kinds of things would be clear to you if you just read a good
> > book on the subject. I don't see why you don't have the intellectual
> > self-reliance just to read a book rather than to remain continually
> > confused and only intermittently informed by ad hoc explanations by
> > other posters.
>
> > > > Your own sense about me is irrelevent. We're talking about your object
> > > > therory and your meta-theory. Whether I am or am not mesmerized by
> > > > anything is irrelevent to the fact that you've not provided proofs of
> > > > your claims. Also, I don't claim any speciality "felicity" with
> > > > symbolization beyond, say, the level of an attentive undergraduate.
> > > > Moreover, you're wrong anyway about what you think is an attraction I
> > > > have for the transfinite. I'd be quite happy to see an axiomatization
> > > > that is as easy to use as set theory while still comprehensive enough
> > > > for mathematics for the sciences but without infinite sets. Moreover,
> > > > as to greater and greater infinities, I do NOT consider that to be,
> > > > for me PERSONALLY, a desirable aspect of a theory; however, I
> > > > recognize that it does comes as part of the price of set theory as an
> > > > easy to use axiomatization for mathematics for the sciences.
>
> > > Even if you do not consider transfinites a desirable aspect of a
> > > theory,
> > > you do come out as defending that with some amount of zeal.
>
> > Please site a PARTICULAR quote in which you think I show "zeal" in
> > defense of infinite sets.
Skipped by you. Your previous claim about this was simply a
fabrication.
> > What I am zealous about is refuting specious arguments that the
> > existence of infinite sets is inconsistent and things along those
> > lines. That is quite differerent from a "zealous defense for the claim
> > that there exist infinite sets".
>
> your defense of 'infinite sets' is entirely based on presuming Z-set
> theories
> as the 'gold standard'.
I've never said Z set theories are "the gold standard". Please do not
put words in my mouth.
> One would have liked you referring to other
> possible
> perspectives , say on the questions -does Z theory allow for an
> iterative
> conception of the universe or whether the smallest infinite set
> contains an infinite
> set.
I've not claimed that Z set theories are to be automatically presumed
to be better than any other theory.
As to the iterative conception and that the least infinite ordinal is
Dedekind infinite, I don't know what point you're trying to make.
> > > You also
> > > overstate the case when you that it comes as a price for an easy to
> > > use axiomatization for mathematics for the sciences. Practically,
> > > sciences work on a 'finite precision arithmetic' . Sciences have used
> > > mathematics but also been pragmatic enough
> > > in interpreting mathematical results. The trajectory of a projectile
> > > is not an open interval and when a mason lay tiles, it is not that the
> > > first tile coincides with the
> > > interval [0,1] and the next with (1,2].
>
> > You said "practically". That's different from the question of formal
> > axiomatization. Yes, one could do all kinds of practical calculations
> > without even having a notion of 'axiom'. But AS TO the specific
> > question of axiomatization, Z set theory does provide such an
> > axiomatization.
>
> > Moreover, I did not claim that there does NOT exist other
> > axiomatizations, with only finite sets, that may also be easy. Only
> > that Z set theory happens to be easy and that any other axiomatization
> > with only finite sets may be compared, whether favorably or
> > unfavorably, to Z set theory on criteria such as ease, simplicity,
> > intuitiveness, faithfulness to reasonable foundational conceptions,
> > etc.
>
> Well, i hope you do not find a theory, every model for which requires
> you to assume an infinite domain and then another infinite set
> contained in the first assumed set, easy and intuitive .
The axioms are easy to work with and intuitive enough. However, I
myself grant that there are puzzling aspects of meta-theory about set
theory. However, such puzzling aspects have not been shown to prove
that set theory is inconsistent.
> If an
> unending series of meta theories are required to interpret the theory,
> i hope you do not find it 'faithful to reasonable foundational
> conceptions'.
Didn't I already address the matter of escalating meta-theories in
this thread? If not in this thread, then I have in other threads,
touching on such points as (1) the infinite escalation is required
only if one demands there can't be an informal starting point, and if
one demands that there not be an informal starting point, then what
OTHER theory can avoid such an infinite escalation? (2) the escalation
can be, for practical purposes, cut after the first one, as one can
see that each esclation works just like the one before it, in a kind
of "successor operation". (3) That it may not be entirely satisfactory
and be puzzling to have such an infinite escalation, this is not in
itself a demonstration that the theory is inconsistent. (4) It's not a
given that an infinite escalation of meta-theories is in itself
objectionable.
> > > > Moreover, as to different perspectives, I do strive to appreciate them, but I
> > > > don't credit yours since it is confused, uninformed, and dogmatic (you
> > > > merely claim, but not prove).
>
> > > it is the prism through which you see that colours your view.
>
> > You should seriously think about submitting that for one of those
> > "Thought ForThe Day" calendars. Meanwhile, it does nothing to refute
> > anything I've said.
>
> as i said earlier, i see this as a forum for discussion and not
> debate.
But you show almost no willingness to LEARN about the subject.
> naturally,
> some of the thoughts need more working or even given up as one
> progresses through a discussion. if one is exploring, one is more
> likely to be tentative,
> rather than when one is just moving on the beaten path.And, you can go
> through the books, and get all the technique and no insight.
I suppose it is possible to get only technique and no insight by
studying the relevent textbooks, but it would seem difficult to do,
since, if one doesn't make sense of the material, it's difficult even
to see how to formalize and use only technique applied to it. But, you
miss the more important point: Even IF studying a textbook doesn't
ensure understanding, (1) there's no reason to think YOU wouldn't gain
understanding, if you just tried, and (2) WITHOUT a textbook you ARE
virtually ensured NOT to understand the subject.
No reply by you. May I take it that you understand now, not just the
dialectic of our conversation at that point, but also the more
substantive matter of what kinds of things are proven in various
theories? Most specifially that you've not shown that your object
theory defines and proves the kinds of things needed for your object
theory to serve as a meta-theory in which there are models of other
theories and, especially, of itself.
Morevover, again, it is not clear what your main point is. Until you
can show that your object theory can be a meta-theory for itself, in
the meantimve, you stated a particular meta-theory that is stronger
than ZFC, having an axiom added to ZFC (as I recall, or perhaps it was
ZFC minus some axioms but added your own axiom, in which case my
remarks still apply, mutatis mutandis). So whatever you derive from
such a theory (such as, say, a contradiction) isn't something that ZFC
is charged with (such as inconsistency) since all we would be able to
conclude, in the case of inconsistency, is that your axioms added to
ZFC is inconsistent. So what?
MoeBlee
MoeBlee
> On Mar 21, 12:38 am, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Mar 20, 12:24 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > > Then interpreting 'R' to stand for the membership relation, all you
> > > > have to do now is PROVE that <D membership-relation> is a model of
> > > > your object theory. And keep in mind that what 'D' stands for in the
> > > > meta-theory cannot be assumed to be as 'D' is defined in the object
> > > > theory.
>
> > > What makes you think that <D, membership relation> does not satisfy
> > > the
> > > axioms of the theory?
>
> > I didn't say it doesn't. But if you claim it does, then you need to
> > prove it. And then if you prove it, what is the signficance? You will
> > have proven something in your own personal meta-theory. That doesn't
> > prove anything about such ordinary theories as Z set theory and the
> > like.
>
> The proof is obvious.
Later in your post, you changed your object theory. If I have time and
inclination tonight or in the next few days, I'll go through your new
axioms to see whether they are properly formulated and whether it is
obvious that <D membership_on_D> is a model of your revised theory.
MoeBlee
MoeBlee
> Later in your post, you changed your object theory. If I have time and
> inclination tonight or in the next few days, I'll go through your new
> axioms to see whether they are properly formulated and whether it is
> obvious that <D membership_on_D> is a model of your revised theory.
P.S. But in the meantime, would you please tell me what exact
mathematical and/or philsophical point do you have in mind with this
exercise?
MoeBlee
MoeBlee
> 1)extensionality.
> 2) Ax x e U [existence of universal set/def of constant U]
> 3)Ax ~x e 0[existence of null set/def of constant 0]
> 4)Ax Ey [Az(z ey <->z e x or z=x][existence of successor;y=Sx and x
> =Py if x exists,x is the predecessor of y ]
> Def : x is finite <->[ x=0 or (~x=0& x has a finite predessor y and
> ~y=x)]
> 5) Ax [x e U <-> x is finite or x=U]
> Axiom 5) ensures that there is a unique infinite set U that contains
> itself
> Also, the set of all finite x does not exist.
We can leave initial universal quantifiers as tacitly understood:
Primitives:
e - 2-place pred
U - 0-place fcn
0 - 0-placed fcn
Axiom: Az(zex <-> zey) -> x=y
Axiom: xeU
Axiom: ~xe0
Axiom: EyAz(zey <-> (zex v z=x))
Theorem: E!yAz(zey <-> (zex v z=x))
Def: Sx = the_y(Az(zey <-> (zex v z=x)))
Def: Py = the_x(y=Sx)
Problem: if there is no such unique x such that y=Sx, then Py
undefined. Use Fregean method and set Py=0 in such cases?
Def: x is finite <-> (x=0 v (~x=0 & Ey(y is finite & x=Sy & ~y=x)))
NO GOOD. CIRCULAR DEFINITION. 'finite' cannot be defined by a right
side of the biconditional that has an occurrence of 'finite'.
Axiom: xeU <-> (x is finite v x=U)
Simpler (since we already have xeU):
x is finite v x=U
Theorem (claimed): ~U is finite & ((xex & ~x is finite) -> x=U)
Not proven. But can't prove anyway since 'finite' is not properly
defined.
Axiom: ~ExAy(yex <-> y is finite).
So,
(1) Decide about 'Py' when there is no such unique x. Use Fregean
method to set Py=0 when there is no such unique x?
(2) Fix definition of 'finite'.
(3) Make explicit definition of 'infinite' as: x is infinite <-> ~x is
finite.
(4) Prove ~U is finite & ((xex & ~x is finite) -> x=U).
Then we talk about models for this theory.
> As to significance, i wonder what happens to all the 'diagonal
> arguments'
> in such a theory.
Which diagonal arguments in particular? As to Cantor's theorem, you
don't have power set in your theory. As to uncountability of the
reals, you don't have a set of reals in your theory. As to Godel's
incompleteness, you haven't shown how to develop any of its apparatus
in your theory. As to the halting problem, you haven't shown how to
express it in your theory.
> and it would be nice resolution to the question
> 'whether
> the smallest infinite set contains an infinite set' that it depends on
> the axioms you
> choose.
What do you mean by 'contains'? Do you membership or subset?
Anyway, no matter which, we can resolve the question right now: Yes,
it depends on what axioms we choose.
So, I still don't see the point of your exercise.
MoeBlee
apoorv
> Def: x is finite <-> (x=0 v (~x=0 & Ey(y is finite & x=Sy & ~y=x)))
> NO GOOD. CIRCULAR DEFINITION. 'finite' cannot be defined by a right
> side of the biconditional that has an occurrence of 'finite'.
finite; definition is recursive and
not circular.
However, There is need for an axiom which will prevent an unending
chain of predecessors
Axiom: ~Ez Ay [y e z-->(Ex : ~x=y and Sx =y)]
It can be seen that any x other than 0 which has no predecessor is
not finite.Also, if any x has a chain of
predecessors ending with any x not equal to 0 than it is not finite.U
is not finite as SU =U
> Axiom: xeU <-> (x is finite v x=U)
> Simpler (since we already have xeU):
> x is finite v x=U
>
> Theorem (claimed): ~U is finite & ((xex & ~x is finite) -> x=U)
> Not proven. But can't prove anyway since 'finite' is not properly
> defined.
1)~U=0
2)SU =U
So, ~Ey: (~y=U and Sy =U)
~ U is finite.
Also ,
x is finite <--> ~x =Sx <--> `~x e x
x e x<--> ~x is finite .
x ex and ~x is finite <--> ~x is finite.
x is finite or x =U.
So,~x is finite <-->x =U
Also,
Suppose
Ex Ay (yex <--> y is finite)
So, ~x =U and A y (y e x<--> ~y=U) .
So, x e x
But x e x<--> x=U , contradiction
~Ex Ay (y e x <--> y is finite)
> Axiom: ~ExAy(yex <-> y is finite).
That is a theorem .see above .
> So,
>
> (1) Decide about 'Py' when there is no such unique x. Use Fregean
> method to set Py=0 when there is no such unique x?
>
> (2) Fix definition of 'finite'.
>
> (3) Make explicit definition of 'infinite' as: x is infinite <-> ~x is
> finite.
>
> (4) Prove ~U is finite & ((xex & ~x is finite) -> x=U).
See above
> Then we talk about models for this theory.
>
> > As to significance, i wonder what happens to all the 'diagonal
> > arguments'
> > in such a theory.
>
> Which diagonal arguments in particular? As to Cantor's theorem, you
> don't have power set in your theory. As to uncountability of the
> reals, you don't have a set of reals in your theory. As to Godel's
> incompleteness, you haven't shown how to develop any of its apparatus
> in your theory. As to the halting problem, you haven't shown how to
> express it in your theory.
That is the point- there is no uncountability, not uncountably more
subsets of
w than can be defined by predicates .
> > and it would be nice resolution to the question
> > 'whether
> > the smallest infinite set contains an infinite set' that it depends on
> > the axioms you
> > choose.
>
> What do you mean by 'contains'? Do you membership or subset?
The usual usage and the context make the meaning obvious -membership.
> Anyway, no matter which, we can resolve the question right now: Yes,
> it depends on what axioms we choose.
So we agree that there are axiomatisation (s) in which the smallest
infinite
set contains itself and the set of all finite set does not exist.
> So, I still don't see the point of your exercise.
i talked about Zeno's paradox; i mentioned the problem with edges of
adjacent tiles being
[0,1] and (1,2] ; i mentioned the several discussions in this forum
debating the question of the smallest
infinite set containing itself-- but those did not get your attention.
-apoorv
MoeBlee
> > > Def: x is finite <-> (x=0 v (~x=0 & Ey(y is finite & x=Sy & ~y=x)))
> > NO GOOD. CIRCULAR DEFINITION. 'finite' cannot be defined by a right
> > side of the biconditional that has an occurrence of 'finite'.
> 'x is finite' is defined in terms of ' predecessor of x ' being
> finite; definition is recursive and
> not circular.
NO, WRONG. Recursive definitions are justified by "definition by
recursion" theorems. You have no such theorem yet shown in your
theory. Your remark shows a fundamental ignorance and confusion about
recursive definition. This is another good example of your need to
learn the basics of this subject from a textbook.
Now, as to fixing the problem in your system, my guess is that your
system will prove the needed definition by recursion theorem (or even
inductivity, since you don't have unions or intersections, etc. that
are part of the ordinary machinery of induction). So instead you could
take 'finite' as primitive then state your "definition" as an axiom
rather than as a definition. I don't know whether that would empower
your system in the way you want, but at least it would be coherent.
In sum, NO, you cannot define 'finite' by referring to 'finite', and
you are not correct to conflate this situation with one in which we
prove various defintion by recursion theorems then utilize those to
prove the existence of certain "recursively (and/or inductively)
defined" functions and predicates.
> However, There is need for an axiom which will prevent an unending
> chain of predecessors
> Axiom: ~Ez Ay [y e z-->(Ex : ~x=y and Sx =y)]
> It can be seen that any x other than 0 which has no predecessor is
> not finite.
I'm going to cut you off there. You used the predicate 'finite'. Get
back to me when you've given a proper definition of 'finite' from your
primitives.
MoeBlee
MoeBlee
> back to me when you've given a proper definition of 'finite' from your
> primitives.
Or taken 'finite' as primitive and given whatever axioms you like with
'finite' in those axioms.
MoeBlee
apoorv
necsasary. the definition is
perfectly coherent.
0 is finite; [0} is finite, indeed , any of the conventionally finite
ordinals is finite
{{{0}}}} has no predecessor and is not finite (and does not exist), z=
{z} is not finite (and does not exist).
Take any x and the definition allows you to say whether x satisfies it
or not and
that is all that the definition has to be. If it were circular, as you
claim, there should be at
least one instance where the definition reduces to
x is finite <--> x is finite and we have no way of saying whether x is
finite or not.
> > However, There is need for an axiom which will prevent an unending
> > chain of predecessors
> > Axiom: ~Ez Ay [y e z-->(Ex : ~x=y and Sx =y)]
> > It can be seen that any x other than 0 which has no predecessor is
> > not finite.
>
> I'm going to cut you off there. You used the predicate 'finite'. Get
> back to me when you've given a proper definition of 'finite' from your
> primitives.
x is finite <--> x =0 Or [0 e x and ~E z(~z=0and (z ex-->~ Ey Sy =z))
and ~Ez (zex and z=Sz) and ~Ez (z e x and (Ay (yez -->Sy e z))) and
Ez (~Sz e x and Ay (y e x and
~y=z <--> Sy e x))]
x is finite iff x is 0 OR[0 is in x AND no other z which has no
predecessor is in x AND no set which is its own successor is a member
AND no inductive set is a member of x AND there is a z in x whose
successor is not in x and the successor of every y in x which is not
z is in x]
And you are not cutting me off. You are possibly trying hard to cut
yourself off from the realization that there
are axiomatisation where the smallest infinite set contains (belongs
to, for abundant clarity) to itself. And you
are cutting yourself off from the perspective that the set theory you
profess by is just one and possibly not
the best way to model reality.
-apoorv
MoeBlee
Exactly. You don't see, and the reason is that you are not familiar
with the basics of basic mathematical definition and recursion and
induction.
> the need for any of the apparatus that you think is
> necsasary. the definition is
> perfectly coherent.
Ask ANY mathematician. He or she will point out to you that you cannot
define a predicate symbol (here with the nickname 'finite') by
mentioning that predicate symbol in your definiens.
Recursive definitions are not an exception, in the sense that when we
fully unpack our recursive "definition" we find that it is not a
definition but rather a statement that a certain kind of function
exists, as we prove such statements by 'definition by recursion
theorems. In the case of predicates, we move simply via the
characteristic function for some recursively "defined" function.
And inductive definitions are enabled by proving certain closure and/
or intersection properties, et.al.
You're ridiculous. There is no MAGIC WAND by which we write down a
CIRCULAR definition and declare it is "recursive".
Any recursive definition can be unpacked ("reformulated") so that we
can see it is NOT circular.
So, LISTEN CAREFULLY: If your definition is indeed NOT circular, you
should be able to reformulate it so that indeed it does not mention
'finite' in the definens of a definition of 'finite'.
> 0 is finite; [0} is finite, indeed , any of the conventionally finite
> ordinals is finite
> {{{0}}}} has no predecessor and is not finite (and does not exist), z=
> {z} is not finite (and does not exist).
Those are all mere stipulations, while you have not given a proper
definition of 'finite'.
> Take any x and the definition allows you to say whether x satisfies it
> or not
You need to show that 'finite' is ELIMINABLE. With a proper defintion,
we get an algorithm to decode any formula with a defined symbol or
predicate in it to an equivalent formula withOUT that defined symbol
or predicate. That is not possible when the definition is circular.
> and
> that is all that the definition has to be. If it were circular, as you
> claim, there should be at
> least one instance where the definition reduces to
> x is finite <--> x is finite
How did you arrive at that conclusion? Prove it.
Rather, the correct statement is that a circular definition does not
ensure eliminability. And it is circular to define 'finite' in terms
of 'finite'. And, as I've explained too much already, recursive and
inductive definition are not an exception since they unpack to
theorems, not actually definitions, that a certain kind of function
exists (recursion) or that we may take the intersection of a set that
is closed under certain relations (induction), etc.
> and we have no way of saying whether x is
> finite or not.
> However, There is need for an axiom which will prevent an unending
> > > chain of predecessors
> > > Axiom: ~Ez Ay [y e z-->(Ex : ~x=y and Sx =y)]
> > > It can be seen that any x other than 0 which has no predecessor is
> > > not finite.
>
> > I'm going to cut you off there. You used the predicate 'finite'. Get
> > back to me when you've given a proper definition of 'finite' from your
> > primitives.
>
> Let me give another definition of finite.
> x is finite <--> x =0 Or [0 e x and ~E z(~z=0and (z ex-->~ Ey Sy =z))
> and ~Ez (zex and z=Sz) and ~Ez (z e x and (Ay (yez -->Sy e z))) and
> Ez (~Sz e x and Ay (y e x and
> ~y=z <--> Sy e x))]
x is finite
<->
(x=0 v
(0ex &
~Ez(~z=0 & (zex -> ~Ey Sy=z)) &
~Ez(zex & z=Sz) &
~Ez(zex & Ay(yez -> Sy e z)) &
Ez(~Sz e x & Ay(yex & (~y=z <-> Sy e x))))
Right?
At least it's not circular. I have no objection to the above
definition for purposes of your system.
But it seems your symbolization is not actually what you mean:
> x is finite iff x is 0 OR[0 is in x AND
Okay.
> no other z which has no
> predecessor is in x
That's not what you symbolized.
What you just said is symbolized this way:
Az((~z=0 & ~Ey z=Sy) -> ~zex)
That is, "If z is not 0 and z has no predecessor, then z is not in x."
Your symbolization,
~Ez(~z=0 & (zex -> ~Ey Sy=z))
is equivalent to:
Az(z=0 -> (zex & Ey Sy=z))
and says "If z is 0, then (z is in x and z has no predecessor)."
> AND no set which is its own successor is a member
Member of what? Member of x, I suppose?
Anyway, your symbolization
~Ez(zex & z=Sz)
is equivalent to:
Az(zex -> ~z=Sz)
and says "If z in x then z is not its own successor."
which is equivalent to what I think you mean:
"If z is its own successor then z is not in x".
So, okay on this one, I guess.
> AND no inductive set is a member of x
What's your definition of 'inductive set'? Do you mean this?:
z is inductive <-> Ay(yez -> Sy e z)
That's fine, but you don't want to include also that 0 in z?
Anyway, this clause in your symbolization does match what you said:
~Ez(zex & Ay(yez -> Sy e z))
equivalent to:
Az(zex -> ~Ay(yez -> Sy e z))
which says, "If z in x then z is not inductive".
So, okay on that one.
> AND there is a z in x whose
> successor is not in x and the successor of every y in x which is not
> z is in x]
That's not what your symbolism says:
Ez(~Sz e x & Ay(yex & (~y=z <-> Sy e x)))
which makes no demand that there is a z in x.
Rather, yours says "There is a z such that the successor of z is not
in x; and everything is in x; and every y is such that (y is not z iff
Sy in x)."
Then, by "the successor of every y in x which is not z is in x" I take
it you mean:
Ay((yex & ~Sy e z) -> Sy e x).
which is to say, "If y in x and the successor of y is not in z, then
the successor y is in x."
Is that what you mean?
> And you are not cutting me off. You are possibly trying hard to cut
> yourself off from the realization that there
> are axiomatisation where the smallest infinite set contains (belongs
> to, for abundant clarity) to itself. And you
> are cutting yourself off from the perspective that the set theory you
> profess by is just one and possibly not
> the best way to model reality.
You really are an ass. What I am doing is - for no consideration other
than the frustration of attemptiong to communicate with someone who is
determined to remain ignorant of the subject on which he is spoutiung
- giving you good information and indulging your exercises. I'm
correcting your formulas, explaining basics of the subject to you, and
still waiting for you to come up with a coherent formulation NOT
because I have any interest at all in "trying to cut myself off from
the realization that there are axiomatisation[s] where the smallest
infinite set [is a member of] itself."
LISTEN CAREFULLY, smartass, I fully grant that one can devise such an
axiomatization. I have NEVER argued that one cannot. It's even
TRIVIAL. All one has to do is state it as an axiom! Or, if you like,
find some axioms that imply it. Who EVER doubted that?
Rather, the point was a DIFFERENT claim, which was that your theory
(which still now depends on you settling for yourself whether we
should take your symbolization of the definition 'finite', though,
now, thankfully, at least not circular) is modelled by the structure
<D membership_on D> mentioned previously.
After you prove a THEOREM, in your object theory, "The smallest
infinite set is a member of itself" (i.e., "There exists a unique x
such that (x is not finite (with YOUR definition of 'finite') and no
other not-finite set is a proper subset of x) and the aforementioned
unique set x is a member of x", then you STILL need to prove that <D
membership_on D> is a model of your theory (if you still wish to
sustain that claim), which requires not proving the just mentioned
purported theorem in your object theory, but rather proving in your
meta-theory that each axiom of your object theory is true in the
structure <D membership_on D>. Unfortunately, you don't know what is
actually involved in proving that a sentence (such as an axiom) is
true in a given structure. Moreover, the proof of your theorem about a
least infinite set has no apparerent relevance to notions of finite
and infinite in ordinary set theory, since your definition of 'finite'
hasn't been shown to be equivalent with any of the ordinary defintions
of 'finite' in set theory.
Next step: Let me know what you work out as your revised (or not)
definition of 'finite'. Then please list your axioms and definitions
for your object theory altogether again (so that I don't have to
reassemble from various posts). Then please reiterate your meta-
theory? It was ZFC-R+"exists an x whose members are all and only the
natural numbers and x itself", right? Then we define, in your META-
theory, D = the unique x such that Ay(y in x <-> (y a natural number
or y=x), as uniqueness derives from extensionality. Then we may
consider whether <D membership_on D> is a model of your object theory.
Then we'll revist the question to all this, "So what?" And you'll
explain whatever it is you think this shows about infinite sets (as
'infinite' is defined ORDINARILY) and whatever else about Zeno's
paradox.
MoeBlee
P.S.
> And you
> are cutting yourself off from the perspective that the set theory you
> profess by is just one and possibly not
> the best way to model reality.
In what sense do I "profess by" ZFC? Be careful in your answer here,
please do not put words in my mouth.
And, OF COURSE I recognize that there are theories, even foundational
theories, other than ZFC. And NOWHERE will you find a statement by me
that ZFC is the "best way to model reality". You won't even find a
claim by me that ZFC is the best foundational theory for abstract
mathematics. I've simply never made such a claim as to "best".
So, instead of bridling against what you THINK I believe, please just
refer to what I've acutally posted or ask me. Meanwhile, your time is
better spent anyway just reading a good introductory textbook on this
subject of which you have think you have something to say yet you know
virtually nothing about it.
MoeBlee
apoorv
'A recursive definition, sometimes also called an inductive
definition, is one that defines a word in terms of itself, so to
speak, albeit in a useful way. Normally this consists of three steps:
1. At least one thing is stated to be a member of the set being
defined; this is sometimes called a "base set".
2. All things bearing a certain relation to other members of the
set are also to count as members of the set. It is this step that
makes the definition recursive.
3. All other things are excluded from the set
For instance, we could define natural number as follows (after Peano):
1. "0" is a natural number.
2. Each natural number has a distinct successor, such that:
* the successor of a natural number is also a natural
number, and
* no natural number is succeeded by "0".
3. Nothing else is a natural number.
So "0" will have exactly one successor, which for convenience we can
call "1". In turn, "1" will have exactly one successor, which we would
call "2", and so on. Notice that the second condition in the
definition itself refers to natural numbers, and hence involves self-
reference. Although this sort of definition involves a form of
circularity, it is not vicious, and the definition is quite
successful.'
> Recursive definitions are not an exception, in the sense that when we
> fully unpack our recursive "definition" we find that it is not a
> definition but rather a statement that a certain kind of function
> exists, as we prove such statements by 'definition by recursion
> theorems. In the case of predicates, we move simply via the
> characteristic function for some recursively "defined" function.
>
> And inductive definitions are enabled by proving certain closure and/
> or intersection properties, et.al.
>
> You're ridiculous. There is no MAGIC WAND by which we write down a
> CIRCULAR definition and declare it is "recursive".
>
> Any recursive definition can be unpacked ("reformulated") so that we
> can see it is NOT circular.
>
> So, LISTEN CAREFULLY: If your definition is indeed NOT circular, you
> should be able to reformulate it so that indeed it does not mention
> 'finite' in the definens of a definition of 'finite'.
A definition of 'x is finite' in terms of 'Px is finite' is perfectly
valid.
> > 0 is finite; [0} is finite, indeed , any of the conventionally finite
> > ordinals is finite
> > {{{0}}}} has no predecessor and is not finite (and does not exist), z=
> > {z} is not finite (and does not exist).
>
> Those are all mere stipulations, while you have not given a proper
> definition of 'finite'.
If you were to apply yourself with a little more open mind , you
would
se that it clearly flows from thw definition unambiguously and are not
mere stipulations.
> > Take any x and the definition allows you to say whether x satisfies it
> > or not
>
> You need to show that 'finite' is ELIMINABLE. With a proper defintion,
> we get an algorithm to decode any formula with a defined symbol or
> predicate in it to an equivalent formula withOUT that defined symbol
> or predicate. That is not possible when the definition is circular.
>
> > and
> > that is all that the definition has to be. If it were circular, as you
> > claim, there should be at
> > least one instance where the definition reduces to
> > x is finite <--> x is finite
>
> How did you arrive at that conclusion? Prove it.
I do not have to prove it -that is obvious.
With x is finite <-->x is finite, you can never ascertain whether x
is finite or not:
with x is finite <--> y is finite, the definition is not circular-
however it will return a truth value
for 'x is finite' only if you know separately whether y is finite or
not.
I think there is a a simpler definition:
x is finite <--> x=0 or [0 e x And ~x e x And Ez (Sz=x &~z = x) And
Ay ((~y=0 & y e x)-->Ez Sz=y &~z =y)]
> > And you are not cutting me off. You are possibly trying hard to cut
> > yourself off from the realization that there
> > are axiomatisation where the smallest infinite set contains (belongs
> > to, for abundant clarity) to itself. And you
> > are cutting yourself off from the perspective that the set theory you
> > profess by is just one and possibly not
> > the best way to model reality.
>
> You really are an ass.
oh dear ! you loose your cool ! what a pity !
What I am doing is - for no consideration other
> than the frustration of attemptiong to communicate with someone who is
> determined to remain ignorant of the subject on which he is spoutiung
> - giving you good information and indulging your exercises. I'm
> correcting your formulas, explaining basics of the subject to you, and
> still waiting for you to come up with a coherent formulation NOT
> because I have any interest at all in "trying to cut myself off from
> the realization that there are axiomatisation[s] where the smallest
> infinite set [is a member of] itself."
> LISTEN CAREFULLY, smartass, I fully grant that one can devise such an
> axiomatization. I have NEVER argued that one cannot. It's even
> TRIVIAL. All one has to do is state it as an axiom! Or, if you like,
> find some axioms that imply it. Who EVER doubted that?
That is all i said in the beginning. YOU CONCEDE BUT AFTER MUCH
PREVARICATION
AND WITH NO GRACE. ALL I WOULD LIKE TO SEE IS THAT YOU QUALIFY
ALL YOUR ASSERTIONS ABOUT 'THE SMALLEST INFINITE SET' WITH THE
STATEMENT
THAT 'THERE ARE AXIOMATISATIONS IN WHICH THE SMALLEST INFINITE
SET IS A MEMBER OF ITSELF'.
> Rather, the point was a DIFFERENT claim, which was that your theory
> (which still now depends on you settling for yourself whether we
> should take your symbolization of the definition 'finite', though,
> now, thankfully, at least not circular) is modelled by the structure
> <D membership_on D> mentioned previously.
> After you prove a THEOREM, in your object theory, "The smallest
> infinite set is a member of itself" (i.e., "There exists a unique x
> such that (x is not finite (with YOUR definition of 'finite') and no
> other not-finite set is a proper subset of x) and the aforementioned
> unique set x is a member of x",
I already proved that but you cut yourself off.
Now that you have agreed that there are axiomatisations in which the
smallest infinite
set is a member of itself, i will revisit these issue in due course.
> > And you
> > are cutting yourself off from the perspective that the set theory you
> > profess by is just one and possibly not
> > the best way to model reality.
>
> In what sense do I "profess by" ZFC? Be careful in your answer here,
> please do not put words in my mouth.
>
> And, OF COURSE I recognize that there are theories, even foundational
> theories, other than ZFC. And NOWHERE will you find a statement by me
> that ZFC is the "best way to model reality". You won't even find a
> claim by me that ZFC is the best foundational theory for abstract
> mathematics. I've simply never made such a claim as to "best".
. '
See extract from this post of yours 31 may 2006:
'Then if, even before mentioning the axiom of infinity, we had
already
defined 'finite ordinal', which we can do, then we can prove w = the
set of finite ordinals. And we can define w = the set of natural
numbers = the set of finite ordinals. And we can prove, trivially,
that
for any n, if n is a member of w, then n = 0 or there exists a k in w
such that n = k+. And it is trivial also that w is a subset of every
successor-inductive set. So there is nothing in w except 0 and
sucessors from 0. Etc, etc, through proofs of the proof by induction
priniciples and the defintion by recursion theorems, through to the
defintions of the operations such as addition, multiplication,
exponentiation, factorial, divides, through to the proof of the
fundamental theorem of arithemetic, and on...and also to the
construction of the integers, rationals, and reals, and on....It's a
beautiful thing'
That is why i feel you are besotted by this garden of the transfinite.
Anyway, as i said earlier, i do hope that whenever you talk of the
'smallest infinite set' you do refer to alternate axiomatisations
with 'the smallest infinite set containing itself'
-apoorv
MoeBlee
> 'See this from wikipedia writeup on definition:
Ah, yes, the method of learning mathematics by reading various ad hoc
Wikipedia articles.
What you cite in the Wikipedia article is okay, as far as it goes, as
a quite general overview of the notions of recursive and inductive
definitions.
Anyway nothing in your quote contradicts anything I've said and much
of it is what I've said already. Moreover, IF you were to investigate
the matter at any level of technical precision - note that our context
is your FORMAL object theory - beyond general remarks in an
encyclopedia article, you will find that further of my own
elaborations are correct - specifically concerning proof of definition
by recursion theorems.
But if you insist on remaining ignorant of the actual technical
specifics as found in many an ordinary textbook on the subject, then
there's probably nothing I can do to dissuade you from what you want
to believe.
> A definition of 'x is finite' in terms of 'Px is finite' is perfectly
> valid.
If 'P' is a previously defined predicate symbol, then it is fine for
'P' to appear in the definiens of a definition of 'finite'. But it is
not correct to define 'finite' with 'finite' in the definiens. And
I've explained already too many times what is required for a recursive
and/or inductive definition in which form 'finite' would be eliminable
from the definiens.
> > 0 is finite; [0} is finite, indeed , any of the conventionally finite
> > > ordinals is finite
> > > {{{0}}}} has no predecessor and is not finite (and does not exist), z=
> > > {z} is not finite (and does not exist).
>
> > Those are all mere stipulations, while you have not given a proper
> > definition of 'finite'.
>
> If you were to apply yourself with a little more open mind , you
> would
> se that it clearly flows from thw definition unambiguously and are not
> mere stipulations.
One does not lack an open mind merely for not accepting circular
definitions.
Anyway, since then you gave a non-circular definition, so I don't see
why you cling now to arguments about your earlier circuluar
definition.
> > > and
> > > that is all that the definition has to be. If it were circular, as you
> > > claim, there should be at
> > > least one instance where the definition reduces to
> > > x is finite <--> x is finite
>
> > How did you arrive at that conclusion? Prove it.
>
> I do not have to prove it -that is obvious.
So you can't prove it, in other words, I take it. (Anyway, you'd have
to define 'reduces to'.)
> With x is finite <-->x is finite, you can never ascertain whether x
> is finite or not:
Yes.
> with x is finite <--> y is finite, the definition is not circular-
It's not even a proper definition. Not only is 'finite' in the
definiens circular, but 'y' is the incorrect variable to appear in the
definiens.
Really, why don't you just learn something about mathematical
definitions? See Suppes 'Introduction To Logic' for one of the very
best, easy to understand, not-too-technical, but just-technical-enough
chapter on this subject.
> however it will return a truth value
> for 'x is finite' only if you know separately whether y is finite or
> not.
You really need to learn about mathematical definitions. I can't keep
wasting more of my time with you on this matter if you continue to
refuse to inform yourself.
> x is finite <--> x=0 or [0 e x And ~x e x And Ez (Sz=x &~z = x) And
> Ay ((~y=0 & y e x)-->Ez Sz=y &~z =y)]
It seems you're missing parentheses around (Sz=y & ~z=y) near the end,
I guess?
So:
x is finite
<->
(x=0 v
(0ex &
~xex &
Ez(x=Sz & ~z=x) &
Ay((~y-0 & yex) -> Ez(Sz-y & ~z=y))))
MoeBlee
MoeBlee
> On Mar 28, 5:53 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 27, 2:21 pm, apoorv <sudhir...@hotmail.com> wrote:
> > You really are an ass.
>
> oh dear ! you loose your cool ! what a pity !
No, no loss of cool; I'm just noting that you're being an ass.
> > LISTEN CAREFULLY, smartass, I fully grant that one can devise such an
> > axiomatization. I have NEVER argued that one cannot. It's even
> > TRIVIAL. All one has to do is state it as an axiom! Or, if you like,
> > find some axioms that imply it. Who EVER doubted that?
>
> That is all i said in the beginning.
No, it is NOT "all" you said. You've made claims about a certain
structure being a model of a certain theory. THAT is what you haven't
proven.
> YOU CONCEDE BUT AFTER MUCH
> PREVARICATION
I'ts hardly a concession since I never disputed that one can state a
theory in which one can prove such and such. And you've shown no
prevarication on my part.
> AND WITH NO GRACE.
Whatever lack of grace on my part, my corrections of your
misunderstandings are a lot more graceful than your misunderstandings.
> ALL I WOULD LIKE TO SEE IS THAT YOU QUALIFY
> ALL YOUR ASSERTIONS ABOUT 'THE SMALLEST INFINITE SET' WITH THE
> STATEMENT
> THAT 'THERE ARE AXIOMATISATIONS IN WHICH THE SMALLEST INFINITE
> SET IS A MEMBER OF ITSELF'.
Why in the world should I do that? For ANY mathematical statement
there is some theory in which that statement is a theorem. When I'm
speaking in a context, such as ZFC, either as explicitly mentioned or
by tacit understanding, I don't have to preface each remark with a
disclaimer that in some OTHER THEORY we get different results. OF
COURSE WE DO. It doesn't need mentioning!
> > Rather, the point was a DIFFERENT claim, which was that your theory
> > (which still now depends on you settling for yourself whether we
> > should take your symbolization of the definition 'finite', though,
> > now, thankfully, at least not circular) is modelled by the structure
> > <D membership_on D> mentioned previously.
> > After you prove a THEOREM, in your object theory, "The smallest
> > infinite set is a member of itself" (i.e., "There exists a unique x
> > such that (x is not finite (with YOUR definition of 'finite') and no
> > other not-finite set is a proper subset of x) and the aforementioned
> > unique set x is a member of x",
>
> I already proved that but you cut yourself off.
You can't have proven ANYTHING about 'finite' in your theory until you
REPAIRED your circular definition of 'finite'. THAT is why I cut it
off. NOW, though, you've given a non-circular definition of
'finite' (and I suppose 'infinite' is meant by you as 'not finite'),
so NOW you may show me your proof that: There exists a unique x such
that (x is not finite (with YOUR definition of 'finite') and no other
not-finite set is a proper subset of x) and the aforementioned unique
set x is a member of x.
But I NEVER disputed that there are axiomaizations in which there is a
smallest infinite set that is a member of itself.
> > In what sense do I "profess by" ZFC? Be careful in your answer here,
> > please do not put words in my mouth.
>
> > And, OF COURSE I recognize that there are theories, even foundational
> > theories, other than ZFC. And NOWHERE will you find a statement by me
> > that ZFC is the "best way to model reality". You won't even find a
> > claim by me that ZFC is the best foundational theory for abstract
> > mathematics. I've simply never made such a claim as to "best".
>
> See extract from this post of yours 31 may 2006:
>
> 'Then if, even before mentioning the axiom of infinity, we had
> already
> defined 'finite ordinal', which we can do, then we can prove w = the
> set of finite ordinals. And we can define w = the set of natural
> numbers = the set of finite ordinals. And we can prove, trivially,
> that
> for any n, if n is a member of w, then n = 0 or there exists a k in w
> such that n = k+. And it is trivial also that w is a subset of every
> successor-inductive set. So there is nothing in w except 0 and
> sucessors from 0. Etc, etc, through proofs of the proof by induction
> priniciples and the defintion by recursion theorems, through to the
> defintions of the operations such as addition, multiplication,
> exponentiation, factorial, divides, through to the proof of the
> fundamental theorem of arithemetic, and on...and also to the
> construction of the integers, rationals, and reals, and on....It's a
> beautiful thing'
Yes, so? Nowhere there did I say that ZFC is "the best way to model
reality" or that ZFC is the best foundational theory.
Again, I'm asking you, please do not represent as if I've said things
that in fact I haven't.
Anyway, with my help your formulations are now precise. So anytime you
wish to prove, in your meta-theory, your claim that your structure is
a model of your object theory, I'll read it, time (and my patience
with you) permitting. Then, if you prove your claim, with time and
patience I'll hear whatever argument you have as to what you think the
signficance you find comes from the result.
MoeBlee
apoorv
> On Mar 28, 4:33 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > On Mar 28, 5:53 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 27, 2:21 pm, apoorv <sudhir...@hotmail.com> wrote:
> > > You really are an ass.
>
> > oh dear ! you loose your cool ! what a pity !
>
> No, no loss of cool; I'm just noting that you're being an ass.
going to
say that you are being an uncivilised boor with the mindset of a
medieval theocrat-no , i am not going to say that.
> But I NEVER disputed that there are axiomaizations in which there is a
> smallest infinite set that is a member of itself.
Since we are in agreement on this issue, it would be in order to close
this discussion on this amicable note of agreement.
-apoorv
-
MoeBlee
> On Mar 30, 10:44 pm, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 28, 4:33 am, apoorv <sudhir...@hotmail.com> wrote:
>
> > > On Mar 28, 5:53 am, MoeBlee <jazzm...@hotmail.com> wrote:> On Mar 27, 2:21 pm, apoorv <sudhir...@hotmail.com> wrote:
> > > > You really are an ass.
>
> > > oh dear ! you loose your cool ! what a pity !
>
> > No, no loss of cool; I'm just noting that you're being an ass.
>
> Now do we have to loose our manners as well ?
No, you don't get an insurance that allows you to be snide, to act
like an ass, but not have that fact pointed out to you.
> In any case, i am not
> going to
> say that you are being an uncivilised boor with the mindset of a
> medieval theocrat-no , i am not going to say that.
You just did. But in typically snide manner that you would contrive
not to have to support your claim.
It's not uncivilized nor boorish to tell boors, such as you, that they
are boors.
And one doesn't have the mindset of a medieval theocrat merely for not
accepting upon your word alone various claims you make without proof.
Nor does one have the mindset of a medieval theocrat merely for not
accepting such things as circular definitions and for pointing out
that circular definitions are not made non-circular just by waving a
magic wand in the form of merely CLAIMING them to be recursive, but
rather that there are a certain mathematical aspects of recursive
definition and inductive definition that need to be shown to apply.
> > But I NEVER disputed that there are axiomaizations in which there is a
> > smallest infinite set that is a member of itself.
>
> Since we are in agreement on this issue, it would be in order to close
> this discussion on this amicable note of agreement.
Then what was the point of ALL THAT RIGMAROLE of your special object
theory and special meta-theory and formulas and definitions, if all
you wanted to say is something that no informed person in mathematical
logic would disagree with: From different axioms we get different
theorems; if you want a certain theorem then you can always have it by
choosing an axiomatization that yields it.
If you had the least bit of intellectual organization (along with some
basic understanding of the subject matter), you could have saved us a
lot time and bother from the outset.
MoeBlee