Minimal set theory for model theory?
Bill Taylor
the question appropriately precise, and then answer it.
The query is prompted by the recent thread on Skolem's paradox,
but can be considered entirely independently of that.
It is a query that informally comes up here from time to time.
The query is not one that bothers me personally - I'm quite happy with
the situation; but I find myself at a slight loss when trying to
answer it,
when posed by someone else. So perhaps this is a pedagogical query,
really.
- - -
It concerns the matter of treating set theory formally and thoroughly,
in particular using model theory. The thing being, that model theory
itself relies quite a bit on set theortetical concepts,
although rather simpler ones than full set theory itself.
But learners are often bothered by the fact that they see us trying to
"define" or "explain" what sets are, formally, by using the details of
FOL
plus its model theory, which is itself reliant on a form of set
theory.
So they detect a circularity here, at least an informal one, and it is
a complaint that needs addressing.
Now I and others have long noted, that the amount of set theory
required
to do model theory is *very* limited compared with full set theory
itself.
So my query is - if one treats FOL, proof theory & model theory
*formally*,
(as one must in mathematical logic), then WHAT is the minimum amount
of simple set theory one can get away with? It presumably includes
simple ideas of finite unions and intersections, finite cross products
(to deal with multiple arities) and basic ideas of membership.
It presumably doesn't have to deal with power sets, cardinality
beyond the finite/infinite distinction, regularity, well-ordering,
replacement, or even very much separation.
It would be nice to know if this basic amount of set theory has itself
been made the topic of a formal (mathematical) theory. Has it?
In particular, what is the MINIMAL amount of formal set theory
required to support the basic concepts of a formal approach to FOL,
proof theory, and model theory.
- - -
I hope the question is clear enough to admit of some precise answer!
------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
------------------------------------------------------------------------
The math is done right, but is the right math done?
------------------------------------------------------------------------
LauLuna
wrote:
> ------------------------------------------------------------------------
> The math is done right, but is the right math done?
> ------------------------------------------------------------------------
I would say that separation is needed to define models of FOL formulae
since we choose a universe of discourse and we specify within it the
interpretation of predicates by defining subclasses.
I can't say much more about this but this reminds me of the paradox of
Orayen that I think springs from the fact that we use FOL to formalize
set theory and set theory to interpret FOL. I'll take a look on the
literature about that paradox.
Regards
LauLuna
> ------------------------------------------------------------------------
> The math is done right, but is the right math done?
> ------------------------------------------------------------------------
A previous post hasn't appeared.
Well, take into account Orayen's paradox. It is caused by the fact
that we use FOL to formalize set theory and set theory to interpret
FOL formulae. The paradox shows that there is no standard model for
first order set theory and derives this from Cantor's paradox: set
theory is supposed to speak about any set, so that the universe of
discourse of the corresponding interpretation should be the set of all
sets, which does not exist. This we usually prove from Powerset and
Cantor's theorem.
So, contrary to what it might seem, model theory for FOL could require
quite a bit of set theory.
Regards
MoeBlee
wrote:
> So my query is - if one treats FOL, proof theory & model theory
> *formally*,
> (as one must in mathematical logic), then WHAT is the minimum amount
> of simple set theory one can get away with?
I'm not an expert (I'm more of a beginner) and I don't know as to a
strict minimal theory, but I have been working on formalizing
mathematical logic and mathematics in general (including formation of
languages, model theory, proof theory, recursion theory, topology,
algebra, analysis) in a formal set theory. I use Z set theory without
the axiom of regularity. That is: extensionality, schema of
separation, pairing, union, power set and infinity. There may be
weaker theories or other theories that can do the job, but in an
ordinary set theoretic context, I can't see how to do without each of
those axioms.
If formulas are taken as sequences (functions) then the schema of
replacement is not needed except for certain purposes (such as the
metamathematics of set theory itself); but if formulas are taken as
ordered n-tuples, then the set of formulas can't be formed without the
schema of replacement. So, to avoid having to use the schema of
replacement, I take formulas to be sequeneces, not tuples. Also, for
uncountable languages and in certain other cases, the axiom of choice
is needed, so I note that on a case-by-case basis (and use a weaker
form when possible). For much of metamathematics of set theory itself,
the axiom of regularity and the schema or replacement are used, which
I note on a case-by-case basis.
MoeBlee
Aatu Koskensilta
> If formulas are taken as sequences (functions) then the schema of
> replacement is not needed except for certain purposes (such as the
> metamathematics of set theory itself); but if formulas are taken as
> ordered n-tuples, then the set of formulas can't be formed without the
> schema of replacement.
I have no idea what you mean by that. Replacement is not needed in proving
the existence of the set of formulas of the usually considered languages,
even if we represent them as tuples. Such representations occur within the
collection of hereditarily finite sets, the existence of which does not rely
on replacement.
> So, to avoid having to use the schema of
> replacement, I take formulas to be sequeneces, not tuples. Also, for
> uncountable languages and in certain other cases, the axiom of choice
> is needed, so I note that on a case-by-case basis (and use a weaker
> form when possible). For much of metamathematics of set theory itself,
> the axiom of regularity and the schema or replacement are used, which
> I note on a case-by-case basis.
Regularity is always unnecessary, by relativizing to the collection of
well-founded sets. I wonder what results about models of set theory you have
in mind that require replacement?
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Frederick Williams
> It presumably doesn't have to deal with power sets, cardinality
> beyond the finite/infinite distinction,
Why not? Chang and Keisler is full of stuff about Cardinality.
--
Remove "antispam" and ".invalid" for e-mail address.
We have lingered in the chambers of the sea
By sea-girls wreathed with seaweed red and brown
Till human voices wake us, and we drown.
Aatu Koskensilta
> Bill Taylor wrote:
>
>> It presumably doesn't have to deal with power sets, cardinality
>> beyond the finite/infinite distinction,
>
> Why not? Chang and Keisler is full of stuff about Cardinality.
It all depends on what one takes model theory to comprise. Depending on that
a weak fragment of set theory migh suffice, or it might be principles
beyond those in ZFC are needed.
george
I too had a previous post fail to appear.
I wasn't bothered by it at first because I am not
actually competent to answer the question being posed.
But the discussion will have to begin on this lower level
until Someone Who Knows deigns to Speak.
> Well, take into account Orayen's paradox.
That's easier said than done, in English anyway.
If you google "Orayen's Paradox" you will not get any
hits outside this thread. If you Google Orayen and Paradox
disjunctively then you will find that Raul Orayen wrote a lot
of things about paradoxes, but most of them are in Spanish.
> It is caused by the fact that we use FOL to formalize set theory
Right.
> and set theory to interpret FOL formulae.
But not necessarily; the content of my missing post was that
the completeness theorem makes interpretation irrelevant.
FOL formulae simply don't need to be interpreted at all.
Although the typical definition of logical consequence under
this paradigm is via satisfaction "in all models", the Completeness
theorem is a demonstration that we have the (extensionally) identical
notion of consequence via synactic inference rules. In other words,
FOL *has* a model theory, but it doesn't *need* one.
If the model theory gets paradoxical or otherwise embarrassing
then the obvious solution is simply to throw out the bathwater
of model theory and deal thereafter with the baby of proof theory.
> The paradox shows that there is no standard model for
> first order set theory and derives this from Cantor's paradox: set
> theory is supposed to speak about any set,
And, more to the point, logical consequence is defined in terms
of any and every model, so it, too, would need to speak about
every set (the point being that every set COULD be [the domain
of] a model of every theory that didn't force a bigger cardinality).
> so that the universe of discourse of the corresponding
> interpretation should be the set of all sets, which does not exist.
Somebody needs to read Zermelo's original 1908 paper.
It says basically that we have a domain of objects B and
that among these are the sets. It does NOT say that B has to
or needs to be a set; indeed, with the axiom of foundation,
it absolutely cannot be (it would have to contain itself if it were
a set). The rebuttal of this objection is that THE CLASS of all sets
DOES exist, and the general dictum should be that the domain
of a model has to be a CLASS. Whether the class is proper or is a
set is not even important. From the viewpoint of the theory
INTERNALLY,
the universal class will ALWAYS be proper, but this does not prohibit
it from becoming a set in a meta-theory.
> This we usually prove from Powerset and Cantor's theorem.
>
> So, contrary to what it might seem,
> model theory for FOL could require
> quite a bit of set theory.
Well, at a bare minimum it requires a class theory as
opposed to the ZFC conceit that proper classes don't
exist. But ZFC's failure to understand its own whole
domain IS NOT a paradox.
MoeBlee
wrote:
> On 2007-04-24, MoeBlee wrote:
> > If formulas are taken as sequences (functions) then the schema of
> > replacement is not needed except for certain purposes (such as the
> > metamathematics of set theory itself); but if formulas are taken as
> > ordered n-tuples, then the set of formulas can't be formed without the
> > schema of replacement.
>
> I have no idea what you mean by that. Replacement is not needed in proving
> the existence of the set of formulas of the usually considered languages,
> even if we represent them as tuples. Such representations occur within the
> collection of hereditarily finite sets, the existence of which does not rely
> on replacement.
By 'formula' I meant in the sense sometimes used of 'expression'
whether a well formed formula or just a string of symbols that is not
necessarily a well formed formula. Actually, I wish I'd said
'expression' since usually I do like to use 'formula' to mean 'well
formed formula'. Anyway, if 'strings' is taken in the sense of an
ordered n-tuple (in the sense of iterated ordered pairs, not in the
sense of a function), then, as I understand, the set {t | Enew t is an
n-tuple of symbols}) can't be formed without the axiom of replacement.
I don't have a proof of that, but it is what Enderton confirms when I
asked him, and goes with remarks in a more general sense by
Moschovakis in his intro set theory text, and was confirmed several
months ago in a post by another poster (I don't recall who it was
now). The basic reasoning is that, even though for any n, we can form
the set of n-tuples of symbols, the set {t | Enew t is an n-tuple of
symbols} is the union of those sets for each n and to gather those
sets together requires replacement (for each n, there is the set of n-
tuples of symbols, then replacement forms the set of all the tuples
for all n). On the other hand, if we take 'strings' to be finite
sequences, that is, actual functions (not just n-tuples), then we
don't need replacement to form the set of all finite sequences of
symbols. So, if in my post you take 'formulas' to be not 'well formed
formulas' but more generally all formulas (strings of symbols),
whether well formed or not, then do you see my point? Yet, if there is
a way to prove the existence of the set of tuples of symbols (tuples
in the sense of iterated ordered pairs and not in the sense of finite
functions), then I do need to learn about that proof.
> Regularity is always unnecessary, by relativizing to the collection of
> well-founded sets.
That may be so, and I need to learn more about what you mean. Anyway,
I didn't say or necessarily mean that regularity is needed, but rather
just to observe that it is used. I have in mind such things as
Enderton's discussion of natural models of Z set theory as in his set
theory textbook. I don't know whether that is considered model theory
or metamathematics or whether it is to be considered just more set
theory, but at least it is discussion of models of Z set theory, and
the axiom of regularity is used to get the V_alphas that are used to
evince the natural models. Maybe it can be done without the axiom of
regularity, I don't know.
MoeBlee
LauLuna
Ordinarily we understand that the domains of our models are sets. But
if you introduce proper classes into the picture, well, then we surely
want to interpret set theory as being about all classes. So, again we
would have no model for it, unless we admit the existence of a
universal class.
Since Raúl Orayen was Argentinian, you may be right that most
references to his paradox are in Spanish. It seemed that Orayen had
some relationship to Quine; nevertheless I cannot provide any
reference in English, at least no more than what you could find on the
web.
A related difficulty lies in the interpretation of some FOL theorems
like Thomson's theorems: -Ex Ay Rxy <-> -Ryy. Note that this is
closely related to Russell's paradox.
Take R as the 'aboutness' relation among propositions, so defined: p
is about q iff there is a proposition r equivalent to p and the
subject of r either denotes q or a class to which q belongs. The we
interpret the theorem as:
(1) there is no proposition exactly about all propositions not about
themselves and only them.
And consider:
(2) all propositions not about themselves share the property that
there is no proposition exactly about them.
Well, (2) is equivalent to (1) and seems to be exactly about all
propositions not about themselves.
Consequently there is an interpretation of the theorem, namely (1),
that seems to violate the theorem itself!
I think we can only escape this by imposing a restriction on the range
of the quantifiers in the interpreted theorem: the universe of
discourse cannot include the interpreted theorem itself.
So, again, paradoxes can force us to consider sophisticated issues of
set theory in order to account for the interpretation of FOL
formulas.
Regards
LauLuna
Since Raúl Orayen was Argentinian, you may be right that most
references to his paradox are in Spanish. Unfortunately I cannot
provide any reference in English, at least no more than what you could
find on the web.
Ordinarily we understand that the domains of our models are sets. But
if you introduce proper classes into the picture, well, then we surely
want to interpret set theory as being about all classes. So, again we
would have no model for it, unless we admit the existence of a
universal class.
A related difficulty lies in the interpretation of some FOL theorems
like Thomson's theorems: -Ex Ay Rxy <-> -Ryy. Note that this is
closely related to Russell's paradox.
Take R as the 'aboutness' relation among propositions, so defined: p
is about q iff there is a proposition r equivalent to p and the
subject of r either denotes q or a class to which q belongs. Then we
interpret the theorem as:
(1) there is no proposition exactly about all propositions not about
themselves.
And consider:
(2) all propositions not about themselves share the property that
there is no proposition exactly about them.
Well, (2) is equivalent to (1) and its subject seems to denote the
class of all propositions not about themselves.
Aatu Koskensilta
> So, if in my post you take 'formulas' to be not 'well formed
> formulas' but more generally all formulas (strings of symbols),
> whether well formed or not, then do you see my point? Yet, if there is
> a way to prove the existence of the set of tuples of symbols (tuples
> in the sense of iterated ordered pairs and not in the sense of finite
> functions), then I do need to learn about that proof.
I think I had something like a proof in mind, but I'll have to check if I
was just hallucinating. I could very well be mistaken. Anyhow, it's just a
technicality as you yourself note; there is no problem in coming up with a
suitable representation for expressions in context of Zermelo set theory.
> That may be so, and I need to learn more about what you mean. Anyway,
> I didn't say or necessarily mean that regularity is needed, but rather
> just to observe that it is used. I have in mind such things as
> Enderton's discussion of natural models of Z set theory as in his set
> theory textbook. I don't know whether that is considered model theory
> or metamathematics or whether it is to be considered just more set
> theory, but at least it is discussion of models of Z set theory, and
> the axiom of regularity is used to get the V_alphas that are used to
> evince the natural models. Maybe it can be done without the axiom of
> regularity, I don't know.
Regularity is always unnecessary. Whenever regularity is used, it can be
avoided by relativizing all formulas to the class WF of well-founded sets,
i.e. replacing "Ax..." with "Ax((x = 0 \/ Ey in x(x intersect y = 0))
--> ..." and "Ex..." with "Ex((x = 0 \/Ey in x(x intersect y = 0)) & ...".
That the usual axioms of set theory provably hold in WF is one of the easier
inner model consistency proofs.
MoeBlee
> the axiom of regularity is used to get the V_alphas that are used to
> evince the natural models.
Darn. I didn't mean to say that. Of course it is transfinite
recursion, courtesy of replacement (not regularity) that provides us
with the V_alphas. What I meant is that (at least it is my impression
that) the axiom of regularity is used to infer that each set is a
member of some V_alpha. What I have in mind specifically (and I am
rusty on this material) is Enderton's chapter on 'Rank' and then his
comment in his section on 'Natural Models' [in brackets are my
contextual additions]: "Notice that we did not use the regularity
axiom in this proof [the proof that the regularity axiom is true in
V_alpha for all alpha], in contrast to the situation with all the
other axioms [the proofs that the other axioms of Z set theory are
each true in some V_alpha]." So, Aatu, I wonder whether I understand
you correctly: Are you saying that instead of using V_alphas as
universes for models of the axioms we can take those subsets of the
V_alphas that are the V_alphas but with just grounded sets? My
admittedly inexact sense of that is that it makes sense; I don't see
why we can't do that. So maybe Enderton just wanted to use a certain
approach for pedagogical reasons and his comment about the need for
regularity should be taken not generally but with respect to the
particular approach he took?
SEPARATE QUESTION:
What assumptions are needed to show that every finite subset of ZF is
consistent? And is the proof very difficult or long to convey? And,
for historical perspective, who came up with the proof and around what
year?
Thanks,
MoeBlee
Aatu Koskensilta
> Darn. I didn't mean to say that. Of course it is transfinite
> recursion, courtesy of replacement (not regularity) that provides us
> with the V_alphas. What I meant is that (at least it is my impression
> that) the axiom of regularity is used to infer that each set is a
> member of some V_alpha. What I have in mind specifically (and I am
> rusty on this material) is Enderton's chapter on 'Rank' and then his
> comment in his section on 'Natural Models' [in brackets are my
> contextual additions]: "Notice that we did not use the regularity
> axiom in this proof [the proof that the regularity axiom is true in
> V_alpha for all alpha], in contrast to the situation with all the
> other axioms [the proofs that the other axioms of Z set theory are
> each true in some V_alpha]." So, Aatu, I wonder whether I understand
> you correctly: Are you saying that instead of using V_alphas as
> universes for models of the axioms we can take those subsets of the
> V_alphas that are the V_alphas but with just grounded sets?
V_alpha is by definition grounded. My remark that regularity is never
necessary is based on the following observation: whenever we have a proof of
a set theoretical statement P relying on regularity, we can transform it
into a proof of "P holds for well-founded sets". Models of set, reals,
naturals, and all usual objects live within the class of well-founded sets
so a statement P about such objects is equivalent to "P holds for
well-founded sets", which, combined with the previous observation, shows
that regularity is not needed in proofs about statements about ordinary
mathematical objects, such as V_alpha.
> SEPARATE QUESTION:
>
> What assumptions are needed to show that every finite subset of ZF is
> consistent? And is the proof very difficult or long to convey? And,
> for historical perspective, who came up with the proof and around what
> year?
That every finite subset of ZF is consistent is equivalent to the
consistency of ZF by compactness.
That for every finite subtheory A of ZF ZF proves "A is consistent" can be
proved in a number of ways, perhaps the most straightforward of which is
showing that ZF proves reflection, i.e. proves for every statement P,
"exists alpha, P <--> V_alpha |= P"; keeping this in mind we can then
reason in ZF as follows: A, hence there exists a V_alpha such that
V_alpha |= A. By soundness of first order logic we then have for every Q
such that A |- Q, V_alpha |= Q, and since V_alpha |=/= 0 = 1 we have
A |-/- 0 = 1, i.e. A is consistent.
Another way is to prove, using Gentzen's haupsatz, that for every B it is
provable in ZF that if B is logically provable, B is true, and then apply
this to a finite subtheory A of ZF as follows; since A is a subtheory of ZF
we have ZF |- A; by the previous observation, ZF |- ("A --> 0 = 1" is
logically provable --> (A --> 0 = 1)); by a trivial proof ZF |- ~0 = 1 and
hence ZF |- ~(A --> 0=1), and by the application of Haupsatz, ZF |-
("A --> 0 = 1" is not logically provable) which is, provably in ZF,
equivalent to "A is consistent".
The result that ZF is not finitely axiomatizable was first proved by
Mostowski, in the 1940's if I'm not mistaken.
MoeBlee
wrote:
> On 2007-04-25, MoeBlee wrote:
> > So, if in my post you take 'formulas' to be not 'well formed
> > formulas' but more generally all formulas (strings of symbols),
> > whether well formed or not, then do you see my point? Yet, if there is
> > a way to prove the existence of the set of tuples of symbols (tuples
> > in the sense of iterated ordered pairs and not in the sense of finite
> > functions), then I do need to learn about that proof.
>
> I think I had something like a proof in mind, but I'll have to check if I
> was just hallucinating. I could very well be mistaken. Anyhow, it's just a
> technicality as you yourself note; there is no problem in coming up with a
> suitable representation for expressions in context of Zermelo set theory.
Yep, indeed it was my point that as long as we use functions, as
opposed to tuples, we can do it in Z.
> > That may be so, and I need to learn more about what you mean. Anyway,
> > I didn't say or necessarily mean that regularity is needed, but rather
> > just to observe that it is used. I have in mind such things as
> > Enderton's discussion of natural models of Z set theory as in his set
> > theory textbook. I don't know whether that is considered model theory
> > or metamathematics or whether it is to be considered just more set
> > theory, but at least it is discussion of models of Z set theory, and
> > the axiom of regularity is used to get the V_alphas that are used to
> > evince the natural models. Maybe it can be done without the axiom of
> > regularity, I don't know.
>
> Regularity is always unnecessary. Whenever regularity is used, it can be
> avoided by relativizing all formulas to the class WF of well-founded sets,
> i.e. replacing "Ax..." with "Ax((x = 0 \/ Ey in x(x intersect y = 0))
> --> ..." and "Ex..." with "Ex((x = 0 \/Ey in x(x intersect y = 0)) & ...".
> That the usual axioms of set theory provably hold in WF is one of the easier
> inner model consistency proofs.
I hope you'll comment, even if only briefly, on the post I made about
that this morning, and about the other question regarding consistency
of any finite subset of ZF.
Thanks.
MoeBlee
Frederick Williams
>
> On 2007-04-25, Frederick Williams wrote:
> > Bill Taylor wrote:
> >
> >> It presumably doesn't have to deal with power sets, cardinality
> >> beyond the finite/infinite distinction,
> >
> > Why not? Chang and Keisler is full of stuff about Cardinality.
>
> It all depends on what one takes model theory to comprise.
Up and down Lowenheim-Skolem at least, I would think.
> Depending on that
> a weak fragment of set theory migh suffice, or it might be principles
> beyond those in ZFC are needed.
--
MoeBlee
wrote:
> On 2007-04-26, MoeBlee wrote:
> > What I meant is that (at least it is my impression
> > that) the axiom of regularity is used to infer that each set is a
> > member of some V_alpha. What I have in mind specifically (and I am
> > rusty on this material) is Enderton's chapter on 'Rank' and then his
> > comment in his section on 'Natural Models' [in brackets are my
> > contextual additions]: "Notice that we did not use the regularity
> > axiom in this proof [the proof that the regularity axiom is true in
> > V_alpha for all alpha], in contrast to the situation with all the
> > other axioms [the proofs that the other axioms of Z set theory are
> > each true in some V_alpha]." So, Aatu, I wonder whether I understand
> > you correctly: Are you saying that instead of using V_alphas as
> > universes for models of the axioms we can take those subsets of the
> > V_alphas that are the V_alphas but with just grounded sets?
>
> V_alpha is by definition grounded.
Okay, then obviously I am at least a bit confused about this. Okay, I
see that the V_alphas are grounded, so now I'm confused as to what
Enderton means by using regularity in those proofs that the axioms are
true in certain V_alphas. I think I just need to study this more.
> My remark that regularity is never
> necessary is based on the following observation: whenever we have a proof of
> a set theoretical statement P relying on regularity, we can transform it
> into a proof of "P holds for well-founded sets".
Okay, I understand that.
> Models of set, reals,
> naturals, and all usual objects live within the class of well-founded sets
> so a statement P about such objects is equivalent to "P holds for
> well-founded sets", which, combined with the previous observation, shows
> that regularity is not needed in proofs about statements about ordinary
> mathematical objects, such as V_alpha.
Okay, that makes sense given that the V_alphas are grounded.
> > SEPARATE QUESTION:
>
> > What assumptions are needed to show that every finite subset of ZF is
> > consistent? And is the proof very difficult or long to convey? And,
> > for historical perspective, who came up with the proof and around what
> > year?
>
> That every finite subset of ZF is consistent is equivalent to the
> consistency of ZF by compactness.
Right, of course.
> That for every finite subtheory A of ZF ZF proves "A is consistent" can be
> proved in a number of ways,
Wait, I want to be clear here as to what is at stake. Are we talking
about (1) or (2)?:
(1) Which set of assumptions are we invoking to prove "Every finite
subset of ZF is consistent" ?
(2) Which set of assumptions are we invoking to prove "For every
finite subset A of ZF, ZF proves "A is consistent"" ?
I sense that you're talking about (2), right? Then you're inferring
"Every finite subset of ZF is consistent" from "For every finite
subset A of ZF, ZF proves "A is consistent"", right? But, if I recall
correctly, someone asked you something along the lines of "How do we
know ZF is consistent?" and your response was to invoke that every
finite subset of ZF is consistent. But while I see that that could be
a technical answer, I don't see it meeting the more epistemological
concern conveyed by words such as 'know'. Though, I do tend to see
Franzen's point as to such epistemological concerns being misplaced or
misconceived in a more basic way.
> perhaps the most straightforward of which is
> showing that ZF proves reflection, i.e. proves for every statement P,
> "exists alpha, P <--> V_alpha |= P";
And we show that with what assumptions? Only those of ZF itself or
with others too?
> keeping this in mind we can then
> reason in ZF as follows: A, hence there exists a V_alpha such that
> V_alpha |= A. By soundness of first order logic we then have for every Q
> such that A |- Q, V_alpha |= Q, and since V_alpha |=/= 0 = 1 we have
> A |-/- 0 = 1, i.e. A is consistent.
>
> Another way is to prove, using Gentzen's haupsatz, that for every B it is
> provable in ZF that if B is logically provable, B is true, and then apply
> this to a finite subtheory A of ZF as follows; since A is a subtheory of ZF
> we have ZF |- A; by the previous observation, ZF |- ("A --> 0 = 1" is
> logically provable --> (A --> 0 = 1)); by a trivial proof ZF |- ~0 = 1 and
> hence ZF |- ~(A --> 0=1), and by the application of Haupsatz, ZF |-
> ("A --> 0 = 1" is not logically provable) which is, provably in ZF,
> equivalent to "A is consistent".
>
> The result that ZF is not finitely axiomatizable was first proved by
> Mostowski, in the 1940's if I'm not mistaken.
Wasn't it Montague somewhat later than the '40s?
Thanks,
MoeBlee
Alan Smaill
Attribute to Montague 1961 (from a conference in 1959)
by mathworld, FWIW:
mathworld.wolfram.com/Zermelo-FraenkelSetTheory.html
>
> Thanks,
>
> MoeBlee
>
--
Alan Smaill
george
> Ordinarily we understand that the domains
> of our models are sets. But
> if you introduce proper classes into the picture,
> well, then we surely want to interpret set theory
> as being about all classes.
No, of course you do NOT want to do that.
Once you introduce proper classes, the set/class
dichotomy is VERY hard and stark, and you STILL
want to be about all SETS. Being about "all" classes
is just sort of fundamentally impossible, INSIDE the
"same" class theory. Bothering with proper classes is
simply ACKNOWLEDING that universal generalization
forces semantic ascent. I will belabor this point to
unreasonable detail in rehashing Rusell's Vicious
Circle Principle below
> So, again we would have no model for it,
The class theory ITSELF, JUST as in the old/refuted ZFC
case, is supposed to provide the machinery for constructing
the classes-that-are-going-to-be-models. The point here
is that one can now talk about theories as having sets as
models WITHOUT having a meltdown when either 1) generalizing over all
of them, or 2) considering set
theories. In both of those cases, we now have the
option of using the proper class of all sets as a domain.
> unless we admit the existence of a universal class.
In most class theories that is simply impossible, for the
simple reason that one way of distinguishing a proper class
from a set is that proper classes are not members of other
classes. Since No class contains Any proper class, it is
analytic that you are not going to have a class of classes.
BUT YOU CAN STILL have a class of all sets, AND YOU CAN
STILL do most of first-order model theory UNDER THE
RESTRICTION that models HAVE to be sets, or at least
have domains that are sets. Given that everybody was
FORMERLY doing model theory (of everything EXCEPT ZFC
and ITS "supermodels") with domains as ZFC-sets, this CAN'T
POSSIBLY be any worse a state of affairs THAN the one that
prevailed formerly. But you now have the option of creating
a fork in your model-theory between theories whose domains
can be sets those whose domains MUST be proper classes
(you can now assign ZFC to the upper tier). This is analogous
to using 2nd-order logic -- you can ascend from sets to proper
classes as opposed to from 0th-order objects to 1st-order
predicates over them -- a class of sets arguably just IS
a predicate over sets-considered-as-base-objects.
> A related difficulty lies in the interpretation of
> some FOL theorems like Thomson's theorems:
> -Ex Ay Rxy <-> -Ryy.
Calling THIS an FOL theorem is like calling Hurricane
Camille a rainstorm, or WW2 a skirmish!
THIS IS a *2*OL theorem!
The R IS UNIVERSALLY quantified!!
> Note that this is closely related to Russell's paradox.
Indeed, it is Rusell's paradox re-phrased as
a 2nd-order validity; it almost doesn't even sink
to the level of "theorem" since it doesn't even
need any premises or axioms! Without even
an axiom to put R into the signature, you can
derive this FROM NOTHING using only *1st*-order
inference rules!
> Take R as the 'aboutness' relation among propositions,
But the WHOLE point is that R can be ANYthing.
It NEVER MATTERS what R is.
This "x" is ALWAYS going to come up non-existent.
> so defined: p is about q iff there is a
> proposition r equivalent to p and the
> subject of r either denotes q or a class to which q belongs.
> Then we interpret the theorem as:
>
> (1) there is no proposition exactly about
> all propositions not about
> themselves and only them.
For if there were, would it be about itself or not about itself?
It is for this reason that the object-language/meta-language
distinction was invented. But it is an OLD invention by now.
> And consider:
>
> (2) all propositions not about themselves share the property that
> there is no proposition exactly about them.
THIS is a MIS-use of "all". THIS is NOT the "all" of the universal
quantifier, the "all" that translates to "each and every",
singularly, individually, conjunctively. THIS is the "all"
that translates into "all COLLECTIVELY", into "the class of all",
considered as ONE thing. You CAN'T use the "individual"
quantifiers of classical logic to talk about "sharing" a property.
THOSE quantifers will ALWAYS be saying about each ONE
thing (or some definitely-finitely-numbered ordered-n-tuple
of things) that it does or doesn't have some property.
> Well, (2) is equivalent to (1)
No, it isn't.
> and seems to be exactly about all
> propositions not about themselves.
To the contrary, it doesn't even seem to be reasonably or
rationally translatable (or even rightly understood,
frankly) by the very people proposing it.
> Consequently there is an interpretation of the theorem,
> namely (1),
> that seems to violate the theorem itself!
That is preposterous. Strawson's "theorem" CAN'T
be violated: IT IS A LOGICAL VALIDITY. A great
many things may ultimately get violated, but NO
*2nd*-order validity IS EVER going to be one of them.
SOME OTHER error is being made; I just identified one.
If you tried to formalize (2), you would see that your natlang
gloss of it is just wrong.
> I think we can only escape this by imposing
> a restriction on the range
> of the quantifiers in the interpreted theorem: the universe of
> discourse cannot include the interpreted theorem itself.
DUH! IT HAS ALWAYS BEEN the case that you risk paradox
when you try to produce a summary-statistic over a domain
that WILL THEN contain THAT VERY product!
FOR[classic]EXAMPLE: IF you have a blackboard
with two or more positive natural
numerals written on it, then you canNOT,
on THAT board, write the sum of the numerals-written-on
THAT board (think of that as a vicious circle of diameter 1).
If you have 2 blackboards (again with two or more positive
natural numerals written on them), you canNOT arrange
for EACH of them to have its sum-of-the-numerals-written-on-it
occurring on SOME one of the 2 (on EITHER of the 2) boards:
think of that as a vicious circle of diameter 2. You can expand
this to a diameter of 3: You cannot write two or more positive
natnums on 3 boards such that for each board, the sum of the numbers
written on it occurs on one of THOSE
3 boards. You can expand this to a diameter of 4,
or, inductively, to any natural number n. If you want to ensure
that the sums of the numbers written on all n boards occur
on some boards, then ONE of the boards-on-which-a-sum-
occurs must be OUTSIDE the n boards summed over (putting
the nth&-last one "in" would complete the "circle").
This is Russell's "vicious circle principle". It is completely
general. You could certainly think of Strawson's theorem
as one way of saying it. The left or "x" argument of R ranges
over possible values of the summary-statistic and the right or
"y" argument ranges over the domain being summarized.
Classical quantifers quantify over the SAME domain at all
argument positions, so the range of the summary statistic
winds up forced to be in the right-arg domain ("Rxx") and
paradox ensues.
Equivalently you could say that Russell's Paradox IS
a paradox precisely BECAUSE the correct formal translation
of it is a logical contradiction, and the proof that that
contradiction
IS a contradiction is just to notice that it IS THE DENIAL of
Strawson's theorem, WHICH IS A *2ND*-order validity.
The first-order Russell's paradox is what you get by
instantiating R to epsilon, but the point is, predicates and
their extensions are sufficiently tightly connected that
for a predicate to be true of something and for its(class-
like)extension to contain that something as a member
are arguably merely orthographic variations on the same theme.
> So, again, paradoxes can force us
> to consider sophisticated issues of
> set theory in order to account for the interpretation of FOL
> formulas.
The vicious circle principle IS NOT a SOPHISTICATED
issue of set theory! That you cannot have a circle of xK
such that x0 < x1 < x2 < x... < xK < x0 again is a TRIVIAL issue
of linear ordering! To the extent that it is a 2nd-order
validity/contradiction, it is not even specific to set theory
at all, although, again, set theory remains SPECIAL in that
you could insist that extensions ALWAYS exist and that
EVERY instance of a predicate being true of something is
ALSO ALWAYS an instance of that something belonging
to a class. The point is simply that some of these classes
must be AT HIGHER LEVELS than others in order to avoid this
paradox. That is fundamentally what the set-class distinction
is about. That is why it is "irresponsible", HAVING created a
class of all sets, to then ask about the class of all classes.
It goes WITHOUT saying that you can't fit generalizations
about ALL of the framework INTO the framework! Remeber
the blackboards!
LauLuna
> On Apr 26, 11:21 am, LauLuna <laureanol...@yahoo.es> wrote:
>
> > Ordinarily we understand that the domains
> > of our models are sets. But
> > if you introduce proper classes into the picture,
> > well, then we surely want to interpret set theory
> > as being about all classes.
>
> No, of course you do NOT want to do that.
> Once you introduce proper classes, the set/class
> dichotomy is VERY hard and stark, and you STILL
> want to be about all SETS.
No, this is not so. Most teories that distinguish between proper
classes and sets are standardly interpreted as letting their variables
range over all classes, not just about sets. This standard
interpretation cannot be expressed in the usual theory of models.
> > A related difficulty lies in the interpretation of
> > some FOL theorems like Thomson's theorems:
> > -Ex Ay Rxy <-> -Ryy.
>
> Calling THIS an FOL theorem is like calling Hurricane
> Camille a rainstorm, or WW2 a skirmish!
> THIS IS a *2*OL theorem!
> The R IS UNIVERSALLY quantified!!
It is indeed a FOL theorem in the way I wrote it. There is a easy
derivation of it in FOL. And yes, predicate variables in FOL are just
that: predicate variables, but they are not quantified.
> But the WHOLE point is that R can be ANYthing.
> It NEVER MATTERS what R is.
> This "x" is ALWAYS going to come up non-existent.
Oh, yes. But I was interested in speaking about just one
interpretation.
> > (2) all propositions not about themselves share the property that
> > there is no proposition exactly about them.
>
> THIS is a MIS-use of "all". THIS is NOT the "all" of the universal
> quantifier, the "all" that translates to "each and every",
> singularly, individually, conjunctively. THIS is the "all"
> that translates into "all COLLECTIVELY", into "the class of all",
> considered as ONE thing. You CAN'T use the "individual"
> quantifiers of classical logic to talk about "sharing" a property.
> THOSE quantifers will ALWAYS be saying about each ONE
> thing (or some definitely-finitely-numbered ordered-n-tuple
> of things) that it does or doesn't have some property.
Those are just linguistic conventions that cannot change the
underlying logical facts.
> FOR[classic]EXAMPLE: IF you have a blackboard
> with two or more positive natural
> numerals written on it, then you canNOT,
> on THAT board, write the sum of the numerals-written-on
> THAT board (think of that as a vicious circle of diameter 1).
> If you have 2 blackboards (again with two or more positive
> natural numerals written on them), you canNOT arrange
> for EACH of them to have its sum-of-the-numerals-written-on-it
> occurring on SOME one of the 2 (on EITHER of the 2) boards:
> think of that as a vicious circle of diameter 2. You can expand
> this to a diameter of 3: You cannot write two or more positive
> natnums on 3 boards such that for each board, the sum of the numbers
> written on it occurs on one of THOSE
> 3 boards. You can expand this to a diameter of 4,
> or, inductively, to any natural number n. If you want to ensure
> that the sums of the numbers written on all n boards occur
> on some boards, then ONE of the boards-on-which-a-sum-
> occurs must be OUTSIDE the n boards summed over (putting
> the nth&-last one "in" would complete the "circle").
I like this example. Thanks.
>
> This is Russell's "vicious circle principle".
Well, I'm not sure. It's something akin or related to it, but perhaps
not just it.
I have never written that Thomson's theorem fails, I just wrote that
it seems to fail; of course it does not fail. The restriction on the
universe of discourse fixes it all. But I would say the question of
the logical levels (whose necessity we both admit) is one of the most
sophisticated; a hint at this is how hard paradoxes are to solve.
>The point is simply that some of these classes
> must be AT HIGHER LEVELS than others in order to avoid this
> paradox. That is fundamentally what the set-class distinction
> is about. That is why it is "irresponsible", HAVING created a
> class of all sets, to then ask about the class of all classes.
> It goes WITHOUT saying that you can't fit generalizations
> about ALL of the framework INTO the framework! Remeber
> the blackboards!
I quite agree but, again, the issue of what exactly those levels are
is far from clear.
I am having poblems too with mu posts and that's why some of them
appear more than once. Sorry.
Regards
Aatu Koskensilta
> Okay, then obviously I am at least a bit confused about this. Okay, I
> see that the V_alphas are grounded, so now I'm confused as to what
> Enderton means by using regularity in those proofs that the axioms are
> true in certain V_alphas. I think I just need to study this more.
I think Enderton's point was rather that in proving that an axiom A holds in
V_alpha (for a suitable alpha) relies on A itself for all axioms except
regularity.
> Wait, I want to be clear here as to what is at stake. Are we talking
> about (1) or (2)?:
I'm talking about "for every finite subtheory A of ZF, ZF proves 'A is
consistent'", since, as noted, "every finite subtheory of ZF is consistent"
is just another way of saying "ZF is consistent". To prove that ZF is
consistent one needs to invoke principles not contained in ZF, e.g. basic
principles concerning set theoretic truth or the existence of an
inaccessible, or more trivially, "ZF is consistent" (and "ZF is consistent if
every of its finite subtheories is consistent", which is provable in ZF).
> I sense that you're talking about (2), right? Then you're inferring
> "Every finite subset of ZF is consistent" from "For every finite
> subset A of ZF, ZF proves "A is consistent"", right?
We can of course conclude "ZF is consistent" from "For every finite
subtheory A of ZF, ZF proves 'A is consistent'", relying on global
reflection for ZF, i.e. the principle "if for every natural n, ZF |- P(n),
then P(n) for all n" (and coding of formulas of set theory as naturals).
While there is nothing wrong with this argument, it is rather pointless and
carries no epistemological weight, in the sense thay if someone doubts the
consistency of ZF their doubts will not evaporate as a result of being
presennted this reasoning.
However, I thought you were interested in the result that ZF is reflexive,
i.e. proves for every of its finite subtheories that they're consistent.
This result is purely finitistic, and provable in e.g. primitive recursive
arithmetic -- this is because we can give a proof theoretical argument that
provides a method that for every finite subtheory A, or equivalently for
every A provable in ZF, provides a proof of consistency of A in ZF.
> And we show that with what assumptions? Only those of ZF itself or
> with others too?
The proof of reflection in ZF relies on replacement. Proving that reflection
is provable in ZF does not, however, and as noted can be done in primitive
recursive arithmetic.
> Wasn't it Montague somewhat later than the '40s?
Uh, yes, as I have myself said earlier. I don't know what I was thinking!
george
> > > But
> > > if you introduce proper classes into the picture,
> > > well, then we surely want to interpret set theory
> > > as being about all classes.
>
> > No, of course you do NOT want to do that.
> > Once you introduce proper classes, the set/class
> > dichotomy is VERY hard and stark, and you STILL
> > want to be about all SETS.
>
> No, this is not so. Most teories that distinguish between proper
> classes and sets are standardly interpreted as letting their variables
> range over all classes,
In EVERY first-order theory, the variables range over
everything that exists in the domain.
> not just about sets.
Obviously, in a class theory, some of the things in the domain
are classes and not sets. But since EVERY ELEMENT of
a class is a set, EVERY class is always going to be a class
OF SETS. So EVERYTHING you are saying, you are saying
ABOUT sets! Introducing classes is basically just 2nd-order
set theory. It would obviously be odd to insist that 2nd-order
PA is about something OTHER than the natural numbers,
when it is categorical and its ONLY model has a domain that
MUST BE [isomorphic to] the natural numbers. The first-order
predicates in any 2nd-order model HAVE to be predicates OVER
the 0th-order individuals. Something very very analogous is
going on with classes over set theory.
> This standard interpretation
> cannot be expressed in the
> usual theory of models.
THIS is NOT any sort of standard interpretation!
> > > A related difficulty lies in the interpretation of
> > > some FOL theorems like Thomson's theorems:
> > > -Ex Ay Rxy <-> -Ryy.
>
> > Calling THIS an FOL theorem is like calling Hurricane
> > Camille a rainstorm, or WW2 a skirmish!
> > THIS IS a *2*OL theorem!
> > The R IS UNIVERSALLY quantified!!
>
> It is indeed a FOL theorem in the way I wrote it.
Oh, SHUT UP.
By that logic, the proof of a simple 1st-order universal
generalization
looks like a propositional proof AND NOT a 1st-order one, since
the letter upon which you will apply the UG inference rule in the
final step just LOOKS LIKE a constant. It is a variable-NAME
but you simply CAN'T TELL THE DIFFERENCE between a variable
and a constant in that context. The only thing that matters,
contextually, is that the constant is not mentioned in any
axiom of premise from which the theorem is being proven.
That sort of MAKES the constant a "variable" whether anyone
otherwise wanted it to be or not. My point is simply that the
syntax is misleading.
> There is a easy derivation of it in FOL.
There IS NOT, dumbass: There IS NO *R* in your signature!
> And yes, predicate variables in FOL are
ABSOLUTELY NON-EXISTENT.
Seriously, NO MATTER WHAT AUTHORITY EVER told
you otherwise, THERE IS NO SUCH THING as a predicate
VARIABLE in FIRST-order logic!
Keith Ramsay
On Apr 23, 11:13 pm, Bill Taylor <w.tay...@math.canterbury.ac.nz>
wrote:
|It presumably doesn't have to deal with power sets, cardinality
|beyond the finite/infinite distinction, regularity, well-ordering,
|replacement, or even very much separation.
|
|It would be nice to know if this basic amount of set theory has
itself
|been made the topic of a formal (mathematical) theory. Has it?
My first reply seems to have disappeared.
Look at the Wikipedia page for "reverse mathematics" and you'll see
that Goedel's completeness theorem for a countable language (only
countably many predicate, function, and constant symbols) is
equivalent in some weak system to WKL: weak Koenig's lemma, the claim
that any infinite binary tree has an infinite branch. The system based
on WKL has been studied quite a bit.
I was going to suggest that was all you probably wanted. Uncountable
models are studied, of course, but in this context, you don't really
care whether you can prove that they exist, do you?
Now someone has suggested you probably would want enough to prove
upward and downward Lowenheim-Skolem. I don't know how much one needs
to prove either of those. I still think you probably don't really need
more than countable models for your purposes, however, and so you can
stick to subsystems of second-order arithmetic.
Keith Ramsay
george
> Look at the Wikipedia page for
> "reverse mathematics" and you'll see
> that Goedel's completeness theorem
> for a countable language (only
> countably many predicate, function,
> and constant symbols) is
> equivalent in some weak system to
> WKL: weak Koenig's lemma, the claim
> that any infinite binary tree has an infinite branch.
Can somebody tell me what the "foundational"
status of WKL is? In these weak reverse systems,
it seems to be being used as an axiom, which
implies that there could be nonstandard systems
with this axiom's denial used as an axiom instead.
But it just seems to me to go without saying that
if every node of the tree only has finitely many immediate
branches then surely WKL *must* be true.
In what context is it ever deniable enough to need
to be an axiom?? What would a structure in which
WKL was false even look like (other than a tree with
infinite BREADTH of branching)??
MoeBlee
wrote:
> On 2007-04-26, MoeBlee wrote:
> I think Enderton's point was rather that in proving that an axiom A holds in
> V_alpha (for a suitable alpha) relies on A itself for all axioms except
> regularity.
Aha, now that would make sense and it clears up the overall issue too.
I thought he meant that regularity was used (and I thought therefore
it must have been used earlier in all the stuff about rank) to prove
the existence of models of each of the axioms except the axiom of
regularity itself. But you're suggesting that each of the axioms is
used to prove that it has a model except that regularity is not used
to prove that it has a model. And I see that your version does make
more sense!
So, good, now I see that we can scratch my original remark about
regularity being used for certain results about models of set theory.
> I'm talking about "for every finite subtheory A of ZF, ZF proves 'A is
> consistent'", since, as noted, "every finite subtheory of ZF is consistent"
> is just another way of saying "ZF is consistent". To prove that ZF is
> consistent one needs to invoke principles not contained in ZF, e.g. basic
> principles concerning set theoretic truth or the existence of an
> inaccessible, or more trivially, "ZF is consistent" (and "ZF is consistent if
> every of its finite subtheories is consistent", which is provable in ZF).
So, as I understand (and this makes sense to me) to prove "For every
finite subtheory A of ZF, ZF proves 'A is consistent'" we need some
principles other than those of ZF. So, if I understand correctly, when
you say that every finite subset of the axioms of ZF is consistent, as
far as proving that, you'd rely on, say, the existence of an
inaccessbile cardinal or something along those lines? That is what I
mean when I say that, epistemologically, it's not persuasive just to
say that every finite subset is consistent, since that is either
begging the question (taking ZF to be consistent anyway, possibly by
some other "epistemological" argument such as ZF being true of the
cumulative hierarchy) or appealing to not just ZF itself, but a
principle added to ZF (such as the existence of an inaccessible
cardinal). So, if I understand all of this correctly, I don't think
the finite subset argument is pertinent to answering the skeptic.
(And, again, I want to emphasize that I do appreciate Franzen's
argument as to why the skepticism is ill-conceived anyway.)
> > I sense that you're talking about (2), right? Then you're inferring
> > "Every finite subset of ZF is consistent" from "For every finite
> > subset A of ZF, ZF proves "A is consistent"", right?
>
> We can of course conclude "ZF is consistent" from "For every finite
> subtheory A of ZF, ZF proves 'A is consistent'", relying on global
> reflection for ZF, i.e. the principle "if for every natural n, ZF |- P(n),
> then P(n) for all n" (and coding of formulas of set theory as naturals).
> While there is nothing wrong with this argument, it is rather pointless and
> carries no epistemological weight, in the sense thay if someone doubts the
> consistency of ZF their doubts will not evaporate as a result of being
> presennted this reasoning.
I'm not up on the technicals regarding reflection and coding, but, in
general, I think you're saying yourself what I said above. So, what I
don't understand is why you present the skeptic with the finite subset
argument while, if I understand you correctly, you yourself recognize
that it has no epistemological clout.
> However, I thought you were interested in the result that ZF is reflexive,
> i.e. proves for every of its finite subtheories that they're consistent.
> This result is purely finitistic, and provable in e.g. primitive recursive
> arithmetic -- this is because we can give a proof theoretical argument that
> provides a method that for every finite subtheory A, or equivalently for
> every A provable in ZF, provides a proof of consistency of A in ZF.
Okay, now I am seriously confused (due to my not yet having
systematically studied into this part of set theory), as what you just
said seems to me to say that ZF proves its own consistency (stronger,
primitive recursive arithmetic does), which I know must be a wildy
incorrect inference from what you said. I suggest that rather than
wear you out trying to explain this to me when I'm not properly
prepared techncially, instead I should wait until I'm better prepared
and perhaps we can resume then.
> > And we show that with what assumptions? Only those of ZF itself or
> > with others too?
>
> The proof of reflection in ZF relies on replacement. Proving that reflection
> is provable in ZF does not, however, and as noted can be done in primitive
> recursive arithmetic.
Same as above.
MoeBlee
Keith Ramsay
On Apr 30, 12:15 am, george <gree...@cs.unc.edu> wrote:
Some of the more natural models of these systems consist of the
natural numbers and some family of subsets of the natural numbers.
This kind of model can be made to satisfy the weaker axioms while
failing the stronger ones. In this case, for example, the model which
has only computable subsets of the natural numbers in it fails
WKL. There are computable infinite binary trees in which each
computable branch is finite.
In general, WKL is regarded as both very intuitive, and reasonable
powerful in the sense that a lot of ordinary mathematics requires
nothing stronger than it.
In case anybody is curious, while WKL in its usual formulation is
nonconstructive (due to the above-mentioned fact), it has a natural
constructive counterpart that was studied by Brouwer, known as
the Fan theorem. So if by Brouwerian mathematics we think only
of mathematics based on principles up to the Fan theorem, then
it's close to the same strength as WKL. (I seem to remember
some minor technical difference....)
Keith Ramsay
Aatu Koskensilta
> So, as I understand (and this makes sense to me) to prove "For every
> finite subtheory A of ZF, ZF proves 'A is consistent'" we need some
> principles other than those of ZF.
No, to prove "For every finite subtheory A of ZF, ZF proves 'A is
consistent'" we need nothing but basic finitistic principles. On the other
hand, to prove "Every finite subtheory of ZF is consistent" we need
principles not contained in ZF.
> So, if I understand correctly, when
> you say that every finite subset of the axioms of ZF is consistent, as
> far as proving that, you'd rely on, say, the existence of an
> inaccessbile cardinal or something along those lines? That is what I
> mean when I say that, epistemologically, it's not persuasive just to
> say that every finite subset is consistent, since that is either
> begging the question (taking ZF to be consistent anyway, possibly by
> some other "epistemological" argument such as ZF being true of the
> cumulative hierarchy) or appealing to not just ZF itself, but a
> principle added to ZF (such as the existence of an inaccessible
> cardinal). So, if I understand all of this correctly, I don't think
> the finite subset argument is pertinent to answering the skeptic.
> (And, again, I want to emphasize that I do appreciate Franzen's
> argument as to why the skepticism is ill-conceived anyway.)
Skepticism is indeed pointless unless based on some specific concern or
doubt. The point of pointing out that ZF proves the consistency of each of
its finite subtheories is not to convince the skeptic, however. Rather, it's
to point out that if one accepts stuff proved in ZF as true, in the sense of
concluding P on basis of P's provability in ZF, say, when P is some number
theoretic statement, one should also accept the consistency of ZF since,
provided one accepts the principle "if ZF |- P then P", it follows from the
consistency of every finite subtheory of ZF. Of course, this is a needlessly
indirect argument, since the consistency of ZF follows from reflection for
ZF ("if P is provable in ZF, then P") simply by noting that "0=1" is not true,
and hence it's not the case that ZF |- 0=1. Sometimes people find the argument
concerning finite subtheories more illuminating than this trivial observation.
PS. I believe it was Torkel and not me who provided the observation about
finite subtheories of ZF as a somewhat facetious reason for concluding that
ZF is consistent.
LauLuna
>
> Oh, SHUT UP.
> There IS NOT, dumbass: There IS NO *R* in your signature!
>
No matter how original your ideas might seem, I do not to like your
manners. So, excuse that I do not go on with the discussion.
Regards