Axiom schema of Replacement naturalness
Marc Alcobé García
"The axiom schema is motivated by the idea that whether a class is a
set depends only on the cardinality of the class, not on the rank of
its elements. Thus, if one class is "small enough" to be a set, and
there is a bijection from that class to a second class, the axiom
states that the second class is also a set. However, because ZFC only
speaks of sets, not proper classes, the schema is stated only for
definable bijections, which are identified with their defining
formulas."
Please, forgive my insistence. I asked a similar question some time
ago, but I still have not been able to understand the "naturalness" of
this axiom scheme.
Clearly, if the image of the bijection is to be a set, then it must
have a rank (since every set must have one), higher than any rank of
any of its elements. Therefore, one cannot simply forget about ranks
and put all the weigth of "setness" on size (cardinality).
How can one be confident that no "defining formula", a first-order
one, is going to biject the elements of a set to sets of higher and
higher rank (unboundedly)?
When I asked this question some time ago, I became in part convinced
by the fact that there are regular cardinals (cardinals equal to their
own cofinalities), but this might be a consequence of the assumption
of the replacement scheme, therefore not being of much help to see why
the axiom schema "must be true".
Marc Alcobé García
Let me restate my question: is there any kind of relative consistency
proof that shows that if we could find a contradiction in ZFC then ZFC
without Replacement would also be inconsistent?
David C. Ullrich
<mal...@gmail.com> wrote:
>In the wikipedia article for Axiom schema of replacement it is read:
>
>"The axiom schema is motivated by the idea that whether a class is a
>set depends only on the cardinality of the class, not on the rank of
>its elements. Thus, if one class is "small enough" to be a set, and
>there is a bijection from that class to a second class, the axiom
>states that the second class is also a set. However, because ZFC only
>speaks of sets, not proper classes, the schema is stated only for
>definable bijections, which are identified with their defining
>formulas."
>
>Please, forgive my insistence. I asked a similar question some time
>ago, but I still have not been able to understand the "naturalness" of
>this axiom scheme.
>
>Clearly, if the image of the bijection is to be a set, then it must
>have a rank (since every set must have one),
Possibly an awesomely stupid question: How do you show that
every set _has_ a rank without replacement?
Marc Alcobé García
> every set _has_ a rank without replacement?
From the standpoint of looking for good axioms that actually describe
some abstraction such as von Newmann's universe, the question would
better be if that can be stated rather than shown.
Anyway, I am not asking how would be a theory without replacement.
What I ask is how can we be so confident that it won't give rise to
contradictions of the kind that the (unrestricted) comprehension
scheme does give rise. I have been told that bounding quantifiers it
is possible to prove the consistency of ZFC by means of L (the
constructible universe), but that it is not possible to prove the
consistency of ZFC from the consistency of ZFC - Replacement.
It is particularly annoying too the fact that one must first of all be
able to check that a certain subformula is a class function (i. e.
that it is a theorem of the theory a formula that states that so it
is), in order to know if a formula is or is not an instance of the
axiom scheme. Is this kind of condicion effectively decidable? How
could this be proved?
David C. Ullrich
<mal...@gmail.com> wrote:
>> Possibly an awesomely stupid question: How do you show that
>> every set _has_ a rank without replacement?
>
>From the standpoint of looking for good axioms that actually describe
>some abstraction such as von Newmann's universe, the question would
>better be if that can be stated rather than shown.
I honestly can't figure out what that sentence means. If _what_
can be stated rather than shown?
>Anyway, I am not asking how would be a theory without replacement.
>What I ask is how can we be so confident that it won't give rise to
>contradictions of the kind that the (unrestricted) comprehension
>scheme does give rise. I have been told that bounding quantifiers it
>is possible to prove the consistency of ZFC by means of L (the
>constructible universe), but that it is not possible to prove the
>consistency of ZFC from the consistency of ZFC - Replacement.
>
>It is particularly annoying too the fact that one must first of all be
>able to check that a certain subformula is a class function (i. e.
>that it is a theorem of the theory a formula that states that so it
>is), in order to know if a formula is or is not an instance of the
>axiom scheme.
Again, I can't quite parse the syntax here. But as of the last
few words I believe I follow what's bothering you: In order
to apply replacement we need first to prove that a certain
formula defines a class function.
You had me worried for a second - if that were so it would
not be at all clear that the axioms were decidable, and that
would be bad. But the axiom of replacement does not
work the way you seem to think it does! I looked it
up just to make sure:
The scheme is not "For every phi such that
(ZFC |- phi is a class function), such and such
is a set".
The scheme is: "For _every_ phi with the
appropriate signature,
(phi is a class function) -> (such and such is a set)
is an axiom".
There's a big difference.
>Is this kind of condicion effectively decidable? How
>could this be proved?
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Rupert
>
> - Show quoted text -
No, definitely not. ZC is ZFC without replacement and the consistency
strength of ZFC is much greater than that of ZC. We can prove in ZFC
that the set V_(omega+omega) exists and that this is a model of ZC. So
ZFC can prove that ZC is consistent.
What the axiom of replacement is saying is that if a first-order
formula defines a function whose domain is a set, then there must
exist an ordinal greater than the ranks of all the sets in the range
of the function.
Rupert
> On Wed, 9 Dec 2009 08:49:26 -0800 (PST), Marc Alcobé García
>
>
>
>
>
> <malc...@gmail.com> wrote:
> >In the wikipedia article for Axiom schema of replacement it is read:
>
> >"The axiom schema is motivated by the idea that whether a class is a
> >set depends only on the cardinality of the class, not on the rank of
> >its elements. Thus, if one class is "small enough" to be a set, and
> >there is a bijection from that class to a second class, the axiom
> >states that the second class is also a set. However, because ZFC only
> >speaks of sets, not proper classes, the schema is stated only for
> >definable bijections, which are identified with their defining
> >formulas."
>
> >Please, forgive my insistence. I asked a similar question some time
> >ago, but I still have not been able to understand the "naturalness" of
> >this axiom scheme.
>
> >Clearly, if the image of the bijection is to be a set, then it must
> >have a rank (since every set must have one),
>
> Possibly an awesomely stupid question: How do you show that
> every set _has_ a rank without replacement?
>
It follows from the axiom of regularity, using well-founded recursion.
This is discussed in Chapter 6 of Jech's "Set Theory".
David C. Ullrich
<rupertm...@yahoo.com> wrote:
>On Dec 12, 4:23�am, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>> On Wed, 9 Dec 2009 08:49:26 -0800 (PST), Marc Alcob� Garc�a
>>
>>
>>
>>
>>
>> <malc...@gmail.com> wrote:
>> >In the wikipedia article for Axiom schema of replacement it is read:
>>
>> >"The axiom schema is motivated by the idea that whether a class is a
>> >set depends only on the cardinality of the class, not on the rank of
>> >its elements. Thus, if one class is "small enough" to be a set, and
>> >there is a bijection from that class to a second class, the axiom
>> >states that the second class is also a set. However, because ZFC only
>> >speaks of sets, not proper classes, the schema is stated only for
>> >definable bijections, which are identified with their defining
>> >formulas."
>>
>> >Please, forgive my insistence. I asked a similar question some time
>> >ago, but I still have not been able to understand the "naturalness" of
>> >this axiom scheme.
>>
>> >Clearly, if the image of the bijection is to be a set, then it must
>> >have a rank (since every set must have one),
>>
>> Possibly an awesomely stupid question: How do you show that
>> every set _has_ a rank without replacement?
>>
>
>It follows from the axiom of regularity, using well-founded recursion.
>This is discussed in Chapter 6 of Jech's "Set Theory".
Well of course it does. I don't have the book - are you certain
that replacement is not used as well?
By well-founded recursion it's enough to show that if every
element of x has a rank then x has a rank. How does one
show _that_ without replacement? Seems clear enough
_with_ replacement - we simply consider the set
{rank(y) : y in x}.
Elsewhere in the thread you say
"What the axiom of replacement is saying is that if a first-order
formula defines a function whose domain is a set, then there must
exist an ordinal greater than the ranks of all the sets in the range
of the function."
Precisely. Now isn't that exactly what's needed to apply
well-founded recursion to show every set has a rank?
Rupert
> On Sat, 19 Dec 2009 06:41:01 -0800 (PST), Rupert
>
>
>
>
>
> <rupertmccal...@yahoo.com> wrote:
> >On Dec 12, 4:23 am, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>
> - Show quoted text -
I apologise, it looks as though I misread your question, or didn't
think my answer through properly, or some such stupid thing.
That cannot be proved, no. It is consistent with Z that sets of rank
omega times two exist but the ordinal omega times two does not.
Marc Alcobé García
> On Thu, 17 Dec 2009 07:25:03 -0800 (PST), Marc Alcobé García
I should have figured it out.