ZFC+~Con(ZFC)
Richard Kennaway
mcc...@darwin.math.usf.edu. writes:
>The standard (ie, wellfounded) models all satisfy Con(ZFC)
Can someone clarify for me the term "standard model"? Is this a concept
with a formal definition, or is a "standard model" of a theory, simply a
model which satisfies the intuitions which inspired the axioms of the
theory?
--
Richard Kennaway SYS, University of East Anglia, Norwich NR4 7TJ, U.K.
Internet: j...@sys.uea.ac.uk uucp: ...mcsun!ukc!uea-sys!jrk
Colin Mclarty
I do not know what all "standard" might mean, but in a lot of
contexts including this one it means "not non-standard" where "non-
standard" is used as in "non-standard analysis". In short, a model is
"standard" in this sense if all the sets it takes to be finite are
actually finite.
If a model is standard in this sense then all the sets it takes
as codes for proofs of ZF actually are such codes, and so there is no set
taken as coding a proof ofa contradiction, so the model satisfies Con(ZF).
Of course all of this assumes there are models in the first place, so ZF
is consistent. As a previous poster said, models of ZF+not-Con(ZF) are
just models that recognize some sets as codes for proofs of contradictions.
But what these sets code are not really finite proofs--i.e. are not
proofs at all.
Robert M. Solovay
>>In article <1992Aug20.1...@ariel.ec.usf.edu> Gregory McColm,
>>mcc...@darwin.math.usf.edu. writes:
>>>The standard (ie, wellfounded) models all satisfy Con(ZFC)
>>
>>Can someone clarify for me the term "standard model"? Is this a concept
>>with a formal definition, or is a "standard model" of a theory, simply a
>>model which satisfies the intuitions which inspired the axioms of the
>>theory?
>>
>>
>
>contexts including this one it means "not non-standard" where "non-
>standard" is used as in "non-standard analysis". In short, a model is
>"standard" in this sense if all the sets it takes to be finite are
>actually finite.
>
The usual terminology in set-theory is as follows. A model is
standard if it is a transitive set (i.e. every member of it is a
subset of it) and it's epsilon relation is just the restriction of the
usual epsilon relation.
By the Mostowski collapse theorem, a model of ZFC is
isomorphic to a standard model iff its ordinals are well-ordered.
A model of ZFC is "correct for finiteness" iff it has no
non-standard integers. Such models are precisely those which are
isomorphic to models whose integers are literally the usual integers.
Models of this latter type are known as omega models.
Every standard model of ZFC is an omega model. If there is an
inaccessible cardinal, then
(a) there is a countable standard model of ZFC.
(b) there is an omega model of ZFC whose ordinals are not [externally]
well-ordered.
I haven't bothered to track down references for this (it's all
standard stuff) but I suspect this material is somewhere in Kunen or
Jech's opuses on set-theory.
DUEN...@dosuni1.rz.uni-osnabrueck.de
question is "Yes":
Each Boolean algebra has a standard model as an algebra of sets via
Stone's theorem, each group has a standard model as a permutation group
via Cayley's theorem. Standard models of relation algebras are algebras
of binary relations, and there are non-standard models (i.e. non -
representable relation algebras) by a theorem of Lyndon.
Ivo Duentsch
Rechenzentrum
Universitaet Osnabrueck
duen...@dosuni1.rz.uni-osnabrueck.de