[FOM] The Minimal Model of ZF

joeshipman@aol.com joeshipman at aol.com
Wed Jan 2 01:54:27 EST 2008

To sharpen the "paradox" I refer to at the end of my previous post:

If ZF has a standard model, then M, the minimal model, exists as a set, 
and Th(M) is a well-defined set of sentences that includes the axioms 
of ZFC, "V=M", and lots of other things. If M does not exist as a set, 
then Th(M) is no more definable than Th(V); but what goes wrong with 
the previous definition, exactly?

Another way to put it: assuming M exists as a set, it is a unique and 
well-defined set, but is there any sentence that we can prove M 
satisfies that is not already a logical consequence of "ZFC+V=M"? 
Assuming Con(ZF), of course Con(ZF) is such a sentence, because M 
satisfies the true sentences of arithmetic which are absolute and if M 
exists as a set then ZF is consistent; but I'd like to see how much 
further this can be taken.  ("Taking it further" corresponds to 
extending the axiom system "ZFC + V=M" with the scheme "Anything which 
'ZFC + M exists as a set' proves is true in M is true in V".)

-- JS

-----Original Message-----
From: joeshipman at aol.com
To: fom at cs.nyu.edu
Sent: Tue, 1 Jan 2008 12:49 am
Subject: [FOM] The Minimal Model of ZF

The following paragraph is adapted from Cohen's book "Set Theory and
the Continuum Hypothesis".

If ZF has a standard model (that is, a model which, when viewed
externally, has no infinite descending membership chains), then there
is a minimal such model M. M is countable, and satisfies V=L (and M
does NOT satisfy "ZF has a standard model", though it satisfies all
true statements of arithmetic including Con(ZF)). M is the intersection
of all standard models, and for every element x in M there is a formula
A(y) in the language of set theory such that x is the unique element of
M satisfying A. Thus in M every element can be "named".

M (which was first studied by J.C. Shepherdson in 1951) can also be
defined by a transfinite recursion in the same way L is, except that
the operations are simpler: start with the empty set and omega  and
close transfinitely under pairing, sumset, internal powerset (for X the
current partial universe and x in X, the set of y in X which are
subsets of x), and taking ranges of functions. If there is a standard
model, this construction stops adding new sets after a certain
countable stage; otherwise, it goes on forever and simply gives L.
Thus, we can treat M as a class and avoid the assumption that a
standard model exists.

The usual proof of the consistency of V=L, GCH, and AC can be carried
out in this setting. But what I am interested in is what happens if we
strengthen V=L to a new axiom "V=M" (this is equivalent to the
conjunction of V=L and "there is no standard model").  V=M means that
not only is there no SET that can serve as a standard model, the only
CLASS that can do so is V; in other words, the only sets that exist are
the ones that MUST exist. Once you start building V, there will never
be any place you can stop with a model of ZF until you have got

This is as parsimonious as a set theory can be. But there is a
confusing issue. When there IS a standard model, so that M is a set,
every element of M has a name. If there is no standard model, M is a
proper class, so there are way too many things in M to correspond to
countably many formulas. I'd like to ask what is the first set in M
which does not have a name, but this runs into the usual paradoxes. On
the other hand, if M is a set, then internally M satisfies V=M, and
there IS a first set in M which, externally, we can name, but which M
does not "know" has a name.

What else can be said about this set? And can anyone clarify the
situation philosophically? It seems to be a more vicious form of
Skolem's paradox.

-- JS
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