[FOM] Paleolithic Model Theory and Non-standard models.
steve newberry
stevnewb at ix.netcom.com
Mon Mar 10 15:34:26 EST 2003
[The following is a reply to Neil Tennant's crushing refutation of my
inadequately formulated problem in a previous post. I present it here,
somewhat corrected and extended beyond its earlier posting on FOM, in hopes
that it may draw responses from others.]
==================================================================================
Of course you are right. I WAS thinking in terms of General Models-FOL
semantics, and should have made that explicit. Clearly, I *must* be more
explicit, and herewith shall do my best in that regard.
[read '@' as the epsilon of membership, ' /@ ' as its negation, and ' /= '
as "not equal".]
The logic is classical, simply-typed, of finite order, FOL-General Models
syntax and semantics, and the ground domain is fixed, discrete and
countable. It either IS omega, or its elements are indexed by omega.
Given those preconditions, I invite you to consider the partition of all
cwffs [of the above stipulated logic] into two negation-invariant blocks: U
and NK, where a cwff B is said to be *absolute* iff it is either a
contradiction or a tautology [and hence in U]; and B is said to be
*contingent* iff it is not absolute, and hence in NK, the contingent
block. But NK splits very nicely in two, to give N + K, so we end up with
three negation-invariant blocks: U + N + K, [where by "negation-invariant"
I mean just that if B @ a block, then ~B also @ that block.] To summarize:
B @ U <=> B is absolute, i.e., ~B is a contradiction and has no models of any
cardinality, and B is the negation of a contradiction, which, for lack of
a better
word, we'll call a tautology. B is "u-valid" [universally valid].
B @ N <=> ~B has an infinite, but no finite models, and B is
valid on all and
only finite domains; ~B is asymptotically close to being a
contradiction, and B
is asymptotically close to being a tautology. B is "n-valid"
[valid on all and
only domains of cardinality n, where n @ omega.] N is not r.e.,
and is the
reason for the "undecidability" of classical logic. Tarski said
it better: Classical
logic is omega-incomplete. [N is a very interesting class of
propositions!]
B @ K <=> both B, ~B have finite models. The models of cwffs in K are
either finite and co-finite, or infinite and infinite, which I call
"co-infinite".]
I claim that this partition , negation invariant and defined by the
cardinality of domains of realization, is complete, i.e., every cwff is
accounted for, and falls into one and only one of the blocks. [I call this
*view* of the universe, and the theorems which follow from it,
"Paleolithic Model Theory", or PMT.]
Are we okay on that? [GM-FOL semantics being now understood.] If not, shoot
me down. If so, then where do you stand on the paradox?
The "Paradox": [This is not really a paradox, but merely, to myself, a bit
surprising.]
Suppose we have a cwff ~B which has an infinite, but no finite models.
[It can be the conjunction of the axioms for any theory that has no finite
realizations.]
Then B is n-valid. If B is *n-valid* then B *has models on all
finite domains*. This class of models comprise what Keisler ["Fundamentals
of Model Theory", in HANDBOOK OF MATHEMATICAL LOGIC, ed. Barwise] terms an
'Elementary Chain' and by the associated 'Elementary Chain Theorem', the
union of this chain is an elementary extension of each structure in the
chain. "It follows that for any theory T , the union of an elementary chain
of models of T is a model of T ." [Up to this point I would include the
foregoing in PMT. Note that no reference is being made to compactness
here, merely the existence of unions.]
In my earlier post on FOM, I remarked that the union of models on an
n-valid cwff "would have to be a model", and you responded:
NT Question: The union of those models would have to be an infinite model
of *what*, exactly?
SN Answer: An infinite model of the n-valid cwff B of which I was
speaking. I'm suggesting that an n-valid cwff is a cwff which, *by merely
existing*, in effect says, "the cwff of which I am the negation is
satisfiable-but-not-valid on omega, hence by the *existence* of the union
of all my finite models, taken together with the fact that such a union is
necessarily infinite, I *must* have an infinite model, despite all my
protestations to the contrary."
A reasonable reaction might be, "OK, sure, but why should I be interested
?" To which my response would be, "Perhaps it is NOT interesting, but it
brings up a question which is interesting to me, because I have only a very
rudimentary [if that!] understanding of the concept of "non-standard"
models. In the scenario which I have outlined above, we find that *every*
n-valid cwff B of classical logic possesses an infinite model, and by
definition, ~B has an infinite model, and I believe that one of these
models is generally held to be "non-standard"; *which one* is the
non-standard model? And if it is the model of ~B which is the
non-standard model, then where does that leave the entire class of theories
which have no finite models?
I really do need some clarification on this question
More information about the FOM
mailing list