Re: Are Hilbert's Axioms independent?

From: KRamsay (kramsay_at_aol.com)
Date: 11/20/04 

Date: 20 Nov 2004 08:16:37 GMT

In article <y93vfc5g1n3.fsf@nestle.csail.mit.edu>, Allan Adler
<ara@nestle.csail.mit.edu> writes:
|One feature of Hilbert's book is that the axioms are not first order
|axioms. So, the completeness theorem for first order logic doesn't apply
|and it is at least conceivable that even if one axiom is independent of
|the others, there might not exist a model in which it is false. There
|might exist a model of set theory in which there is a model of geometry
|in which it is false, but that is a much bigger problem than one normally
|expects to have to deal with in confronting independence of axioms for
|geometry.

I don't think it's customary to use the term "independent" to mean
what you're using it to mean, for second-order axioms.

Second-order logic has an associated "satisfaction" relationship
between sentences and structures. We can define a consequence
relation between sets of sentences and sentences by saying that
a sentence A is a consequence of a family {B_i} of sentences if
every structure satisfying all of the B_i also satisfies A. Trivially,
if one of the axioms is a consequence in this sense of the others
then it can be omitted without affecting which structures satisfy
the axioms and conversely.

The failure of the compleness theorem to apply means that there
isn't a formal deductive system for second-order logic that
suffices to deduce all the consequences of an arbitrary set
of second-order axioms. As far as I know, in fact, there is no
formal deductive system that is treated as "standard" for it.
Consequently I would assume that there is no canonical
notion of "this sentence cannot be deduced in the canonical
way from these other sentences", as distinct from "this
sentence is satisfied in all the structures where these other
sentences are satisfied".

It appears that you mean by "independent" that an axiom
cannot be deduced (or refuted) from the remaining axioms
in (some kind of) set theory. It's true that this entails only that
there exists a model of that kind of set theory in which there
is a structure satisfying the remaining axioms but not satisfying
the "independent" one. But, correct me if I am wrong, I don't
think the term "independent" is ordinarily used for that.

|Incidentally, Marvin Greenberg's book on non-euclidean geometry makes
|a fundamental mistake on this point and I think that greatly impairs its
|usefulness as a textbook, in spite of its many admirable features.
|Specifically, he claims that because of the Godel completeness theorem,
|one is entitled to have a model of non-Euclidean geometry, including the
|continuity axiom, if it is consistent, and organizes a certain amount of
|the presentation around the belief that is true. Maybe this problem was
|fixed in a later edition, but I'm not aware of it.

That's too bad! If he meant to claim that the parallel axiom[*] is not a
consequence of the others in the sense I defined above, then it's
tautological to say that there are non-Euclidean geometries; to show
that there are non-Euclidean geometries is what one would need to
do to prove that the parallel axiom is not a consequence of the
other axioms. The Goedel completeness theorem doesn't enter
into it.

It seems like sort of an odd statement anyway. Even if it were only
a question of the first-order axioms, the easiest way to show that
there is no proof (in a standard deductive system for first-order logic)
of the parallel axiom[*] from the remaining axioms is to exhibit a
model, isn't it?

[*] Or whatever distinguishes Euclidean from non-Euclidean in
his system.

Keith Ramsay

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