[net.math] Loweheim Skolem

cbostrum (02/01/83)

While we are on the subject of infinity, what opinions are there
on the theorem that says every set of first order axioms in a 
countable language has a countable model?? This makes the people
with big infinities look silly since even if they write down their
set theory axioms and prove that there is an "uncountable set" (so
there are uncountably many objects in their universe), that set
of axioms from which they proved their wondrous result has a countable
model. So where are the uncountable sets, really. (Of course, all they
have proved is that there are infinite sets such that there exists no
1-1 onto corrspondence. THIS DOES NOT ESTABLISH THAT THERE ARE 
UNCOUNTABLE SETS BY ANY MEANS! It just says that what it says.
(I am what I am?)). Comments?

leichter (02/02/83)

You're ignoring the assumptions of Lowenheim-Skolem:  "Every theory WITH
FIRST ORDER AXIOMS ... ".  There does not exist a theory with first-order
axioms that defines a complete ordered field, for example.  (Proof:  One
shows that all complete ordered fields are isomorphic to the reals, and
that the reals are uncountable.  QED)  It IS possible to build countable
models of the reals that have the same first-order properties; that's one
kind of "non-standard" analysis.

For those who are wondering what this is about:  A first order statement is
one in which one can have objects and sets of those objects, but not sets
of sets of those objects.  (This is informal but I don't remember the formal
definitions any more.)  What you consider the "objects" to be is arbitrary,
but if you want a set of first-order axioms, you have to make a consistent
choice once and for all, such that all the variables in ALL your axioms
represent either objects, or sets of those objects - but nothing nested deeper.
The problem with the "complete ordered field" is stating the "every set with
an upper bound has a least upper bound" axiom.
							-- Jerry
						decvax!yale-comix!leichter