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