NSA: Robinson's "overspill" lemma

This lemma presents an attractive packaging to the weak saturation condition which appears in NSA: Introduction and Construction, in the setting of an enlargement of ZF set theory.

So let M be a model of ZF set theory, and let *M be its enlargement (the extension described in NSA: Introduction and Construction). L, the language of M, has a constant name, say s, for every set of M. *M also assigns an interpretation to L, so in particular it assigns some object to s.

Lemma.

Let s be a name of a set in M. Then s *contains non-standard *elements in *M iff s is infinite in M.

Proof.

For any element a of s in M, we may form the sentence "a is an element of s" in L, and by the transfer principle a is an element of s in *M, too; applying the same principle to a's which are not elements of s in M, we see that the only standard *elements of s in *M are those which are elements of s in M.

What about non-standard elements?

• Suppose s is finite in M. Enumerate (the names of) its members as a₁,...,a_k. Now consider this sentence S of L: "for all x in s, x=a₁ or ... or x=a_k" (note that this sentence can be very long, when k is very large; but we can always write it down once and for all, enumerating each element a_j of s; this is where we use the finiteness of s).

  The sentence S is true in M, therefore it's true in *M. Thus we conclude that these k elements are the only elements of s in *M, too -- s *contains no non-standard elements.

• Now suppose s in infinite. For every a in s, create the formula f^a(x) "x is an element of s, and x != a". Let F be the set of all such f^a. Since s is infinite, F is finitely satisfiable (for instance, to satisfy {f^a₁,f^a₂,f^a₃}, pick x to be an element of s (in M) other than a₁,a₂,a₃; this is possible since s is infinite, and therefore contains more than 3 elements).

  By the weak saturation condition, F is satisfiable in *M. That is, there exists some object *p of *M, such that for every element a in s (in M!) *p != a, and yet *p is a *element of s (in *M). In other words, *p is non-standard (hence has no name in L; if it had a name in L, it would have been in s in M too, and therefore been included in F), and a *element of s in *M. Thus s *contains non-standard *elements.