First, two exercises to work some with the notion of ultrapower: Check that whenever
is a nonprincipal ultrafilter on the natural numbers, and
for all
or
Our argument for compactness required the existence of nonprincipal ultrafilters. One might wonder whether this is a necessity or just an artifact of the proof. It is actually necessary. To see this, I will in fact show the following result as a corollary of compactness:
Theorem. If is a nonprincipal filter on a set
then there is a nonprincipal ultrafilter on
that extends
(Of course, this is a consequence of Zorn’s lemma. The point is that all we need is the compactness theorem.)
Proof. Consider the language Here, each
is a constant symbol,
is another constant symbol, and
is a symbol for a binary relation (which we will interpret below as membership).
In this language, consider the theory A model
of this theory
would look a lot like
except that the natural interpretation of
in
namely,
is no longer nonprincipal in
, because
is a common element of all these sets.
Note that there are indeed models of
thanks to the compactness theorem.
If let
and note that
is a nonprincipal ultrafilter over
that contains
Ha! Yes, sorry. (It is fixed now.)
I think there is a typo in the assignment. You say that
is an ultrapower. Did you mean ultrafilter here?