## 502 – Ultraproducts of finite sets

October 2, 2009

I want to sketch here the proof that if ${(M_n\mid n\in{\mathbb N})}$ is a sequence of finite nonempty sets, and ${\lim_n |M_n|=\infty,}$ then ${\prod_nM_n/{\mathcal U}}$ has size ${|{\mathbb R}|}$ for any nonprincipal ultrafilter ${{\mathcal U}}$ on ${{\mathbb N}.}$

The argument I present is due to Frayne, Morel, Scott, Reduced direct products, Fundamenta Mathematica, 51 (1962), 195–228.

The topic of the size of ultraproducts is very delicate and some open questions remain. For ultraproducts of finite structures, this is continued in Keisler, Ultraproducts of finite sets, The Journal of Symbolic Logic, 32 (1967), 47–57, and finally in Shelah, On the cardinality of ultraproduct of finite sets, The Journal of Symbolic Logic, 35 (1) (Mar., 1970), 83–84. Shelah shows that if an ultraproduct of finite sets is infinite, say of size ${\kappa,}$ then ${\kappa^{\aleph_0}=\kappa.}$ His argument is a very nice application of non-standard analysis. The case that interests us is easier.

Clearly,

$\displaystyle |\prod_nM_n/{\mathcal U}|\le|\prod_n M_n|\le|{\mathbb N}^{\mathbb N}|=|{\mathbb R}|,$

so it suffices to show that ${|{\mathbb R}|\le|\prod_nM_n/{\mathcal U}|.}$

## 502 – Notes on compactness

October 1, 2009

The goal of this note is to present a proof of the following fundamental result. A theory ${T}$ is said to be satisfiable iff there is a model of ${T.}$

Theorem 1 (Compactness) Let ${T}$ be a first order theory. Suppose that any finite subtheory ${T_0\subseteq T}$ is satisfiable. Then ${T}$ itself is satisfiable.

As indicated on the set of notes on the completeness theorem, compactness is an immediate consequence of completeness. Here I want to explain a purely semantic proof, that does not rely on the notion of proof.

The argument I present uses the notion of ultraproducts. Although their origin is in model theory, ultraproducts have become an essential tool in modern set theory, so it seems a good idea to present them here. We will require the axiom of choice, in the form of Zorn’s lemma.

The notion of ultraproduct is a bit difficult to absorb the first time one encounters it. I recommend working out through some examples in order to understand it well. Here I confine myself to the minimum necessary to make sense of the argument.