## 502 – Compactness

First, two exercises to work some with the notion of ultrapower: Check that $|\prod_n M_n/{\mathcal U}|=|{\mathbb R}|$ whenever ${\mathcal U}$ is a nonprincipal ultrafilter on the natural numbers, and

1. $M_n={\mathbb N}$ for all $n,$ or
2. $\lim_{n\to\infty}|M_n|=\infty.$

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 ${\mathcal F}$ is a nonprincipal filter on a set $I,$ then there is a nonprincipal ultrafilter on $I$ that extends ${\mathcal F}.$

(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 ${\mathcal L}=\{\hat X\mid X\subseteq I\}\cup\{c,\hat\in\}.$ Here, each $\hat X$ is a constant symbol, $c$ is another constant symbol, and $\hat\in$ is a symbol for a binary relation (which we will interpret below as membership).

In this language, consider the theory $\Sigma=\{c\hat\in\hat X\mid X\in {\mathcal F}\}\cup{\rm Th}(I\cup{\mathcal P}(I),\in,X\mid X\subseteq I).$ A model $M$ of this theory $\Sigma$ would look a lot like $I\cup{\mathcal P}(I),$ except that the natural interpretation of ${\mathcal F}$ in $M,$ namely, $\{\hat X^M\mid X\in{\mathcal F}\}$ is no longer nonprincipal in $M$, because $c^M$ is a common element of all these sets.

Note that there are indeed models $M$ of $\Sigma,$ thanks to the compactness theorem.

If $M\models\Sigma,$ let ${\mathcal U}=\{X\subseteq I\mid M\models c\hat\in \hat X\},$ and note that ${\mathcal U}$ is a nonprincipal ultrafilter over $I$ that contains ${\mathcal F}.$ $\Box$

### 2 Responses to 502 – Compactness

1. andrescaicedo says:

Ha! Yes, sorry. (It is fixed now.)

2. Andrew Misseldine says:

I think there is a typo in the assignment. You say that $\mathcal{U}$ is an ultrapower. Did you mean ultrafilter here?