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

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2 Responses to 502 – Compactness

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

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

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