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so I'm studying for my exam in the Model Theory course I take.

I just can't really wrap my head around ultraproducts. It's so abstract and complex. Is there a place where I can be shown with an explicit example of an ultraproduct, or anything close to any explicit example?

Thank you in advance for any assistance!

Reyo
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    It’s impossible to give an explicit example of a free ultrafilter, so it’s also impossible to give a completley explicit example of a non-trivial ultraproduct. – Brian M. Scott Jul 04 '15 at 20:24
  • So maybe an online resource explaining it a bit more thoroughly? I can't understand the building of the structure from the ultraproduct... Or is the only solution reading it over and over until I get it? – Reyo Jul 04 '15 at 20:25
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    The second PDF linked from this answer goes into the specific example of constructing the hyperreals as an ultrapower of $\Bbb R$ in quite a bit of detail; you might find it helpful. – Brian M. Scott Jul 04 '15 at 20:33
  • You’re welcome. Good luck! – Brian M. Scott Jul 04 '15 at 20:53

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I like to think if it in terms of finitely additive measures. A non-principal filter on $\{ 0,1,2,3,\ldots \}$ contains every cofinite set, so think of a measure that behaves like this $$ m(A) = \begin{cases} 0 & \text{if $A$ is finite}, \\ 1 & \text{if the complement of $A$ is finite}, \\ \text{???} & \text{otherwise}. \end{cases} $$ Zorn's lemma implies the domain of this measure can be extended to the whole set of all subsets of $\{0,1,2,3,\ldots\}$, so you have a function $m$ assigning $0$ or $1$ to every subset, and the measure of a set is $1$ iff the measure of its complement is $0$, and this measure is finitely additive, i.e. $m(A\cup B)=m(A)+m(B)$ if $A\cap B=\varnothing$. The ultrafilter is the set of all sets of measure $1$.

A first-order sentence is true in an ultraproduct if and only if it is true in almost every factor, and that means the measure of the set of factors in which is is not true is $0$. That statement is a theorem rather than a definition, called Łoś's theorem. ("Ł" is pronounced like the "W" in "Washington", and "ś" like "sh" in "Washington". It's a Polish name.)

So you have an indexed family of structures $B_i,\ i\in\{0,1,2,3,\ldots\}$ (or the index set could be some other set). A member of the product $\prod_{i=0}^\infty B_i$ is a sequence $b_0,b_1,b_2,\ldots$, or if the index set is something other than $\{0,1,2,3,\ldots\}$, then an indexed family $(b_i)_{i\in\text{index set}}$. Each structure $B_i$ has not only an underlying set but also some "structure", which can be some relations or functions on $B_i$. Let's set you have a binary relation $R_i$ among members of $B_i$. You then have a product relation $R$ according to which $x$ is related to $y$, i.e. $x\,R\,y$, if and only if for every index $i$ you have $x_i\,R_i\,y_i$.

In the ultraproduct rather than the product, you have $x\,R\,y$ if and only if for almost every index $i$ you have $x_i\,R_i\,y_i$ (with "almost" defined as above).

If $F$ is the ultrafilter --- the set of subsets of the index set whose complements have measure $0$ --- then the ultraproduct is denoted $\prod_{i=0}^\infty B_i/F$.

Michael Hardy
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