# Ultraproducts and the compactification of Lindenbaum algebras

I'm trying to understand the relationship between ultraproducts and the compactification of Lindenbaum algebras and I wanted to check if I'm getting thing right (so far).

We start with a theory $T$ and form the Lindenbaum algebra of sentences equivalent modulo $T$. This algebra will contain elements of the form $[\phi]_T$ for some sentence $\phi$. These elements will be like finite theories; so when we take filters of these elements, these filters may converge to an ideal element not in the algebra, namely a complete theory. So we add ultrafilters to obtain the necessary ideal points, compactifying (is that a verb?) the space, which is now $S(T)$, the space of complete theories in the language of $T$ and corresponds to the Stone space of the original algebra (this also explains why complete types can be considered as $\{0, 1\}$-valued measures: they are ultrafilters, which are such measures).

On the other hand, if we consider the models associated to the theories, we see that if $\mathcal{A}_i$ is the model of a theory $i$ in a filter $F$ from the original Lindenbaum algebra, then we see that to the ultrafilter $U$ extending $F$ there will correspond an ultraproduct $\mathcal{A}$. This ultraproduct will be in a certain sense (which?) the limit of the structures $\mathcal{A}_i$, and its theory will correspond to the ultrafilter of the Lindenbaum algebra, i.e. a point in the Stone space $S(T)$.

Is the above account correct? Am I missing something? Are there other things that I should pay attention to in filling out the details of the above?

I think there's some confusion here between ultrafilters on sets and Boolean algebras. An ultrafilter on a Boolean algebra $B$ is a subset $U\subseteq B$. An ultrafilter on a set $X$ is a set of subsets $U\subseteq\mathcal{P}(X)$; in particular, it is an ultrafilter on the powerset Boolean algebra $\mathcal{P}(X)$.

In what sense is $S(T)$ a compactification of the Lindenbaum algebra? Usually, to compactify means to embed a topological space into a compact space. But the Lindenbaum algebra is a Boolean algebra, not a space, and it doesn't naturally embed in $S(T)$. Instead of the compactification of the Lindenbaum algebra $L(T)$, $S(T)$ is the space corresponding to $L(T)$ under Stone duality.

On the other hand, you can view certain Stone spaces as compactifications of sets. Given a set $X$, the Stone-Čech compactification $\beta X$ is the space of ultrafilters on the set $X$, i.e. ultrafilters on the powerset Boolean algebra of X. That is, $\beta X = S(\mathcal{P}(X))$. And $\beta X$ is a compactification of $X$ when we view $X$ as a discrete space, by the embedding sending $x\in X$ to the principal ultrafilter generated by $x$.

A possible confusion between ultrafilters on sets and Boolean algebras is also reflected in your last paragraph, where you appear to be trying to take an ultraproduct of structures $(\mathcal{A}_i)_{i\in I}$ by an ultrafilter on a Boolean algebra. Usually, the ultraproduct is defined using an ultrafilter on the index set $I$.

• "I thought that the Stone space was the precisely the Stone-Čech compactification of a Lindenbaum algebra with the discrete topology. Is this mistaken?" Yes, that's the source of your confusion. An ultrafilter on a Boolean algebra $B$ is a subset of $B$. If you instead let $\overline{B}$ be the underlying set of the boolean algebra, then an ultrafilter on $\overline{B}$ (i.e. a point of the Stone-Čech compactification of $\overline{B}$) is a set of subsets of $B$, i.e. a subset of $\mathcal{P}(\overline{B})$. These are totally different things. – Alex Kruckman Sep 19 '18 at 18:06
• Similarly, if $F\subseteq B$ is a filter, then an ultrafilter on $F$ (viewed as a set) is a totally different kind of thing than than ultrafilter on $B$ (viewed as a Boolean algebra) extending $F$. The first is $U\subseteq \mathcal{P}(F)$, while the second is $F\subseteq U\subseteq B$. – Alex Kruckman Sep 19 '18 at 18:08
• Being pedantic: If $B$ is the Lindenbaum algebra (whose elements are sentences up to equivalence mod $T$), we usually call ultrafilters over $B$ complete theories (completions of $T$), not complete types. A complete type (in variable context $x$) is an ultrafilter on the Boolean algebra of formulas with free variables from $x$ up to equivalence mod $T$. So completions of $T$ are complete types in the empty variable context. – Alex Kruckman Sep 19 '18 at 19:56
• "Is there any relation then between $S(T)$ and $\beta\overline{B}$?" Here's one relationship. Stone duality say that the powerset algebra of $\overline{B}$ is isomorphic to the algebra of clopen sets of $\beta\overline{B}$. That is, a subset $Y\subseteq B$ corresponds to a clopen set in $\beta\overline{B}$, namely $\{U\in \beta\overline{B}\mid Y\in U\}$. In particular, since ultrafilters on $B$ are subsets of $B$, they correspond to certain clopen sets in $\beta\overline{B}$. But I don't see any natural way to make them correspond to points of $\beta\overline{B}$. – Alex Kruckman Sep 19 '18 at 20:05
• The point is that when you pass to $\overline{B}$, you forget about the Boolean algebra structure on $B$. Now it's just a set. So from the point of view of $\beta\overline{B}$, there's no difference between an ultrafilter on $B$ and any other subset of $B$. – Alex Kruckman Sep 19 '18 at 20:07