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Let $X$ be a set and let $\mathcal A$ be a non-empty collection of subsets of $X$.

For every $x\in X$ define $\mathcal A_x:=\{A\in\mathcal A\mid x\in A\}$.

Now let it be that collection $\{\mathcal A_x\mid x\in X\}\subseteq\mathcal P(\mathcal A)$ is an algebra on $\mathcal A$ in the sense that the collection is closed under intersection and complementation.

Also let it be that the function $\phi:X\to\mathcal P(\mathcal A)$ prescribed by $x\mapsto \mathcal A_x$ is injective.

Then $(X,\leq)$ where by definition $x\leq y\iff \mathcal A_x\subseteq\mathcal A_y$ can be recognized as a Boolean algebra and function $\phi$ mentioned above as an isomorphism. Also every $A\in\mathcal A$ appears to be an ultrafilter of $(X,\leq)$.

Now my question:

Is $\mathcal A$ necessarily the collection of all ultrafilters of $(X,\leq)$?

Remark1: it is common to start with a Boolean algebra in order to show a correspondence between this algebra and its set of ultrafilters. In the setup above things are turned around.

Remark2: I added tags "logic" and "predicate-logic" because actually this question arose when I tried to construct a Boolean algebra on (equivalence classes of) $\mathcal L$-formulas by the use of functions that split up all $\mathcal L$-formulas in true formulas and false formulas. These functions correspond with pairs $(\mathfrak A,\sigma)$ where $\mathfrak A$ is an $\mathcal L$-structure and $\sigma$ is an assignment. They provide ultrafilters but I want to know: do they provide all ultrafilters?

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  • $\begingroup$ The statement that all ultrafilters on $\mathcal{L}$-formulas come from a pair $(\mathfrak A,\sigma)$ is just a rephrasing of the completeness theorem. (It's just saying every consistent theory has a model, where you enlarge the language by adding constants for each variable you use in your formulas.) $\endgroup$ – Eric Wofsey Aug 30 '20 at 14:24
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No. For instance, let $X$ be any Boolean algebra, let $S$ be the Stone space of $X$, and let $\mathcal{A}\subseteq S$. Your map $\phi:X\to\mathcal{P}(\mathcal{A})$ is then a Boolean homomorphism, and identifying $X$ with the algebra of clopen subsets of $S$, it is just the map that takes a clopen subset of $S$ and intersects it with $\mathcal{A}$. This makes it clear that $\phi$ is injective iff $\mathcal{A}$ is dense in $S$, so that it intersects every nonempty clopen set. So, $\mathcal{A}$ need not be all the ultrafilters on $X$, but only a dense subset of them.

(Note that every example is in fact isomorphic to one of the form above, since instead of starting with $\mathcal{A}$, you can start with the induced Boolean algebra structure on $X$ and then take $\mathcal{A}$ as a set of ultrafilters on $X$.)

For a very explicit example, you could take $X$ to be the power set of an infinite set $Y$, and $\mathcal{A}$ to be just the principal ultrafilters on $Y$. Then $\phi:\mathcal{P}(Y)\to\mathcal{P}(\mathcal{A})$ is just the isomorphism induced by the obvious bijection between $Y$ and $\mathcal{A}$.

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