# Are - in the situation sketched below - all ultrafilters of $X$ in $\mathcal A$?

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?

• 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.) – Eric Wofsey Aug 30 '20 at 14:24

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 $$\mathcal{A}$$.