# Ordering a subset of the clopens of a Stone space

Let $$P$$ be a countably infinite set of propositional variables and $$\mathcal{L}_P$$ be the propositional language generated from $$P$$ and the usual connectives $$\wedge$$, $$\neg$$, $$\vee$$. The set $$\mathcal{W}_P$$ denotes the set of all propositional interpretations on $$P$$. Given a set of formulae $$\Phi \subseteq \mathcal{L}_P$$, $$[\Phi]$$ denotes the set of interpretations that are models of all formulae from $$\Phi$$.

We consider the Stone topology $$S$$ on $$\mathcal{W}_P$$ defined as the topology with the basis $$\{[\{\varphi\}] \mid \varphi \in \mathcal{L}_P\}.$$

Remark 1: $$x \in S$$ is closed iff $$x = [\Phi]$$ for some $$\Phi \subseteq \mathcal{L}_P$$

Remark 2: $$x \in S$$ is closed and open iff $$x = [\Phi]$$ for some finite $$\Phi \subseteq \mathcal{L}_P$$.

Let $$Clop(S)$$ be the set of clopen sets of $$S$$, and $$(X, \subset)$$ be a subset of $$Clop(S)$$ strictly ordered by inclusion.

I am interested in the following property:

Property $$P$$. For each $$z \in Clop(S)$$, if there exists $$x \in X$$ such that $$x \cap z \neq \emptyset$$, then there also exists a smallest $$y \in X$$ (smallest w.r.t. $$\subset$$) such that $$y \cap z \neq \emptyset$$.

Is there a condition on $$(X, \subset)$$ which captures property $$P$$ above? That is, a condition on $$(X, \subset)$$ whose statement does not mention any element outside of $$X$$?

For instance, $$(X, \subset)$$ being a well-order is a too strong condition.

I am not an expert in topology, and in absence of a definite answer I would be glad to hear suggestions on where to look at.

• It seems you're introducing the Stone space only to redefine $B=Clop(S(B))$!
– YCor
Nov 7, 2019 at 10:42
• Thank you for your comments, I have now realized the Stone space I am interested in is not "the Stone space generated from a boolean algebra". I revised the post accordingly. Nov 20, 2019 at 9:28

If $$B$$ is a complete boolean algebra (i.e. if $$S(B)$$ is extremally disconnected) then Property $$P$$ is equivalent to "$$(X,\subseteq)$$ is a well-order". On the other hand if $$B$$ is countable (and I suppose in many other cases) we can easily find an $$(X,\subseteq) \cong \omega + \omega^*$$ with property $$P$$ and we can also find $$(X,\subseteq) \cong \omega + \omega^*$$ without property $$P$$. So in many cases it will be impossible to characterize $$P$$ without mentioning elements outside of $$X$$.
• Could you please explain what "$(X, \subseteq) \cong \omega + \omega^*$" means? Nov 20, 2019 at 11:03
• It means that $(X,\subseteq)$ is isomorphic (as a linear order) to the order $\omega+\omega^*$, where $\omega$ is the first infinite ordinal and $\omega^*$ is $\omega$ with the order reversed. Nov 20, 2019 at 12:56