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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.

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  • $\begingroup$ It seems you're introducing the Stone space only to redefine $B=Clop(S(B))$! $\endgroup$
    – YCor
    Nov 7, 2019 at 10:42
  • $\begingroup$ 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. $\endgroup$
    – user109711
    Nov 20, 2019 at 9:28

1 Answer 1

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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 $X$.

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  • $\begingroup$ Could you please explain what "$(X, \subseteq) \cong \omega + \omega^*$" means? $\endgroup$
    – user109711
    Nov 20, 2019 at 11:03
  • $\begingroup$ 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. $\endgroup$ Nov 20, 2019 at 12:56

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