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For some infinite set $S$, let

$W:=\mathcal{P}(S)$

$B:=\mathcal{P}(W)$

$F:= \{p\in B: \exists s\in S\text{ s.t. }p=\{w\in W:s\in w\}\text{ or }p=\{w\in W:s\not\in w\}\}$

$A:= \{p \in B: \forall X\subseteq F(\bigcap X\neq\emptyset\text{ and }\bigcap X\subseteq p \Rightarrow \exists Y\subset X\text{ s.t. }\bigcap Y\subseteq p)\}$.

(We might think of $S$ as a set of independent possible events, $W$ as the set of possibilities (one for each set of events, in which all and only those events obtain), $B$ as the set of propositions (with a proposition identified with the set of possibilities in which it is true), $F$ as the set of fundamental propositions (those saying that some given event either does or does not obtain), and $A$ as the set of amorphous propositions (those which, when true, lack any minimal basis among the fundamental propositions which implies their truth.))

My question is:

Is $A\cup\{W,\emptyset\}$ a Boolean subalgebra of $B$ (under the natural set-theoretic operations)? If so, is it complete and/or atomic?

Jeremy
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1 Answers1

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No. For instance, take a countably infinite proper subset $T\subset S$, and topologize $\mathcal{P}(T)$ by identifying it with $\{0,1\}^T$ with the product topology. Say two elements of $\mathcal{P}(T)$ are equivalent if their symmetric difference is finite. By transfinite recursion, you can construct a subset $p\subset\mathcal{P}(T)$ which is a union of equivalence classes such that $p$ intersects every uncountable closed subset of $\mathcal{P}(T)$ but does not contain any uncountable closed subset of $\mathcal{P}(T)$. (Sketch: there are $\mathfrak{c}$ uncountable closet subsets and each has cardinality $\mathfrak{c}$, so you can go through them one by one and put an equivalence class which intersects each one in $p$ and another such equivalence class in the complement of $p$.)

I claim then that $p\in A$. Indeed, suppose $X\subseteq F$, $\bigcap X$ is nonempty, and $\bigcap X\subseteq p$. If $\bigcap X$ is finite, then $X$ must contain either $\{w\in W:s\in w\}$ or $\{w\in W:s\not\in w\}$ for all but finitely many $s\in S$. In particular, there is some such element of $X$ for which $s\in T$; let $Y$ be $X$ with that element removed. Then $\bigcap Y$ will still be contained in $p$ since $p$ is a union of equivalence classes. If $\bigcap X$ is infinite, then it is an uncountable closed subset of $\mathcal{P}(T)$, so it cannot be contained in $p$.

So $p\in A$, and by the same reasoning, $\mathcal{P}(T)\setminus p\in A$. However, the union $p\cup(\mathcal{P}(T)\setminus p)=\mathcal{P}(T)$ is not in $A\cup\{W,\emptyset\}$.

Eric Wofsey
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  • Thank you! I don't know what the product topology on ${0,1}^T$ is. Is there another characterization of the open sets of $\mathcal{P}(T)$ so topologized? – Jeremy Aug 25 '20 at 20:51
  • The topology here isn't important and was just a lazy way for me to avoid some circumlocutions. Read "uncountable set of the form $\bigcap X$" everywhere I say "uncountable closed set" and the argument works the same. – Eric Wofsey Aug 25 '20 at 21:11
  • I see. So intuitively $p$ is a big disjunction of continuum many specifications modulo-finite-difference of which members of $T$ do and don't obtain, chosen in such a way that $p$ is logically independent of every proposition that specifies exactly (i.e., not modulo finite difference) the obtaining or not of the members of some $U\subset T$ while remaining neutral on infinitely members of $T$ (i.e., such that $U\backslash T$ is infinite), and also implying that only members of $T$ obtain. – Jeremy Aug 25 '20 at 21:36
  • This was very helpful. I've posted a follow-up question here https://math.stackexchange.com/questions/3803400/another-question-about-subalgebras-of-22s – Jeremy Aug 25 '20 at 23:50