This is a follow up to my earlier question Is this a complete and/or atomic subalgebra of $2^{2^S}$?

For some infinite set $S$, let



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

$C:= \{p \in B: \forall X\subseteq F\text{ s.t. }\bigcap X\subseteq p, \exists Y\subseteq X\text{ s.t. }\bigcap Y\subseteq p\text{ and }\forall Z\subset Y\bigcap Z\not\subseteq p))\}$.

(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 $C$ as the set of crisp propositions (those which, when true, have a minimal basis among the fundamental propositions that implies their truth.))

My question is: Is $C$ a Boolean subalgebra of $B$ (under the natural set-theoretic operations)?


I will show that $C$ is not a Boolean subalgebra. In particular, I will show that it is not closed under complement.

For convenience, let $S = \{1, 2, 3, \ldots \}$, so that possible worlds are subsets of the natural numbers. Also for convenience, denote by $p_i$ the fundamental proposition that $i$ obtains, and by $\lnot p_i$ the fundamental proposition that event $i$ does not obtain (i.e. $p_i = \{w \in W: i \in w\}$ and $\lnot p_i = \{w \in W : i \notin w\}$), so that $F = \{p_1, \lnot p_1, p_2, \lnot p_2, p_3, \lnot p_3, \ldots\}$.

Let $w_k \in W$ be the subset of the first $k$ natural numbers $\{1, 2, 3, \ldots, k\}$. Let $c = \{w_1, w_2, w_3, \ldots\}$; this is the proposition that some finite initial segment of events obtain, and the rest do not. I claim that $c$ is crisp ($c \in C$), but its complement is not.

  • First we show $c$ is crisp. Consider any subset $X$ of $F$ which implies $c$. What does this look like? First there is the case that $X$ is inconsistent (contains both $p_i$ and $\lnot p_i$ for some $i$), in which case a minimal basis is just $p_i$ and $\lnot p_i$ for that $i$. Otherwise, $X$ has to be almost maximal, by which I mean that it must contain $p_i$ or $\lnot p_i$ for all but at most one $i$. If not, then say it doesn't contain $p_i$ or $\lnot p_i$, and also doesn't contain $p_j$ or $\lnot p_j$, for some $i < j$. So it doesn't say anything about event $i$ or about event $j$. Then it is consistent with $X$ that $i$ does not occur, and $j$ occurs. But that can't happen in any initial segment of the natural numbers, so $X$ does not imply $c$, contradiction. So, $X$ is almost maximal. Because all possible $X$ are almost maximal, a minimal basis for a given $X$ is always either $X$ itself, or $X$ minus one element. (Concretely, the minimal bases are of the following form: $\{p_1, p_2, p_3, \ldots, p_{k-1}, \lnot p_{k+1}, \lnot p_{k+2}, \lnot p_{k+3}, \ldots\}$. The basis states that events $1$ through $k - 1$ obtain, and events $k+1$ and beyond do not obtain.)

  • Next we show that $c^C$ is not crisp. To do so, we need to exhibit a subset $X$ of $F$ which implies $c^C$, but has no minimal basis which implies $c^C$. Take $$ X = \{p_1, p_2, p_3, \ldots \}. $$

    This implies $c^C$ because clearly, if all events obtain, then the set of events which obtain cannot be a finite initial segment of the natural numbers. On the other hand, there can be no minimal subset which implies $c^C$. To see this, note that any subset of $X$ is either finite or infinite. If finite, then it contains only finitely many propositions, say for example $\{p_3, p_6, p_{13}\}$, in which case it doesn't imply $c^C$ because it is consistent with $c$ (for example consistent with the world where $1$ through $15$ obtain). If infinite on the other hand, then the subset asserts that infinitely many events obtain (so it implies $c^C$), but we can always obtain an even smaller subset by removing some $p_i$, while still having an infinite set. Thus we obtain smaller and smaller infinite sets which nevertheless all assert that infinitely many events obtain, and thus all imply $c^C$.

  • $\begingroup$ Great example! It shows that I hadn't been careful to distinguish the notion of crispness from a related notion. Say that $p$ is weakly crisp iff, for any $w$, if $p$ is true at $w$, then there is some minimal set $X$ of fundamental propositions that jointly entail $p$ and all of which are true at $w$. Your example of $\neg c$ is not crisp but it is weakly crisp. Do you know whether the weakly crisp propositions form a BA? $\endgroup$ – Jeremy Sep 3 '20 at 0:59
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    $\begingroup$ @Jeremy Perhaps I am misunderstanding, but it seems to me $\lnot c$ isn't weakly crisp either. The example $X$ exhibited corresponds to the single world $w = \{1, 2, 3, \ldots \}$ where all $p_i$ are true, and as the answer argues, there is no minimal subset of fundamental propositions $p_i$ that implies $X$. $\endgroup$ – 6005 Sep 3 '20 at 1:14
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    $\begingroup$ You're right, I was confused. (I was thinking about $v = \{2,3,4,\dots\}$, and seeing that, although $\{p_i:i>1\}$ entails $\neg c$ without having a minimal subset that does, $\{\neg p_1,p_2\}$ also entails $\neg c$ and is minimal in doing so. But that doesn't work at $w$.) $\endgroup$ – Jeremy Sep 3 '20 at 1:18

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