# Another question about subalgebras of $2^{2^S}$

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

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

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

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

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

(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 $$c^C$$.

• 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? Sep 3, 2020 at 0:59
• @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 $X$.
– 6005
Sep 3, 2020 at 1:14
• 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 $w$.) Sep 3, 2020 at 1:18