This question is inspired by modal logic, but reduces to a basic set theory problem. David Lewis in Counterfactuals claims that the answer to my question is "easily verified" but I can't figure it out.
Suppose you have a set of points $W$ and a function $f\colon \wp(W) \rightarrow \wp(W)$ such that:
- there's a $x \in W$ such that if $x \in A$, then $f(A) = \{x\}$
- $f(A) \subseteq A$
- if $f(A) = \varnothing$, then $A = \varnothing$
- if $A \subseteq B$ and $f(A) \neq \varnothing$, then $f(B) \neq \varnothing$
- if $A \subseteq B$ and $A \cap f(B) \neq \varnothing$, then $f(A) = A \cap f(B)$.
Intuitively, $f$ is a selection function, that picks out the "relevant" points of evaluation. Notice that $f(f(A)) = f(A)$ by 1 and 4 (and 2 for the empty case).
Now, let us say a sphere is a subset $S \subseteq W$ such that:
- for all $x \in S$, there is an $A \subseteq W$ such that $x \in f(A)$
- for all $A \subseteq W$, if $A \cap S \neq \varnothing$, then $f(A) \subseteq S$.
It's easy to verify that the set of spheres is closed under arbitrary unions and nonempty intersections. It's also easy to see that $\{x\}$ is a sphere.
Question: Must the set of spheres be nested? That is, if $S,T$ are spheres, must it be that either $S \subseteq T$ or $T \subseteq S$?
Lewis says the answer should be yes and that it's "easily verified" (p. 59). I tried to prove this by reductio, but I can't get a contradiction out. Any thoughts?
Suppose that $S$ and $T$ are both spheres, and that neither is a subset of the other. Then $S' = S\setminus T$ and $T'=T\setminus S$ are both nonempty.
Setting $A=S' \cup T'$ in the second condition for being a sphere we learn that $f(S'\cup T')\subseteq S$ and $f(S'\cup T') \subseteq T$. Together with axiom (2) this means that $$ f(S'\cup T') \subseteq S \cap T \cap (S' \cup T') $$ But the right-hand side of this is empty, which contradicts axiom (3).