Limits of Logic #3: Kaplan's Paradox

15 Jul, 2024

1.5.6 Exercise: Kaplan's Paradox

Let $P$ be a set of propositions, and let $W$ be a set of possible worlds.

We’ll consider two relations between propositions and possible worlds.

First, a proposition can be true at a possible world.
Second, a proposition $p$ can be the only proposition that anyone believes at $w$ ; in this case we say that $w$ singles out $p$ .

We’ll make two assumptions about these relations.

First, for any set $X$ of possible worlds, there is some proposition $p X$ which is true at each possible world in $X$ , and which is not true at any possible world which is not in $X$ .
Second, no world singles out more than one proposition.

Given these assumptions, show that there is at least one proposition which is not singled out by any possible world.

In other words, some proposition cannot possibly be uniquely believed.

Hint. Consider the function that takes each world $w$ that singles out some proposition $p$ to the set of worlds at which $p$ is true.

Proof:

Consider the function $f : W \to P W$ such that:

If $w \in W$ singles out some proposition $p$ , then we have that $p$ is true at all and only the worlds in $f (w)$ .

If $w \in W$ does not single out any proposition $p$ , then $f (w) = \emptyset$ .

Note that by Cantor's theorem, the function $f$ cannot be onto. Therefore, there is some set of possible worlds $X \subseteq P W$ that is not in the range of $X$ .

This means that no world $w \in W$ singles out the proposition $p X$ i.e. there is some proposition that is not singled out by any world. In other words, some proposition cannot possibly be uniquely believed. $∎$