# Speak with the vulgar.

Think with me.

## Oops

with one comment

I’ve been cleaning up my “Indefinite Divisibility” paper from last year. One of my arguments in it concerned supergunk: X is supergunk iff for every chain of parts of X, there is some y which is a proper part of each member of the chain. I claimed that supergunk was possible, and argued on that basis against absolutely unrestricted quantification. I even thought I had a kind of consistency proof for supergunk: in particular, a (proper class) model that satisfied the supergunk condition as long as the plural quantifier was restricted to set-sized collections. Call something like this set-supergunk.

Well, I was wrong. I’ve been suspicious for a while, and I finally proved it today: set-supergunk is impossible. So I thought I’d share my failure. In fact, an even stronger claim holds:

Theorem. If $x_0$ is atomless, then $x_0$ has a countable chain of parts such that nothing is a part of each of them.

Proof. Since $x_0$ is atomless, there is a (countable) sequence $x_0 > x_1 > x_2 > \dots$. For each positive integer $k$, let $y_ k$ be $x_{k-1} - x_ k$. Then let $z_ k$ be the sum of $y_ k, y_{k+1}, y_{k+2}, \dots$. Note that the $z_ k$’s are a countable chain. Note also that each $z_ k$ is part of $x_{k-1}$.

Now suppose that $z_\omega$ is a part of each $z_ k$. In that case, $z_\omega$ is part of each $x_ k$. But since $x_ k$ is disjoint from $y_ k$, this means that $z_\omega$ is disjoint from each $y_ k$, and so by the definition of a mereological sum, $z_\omega$ is disjoint from each $z_ k$. This is a contradiction.