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 is atomless, then has a countable chain of parts such that nothing is a part of each of them.
Proof. Since is atomless, there is a (countable) sequence . For each positive integer , let be . Then let be the sum of . Note that the ’s are a countable chain. Note also that each is part of .
Now suppose that is a part of each . In that case, is part of each . But since is disjoint from , this means that is disjoint from each , and so by the definition of a mereological sum, is disjoint from each . This is a contradiction.
Subscribe to comments with RSS.