Speak with the vulgar.

Think with me.

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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.

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Written by Jeff

January 15, 2010 at 6:22 pm

One Response

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  1. Thanks for writing this.

    Perhaps another definition of hypergunk that might be in a similar ballpark:

    a is a piece of hypergunk iff whenever the xx form a well-ordered chain of a’s parts there are yy which form a well-ordered chain of a’s parts that are isomorphic to the successor of the xx.

    You can state this rigorously in second order logic. (Or even plural logic, I think, if you look at the appendix to parts of classes.)

    It’s also inconsistent if you assume the plural quantifiers are ranging over all pluralities (you get a kind of Burali-Forti paradox) but not if they only range over set sized pluralities so it might tie in with the indefinite extensibility stuff.

    Andrew Bacon

    January 18, 2010 at 4:57 am


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