Thursday, June 21, 2007

Hypergunk!

A few definitions of Hypergunk were given in Nolan's paper. Two on page 6:



(S) Something is Hypergunk iff it is atomless, and for every set containing only its parts, there is a strictly larger set containing only its parts



(P) Something is Hypergunk iff it is atomless, and whenever there are some of its parts, there are some others of its parts such that there are more of the second than there are of the first



The second is a paraphrase of the first meant to eliminate set-theory talk, for fear that plural quantification can only go as far as set-theory. He mentions breifly that if plural quantification over more than set many objects is possible then (P) is inconsistent.



However, I think there's something I don't understand here. He glosses over that argument pretty quickly. Assuming one can quantify over more than set many things, he argues against (P).



1) We can quantify over more than set many things

2) if (1) then (3)

3) We can refer to ALL of x's parts (where x is a piece of hypergunk)

4) (P)

5) (3)&(4)

6) If (5) then (7)

7) There are some parts of x such that there are more of these than ALL of x's parts



Reductio, ~(P) voila. Hence, if we can quantify over more than set many things we should stick to (S). That way, we can speak of all of x's parts, we just can't speak of the set of all of x's parts. If we can't quantify over more than set many things, we can't speak of all of x's parts at all (it seems). That would have bad consequences, so let's assume we can.

I'm not going to argue that the concept of hypergunk is incoherent, rather that it has a couple veeery strange properties. Consider thesis (H).

(H) Every piece of hypergunk has a proper part which is itself hypergunk


Not going to prove this here, but intuitively this is very plausible. A quick reductio. Assume not for hypergunk chunk x. Each part of x has a largest set of parts. For any two sets, the union of those sets forms a set. Union up all the largest sets and you have a set of all of x's parts. That makes x set-sized, but x can't be set-sized (I'm pretty sure this goes through, but I have to brush up on my set-theory).

Now suppose you have a finite chunk of something (matter - finite mass, time - finite seconds, whatever you like). This cannot be the fusion of only hypergunk parts. Consider thesis (M).

(M) No amount of massless (I'll use mass for simplicity, take any unit you like) hypergunk can form an object with finite mass.

This I'll need to prove.

1) suppose for reductio: It takes cardinality C many pieces of massless hypergunk to form a piece of hypergunk X with finite mass.

2) Each part of X has at least C many parts.

3) Each part of X is massless.

4) if (2) then (5)

5) ~(3)

6) (3)&~(3)

Notice this argument won't work for regular gunk. Not matter how divisible regular gunk is, the buck stops somewhere. If a piece of regular gunk has cardinality aleph67 many parts, you can just say aleph67 pieces of massless gunk form a piece of gunk with finite mass. It's hard to see how massless things can constitute a thing with mass anyway, but we do this all the time. A line has length, and can be seen as a union of points. Yet, no point has a length. And it's also intelligble to say that continuum many points can form a line with a finite length. Moreso, it seems like we need to do something like this if we're to say something is infinitely divisible in any sense. If every point in a line had length, and there were continuum many points, the line would be infinitely long.

So, if there's hypergunk around, it's either devoid of any qualitative unit of measurement, or it has that measurement to infinity. If it does have a finite amount of some measurement, that is in virtue of some part that is not hypergunk. It seems like a bad thing for hypergunk, although I'm not sure exactly how damaging it is (or helpful for GR and AR). It's a kind of out-there sketch.

4 comments:

Chris Tillman said...

I think Nolan's second formulation is inconsistent since the expressive power of plural quantification outstrips set theory. Thus, Ludicrousgunk:

Something is Ludicrousgunk iff it is atomless, and for any xxs, if the xxs are part of it and it’s not the case that for all yys, if the yys are part of it then the yys are among the xxs and the xxs are among the yys, then there are some zzs such that the zzs are part of it and there are more of the zzs than there are of the xxs.

(The uses of ‘it’ throughout should be read as plurally referring expressions, which refer in the plural to the pieces (parts) of LG, so that the formulation does not presuppose that the parts of LG have a fusion.) To mitigate worries of “metaphysicalness” that Nolan mentions, the formulation may be restated using only plural demonstratives and anaphora, though this would impede readability.

So LG has the virtues Nolan claimed for HG but no worries about consistency.

Dan said...

let me paraphrase ludicrousgunk to see if I'm understanding this right:
Something is LG iff it's atomless and for any number of parts that doesn't consitute the whole, then there is some parts that are more numerous than the first.
Why should we be careful about the parts forming a fusion? Are there problems associated with that as well?

Chris Tillman said...

Technically, LG does not imply that any of the relevant parts form a whole. But no, I do not believe there are similar problems with fusions.

Helenski said...

Hi, I am somewhat confusd as to your definition of gunk, specifically "regular gunk" to hypergunk.

Your "regular gunk" doesn't seem to sound like gunk at all. For me, gunk always has parts which themselves have parts, so there is no "ultimate part". So I do not see how we can say:

"Not matter how divisible regular gunk is, the buck stops somewhere. If a piece of regular gunk has cardinality aleph67 many parts, you can just say aleph67 pieces of massless gunk form a piece of gunk with finite mass."

Gunk shouldn't have aleph67 indivisible parts. Gunk shouldn't have any ultimate parts at all, no matter whether or not there are a certain infinite amount or not. Otherwise we would have a point-continuum, which runs contrary to the atomless gunk theory.