Not Countably Many

26 04 2005

This post has nothing to do with my qual topics, except that it uses a little bit of set theory. However, I have a bit of a soft spot for mereology.

I was browsing through Brian Weatherson’s archive last night and stumbled across a post wondering how many things* there could be in an infinite universe if all mereological fusions exist. Daniel Nolan answered that if there is no atomless gunk (ie, if every object has a part with no proper part), then there must be 2^k objects for some cardinal k. This only obviously rules out inaccessible cardinalities. Gabriel Uzquiano then mentioned a result that said there are complete atomless boolean algebras of every inaccessible cardinality, and since a world composed entirely of atomless gunk is basically just a complete atomless boolean algebra, this seems to suggest that every infinite cardinality is at least possibly the cardinality of some universe of unrestricted mereological fusion.

However, the countable cardinality (I’ll use “A” instead of aleph_0 for typographical considerations) isn’t obviously inaccessible in the same way as uncountable inaccessibles, and I conjectured that it might not have a complete atomless boolean algebra. Today, after consulting Thomas Jech’s Set Theory (3rd Millenium Edition), I found an exercise on pg. 88 that proves my conjecture. In fact, any universe of unrestricted mereological fusion has to be at least the size of the continuum. If there is no atomless gunk, then the size of the universe is the size of the powerset of the set of atoms. If there is some object composed entirely of atomless gunk, then it’s clear that this object has infinitely many disjoint parts (because if n was the greatest natural number such that it had n disjoint parts, then each of those parts would have to be atoms). But then every collection of these parts forms an object, so the universe must have at least as many objects as this powerset, which is at least the size of the continuum, QED.

In fact though, the result mentioned in Jech is substantially stronger. It says that if K is the cardinality of some complete boolean algebra, then K=K^A, where again A is the countable cardinality. Exactly which class of cardinalities satisfy this property is unclear, because of things like the Singular Cardinals Hypothesis, and Easton’s Theorem (which states that it is consistent that the cardinalities of the powersets of each regular cardinal be anything, as long as they are non-decreasing, and the powerset of each K has cofinality strictly greater than K). It is clear that 2^K satisfies it (since (2^K)^A=2^(KA)=2^K, by simple cardinal arithmetic). And the result Gabriel Uzquiano cites suggests that every (uncountable, strong) inaccessible satisfies it as well.

However, it is clear that nothing with cofinality A is a possible size of the universe, which rules out A itself (so the universe must be uncountable, answering Brian Weatherson’s first question negatively), and aleph_A, and aleph_(aleph_A), and aleph_(A+A), and epsilon_0 (the first fixed point of the aleph function, ie aleph_(aleph_(…))).

In addition, assuming there aren’t a proper class of objects, the universe can be separated into a part consisting of all the atoms and a part consisting of all the atomless gunk. Since every object can be partitioned the same way, the cardinality of the universe is going to be the product of the cardinalities of these two parts, which is just the larger of the two cardinalities, by a basic result of cardinal arithmetic. The atomic part has cardinality 2^K, which is fairly restrictive, since this means that it can’t have cofinality A and can’t be inaccessible. In addition, the only 2^K that can have cofinality aleph_1 is 2^A, so either the atomic part is required to have cofinality at least aleph_2, or 2^A is at least aleph_(aleph_1), in which case the universe is required to have at least that cardinality, which rules out uncountably many cardinalities. Similar results obtain for every cofinality, so that if the atomic part of the universe has size at least 2^K, then it has cofinality strictly greater than K, which rules out most limit cardinals. GCH seems to be the way to make this the most permissive, since allowing any one limit cardinal with cofinality L requires increasing 2^K to be that cardinal, which rules out uncountably many lower cardinals. So intuitively, using forcing to make every powerset as small as possible is the way to rule out the fewest cardinals, so we get GCH, which says that the powerset carinalities are exactly the successors, so under that hypothesis, the atomic part of the universe has a successor cardinality.

Now, the atomless part of the universe is less clear. Gabriel Uzquiano says that every inaccessible is possible as a cardinality for this part, and this result sounds quite plausible. The appropriate chapter in Jech didn’t seem to give any sufficient conditions on a cardinal for it to have a complete atomless boolean algebra, just necessary conditions, so all I can say is that the atomless part could have inaccessible cardinality, and it certainly has cardinality such that K^A=K, so it does not have countable cofinality.

So an answer to Brian Weatherson’s second question requires a solution to the generalized continuum problem for the atomic part of the universe (ie, what cardinals are the sizes of powersets), and some more knowledge of complete atomless boolean algebras than I have for the atomless part.

But at any rate, the universe is not countable, and does not have countable cofinality, and is also at least the size of the continuum.

*This entire discussion assumes that the sets aren’t part of the universe, so that we can properly talk about the cardinality of the universe. One way to accomplish this is to be a nominalist and think that there just aren’t any sets, but allow ourselves to use sets fictionally to talk about cardinalities. Another way is to consider the universe from the perspective of some “bigger universe” containing more sets, so that the actual universe (including all its sets) forms a set and not a proper class. However, if the universe satisfies ZFC, then we can’t be total mereological universalists, because not every collection of sets has a fusion – otherwise, the universe would form a set, since we could take the fusion of all the singletons. To address the case where the universe contains sets, I think you have to be careful just how you phrase the complete mereology axiom. The discussion here seems like it will be quite useful.




One response

23 01 2007

Since we wish to avoid metaphysical claims that are too strong, note that it is also possible (and arguably not especially unlikely) that such a universe would contain # things, where # is 1/0 (more or less, see, a possible number of points in physical continua.

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