## How big is the universe?

December 27, 2007

I’ve been reading some interesting old posts from Kenny Easwaran’s blog (and this one from Brian Weatherson) recently, so I’m going to take a break from topless mereology for a bit and consider what happens when unrestricted composition does hold. In particular, I want to know how many things there can be in the presence of UC. (I shall use the expression “there can be kappa many things” to mean there is a kappa sized model of plural mereology.)

The first observations are that the universe will be of size $2^\kappa$ or $2^\kappa-1$ when kappa is infinite and finite respectively, and the the world contains exactly kappa many atoms, and no gunk. If there is gunk there are infinitely many disjoint things (suppose there were n: $a_1 \ldots a_n$. $a_1$ would have a proper part b. $a_1 - b$, and b would be distinct from $a_1 ... a_n$ and disjoint so there would $n+1$ disjoint things, contradiction) and thus uncountably many fusions of them. For mixed gunky/pointy universes their size is just the max of the number of gunky things and the number of non-gunky things. So we know there cannot be any countable models. (This is why we didn’t use a first order theory of mereology.)

Kenny cites the result that every infinite complete boolean algebra has size $\kappa$, where $\kappa^{\aleph_0} = \kappa$. The sufficient conditions for $\kappa$ to be a size of the universe were left unclear, although he mentioned that if $\kappa$ was inaccessible, then you can find a model of that size. I think you can do a bit better than this: if $\kappa^{\aleph_0} = \kappa$ then you can have a $\kappa$ sized model, which combining these results gives us:

$\kappa$ is a possible size of the universe iff $\kappa = 2^n - 1$ for some finite n, or $\kappa^{\aleph_0} = \kappa$.

To see the the right-left direction, let $\kappa = \kappa^{\aleph_0}$ and take the regular open sets in $2^\kappa$ (the Tychonoff product of the discrete topology on 2.) There are $\kappa$ regular open sets in $2^\kappa$, which is slightly surprising. I’m not going to write the proof out here, but it helps to know that $2^\kappa$ has the Suslin property. The interesting thing is that this construction is always gunky, so all the possible sizes are exhausted by gunky worlds alone – considering atoms does not change anything size wise.

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### 2 comments

1. Interesting — some readers may like to see the proof that there are 2^k objects in the finite case, which they can get from here: http://mathworld.wolfram.com/k-Subset.html (NB k there is the number of atoms in the object, n the number of atoms in the universe). I assume you subtract 1 from this because you don’t want to count the “null object”.

When thinking about infinite cases, doesn’t it simplify matters if one sees fusions as sets of atoms, so that the set of all fusions is just the power set of the set of all atoms? Cardinalities of power sets are pretty well-understood. Do you see any downside to that approach?

2. Hi Rich. Reasoning as you suggested is fine if you’re in a completely atomic world. The size of the universe is then $2^\kappa$ or $2^\kappa - 1$ when the universe is infinite and finite respectively. The difficulty comes when you want to work out what the size of the universe is when you allow for gunk (something such that all of its parts have proper parts.) It turns out that if the universe is infinite, then it is of size $\lambda$ if and only if $\lambda^{\aleph_0} = \lambda$.