I read that there are some non-standard versions of set theory that allow for the existence of a universal set. My first question is: what (if anything) can be said about the cardinality of the universal set in such theories? Is it greater than the cardinality of every other set? I'm guessing Cantor's theorem fails in those theories, so I'm wondering if there is a greatest cardinality according to them (and, in particular, if that greatest cardinality, if it exists, is associated with the universal set). More generally, are there indeed set theories according to which there is a greatest cardinality?
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Note that any halfway decent set theory with a universal set will allow you to form the function $x\mapsto x$ restricted to any set. This is certainly the case for $\mathsf{NF(U)}$ or positive set theory. This provides an injection from any set into the universe, which means that every set has cardinality $\leq V$.
As for what can be said about this cardinality, that depends. Here I'll speak only for $\mathsf{NF(U)}$ since I'm not familiar enough with positive set theories. Choice is inconsistent with $\mathsf{NF}$, so it's hard to say anything in particular about the cardinality of $V$ using standard terminology. In this setting, Cantor's theorem fails and not only is $|\mathcal{P}(V)|=|V|$, but $\mathcal{P}(V)=V$. A form of Cantor's theorem holds, however: $|\{\{x\}:x\in X\}|<|\mathcal{P}(X)|$ for all sets $X$. Since there's a largest cardinality, it shows that $V$ cannot be the same size as any set of singletons---this is really sort of the fundamental result on the cardinality of $V$ in $\mathsf{NF(U)}$.
In $\mathsf{NFU}$, where we can restore choice, assertions about what kind of cardinal the universe is seem to be fairly non-trivial; it needn't be, say, inaccessible, but it is consistent (modulo some large cardinals in the metatheory) to assume that it is. Interestingly, the cardinalities of the set of singletons, of singletons of singletons, etc. will also be smaller inaccessible cardinals. In $\mathsf{NFU+Choice}$, Cantor's theorem also fails, but it fails even more because now $|\mathcal{P}(V)|<|V|$! However, the relation between the set of singletons of members of $X$ and the powerset of $X$ is still exactly the same as in $\mathsf{NF}$.
Malice Vidrine
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Thanks! Very interesting. – user171348 Jul 15 '15 at 04:06
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1The nice thing about the failure of Cantor's theorem in $\sf NF$ is that ${x\mid x\notin f(x)}$ does not define a set to begin with! :-) – Asaf Karagila Jul 17 '15 at 09:52
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Hereditarily finite NF [1] supports a simple model in which the universal set has countably many elements, but hereditarily finite NF(U) allows arbitrarily many urelements to be added to the universal set, so the only thing that can be said is that the universal set has some infinite cardinality.
[1]: This is a theory introduced by Thomas Forster whose sets are the hereditarily finite sets of ZF plus the sets that it is possible to get from these by replacing the 'leaves' (i.e., empty sets in the membership tree) by the universal set. Forster showed that this structure is just the hereditarily finite part of NF.
Charles Stewart
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I didn't know that $\sf NF$ has a model. I thought its consistency was an open problem. – Asaf Karagila Jul 17 '15 at 09:51
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@Asaf: Oops - both NF and NF(U) in my answer should be qualified with 'hereditarily finite'. I'll change that in a moment. – Charles Stewart Jul 17 '15 at 09:57
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@AsafKaragila - IIRC, it is only NF plus the infinity axiom (i.e., the existence of infinite sets other than the universal set) whose consistency is an open problem. – Charles Stewart Jul 17 '15 at 10:04
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No, NF itself is not known consistent, though Randall Holmes has a promising looking proof of Con(NF). Infinity is a theorem of NF, proved by Specker quite some time ago. – Malice Vidrine Jul 17 '15 at 11:24
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@AsafKaragila et al: Actually, there have been two relatively recent claimed proofs of the consistency of NF: one by Randall Holmes (http://math.boisestate.edu/~holmes/holmes/nfisconsistentbytangledtypes.pdf) and one by M. Gabbay (http://arxiv.org/abs/1406.4060). I am not sure how much either proof has been vetted yet. Perhaps confusingly, Holmes' own page on NF (http://math.boisestate.edu/~holmes/holmes/nf.html#Problem) says that the consistency problem is still open. :P – Noah Schweber Jul 28 '15 at 19:56
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@Noah: Yeah, that's from last year. I actually had my hand in trying to understand some of those proofs, but time allocation dictated otherwise. :-) – Asaf Karagila Jul 28 '15 at 20:20
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How about Cantor's Absolute Infinite?
It's not a rigorously defined mathematical object but it serves the simple definition that it is a quantity bigger than all others, even cardinal numbers, and that fits the cardinality of the universal set pretty well in my opinion.
SMMH
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