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Can all the large cardinal axioms not presently known to be inconsistent with ZFC be arranged in a strict linear order based on the following criterion. Let A(1) and A(2) be any distinct pair of these axioms and let both be adjoined to ZFC. We will say that $A(1)\lt A(2)$ if and only if the smallest cardinal number satisfying A(1) is smaller than the smallest cardinal number satisfying A(2). Should such a strict linear ordering of these axioms be possible, which of them ranks the largest?

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@Andres: I suppose that would be an interesting question, but how did you get that from what Garabed wrote? –  Ricky Demer May 2 '11 at 20:07
    
I looked at what was written and fixed the typo (a $\lt$ sign). You can see this by looking at the edit history. –  Andres Caicedo May 2 '11 at 20:11
    
It's not just the edit history that would need to be looked at, it's the source of the original post. –  Ricky Demer May 2 '11 at 20:18

2 Answers 2

up vote 9 down vote accepted

Garabed:

For many large cardinals, it is indeed true that the order we usually use (consistency strength over ZFC) coincides with the ordering you suggest: The least inaccessible is strictly weaker than the least Mahlo, which is strictly weaker than the least weakly compact, for example. But this ordering does not behave well for all large cardinals.

Here is an example illustrating the problem:

Magidor showed that it is consistent (relative to the existence of supercompact cardinals) that the smallest strongly compact cardinal is also the smallest measurable cardinal. He also showed that it is consistent that the smallest strongly compact cardinal is the smallest supercompact cardinal. Now: Supercompact cardinals are limit of measurable cardinals, so we have a problem if we use your ordering, as the position of "strong compactness" would depend on the model we are considering, which is not what one would like, namely, that the ordering depends only on the axioms and the theories they imply.

Another example is that we may have a model with Wodin cardinals, and no strong cardinals, and models with many strong cardinals but no Woodin cardinals.

And many other examples can be produced with other pairs of large cardinal assumptions.

The ordering in terms of consistency strength is more subtle. It coincides (within the region where we can currently prove these things) with a natural ordering of certain inner models. Independently of whether the least strongly compact cardinal is the least measurable cardinal or not, if there is a strongly compact cardinal, there are inner models with many measurable cardinals (and much more). On the other hand, we can identify a canonical smallest inner model with a measurable cardinal (an analogue of $L$), and there can be no strongly compact cardinals in any of its submodels. This shows that in a clear sense, "strong compactness" is a stronger assumption.

The issue with Woodin cardinals and strong cardinals is similar, but there is an additional subtlety, namely, (simplifying matters a tiny bit) to be a Woodin cardinal is a "local" assumption, we just need a large enough initial segment of $V$ to verify the Woodinness of a given cardinal. However, to be a strong cardinal is a "global" assumption", that requires that we check the whole universe. Something similar occurs with supercompactness (global) and other assumptions, like the existence of rank into rank embeddings(local). What is true is that, independently of size, if there is a Woodin cardinal, then there are inner models with many strong cardinals (but, just as before, there is a canonical smallest inner model with a strong cardinal, and no cardinals are Woodin in any of its submodels), so the assumption that there is a cardinal that is Woodin is strictly stronger.

Moreover, we are more interested in "large cardinal assumptions" than in "large cardinals" per se, and for many assumptions, largeness in terms of cardinality is not a good measure. For example: The existence of $0^\sharp$ is a large cardinal assumption. It is significantly stronger than the existence of a weakly compact cardinal (for example, if $0^\sharp$ exists, then there is a proper class of weakly compact cardinals in $L$). However, $0^\sharp$ is a countable object, while weakly compact cardinals are not.

The follow up question is what large cardinal assumption is the largest (in whichever appropriate ordering we end up using). At the moment, such an assumption does not seem to exist. We have a template that generates strong large cardinal axioms, namely describing properties of elementary embeddings. We know not everything is possible: Kunen showed there are no Reinhardt cardinals, i.e., no embeddings $j:V\to V$. This gives us an upper bound on what we can have. We have been considering assumptions that in a sense look very close to this boundary (rank into rank embeddings, for example), but it may well be that closeness is a feature of the resolution of our current telescopes. Years ago, it was expected that supercompactness was an assumption just slightly above the existence of strong cardinals. Nowadays we see a huge chasm in between.

I feel it would be slightly naive to go an extra step and say that there can be no largest axiom, but currently this feels close to the truth. Now, of course, among the large cardinal assumptions studied so far, there is one that is the largest, but that seems to me to be simply a historical accident, and bound to change within a few years, so I do not think it would be a good answer to simply point at a book and say "look at the assumption in such and such page" (though I can think of a good candidate for the current record).

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Wouldn't it be Woodinness? :) –  Mariano Suárez-Alvarez May 3 '11 at 1:12
    
I knew there was something weird in that paragraph. But I was worried about whether to write "strength" or "strongness" (as Joel does), and ended up using neither and not fixing the typo. Thanks! –  Andres Caicedo May 3 '11 at 1:22
    
Thanks to you all for your very illuminating answers. Even though the relative size of the cardinals involved seems not to be a good way of ranking these axioms, it appears to me that the more we know about these relative sizes, the sharper becomes our intuitive picture of the universe (or the "possible universes") of set theory. –  Garabed Gulbenkian May 3 '11 at 17:20
    
Oh, sure, I agree. And even on a purely pragmatic level, it is important to study both orderings, to be aware that they differ. –  Andres Caicedo May 3 '11 at 17:24

Just to give a short, explicit example of the local-versus-global issue that Andres mentioned, consider supercompactness (global) and hugeness (local). In terms of consistency strength, hugeness is far stronger than supercompactness. Yet, if there exist both a supercompact cardinal and a huge cardinal, then the first supercompact cardinal is larger than the first huge cardinal.

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But Andreas, does this actually show non-linearity in the proposed order? I don't think so. After all, from the existence of both a supercompact and a huge, we can prove that the least huge is smaller than the least supercompact, and so Huge $\lt$ supercompact in this order. The point instead is that $\lt$ is not the right order, since it differs so strongly from the consistency strength order. A more extreme example would be that Huge $\lt$ $\Sigma_2$-reflecting, since $\Sigma_2$ reflecting cardinals will reflect hugeness, but this is a strong reversal of the consistency strength order. –  Joel David Hamkins May 3 '11 at 0:33
    
Similarly, Supercompact $\lt$ $\Sigma_3$-reflecting on the proposed order, even though $\Sigma_3$-reflecting cardinals are very weak in consistency strength (below Mahlo, or even just ZFC, if you drop inaccessibility requirement). –  Joel David Hamkins May 3 '11 at 0:35
    
@Joel: I didn't intend to imply that this is a nonlinearity in either of the orders (size or consistency strength). It's just a disagreement between the two orders. –  Andreas Blass May 3 '11 at 2:17
    
Does there or can there exist a large cardinal axiom A such that (1) A is local (2) "Supercompact"<A (3) A is stronger than "Supercompact" in consistency strength? –  Garabed Gulbenkian May 13 '11 at 18:18

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