Talk:List of large cardinal properties

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Note on history of this page[edit]

This page was created on 6 November 2005 by cutting and pasting the list part of the article Large cardinal property. To see the history of the list before that date, see the history of Large cardinal property. --Trovatore 05:34, 7 November 2005 (UTC)[reply]

Problem with the list[edit]

The consistency strength of zero-dagger is strictly between the existence of one measurable cardinal and the existence of two measurable cardinals. I've modified the list to reflect this. But now of course it's not strictly a list of large cardinal properties anymore; more a list of large cardinal axioms. We should figure out what to do about this. It's the axioms that are linearly ordered by consistency strength. But it's the properties that correspond to articles we want to link to. What's more important, the ordering or the links? We need to decide, or else figure out how to reword things. (Note that this problem is not by any means unique to zero-dagger; for example the λ-supercompact and λ-extendible cardinals interleave, for different values of λ and similarly for n-huge and n-superstrong.) --Trovatore 00:27, 30 November 2005 (UTC)[reply]

By the way, apparently it's open (or was open in 2000) whether the consistency strength of the existence of a supercompact is strictly above that of a strong compact. See The Bulletin of Symbolic Logic 6(1):89 (March 2000). --Trovatore 01:20, 30 November 2005 (UTC)[reply]
all the axioms should definitely be included. maybe rename the page? Sam Coskey 19:24, 26 October 2006 (UTC)[reply]

the 1=0 joke[edit]

I'm certainly not going to revert R.e.b. but I wonder why the explanation of the joke was removed.[1] Was it incorrect?

Just to save space: I was trying to get the section small enough to fit into one browser page so that you can see all properties on one screen. R.e.b. (talk) 14:49, 22 November 2007 (UTC)[reply]
I'm not sure I'd call 1=0 a joke -- rather it's just the usual way of succinctly giving a false mathematical statement. Putting 0=1 at the top of the large cardinal hierarchy captures the idea that the large cardinals are increasing in consistency and eventually reach inconsistency if pushed far enough (i.e. to a Reinhardt cardinal).Norman314 (talk) 08:14, 13 November 2010 (UTC)[reply]

Absolute Infinite[edit]

I replaced "0=1" with Absolute Infinite which maintains the joke, but seems more like a large cardinal number to me. However, Trovatore reverted me saying "I don't know whether 0=1 should be there, but "absolute infinite" is problematic (not really a property of cardinals).". Well yes, strictly speaking neither the "small cardinals" nor "0=1" belong on this list. However, the idea of a cardinal number such that "every property of the Absolute Infinite is also held by some smaller object" (as the article on it says) certainly sounds like a large cardinal property (except for being inconsistent, since it would have to be bigger than itself). JRSpriggs (talk) 08:43, 12 January 2010 (UTC)[reply]

OK, I'm having trouble coming up with a good way of expressing why I don't think this is a good idea. Partly it's that the absolute infinite seems more like a philosophical concept, not a description of a mathematical object.
Maybe more to the point is that the absolute infinite per se is not contradictory, just its existence as a completed totality (an object, say, rather than a predicate). The existence of predicates that apply to absolutely infinitely many things has no large-cardinal strength at all.
I think maybe the best solution is to remove the entry entirely. Sort of cute, maybe, but it's not clear that "cute" is desirable in an encyclopedia article. --Trovatore (talk) 08:59, 12 January 2010 (UTC)[reply]

Link from other articles?[edit]

I added a link to this page on the ineffable cardinal page. Originally, when I read that page, I thought wikipedia did not have information on the consistency strength of ineffables. Perhaps we should link this page explicitly as a source for consistency strength info on all large cardinal pages, or at least make sure that all large cardinal pages contain info about consistency strength.

Wishlist[edit]

  • Indications of the knowledge gaps in the alleged linear order of proof-theoretic strength.
  • The ordering by size as far as is known.
  • The ordering of implications as far as is known.

Sadly I am not volunteering, as this is currently way over my head.

I would also guess that ordering by consistency also happens to be ordering by arithmetic soundness, and even soundness (which is true in the realm of ordinal analysis). If so I figure that would be worth stating. It might even be true for all sentences of 2nd-order or nth-order arithmetic. Maybe not though since as I understand V=L has consequences and some large cardinals contradict V=L while others don't. Again, over my head. --174.119.182.107 (talk) 01:53, 24 November 2010 (UTC)[reply]

I'm not quite sure what you mean by the ordering by Γ-soundness (for the various Γ you mention). Of course they're all sound, because they're all true. Maybe you mean the collection of their Γ consequences? I would expect that order to be the same. Sure, V=L has consequences that are contradicted by (say) the existence of a measurable, but V=L is not a large-cardinal hypothesis. --Trovatore (talk) 02:04, 24 November 2010 (UTC)[reply]
The information for which you are wishing is too much to put into this article. It would need to be dispersed among the articles on the various large cardinals. In some cases, it already is.
Let me rewrite your list — assuming the existence of Κ a large cardinal of some specified type (e.g. measurable): (1) what other types must it also be among (e.g. Κ must also be a Ramsey cardinal); (2) what other types must exist below Κ (e.g. some (indeed a stationary set of) inaccessible cardinals exist below Κ); (3) what other types' existence must be consistent with ZFC, if those are not among the ones listed under #1 or #2; (4) what other types are neither known to be weaker or stronger than Κ's type.
It would be nice if some expert person would go through all the articles and make sure they contain complete information of this kind, but that is probably too much to expect. JRSpriggs (talk) 08:35, 24 November 2010 (UTC)[reply]
I don't see why it would be too much for this article, other than the problem of collecting the info in the first place. See for example list of finite simple groups which is quite detailed.
I do mean ordering by consequences, and I was, in fact, expecting that the order would be the same up to . Why I wasn't sure beyond this, and why I mentioned V=L, is this: the existence of (say) a measurable implies V=/=L (which has certain second-order consequences), but the consistency of the existence of a measurable, by itself, does not. The list is ordered by consistency strength. So the little knowledge I have doesn't lead me to expect that the entries higher on the list all imply V=/=L, and so I am not sure they have all those second-order consequences.
I guess Trovatore's position is that large cardinal axioms are true, and obviously truth would imply the various forms of soundness. For disclosure purposes, then, I will state that yes, I doubt large cardinal axioms. It's not the for the "typical" reasons, if that's what you're reacting to: I already doubt ZF, and Z, and less. I'm certainly not claiming V=L. I brought it up for a different reason entirely. --174.119.182.107 (talk) 03:42, 27 November 2010 (UTC)[reply]
Oh, OK, I think I see. Yes, all the ones above measurable on the list imply V!=L, and that in particular there are lots of sharps. I don't know any counterexample to the claim that if large-cardinal axiom Bob has higher consistency strength than large-cardinal axiom Joe, then the consequences of Bob (for any n you like) include those of Joe. As Steel puts it, LCAs are "natural markers of consistency strength" — they're also stronger in most ways you care about.
The only thing I know about that seems to violate that general pattern (though not the specific claim about consequences) is that the order of consistency strength is not always the same as the order of the first representative. For example the existence of a huge cardinal is much greater has much greater consistency strength than the existence of a supercompact, but the first huge is below the first supercompact. This has to do with the fact that supercompactness is not a "local" property. In some sense that's sort of a pathology. Woodin's abstract definition of large-cardinal axiom requires locality (and so excludes "there exists a supercompact" per se, but not local variants of it which AFAIK are adequate for anything you want). --Trovatore (talk) 06:12, 27 November 2010 (UTC)[reply]


  • Oh, one technical nit: By "second-order consequences" you mean second-order arithmetic — that is, for some n? Actually, V!=L per se has no such consequences beyond those of ZFC. However, all large cardinal axioms that imply V!=L do have such consequences (specifically, they all imply the existence of 0#, which is a assertion). --Trovatore (talk) 07:00, 27 November 2010 (UTC)[reply]
I see. Assuming it can be cited I would think it's worth mentioning (here or at large cardinal). Thanks for clearing that up (and also for clearing up V=/=L versus 0#). --174.119.182.107 (talk) 18:34, 27 November 2010 (UTC)[reply]

Reinhardt cardinals[edit]

If anyone has shown that ZF + exists j:V->V has stronger consistency strength than rank-into-rank, I'm not aware of it. Would be interesting to know, certainly, if true.

My understanding is that the general sense is that j:V->V "should" be inconsistent, even without AC. I vaguely recall that there was a paper in the Bulletin a few years ago about it. --Trovatore (talk) 06:42, 27 November 2010 (UTC)[reply]

Well, that hardly justifies removing it entirely now does it. --174.119.182.107 (talk) 18:34, 27 November 2010 (UTC)[reply]
Well, it's in any case one of these things that's not like the others. I don't really see it as belonging in the same category (AC, after all, is true, so being inconsistent with AC means being false). But if one wants to mention it, its distinct status, and (possible) exemption from the consistency-strength order, should also be made clear. --Trovatore (talk) 22:20, 27 November 2010 (UTC)[reply]
It clearly belongs, and I've restored it. The article already states its status with respect to AC, and there are other LCAs whose relative strength isn't known exactly (bringing us back to the wishlist). --174.119.182.107 (talk) 14:38, 28 November 2010 (UTC)[reply]
The notion probably deserves mention of some sort. It should absolutely not be a bullet point in the same list; that's completely misleading. I'll think about how to reword, but can't stay in the current form. --Trovatore (talk) 19:32, 28 November 2010 (UTC)[reply]
  • How is it the least bit misleading? Sure, some set theorists, maybe even most, don't believe it. That seems to be entirely irrelevant. It's a large cardinal axiom, this is a list of large cardinal axioms. Incidentally, saying AC is true therefore this is false is not going to be the least bit convincing to me, and I thought that much was implicit in what I said earlier. It just makes me feel you're "up to something". --174.119.182.107 (talk) 03:32, 29 November 2010 (UTC)[reply]
    • It's misleading because the others fit together into a logical hierarchy, and this one doesn't. --Trovatore (talk) 04:01, 29 November 2010 (UTC)[reply]
    • Also I don't agree that it's a large-cardinal axiom at all, unless you mean to include ones that have been refuted. The existence of Reinhardt cardinals has been refuted. --Trovatore (talk) 04:04, 29 November 2010 (UTC)[reply]
Putting Reinhardt cardinal at the end of the list would be justified, if you could show that ZF+RC implied the existence of an inner model of ZFC+I0 (and that ZF+RC was not known to be inconsistent). I am not aware that this has been shown. JRSpriggs (talk) 09:51, 29 November 2010 (UTC)[reply]
  • Does ZF+RC imply the existence of an inner model of something somewhat further down the list? (Is there an instance of implying consistency other than implying the existence of an inner model?) --174.119.182.107 (talk) 05:43, 30 November 2010 (UTC)[reply]
I am not familiar with the literature, so I do not know what may have been proven or disproven. However, if Κ is a Reinhardt cardinal (i.e. the critical point of an elementary embedding of the universe into itself), then it should also be a measurable cardinal for some measure μ. But whether L(μ) (see constructible universe#Relative constructibility) satisfies the axiom of choice is not clear to me. JRSpriggs (talk) 07:53, 30 November 2010 (UTC)[reply]
Its consistency strength is not my main objection. I would be interested to know what is known about that. I think the general expectation is that it probably implies everything, because it's probably inconsistent, but (barring recent developments that I haven't heard of) that isn't known.
The main problem is that, unlike the others, it's not a candidate axiom for set theory, because it has been refuted in ZFC.
Now, it's true that there are other things that could be mentioned, that are like large-cardinal axioms except for being inconsistent with AC. For example, the existence of a strong partition cardinal. It might be worth mentioning those in a separate section. However that one is also not quite the same thing, because there are well-behaved inner models that allow partition cardinals, whereas none is known for Reinhardt cardinals. --Trovatore (talk) 08:42, 30 November 2010 (UTC)[reply]
I see that strong partition cardinal is in Category:Cardinal numbers rather than in its subcategory Category:Large cardinals where Reinhardt cardinal is. Perhaps it should be moved.
Perhaps we need a notion of semi-consistency with the axiom of choice. A hypothesis H would be semi-consistent with AC, if ZF+H has a standard model (a transitive class with the usual element relation restricted to it) in V (which satisfies ZFC), i.e. if H relativized to the class is consistent with ZFC. JRSpriggs (talk) 18:27, 30 November 2010 (UTC)[reply]

Pronunciation of Mahlo[edit]

Can anybody please give the correct pronunciation of Mahlo in IPA? VladimirReshetnikov (talk) 23:04, 10 February 2012 (UTC)[reply]