Timeline for Large cardinal axioms and Grothendieck universes
Current License: CC BY-SA 2.5
22 events
| when toggle format | what | by | license | comment | |
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| Aug 27, 2013 at 20:44 | comment | added | Baby Dragon | One might say that the axiom of infinity is the smallest large cardinal axiom. | |
| Jan 24, 2010 at 14:57 | comment | added | François G. Dorais | Extendibles and superhuge cardinals are much stronger than UA. There is no single large cardinal whose existence is precisely equivalent to UA. | |
| Jan 24, 2010 at 8:08 | comment | added | algori | F.G. -- re the updated version: you say "Although UA is indeed a large cardinal axiom, there is no way to formulate UA as the existence of a single large cardinal." I'm again confused. In the comments to his answer, Joel says that for some axioms close to the top of the list the existence of a single cardinal with these properties directly implies UA (or AU if you like:)). He mentions extendible and superhuge. I've thought about it a little and I think I sort of understand how to deduce UA from superhuge. Do I understand correctly that you're saying that this is impossible? | |
| Jan 24, 2010 at 6:20 | history | edited | François G. Dorais | CC BY-SA 2.5 |
grammar
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| Jan 24, 2010 at 3:23 | comment | added | Harry Gindi | @JDH, I would upvote your single comment, but then it would pop up in the wrong spot. | |
| Jan 24, 2010 at 3:22 | comment | added | algori | Joel, F.G. -- thanks, I think I've got it. I'm just not used to think in terms of consistency strength. So when you said "weaker", I took it as "implies", while you meant "weaker in consistency strength". | |
| Jan 24, 2010 at 3:11 | comment | added | Joel David Hamkins | @Harry: Every infinite kappa is like that, since kappa^2=kappa in infinite cardinal arithmetic. | |
| Jan 24, 2010 at 2:50 | comment | added | Harry Gindi | Forgive my unprofessional phrasing of this: but is there a cardinal, call it k, so big that the existence of k implies that there exists a set of cardinality k of elements, each of which is of size k. | |
| Jan 24, 2010 at 2:49 | history | edited | François G. Dorais | CC BY-SA 2.5 |
addendum
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| Jan 24, 2010 at 2:44 | comment | added | François G. Dorais | @algori: No it has stronger consistency strength, which is different. I'll add something to clarify that. | |
| Jan 24, 2010 at 2:43 | comment | added | Joel David Hamkins | @Algori. It implies it in consistency strength rather than direct implication, as I explain in my answer. If kappa is 2-inacessible, then V_kappa is model of AU. | |
| Jan 24, 2010 at 2:40 | comment | added | algori | F.G. -- ok, but then this does not imply the Universe axiom (in the form stated in the posting: for every cardinal there is a strictly greater (strongly) inaccessible cardinal). Correct me if I'm wrong. | |
| Jan 24, 2010 at 2:39 | history | edited | François G. Dorais | CC BY-SA 2.5 |
correction
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| Jan 24, 2010 at 2:37 | comment | added | François G. Dorais | @Joel: I think it's pretty funny, I'm having fun! People must be thinking we're racing or something. (You write better than I do, so I always learn something by comparing our answers.) | |
| Jan 24, 2010 at 2:34 | comment | added | François G. Dorais | @algori: If $\kappa$ is 2-inaccessible, then it is a limit of inaccessibles, so for every $\mu<\kappa$ there is an inaccessible $\lambda$ with $\mu \leq \lambda < \kappa$. (There may not be any inaccessibles above $\kappa$ at all! This is one of the strange things about large cardinal axioms which takes some getting used to...) | |
| Jan 24, 2010 at 2:28 | comment | added | algori | What I meant was: how does one deduce that there is an inaccessible greater than a given $\lambda$ from the existence of a 2-inaccessible $\kappa$? | |
| Jan 24, 2010 at 2:27 | comment | added | Joel David Hamkins | Francois, we're stepping on each other's toes this evening, aren't we? Sorry! | |
| Jan 24, 2010 at 2:27 | comment | added | Harry Gindi | books.google.com/… just for people who wanted to see the chart. | |
| Jan 24, 2010 at 2:23 | comment | added | François G. Dorais | You don't "construct" them, they're too big for that! The ordinals are wellordered, so you just find the first inaccessible which is larger than $\lambda$. | |
| Jan 24, 2010 at 2:18 | comment | added | algori | Thanks, F.G.! Sorry, I'm not sure I get it. Suppose $\kappa$ is 2-inaccessible and $\lambda$ is arbitrary. How does one construct a 1-inaccessible cardinal greater than $\lambda$? | |
| Jan 24, 2010 at 2:09 | history | edited | François G. Dorais | CC BY-SA 2.5 |
grammar
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| Jan 24, 2010 at 2:00 | history | answered | François G. Dorais | CC BY-SA 2.5 |