Timeline for What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 23, 2012 at 20:42 | comment | added | Mike Battaglia | Great. Much thanks for the clear explanation. I guess I can abbreviate "ZFC+Grothendieck universes exist" to "TG" from now on. | |
| Jul 23, 2012 at 20:41 | vote | accept | Mike Battaglia | ||
| Jul 23, 2012 at 16:34 | comment | added | Trevor Wilson | @Mike: regarding your first comment, modulo ZFC all three axioms are equivalent by what I wrote combined with what you wrote (that ZFC+U and ZFC+Ca are equivalent.) It does seem to be known that Tarski's Axiom A is not equivalent to the other two under ZF for the reason you said. | |
| Jul 23, 2012 at 14:25 | comment | added | Trevor Wilson | @Mike: I'll respond to your second comment first, regarding my other attempt at answering this question, which I deleted quickly but it looks like you were still notified of. If there is a single inaccessible cardinal then for every $x$ there is a Tarski set $y$ with $\lbrace x \rbrace \in y$, and this implies that there is a proper class of Tarski sets, but it does not imply Tarski's axiom that for every $x$ there is a Tarski set $y$ with $x \in y$. | |
| Jul 23, 2012 at 7:14 | comment | added | Mike Battaglia | Also, I'm slightly confused - I got a note in my MathOverflow email update for this question (from you?) telling me to look at the accepted answer to this question: mathoverflow.net/questions/28389/… It looks like ZFC+Ca is equivalent to ZFC+Tarski, but ZFC+"There is a proper class of Tarski sets" doesn't imply ZFC+Ca, because Tarski sets aren't transitive and hence ZFC+"there is exactly one inaccessible cardinal" would also be a model of ZFC+"There is a proper class of Tarski sets." And that seems consistent with what you wrote. | |
| Jul 23, 2012 at 7:14 | comment | added | Mike Battaglia | OK Trevor, that makes sense. One lingering question I have, though, is - within the context of ZFC, do the existence of Tarski sets imply the existence of Grothendieck universes and vice versa? The former don't have to be transitive, but the latter do. Is the idea that they're equivalent in ZFC but not in ZF or something? I've seen some reference to the idea that ZF+Tarski's implies AC but that ZF+U doesn't, and that implying AC from the getgo causes the additional assertion of the existence of either Tarski and Grothendieck universes to imply one another. | |
| Jul 22, 2012 at 9:53 | comment | added | Asaf Karagila♦ | Stefan, yes. Indeed if $\kappa$ is $2$-inaccessible then $V_\kappa$ is a model of ZFC+proper class of inaccessible. I suppose that I get the minor difference. | |
| Jul 22, 2012 at 8:12 | comment | added | Stefan Geschke | Is a 2-inaccessible an inaccessible limit of inaccessibles? In this case, if $\kappa$ is 2-inaccessible, then $V_\kappa$ is a model of TG. | |
| Jul 22, 2012 at 7:13 | comment | added | Asaf Karagila♦ | I always thought that it was $2$-inaccessible (which is slightly stronger than a proper class of inaccessibles) that was equivalent to the TG axioms. | |
| Jul 22, 2012 at 2:12 | history | edited | Trevor Wilson | CC BY-SA 3.0 |
minor simplification and clarification
|
| Jul 22, 2012 at 1:51 | history | edited | Trevor Wilson | CC BY-SA 3.0 |
Fixed some typos
|
| Jul 22, 2012 at 1:10 | history | answered | Trevor Wilson | CC BY-SA 3.0 |