Timeline for How much Replacement does this axiom provide?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 27, 2021 at 16:53 | vote | accept | Tom Leinster | ||
| Jul 21, 2021 at 22:46 | comment | added | David Roberts♦ | @AsafKaragila and indeed, in really only needs $\beth_\alpha$ for all countable $\alpha$, not even $\beth_{\omega_1}$! | |
| May 13, 2019 at 13:35 | comment | added | Asaf Karagila♦ | @Gro-Tsen: See my comment on the question. | |
| May 13, 2019 at 12:53 | comment | added | Gro-Tsen | I'm sure everyone here is well aware of this, but for the completeness of MO it might be worth pointing out that if $δ$ is a fixed point of $α\mapsto\beth_α$ then $V_δ=H(δ)$ (sets hereditarily of cardinal $<\delta$), and $H(\delta)$ is a model of $\Sigma_1$-Replacement (plus all ZFC axioms besides Replacement). Indeed, in this context it follows from $\Delta_0$-Collection, and by a theorem of Lévy, $H(\delta)$ is a $\Sigma_1$-elementary submodel of the universe. | |
| May 13, 2019 at 10:03 | comment | added | Asaf Karagila♦ | @Tom: Even the famous Borel determinacy result (which is the usual appeal of Replacement outside of set theory, although that is too debatable) only requires $\beth_{\omega_1}$ which is far, far below the least fixed point. | |
| May 13, 2019 at 9:52 | comment | added | Tom Leinster | @B2C thank you for this; much appreciated. I asked for statements either inside set theory or outside of it, and you gave me one inside. But still, I can't help wondering: if you had to convince a non-set-theorist that it was important to assume Replacement rather than just "my" weak version of it, what would you say? E.g. do you know of parts of math outside set theory that require the existence of $\beth$ fixed points? | |
| May 13, 2019 at 9:47 | comment | added | Asaf Karagila♦ | @David: Well, you still have the power objects. And what are the $V_\alpha$ if not these? (Yes yes, you don't have the recursion needed for these to exist. Long live Replacement, have I said that already? :)) | |
| May 13, 2019 at 6:45 | comment | added | David Roberts♦ | Of course, but to say that beth_a = a naively, where a is an ordinal, gives well-orderable powersets. And in a structural set theory like ETCS (which I do realise Tom was not necessarily asking for, but which I was thinking of in the context of the question), you don't have the von Neumann hierarchy, at least, not in an obvious way. | |
| May 13, 2019 at 4:59 | comment | added | Asaf Karagila♦ | @David: You don't need choice to define $\beth$ numbers. In effect, $\beth_\alpha$ can be defined as simply the cardinality of $V_{\omega+\alpha}$ (or just iterated power sets from a fixed set). So you don't actually need choice there. Now to get the equivalent of a fixed point, use the Lindenbaum number of $V_{\beta_n}$ to define $\beta_{n+1}$ instead. Some people would argue that the Lindenbaum numbers are the $\beth$ numbers, by the way, which is another way of looking at it. | |
| May 13, 2019 at 3:11 | comment | added | David Roberts♦ | A natural generalisation of ETCS is to constructive models of set theory, or at the least, choice-agnostic. So knowing power sets are well-orderable (as beth-fixed-points are), while part of ETCS, may not be one desired generalisation. Interesting! | |
| May 13, 2019 at 2:11 | history | answered | B2C | CC BY-SA 4.0 |