Timeline for I can't believe it's not Replacement!
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 18 at 2:58 | vote | accept | David Roberts♦ | ||
| Mar 8, 2022 at 1:04 | answer | added | David Roberts♦ | timeline score: 4 | |
| Jun 17, 2021 at 7:42 | comment | added | David Roberts♦ | @Wojowu thanks for the comments. I should have realised that no theorem of ZFC requires more than finitely many instances of Replacement! Someone else pointed my to Don Martin's book on Determinacy, and I see rather precise statements of the form you outline. In particular, he notes that $\Sigma_1$ replacement can be removed, at the cost of using different sets for ordinals (as opposed to von Neumann ordinals), and then $\alpha$-iterated power sets for all countable $\alpha$ is enough for BD. | |
| Jun 16, 2021 at 10:35 | comment | added | Wojowu | Also, although this may be a slightly silly point, it should not at all be surprising that BD doesn't require full replacement - BD is a single formula in the lanuage of ZFC, so it can't depend on an entire axiom schema. | |
| Jun 16, 2021 at 10:25 | comment | added | Wojowu | I believe it is not so much replacement that is needed, but rather existence of suitably large power iterated power sets that is. A reasonably precise statement of strength is given in Theorem 1.1 here - over set theory without power set, existence of $P^\alpha(\omega)$ is sufficient for $\Sigma^0_{1+\alpha+2}$-determinacy, but not $\Sigma^0_{1+\alpha+3}$ determinacy, at least in the presence of $\Sigma_1$-replacement. I'm not sure to what extent this answers your question. | |
| Jun 16, 2021 at 4:32 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
Clarification of question in response to comments
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| Jun 16, 2021 at 4:29 | comment | added | David Roberts♦ | @NoahSchweber Ah, thanks. And apologies for the mix-up. Hopefully people can see through my misunderstanding to what I'm hinting at, and what François' comment seems to answer. | |
| Jun 16, 2021 at 4:25 | comment | added | François G. Dorais | That's exactly right: BD is a fixed target so general axioms are too general for that. | |
| Jun 16, 2021 at 4:25 | comment | added | Noah Schweber | @DavidRoberts There's an odd mix-up in this question between theories and structures here. The $V_{\omega+\omega}$ (for example) of any model $V\models\mathsf{ZFC}$ will satisfy BD, since BD is essentially just a statement about reals. But this has nothing to do with the amount of replacement needed for BD. Basically, "Which $V_\alpha$s (or similar) satisfy BD?" is just not the right question to ask. | |
| Jun 16, 2021 at 4:21 | comment | added | David Roberts♦ | @FrançoisG.Dorais ah, ok. That a nice upper bound, at least. It's a bit like having DC or countable choice, in that there is probably an argument that give a level of justification not available for the an unrestricted axiom. | |
| Jun 16, 2021 at 4:18 | comment | added | François G. Dorais | Sorry for the wording, countable replacement is enough but it's a bit too much for reasons I explained. (Though the "too much" part is mostly visible in combination with other axioms.) | |
| Jun 16, 2021 at 4:12 | comment | added | David Roberts♦ | Hmm, that makes this even less clear! Weasel words (sorry!) like "basically" are what contribute to the hazy picture, in my head,at least. But if Replacement in the sense of having a function from a countable set is all that's needed, that's awesome! | |
| Jun 16, 2021 at 4:01 | comment | added | François G. Dorais | What is necessary over ZC is basically countable replacement, which is enough to get $V_\alpha$ for $\alpha<\omega_1$. (That takes more work than you might think!) But you only need the instances of countable replacement used in the proof of BD, though. | |
| Jun 16, 2021 at 3:55 | comment | added | David Roberts♦ | @FrançoisG.Dorais I see what you are saying, but $V_{\omega+\omega}$ is the poster child of models of BZC (say), and I was led to believe this is not enough to prove BD! Maybe I don't mean models, but I'm trying to get a handle on exactly how to describe the set theory that has BD, partly by looking at models where it does that aren't models of ZFC. | |
| Jun 16, 2021 at 3:44 | comment | added | François G. Dorais | Seriously though, the language you're looking for is $n$-th order arithmetic, often denoted $Z_n$ (but I personally think this historical notation should be revised). | |
| Jun 16, 2021 at 3:31 | comment | added | François G. Dorais | BD holds in all $V_\alpha$ in which all relevant parameters exist: so $\alpha \geq \omega+5$ is enough for most encodings and $V_{\omega+\omega}$ is more than enough for everyone who is not deliberately acting silly. Similar (but slightly different) situation for the $H$ hierarchy are true. This is because these are defined in ZFC and $\beth_{\omega_1}$ does exist; these are not predicative definitions! | |
| Jun 16, 2021 at 2:47 | history | asked | David Roberts♦ | CC BY-SA 4.0 |