Timeline for answer to Axiom of choice, Banach-Tarski and reality by user13113
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| when toggle format | what | by | license | comment | |
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| Jun 18, 2021 at 23:20 | comment | added | Anixx | Is not it evident that there are sets that cannot be constructed? | |
| Jan 25, 2017 at 15:44 | comment | added | Timothy Chow | @MattF. : Regardless of what "foundations" means, it is clear from context that when Hurkyl is saying that $ZFC$ is a "better choice", the intended assertion is that $ZFC$ is a better choice than $ZF + \neg AC$. Dragging $PRA$ into the picture out of left field is the non sequitur. | |
| Jan 23, 2017 at 20:24 | comment | added | user44143 | @TimothyChow, the question is what concept of foundations is used here. (Note that the higher-voted answers do not rely on that word.) Until I see that articulated a bit more, it's Hurkyl's argument that looks like a non-sequitur to me. | |
| Jan 23, 2017 at 16:17 | comment | added | Timothy Chow | @MattF. : Again, I'm not defending Hurkyl's argument, but the thesis isn't that $ZFC$ is better than $PRA$; rather, the thesis is that $ZFC$ is better than $ZF + \neg AC$, because $ZF + \neg AC$ implies something that seems unjustified (namely, that $V\ne L$). The fact that this argument cites a fact that happen to be provable in $PRA$ doesn't at all suggest that we should use $PRA$ for foundations; that's a non sequitur. | |
| Jan 22, 2017 at 22:10 | comment | added | user44143 | You say $ZF + (V=L) \vdash AC$. But that claim of probability can be proved in PRA. Why then do you say that the better choice for foundations is something like ZFC rather than something like PRA? | |
| Jan 22, 2017 at 21:31 | comment | added | Timothy Chow | @Burak : Although I'm not convinced by Hurkyl's argument, I don't think that you can replace V=L by GCH. The point is that L is essentially the minimal model for ZF. Hurkyl's point is that AC is true if you take the smallest possible universe of sets that are forced on you by ZF. | |
| Jan 21, 2017 at 9:35 | comment | added | Burak | @Hurkyl: I still don't understand why this is supposed to be an argument for $AC$. Do you think the following is an argument for $AC$? "We have $ZF+GCH \vdash AC$ and hence in order to assert $\neg AC$ one should posit the existence of additional structure on cardinal numbers other than expected. Also, $ZF+GCH$ as a starting point is rich enough to allow constructions of other interesting universes. etc." Your whole argument can be recast with $V=L$ replaced by $GCH$. | |
| Jan 21, 2017 at 6:08 | comment | added | David Roberts♦ | Who knows? I think Timothy Chow is making some valid points, however. I'm generally agnostic regarding foundations (the best mathematics is generally foundation-independent, give or take) but Maddy's arguments about V=L being an impoverishment of the universe of sets are quite striking. | |
| Jan 21, 2017 at 5:40 | comment | added | user13113 | @Timothy: Okay. In ZFC, I construct the arrow category $X \to Y$. Since that arrow is an epimorphism, "AC holds, period" implies it is split. But I don't see any arrows from $Y$ to $X$ in the arrow category.... Having AC in the ambient set theory does not imply that other structures you construct satisfy their internal versions of AC. | |
| Jan 21, 2017 at 5:33 | comment | added | Timothy Chow | @Hurkyl : No, ZFC says that AC holds, period. That AC holds in models of ZFC is practically a tautology. | |
| Jan 21, 2017 at 5:30 | comment | added | user13113 | @Timothy; ZFC only says that AC holds in models of ZFC. | |
| Jan 21, 2017 at 5:28 | comment | added | Timothy Chow | @Hurkyl: I think you misunderstand my point. That $ZF + (V=L) \vdash AC$ tells us only that $AC$ holds if we restrict ourselves to constructible sets. That (to me) does not suggest that $AC$ holds for all sets, which is what $ZFC$ says. If we accept your line of argument, then it would also be an argument for the generalized continuum hypothesis, and are you really going to argue that? This only seems convincing to me if you're going to argue that $V=L$ is true. | |
| Jan 21, 2017 at 5:00 | comment | added | user13113 | @David: Can I study universes of sets satisfying, say, ZFC, if I take ZF-I as foundations? My impression is no. Am I wrong? Taking ZFC+(V=L) as a starting point, on the other hand, is still rich enough to allow construction of lots of other interesting universes. | |
| Jan 21, 2017 at 4:57 | comment | added | user13113 | @Timothy: That the universe of sets we take for foundations satisfies AC by no means implies the other universes we may wish to study must also satisfy AC. | |
| Jan 21, 2017 at 3:16 | comment | added | Timothy Chow | I agree with David Roberts. This is an unconvincing argument for AC. The argument amounts to this: If we restrict ourselves to a very limited notion of set, namely constructible sets, then there is always a choice function for a constructible family of constructible sets, and in fact the choice function is constructible. Therefore (?!) choice functions surely (?!) exist for arbitrary families of sets. | |
| Jan 21, 2017 at 0:39 | comment | added | David Roberts♦ | Every model of ZF contains a model of ZF - Infinity. Is that a good argument for finitism? | |
| Jan 20, 2017 at 23:21 | comment | added | user13113 | @Burak: Same reason: every model contains an $L$. Assuming $V \neq L$ is assuming extra structure built atop of $L$, so taking $L$ is better for foundations. | |
| Jan 20, 2017 at 23:21 | history | edited | user13113 | CC BY-SA 3.0 |
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| Jan 20, 2017 at 23:11 | history | answered | user13113 | CC BY-SA 3.0 |