Timeline for Relationship between fragments of the axiom of choice and the dependent choice principles
Current License: CC BY-SA 4.0
43 events
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| Aug 23, 2022 at 14:13 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
Corrected a problem with the original argument.
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| Aug 22, 2022 at 15:25 | comment | added | Asaf Karagila♦ | @Lorenzo: Yes, that looks fine. I came back today from the conference and tomorrow I'll have the time to fix my answer. | |
| Aug 22, 2022 at 15:17 | comment | added | Lorenzo | @AsafKaragila Could you give a look at the answer I posted when you find the time? Thanks! | |
| Aug 9, 2022 at 14:47 | comment | added | Asaf Karagila♦ | @Lorenzo: I think you're right, but I have quite a lot on my head over the next few days so I'll have to try and check this only in a bit. Regardless of that, this can be easily solved by adding $\kappa^+$ subsets to $\kappa$ and taking supports of size $\kappa$; this will give us $\sf DC_\kappa$, rather than $\sf DC{<\kappa}$, but we already agreed that we are dealing with a successor cardinal. Again, I will need to think about it and it will take a few days. But thanks again for the tenacity. | |
| Aug 8, 2022 at 8:45 | comment | added | Lorenzo | hence there is none and if we let $\dot{A}_E$ be the canonical name for the set $\{a_\alpha \mid \alpha \in E\}$ then $p' \Vdash \text{ran}(\dot{f}) \subseteq \dot{A}_E$. But then $p'$ would force $\text{ran}(\dot{f}) \cap \dot{[\gamma]} = \emptyset$, contradiction. Hence in $N$ there is no choice function for $\mathcal{B}$. | |
| Aug 8, 2022 at 8:45 | comment | added | Lorenzo | Let $E\in I$ be the support for $\dot{f}$ and assume wlog that $s(p)\subseteq E$. As $|E| < \kappa$ there exists $\gamma \in \kappa$ such that $(\kappa \times \{\gamma\}) \cap \text{dom}(p) = \emptyset$, fix one and let $p' = p \cup \{((\alpha, \gamma), 0) \mid \alpha \in E\}$. Clearly $p' \le p$, assume now that $\exists q \le p', \beta \in \kappa$ and $\alpha\in \kappa\setminus E$ such that $q\Vdash \dot{f}(\check{\beta}) = \dot{a_\alpha}$, this would be a contradiction (for the same reason $W_\kappa$ fails) | |
| Aug 8, 2022 at 8:44 | comment | added | Lorenzo | @AsafKaragila I think that $\text{AC}_{\kappa}$ actually fails in the symmetric model defined in your proof of Theorem I: to see it consider in $N$ the family $\mathcal{B} = \{[\beta] \mid \beta \in \kappa\}$ with $[\beta] = \{a_\alpha \mid \beta \in a_\alpha\}$. Let $\dot{[\beta]},\dot{\mathcal{B}}$ be their canonical names and note that, for each $\beta$, $1_{\mathbb{P}}\Vdash \dot{[\beta]}\neq \emptyset$. Suppose that there is a condition $p \in G$ and a symmetric name $\dot{f}$ such that $p\Vdash ``\dot{f}: \check{\kappa} \rightarrow \dot{A}$ is a choice function for $\dot{\mathcal{B}}$''. | |
| Aug 4, 2022 at 10:53 | comment | added | Lorenzo | Ok, I'm sorry if I'm being repetitive, the point is that I'm not entirely convinced that we've showed that the set of $q$s is dense below $p$, for the reasons I gave briefly 7 comments above. I'll work on it, thanks again! | |
| Aug 4, 2022 at 10:42 | comment | added | Asaf Karagila♦ | @Lorenzo: And that is something you want to do when you want to have atom-like things. We don't need that here. As I keep saying. | |
| Aug 4, 2022 at 10:40 | comment | added | Lorenzo | Because he uses it to prove that the set of $q$s is dense below $p$, after one encodes his argument into ours.. More precisely it seems that he actually proves that $q$ and $\pi q$ (I'm referring to your proof) are compatible | |
| Aug 4, 2022 at 10:33 | comment | added | Lorenzo | Thanks for the insight, what I meant was the automorphism is more complicated even taking into consideration what you just wrote | |
| Aug 4, 2022 at 10:30 | comment | added | Asaf Karagila♦ | @Lorenzo: The point being that in order to get something that "feels" like atoms, one has to begin with sets of sets of Cohen generics, rather than just a set of Cohen generics. And by that virtue alone, everything gets more complicated. When trying to just emulate the original proof via forcing, some arguments get simpler, and some arguments get harder, but the model, as a whole, tends to be simpler in that sense. | |
| Aug 4, 2022 at 10:25 | comment | added | Asaf Karagila♦ | @Lorenzo: Transfer theorems tend to be more complicated than necessary, since they are trying to preserve some "additional structure". If you look at the proof of the Jech–Sochor theorem, they effectively embed an entire initial segment of the permutation model into $\kappa$-Cohen generics for a suitably chosen $\kappa$. We don't have that obligation, and we don't care about that. You can also open my paper about realizability results via classical methods, which extends the proof given here to a more general context. | |
| Aug 4, 2022 at 10:15 | comment | added | Lorenzo | Indeed, maybe I'm wrong, but reading Jech's proof of this result, specifically the transfer argument he sketches, he uses a more complicated automorphism group than the one we are using, and it seems to me that it is necessary to take care of precisely this point.. | |
| Aug 4, 2022 at 10:10 | comment | added | Lorenzo | Ok, but in order to prove that the set of $q$s is dense below $p$, we need to prove that given any $p'\le p$ we can find a condition $q$ below $p'$... With your argument we can show that there exists a condition $q$ below $p'\upharpoonright E$ (which forces the same as $p'$ by homogeneity) but still this $q$ may not be below $p'$... | |
| Aug 4, 2022 at 9:55 | comment | added | Asaf Karagila♦ | @Lorenzo: But homogeneity plays a bit role here, so we can always assume that $s(p)=E$, since homogeneity tells us that if $p$ forced something "reasonable" about $\dot X$, then $p\restriction E$ already forced that. | |
| Aug 4, 2022 at 9:51 | comment | added | Lorenzo | @AsafKaragila mmm didnt' we show that given any $p$ such that $s(p)\subseteq E$ (and which forces etc.) then there exists a $q\le p$ with etc. ? A priori this is not saying that the set of such $q$s is dense below $p$, because there are a lot of conditions $p'$ below $p$ such that $s(p')\not\subseteq E$, and our argument is not considering them (i.e. a priori we cannot find such conditions $q$ below this $p'$) | |
| Aug 4, 2022 at 9:26 | comment | added | Asaf Karagila♦ | @Lorenzo: We effectively showed that given any $p$, there will be some extension which has these desired properties. And so, we effectively show that this set is dense below $p$. | |
| Aug 4, 2022 at 7:45 | comment | added | Lorenzo | in other words: who says that the set $\{q \in \mathbb{P} \mid s(q)\subseteq E\cup E' \text{ and } \exists \dot{x}_\gamma \ q \Vdash \dot{x}_\gamma \in \dot{X}(\check{\gamma}) \text{ with supp}(\dot{x}_\gamma) \subseteq E\cup E'\}$ is dense below p (forcing that $\dot{X}$ has the wanted properties)? | |
| Aug 4, 2022 at 7:38 | comment | added | Lorenzo | @AsafKaragila I read the proof of Theorem 1 and I'm wondering: why is every generic filter intersecting $D_\gamma$ (for each $\gamma$)? You defined $D_\gamma$ to be a maximal antichain of certain conditions $q$ with $s(q)\subseteq E\cup E'$ below some $p\in G$ with $s(p)\subseteq E$... why couldn't we have a condition $p\in G$ such that $s(p)\subseteq E\cup E'$ and $p$ forces $\dot{X}$ has the wanted properties but such that $p$ is incompatible with every condition in $D_\gamma$? Thanks | |
| Nov 15, 2021 at 11:56 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
Corrected an old, but crucial mistake.
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| May 2, 2020 at 9:25 | comment | added | Asaf Karagila♦ | @Taras: I sent you an email with some remarks. Sometimes the emails I send end up in spam folders (it's quite tragic, really, some students missed important announcements because of that in the past). | |
| May 2, 2020 at 4:57 | comment | added | Taras Banakh | @AsafKaragila Dear Asaf, please look at the update at researchgate.net/publication/… Now it contains two (last) chapters devoted to Choice and Global Choice. I would appreciate any comments especially concerning the Clobal Choice. Thank you. | |
| Apr 30, 2020 at 11:12 | comment | added | Asaf Karagila♦ | @Taras: Okay, thanks. | |
| Apr 30, 2020 at 11:09 | comment | added | Taras Banakh | @AsafKaragila I post the current version to researchgate.net/publication/… At the moment the section about Choice is absent but maybe today or tomorrow I will finish writing it and will update the text. | |
| Apr 30, 2020 at 10:56 | comment | added | Asaf Karagila♦ | @Taras: If you'd like, send me the chapter, I can give you more feedback. | |
| Apr 30, 2020 at 10:53 | comment | added | Taras Banakh | @AsafKaragila Thank you for the expert consultation. It was very helpful (at the moment I am writing a textbook for students and wanted to include some known info on versions of AC). | |
| Apr 30, 2020 at 10:48 | comment | added | Asaf Karagila♦ | @Taras: Yes. That is correct. $\sf DC\to AC_\omega\to CUT\to AC_\omega^{fin}$, and none of the implications are reversible. Note that some authors switch the subscript and superscript in the notation. But I prefer this way. | |
| Apr 30, 2020 at 10:46 | comment | added | Taras Banakh | @AsafKaragila Aha, then $AC^\omega_\omega$ is another weakening between $CUT$ and $AC^{<\omega}_\omega$? By $AC^{<\omega}_\omega$ I means the existence of a choice function for countable families of finite sets. It follows from the existence of linear orders (as you wrote in your post on stackexchange). | |
| Apr 30, 2020 at 10:43 | comment | added | Asaf Karagila♦ | @Taras: That is known as the Countable Union Theorem, denoted by CUT. What you're writing is "choice from countable families of countable sets", which is not strong enough to prove the CUT. | |
| Apr 30, 2020 at 10:42 | comment | added | Taras Banakh | @AsafKaragila By the way, do you the name for the statement "The union of a countable family of countable sets is countable", denoted by $AC^\omega_\omega$? Is this notation standard? By the "name" I mean something like "Countable Axiom fo Choice" because the "Axiom of Countable Choice" is already occupied: en.wikipedia.org/wiki/Axiom_of_countable_choice | |
| Apr 30, 2020 at 7:25 | comment | added | Asaf Karagila♦ | @Taras: Not very elementary indeed. The one with the compactness theorem is much easier. But the proof of that equivalence is not trivial in itself, so it's not a good substitute if you don't have it in the bag already. | |
| Apr 30, 2020 at 7:23 | comment | added | Taras Banakh | @AsafKaragila Thank you for the prompt answer. At the moment it suffices for my purposes (I mean no need for "proper cite-able reference"). Anyway, the proof is not so elementary (somehow it escaped from me). | |
| Apr 30, 2020 at 7:14 | comment | added | Asaf Karagila♦ | @Taras: You can find the proof in Jech's "Axiom of Choice", I am sure. Here are two ways to prove it. (1) The Ultrafilter Lemma is equivalent to the compactness theorem for FOL, given a set $A$, look at the language of $\{<,c_a\}_{a\in A}$ and the axioms stating that $<$ is linear and that each $c_a$ is distinct from the others, easily compactness apply here and we have a model. (2) We show the Order Extension Property (then apply it to the discrete order), and you can find a proof of that in math.stackexchange.com/a/271012/622 if you want a "proper cite-able reference", let me know. | |
| Apr 30, 2020 at 6:18 | comment | added | Taras Banakh | @AsafKaragila Dear Asaf, on stachexchange I found your very helpful answer (math.stackexchange.com/a/211621) in which you mention that the Ultrafilter Lemma implies the existence of a linear order on each set. Could you give me a reference to this fact or a hint how it can be proved? Thank you. | |
| Apr 24, 2020 at 6:36 | comment | added | Taras Banakh | @AsafKaragila Thank you, but anyway, it is better to have all notions appearing in formulations of theorem defined, just for convenience of readers. Unfortunately I do not have Jech's book on Axioms of Chice (only his funcdamental "Set Theory"). | |
| Apr 24, 2020 at 6:32 | comment | added | Asaf Karagila♦ | @Taras: That's from Jech's book. It states "Every cardinal is comparable with $\kappa$", that is to say, if a set does not inject into $\kappa$, then $\kappa$ injects into that set. | |
| Apr 24, 2020 at 6:31 | comment | added | Taras Banakh | @AsafKaragila Please add the definition of $\mathsf W_\kappa$ to your answer. | |
| Aug 20, 2014 at 12:09 | comment | added | Victoria Gitman | Thanks for the great answer, Asaf! I am going to try to work out the details of the second proof. | |
| Aug 20, 2014 at 12:07 | vote | accept | Victoria Gitman | ||
| Aug 20, 2014 at 6:15 | comment | added | Asaf Karagila♦ | Yes, maybe I went a bit too far with the length of this answer... :-) | |
| Aug 20, 2014 at 6:04 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |