Timeline for Is there an infinite set $X$ such that for every isotone $f\colon[X]^\omega\to[X]^\omega$ there is a free decreasing sequence?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2017 at 18:48 | comment | added | Asaf Karagila♦ | Okay... I'm confused now... | |
| Dec 16, 2017 at 18:41 | comment | added | Paul Larson | Having looked at it a little more, I think that this is close to Kunen's original proof, including a proof of the Erdos-Hajnal theorem. | |
| Dec 14, 2017 at 17:30 | history | edited | Paul Larson | CC BY-SA 3.0 |
I fixed a few places where outputs of f were confused with their members.
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| Dec 14, 2017 at 10:17 | comment | added | Asaf Karagila♦ | Okay, so $L(V_{\lambda+1})\models\sf DC_\lambda$, which would indeed give us a sequence. But in any case, I don't think that this necessarily contradicts a Reinhardt cardinal. And won't a selector function on $[\lambda]^\omega$ mod. finite with combination of $\sf DC$ give you some semblance of one of the usual inconsistency proofs in $\sf ZFC$ anyway? | |
| Dec 14, 2017 at 7:16 | comment | added | Asaf Karagila♦ | Isn't there a proof that a Reinhardt cardinal implies the consistency of ZFC+I0 that sort of goes through adding bits and odds of choice that should allow this proof to go through? | |
| Dec 13, 2017 at 15:56 | comment | added | Paul Larson | Arguing outside the universe is what absoluteness is for. | |
| Dec 13, 2017 at 15:55 | comment | added | Paul Larson | This seems to give that the existence of a Reinhardt cardinal is inconsistent with ZF + DC plus a selector for the mod-finite equivalence classes of the countable subsets of the first fixed point (above the critical point). I don't know if this was previously known. | |
| Dec 13, 2017 at 15:47 | comment | added | Asaf Karagila♦ | It sounds like this should be doable from ZF+Reinhardt, if it is doable from ZFC+I0. But you argue "outside the universe", which I find a bit odd. | |
| Dec 13, 2017 at 15:44 | comment | added | Paul Larson | You can run the construction in $V$, with each step carried out in $L(V_{\lambda + 1})$. All you need is the output, which is a countable subset of $\lambda$. The whole output is codable by a subset of $\lambda$ anyway. | |
| Dec 13, 2017 at 15:39 | comment | added | Asaf Karagila♦ | But choosing representatives for mod finite from $[\lambda]^\omega$ requires a lot more choice than you'd have in $L(V_{\lambda+1})$ anyway. | |
| Dec 13, 2017 at 15:39 | comment | added | Asaf Karagila♦ | Uh, but why is the choice of $X_n$ canonical in any sort of way? I mean, countable choice fails in $L(V_{\lambda+1})$, if I recall correctly. No? As for the comment by fedja, Pierre commented on this today that probably the idea is to start with any $f$, then for every equivalence class (mod finite) choose a representative $A$ and declare $g(B)=f(A)$ whenever $B$ is equivalent to the representative $A$. Now this property has to fail, since a free dec. seq. gives a constant answer, and we can assume that $f$ satisfied $x\subseteq f(x)$ meaning most of the $x_n$'s are in the corresponding $f$. | |
| Dec 13, 2017 at 15:32 | comment | added | Paul Larson | This comment (from fedja on Dec 7 at 4:45) : 'remove the "isotone" condition or just relax it a tiny bit allowing a finite number of elements to stick out, and the answer is trivially "No".' [None of my attempts to make tags seem to work.] | |
| Dec 13, 2017 at 15:28 | comment | added | Paul Larson | The iterative construction I had in mind was : pick $\alpha_{0}$ and $X_{0}$ by the claim such that $\alpha_{0}$ is not in the image of $[X_{0}]^{\omega}$; since $X_{0}$ has cardinality $\lambda$, we can apply the claim to the restriction of $f$ to $X_{0}$ to find $\alpha_{1}$ in $X_{0}$ and $X_{1} \subseteq X_{0}$ witnessing the claim for this restriction. Continuing in this way, the resulting set of $\alpha_{i}$'s should be as desired. | |
| Dec 13, 2017 at 6:45 | comment | added | Asaf Karagila♦ | When you "build the function iteratively", how do you choose $X$ at each step? In fact, how do you actually construct the family? | |
| Dec 13, 2017 at 6:41 | comment | added | Asaf Karagila♦ | Also, Pierre was interested in models of ZFC. But I will talk to him today and see what he has to say on your answer. | |
| Dec 13, 2017 at 6:32 | comment | added | Asaf Karagila♦ | Which December 7th comment? Can you provide a link to it? (the timestamp functions as a direct link.) | |
| Dec 13, 2017 at 2:01 | history | answered | Paul Larson | CC BY-SA 3.0 |