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?
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| Dec 13, 2017 at 2:01 | answer | added | Paul Larson | timeline score: 5 | |
| Dec 8, 2017 at 21:28 | comment | added | fedja | This is all I know so far. Call two countable sets equivalent if the symmetric difference is finite. The question is equivalent to asking whether for every $X$, there is a mapping $F$ of equivalence classes $q$ to actual countable sets $F(q)\subset X$ such that $q<F(q)$ and $q<r\Longrightarrow F(q)\subset F(r)$ where $q<s$ means that the representatives of $q$ are contained in (the representatives of) $s$ up to finitely many elements. | |
| Dec 7, 2017 at 4:45 | comment | added | fedja | The conditions are set up with devilish cunning: 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". Allow outputs of cardinality $c$ (keeping countable inputs) and you get a trivial "No" again. So everything is just right on the edge,,, Still thinking (though falling asleep slowly...) | |
| Dec 6, 2017 at 21:18 | comment | added | Joel David Hamkins | I had meant that the finitely many points are removed during the free descent, since you remove them all one at a time. But I agree, it might not be useful. | |
| Dec 6, 2017 at 21:13 | comment | added | Asaf Karagila♦ | @Joel: Ah, so you remove some finitely many points, perhaps, to get to a tighter definition. But in any case, if there is a free desc. seq., you can always refine it to have order type $\omega$ anyway. So I'm not sure what the extra assumption that all such sequences have order type $\omega$ gives you. (Not to mention that this is only for one function anyway...) | |
| Dec 6, 2017 at 21:09 | comment | added | Joel David Hamkins | I think it is simplest to use $f(A)=\bigcup_{\alpha\in A'}f_\alpha(A\cap\alpha)$, that is, use only $\alpha$ that are limit points of $A$. This is isotone, and there is no free descending sequence, since after finitely many steps the sup $\alpha$ will stop changing and it would contradict the choice of $f_\alpha$. This also shows that for $\kappa$ of cofinality $\omega$, there is a function all of whose free descending sequences, if any, have order type exactly $\omega$. | |
| Dec 6, 2017 at 20:32 | comment | added | Asaf Karagila♦ | @Joel: Yes. I think that we converged on a fairly correct integration. :-) | |
| Dec 6, 2017 at 20:31 | comment | added | Joel David Hamkins | Indeed, your function doesn't seem to be isotone, since you didn't use the closure of $A$ as in my definition, since you added $\sup(A)$, but not the other limit points of $A$, which might be $\sup(B)$ for some $B\subset A$. (But in my definition, I should have added $A\cap\alpha$ infinite, as you did.) | |
| Dec 6, 2017 at 20:25 | comment | added | Asaf Karagila♦ | @Joel: Yeah, that might also work. I guess that there are many different ways to integrate these counterexamples (and in the case the sequence is of order type $\omega$, these would usually turn up equivalent). | |
| Dec 6, 2017 at 20:24 | comment | added | Joel David Hamkins | But then why don't you just use $f(A)=\bigcup f_\alpha(A\cap \alpha)$ for all $\alpha$ in the closure of $A$. This is now isotone, and there is no free descending sequence, since eventually after removing the finitely many points on top, the sup $\alpha$ won't change, and it would have been free descending for $f_\alpha$. No need for refining. | |
| Dec 6, 2017 at 20:21 | comment | added | Will Brian | Oh, I see now. Writing "lim" instead of "sup" will result in exactly the same function, so it doesn't matter which one you pick. | |
| Dec 6, 2017 at 20:12 | comment | added | Joel David Hamkins | Ah, yes, you are right. You've got to add in all the smaller guys. | |
| Dec 6, 2017 at 20:11 | comment | added | Asaf Karagila♦ | @Joel: But then it's not necessarily isotone, since $f_\omega$ and $f_{\omega+\omega}$ can be very different. You need to somehow integrate the counterexamples, and you're using regularity to ensure that this remains well-defined for every countable set, since it's bounded. My initial thought when I saw this was Magidor filters and Magidor cardinals, since there you're interested in bounded-countable-sets, which in this case you can also integrate over to ensure counterexamples don't appear. | |
| Dec 6, 2017 at 20:10 | comment | added | Joel David Hamkins | Asaf, isn't it easier just to argue like this: let $f(A)=f_\alpha(A\cap\alpha)$ where $\alpha=\sup(A)$. Now, the point is that for any countable $A$, after removing finitely many points, the sup won't change and won't be an element, so there can be no free descending sequence, since eventually those finitely many points on top will be removed, so it will be a free descending sequence for the limsup $\alpha$, contrary to choice of $f_\alpha$. | |
| Dec 6, 2017 at 20:09 | comment | added | Asaf Karagila♦ | @Will: No, I'm pretty sure that I meant to write $\sup$. Why would you think that I would mean $\lim$? (Honestly, Pierre showed me this two weeks ago, and I'm pulling this now out of memory, so I might be mistaken.) | |
| Dec 6, 2017 at 20:07 | comment | added | Will Brian | Thanks -- this makes a lot more sense now. (Although, did you mean to write "$\lim A$" instead of "$\sup A$" to define the set in your comment above?) | |
| Dec 6, 2017 at 19:51 | comment | added | Asaf Karagila♦ | @Will: The refinement goes like this, pick the least element that has infinitely many points from the sequence above it, and now start forming a strictly increasing sequence. By isotony, at each point your sequence will be a subsequence of the original sequence, and thus retain the free-ness property. (And for what it's worth, this is not trivial at all. You're not being stupid for asking, of course!) | |
| Dec 6, 2017 at 19:50 | comment | added | Asaf Karagila♦ | @Will: Suppose that $\kappa$ has uncountable cofinality, and for every $\alpha<\kappa$, we have $f_\alpha$ is a counterexample. Then for a countable set $A\in[\kappa]^\omega$, define $$f(A)=\bigcup\{f_\alpha(A\cap\alpha)\mid\alpha\in A\cup\{\sup A\}, A\cap\alpha\text{ infinite}\}.$$ Then $f$ is indeed isotone; and if $x_n$ is a free dec. seq., by induction we can refine to assume that $x_n$ is a strictly increase set of type $\omega$. But then $f$ applied to the sequence is just $f_\alpha$ applied to it, and $f_\alpha$ had no free dec. seq. | |
| Dec 6, 2017 at 19:29 | comment | added | Will Brian | Sorry if I'm just being stupid, but I don't see how to "glue" counterexamples together. For $\kappa = \omega_1$, I can see that the function $A \mapsto \sup A$ is a counterexample. How do I use this function to build a counterexample for $\omega_2$? | |
| Dec 6, 2017 at 17:44 | comment | added | Asaf Karagila♦ | @fedja: Just a general remark, if you put https or http it will be an actual link, not just a bunch of text. | |
| Dec 6, 2017 at 17:35 | comment | added | fedja | I have the feeling that www-cs.stanford.edu/~jbaek/infinite-ramsey.pdf (especially the very end) may be very relevant here, but have no time to follow this avenue myself now, so I'm just attracting your attention to it :-) | |
| Dec 6, 2017 at 15:48 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
Better title.
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| Dec 6, 2017 at 15:04 | comment | added | Asaf Karagila♦ | @fedja: Okay, I clarified that. | |
| Dec 6, 2017 at 15:03 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
Clarified type issues.
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| Dec 6, 2017 at 15:02 | comment | added | fedja | Indeed. But it wasn't spelled out that it is an element of $X$ either and with all fancy notation I got totally confused. You are right: there is a unique reading that makes everything fit together, but it is not very easy to discern :-) | |
| Dec 6, 2017 at 15:00 | comment | added | Asaf Karagila♦ | @fedja: I don't see where it was implied that $x_k$ is a subset of $X$ and not an element of $X$. If you point me to that location, I will clarify there. | |
| Dec 6, 2017 at 14:59 | comment | added | fedja | What is $f(\{x_k:k>n\})$? $x_k$ are already subsets of $X$, so the set used in the argument of $f$ is an element of $[[X]^\omega]^\omega$ in your notation and we get an argument type mismatch.... | |
| Dec 6, 2017 at 14:55 | comment | added | Asaf Karagila♦ | @Yair: Just the usual definition of $[X]^\omega$. All the countably infinite subsets of $X$. | |
| Dec 6, 2017 at 14:53 | comment | added | Yair Hayut | In the definition you write $[X]^\omega$, do you mean $\omega$-sequence of ordinals, or set of ordinals with order type $\omega$ (in the usual order of the ordinals)? | |
| Dec 6, 2017 at 14:44 | comment | added | Asaf Karagila♦ | @Yair: You can always assume that $x\subseteq f(x)$, by replacing $f$ with $f(x)\cup x$. When I say $f$ is isotone, I just mean that $x\subseteq y$ implies $f(x)\subseteq f(y)$. | |
| Dec 6, 2017 at 14:43 | comment | added | Yair Hayut | Do you mean that $f(x) \subseteq x$ or $x \subseteq f(x)$ (or something else)? | |
| Dec 6, 2017 at 14:36 | comment | added | Asaf Karagila♦ | Inclusion, of course. | |
| Dec 6, 2017 at 14:36 | comment | added | Joel David Hamkins | You mention "isotone" functions, but what is the order here? | |
| Dec 6, 2017 at 14:30 | history | asked | Asaf Karagila♦ | CC BY-SA 3.0 |