Timeline for An infinite hat puzzle variation—if we don't know our place, can we still be almost all correct?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31 at 14:01 | vote | accept | Joel David Hamkins | ||
| Jul 31 at 10:32 | history | became hot network question | |||
| Jul 31 at 4:45 | answer | added | paste bee | timeline score: 12 | |
| Jul 31 at 4:31 | answer | added | bof | timeline score: 9 | |
| Jul 31 at 3:55 | comment | added | bof | Namely, if $f:^\omega\kappa\to\kappa$ is as in Problem 5348, then the algebra $(\kappa,f)$ can't have an infinite descending sequence of subalgebras, and a minimal subalgebra of cardinality $\kappa$ is a Jonsson algebra. | |
| Jul 31 at 3:51 | comment | added | bof | By the way, note that infinitary Jonsson algebras follow immediately from the hat problem for $\omega$ sequences and many colors a.k.a. AMM Problem 5348; this was the subject of an old paper by Galvin and Prikry. | |
| Jul 31 at 3:40 | comment | added | Joel David Hamkins | That of course would require global choice, but this is fine. I look forward to an answer by one of you for this $\omega$ case. Meanwhile, as far as I understand, this doesn't seem yet to answer the question for longer ordinals. | |
| Jul 31 at 3:36 | comment | added | bof | The solution (as I recall) was exactly as outlined in comments by Daniel Weber, i.e., choose a representative from each equivalence class, with special attention to eventually periodic sequences. Any number of colors; as stated by the proposer, the colors could range over the whole universe; i.e., he seems to be talking about a function $f:V^\omega\to V$. | |
| Jul 31 at 3:36 | comment | added | Joel David Hamkins | @bof Could you post the solution as an answer for that case? Does it generalize to higher ordinals? | |
| Jul 31 at 3:19 | comment | added | bof | The "easiest case" (unknown place in an $\omega$ sequence) but with an arbitrary set of colors was American Mathematical Monthly Problem 5348. I recall that they published a faulty solution (not the proposer's) but later published a correct one. The original statement (from memory) was something like this: "Prove that there is a function $f$ such that, for every infinite sequence $\{x_n\}$, we have $x_n=f(x_{n+1},x_{n+2},x_{n+3},\dots)$ for all but finitely many $n$." | |
| Jul 31 at 3:13 | comment | added | Joel David Hamkins | Hmmn. Is it right? If so, this would handle the easy case of $\omega$. Can you generalize to larger ordinals? Please post an answer if the details work out. | |
| Jul 31 at 3:11 | comment | added | Daniel Weber | If they see that the sequence is periodic then it has to be handled separately, by simply continuing it backwards. | |
| Jul 31 at 3:09 | comment | added | Daniel Weber | I think the version with hidden $\theta$ is also solved by this by finding an upper bound for all ordinals you see, then considering the representative amongst the sequences with ordinals up to it | |
| Jul 31 at 3:07 | comment | added | Joel David Hamkins | Perhaps everyone is $0$, but the representative of the eventually zero sequence has a big string of $1$s initially. Why can't it be using your method that everyone guesses $1$? | |
| Jul 31 at 3:01 | comment | added | Daniel Weber | I think it is correct, because if two different suffixes agree then the sequence is eventually periodic, and that can just be handled separately | |
| Jul 31 at 2:57 | comment | added | Daniel Weber | Consider the equivalence relation on sequences defined by $\exists N,M \forall i, a_{N+i} = b_{M+i}$, and choose a representative from each congruence class. All people view sequences in the same class, so from some point their sequence must be a suffix of representative. Now I think everyone can find the first appearance of what they see as a suffix of the representative, and say what appears before it. However, I'm not sure if looking at the first appearance is correct | |
| Jul 31 at 2:44 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 9 characters in body
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| Jul 31 at 2:35 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
edited title
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| Jul 31 at 2:29 | history | asked | Joel David Hamkins | CC BY-SA 4.0 |