Timeline for Stronger negation of AC given by rejecting "infinite hat" puzzles
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3 at 19:09 | comment | added | Ingo Blechschmidt | In the meantime, Elliot has wrote up the strategy for the $\mathcal{P}(\mathbb{R})$ here: arxiv.org/abs/2211.10474 Very insightful! | |
| Feb 2, 2019 at 4:29 | vote | accept | Mike Battaglia | ||
| Feb 1, 2019 at 7:45 | comment | added | Elliot Glazer | Let us continue this discussion in chat. | |
| Feb 1, 2019 at 7:28 | comment | added | Mike Battaglia | For example, if $X_j=X_1$, we could have $s^1 = (\omega, \omega+2, \omega+4, ...)$ and $s^2 = (\omega^2, \omega^2+2, \omega^2+4,...)$, or something like that, as long as $F$, when evaluated at each ordinal in each of those lists, eventually yields identical results for both lists. If that is correct, my question is - why should we ever think there will be $s^1, s^2$ that $F$ agrees on in such a way? (2/2) | |
| Feb 1, 2019 at 7:27 | comment | added | Mike Battaglia | Ok, so it seems that $X_1$ will start with $(0,2,4,6,...)$ and $X_2$ with $(1,3,5,7,...)$, or something like that. Then it seems you want to find the first two sequences $s^1$ and $s^2$ of order type $\omega$, within $X_j$, such that $F \restriction s^1 = F \restriction s^2$ eventually agree after some differing initial segment. (1/2) | |
| Feb 1, 2019 at 7:07 | comment | added | Elliot Glazer | It's Cartesian product. I'm splitting $X$ into two interleaved subsequences, which are in turn split into $|X|$ many $\omega$-sequences. | |
| Feb 1, 2019 at 7:04 | comment | added | Mike Battaglia | Ok, I got tripped up at the part where you say $X \cong 2 \times \omega \times X$. Does $\times$ refer to ordinal multiplication? I thought you were splitting $X$ into two interleaved subsequences, but I may be confused here. | |
| Feb 1, 2019 at 6:42 | comment | added | Elliot Glazer | The variant in the linked post is slightly different. Rather than there being one box for every set of reals (i.e. the index set is $\mathcal{P}(\mathbb{R})$), I have the index set $X$ be the Hartogs number of $\mathbb{R}.$ I was just describing how to construct this set. | |
| Feb 1, 2019 at 6:38 | comment | added | Mike Battaglia | Sorry, writing comments at the same time - I thought about using $\Bbb R$ as an index set instead of $\Bbb N$, or to let there be arbitrary index sets, but wasn't certain how that would make anything stronger. But your version seems to use well-ordered subsets of $\Bbb R$, "modded out by isomorphism" - what do you mean by that last part? | |
| Feb 1, 2019 at 6:31 | comment | added | Elliot Glazer | Is it strong enough to rule out the variant I describe in the second paragraph? My answer isn't about the standard 100 mathematician puzzle. | |
| Feb 1, 2019 at 6:29 | comment | added | Mike Battaglia | Hi Elliot - if you look at the way I use interpolators, it's strong enough to rule out both the usual infinite hat puzzle and the 100 mathematician variant, since both depend on an interpolator $f$ that is guaranteed to make perfect guesses for some missing value eventually as the index goes to infinity. My negation was stronger than just saying no "eventually perfect" interpolator exists, it was saying no interpolator should be guaranteed for all sequences to give correct guesses with natural density greater than 1/|S| for finite S, or produce any correct guesses at all for uncountable S. | |
| Feb 1, 2019 at 6:16 | history | answered | Elliot Glazer | CC BY-SA 4.0 |