Timeline for Unbounded $\omega_1$-sequence in $^*\mathbb{N}$
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Apr 18 at 18:48 | vote | accept | Sergey Grigoryants | ||
| Sep 16, 2020 at 16:03 | answer | added | nombre | timeline score: 4 | |
| Sep 8, 2020 at 12:18 | comment | added | YCor | I'm maybe too hasty. Possibly one should first replace $m\mapsto u_n(m)$ with $\min(u_n(m),m2^{-n})$. | |
| Sep 8, 2020 at 11:46 | comment | added | Sergey Grigoryants | @YCor, Sorry, maybe I don't understand, but I don't think this construction is working. For example, let's take $$u_n(k) = \begin{cases} 2^{2k},\quad k = n \\ \floor{\sqrt{k}},\quad else \end{cases}$$ In this case, $u(m) = 1 + 2^m$, hence $u(m) \neq o(m)$. | |
| Sep 7, 2020 at 21:32 | comment | added | YCor | Let $(u_n)$ be a sequence of such sequences. I guess: set $u(m)=1+\max_{k\le m}\lceil 2^{-k}u_k(m)\rceil$ satisfies $u\ge 2^{-k}u_k$ for all $k$ and $u(m)=o(m)$ when $m\to\omega$, so if $v(m)=m/\sqrt{m/v(m)}$, then it should work. | |
| Sep 7, 2020 at 16:52 | comment | added | YCor | Let $(u_n)$ be a sequence of such sequences. I guess something like $u(m)=\max_{k\le m}\lceil 2^{-k}u_k(m)\rceil$ should work. | |
| Sep 7, 2020 at 16:05 | comment | added | Sergey Grigoryants | @Ycor thank you now I see the whole point. But how to prove, that set of $n \in *\mathbb{N}$ such that $\frac{n(m)}{m} \to 0$ has uncountable cofinality? | |
| Sep 7, 2020 at 15:59 | comment | added | YCor | By "$X$ has uncountable cofinality" I mean "for every countable subset $Y\subset X$ there exists $x\in X$ such that $\forall y\in Y$, $y<x$". | |
| Sep 7, 2020 at 15:51 | comment | added | Emil Jeřábek | You seem to be misreading cofinality as cardinality. | |
| Sep 7, 2020 at 15:45 | comment | added | Sergey Grigoryants | Sorry, maybe I don't understand, but this claim is not true for arbitrary totally ordered uncountable set. For example, there is no strictly increasing $\omega_1$ sequence of reals. | |
| Sep 7, 2020 at 15:25 | comment | added | YCor | If $X$ is a totally ordered set of uncountable cofinality, define (with choice) by transfinite induction $x_\alpha$ as some element $>x_\beta$ for all $\beta<\alpha$, for $\alpha\in\omega_1$. | |
| Sep 7, 2020 at 15:23 | comment | added | Sergey Grigoryants | @Ycor thank you for your first note. I removed the redundant condition. But how uncountable cofinality implies the existence of strictly increasing $\omega_1$ sequence? | |
| Sep 7, 2020 at 15:16 | history | edited | Sergey Grigoryants | CC BY-SA 4.0 |
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| Sep 7, 2020 at 14:39 | comment | added | YCor | The question seems to be whether every sequence is unbounded, which is purportedly negatively answered here. Isn't the question rather whether some sequence is unbounded (which would better fit the title)? | |
| Sep 7, 2020 at 14:35 | comment | added | YCor | The set of $n\in{^*\mathbb{N}}$ such that $\lim_{m\to\omega} n(m)/m=0$ is convex but has uncountable cofinality. So there exists $(n_\alpha)$ satisfying (1) such that each $n_\alpha$ is less than the sequence $m\mapsto m$. | |
| Sep 7, 2020 at 14:29 | comment | added | YCor | I don't think (2) will play any role (if $u$ satisfies (1), it's easy to produce $u'$ satisfying (1) and (2), and such that $u$ is bounded iff $u'$ is bounded). | |
| Sep 7, 2020 at 14:25 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 7, 2020 at 14:08 | history | asked | Sergey Grigoryants | CC BY-SA 4.0 |