Timeline for Bounds for a covering number of the circle group $\mathbb T$ by some its small subgroups
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2021 at 14:24 | answer | added | Will Brian | timeline score: 3 | |
| Sep 18, 2021 at 13:44 | comment | added | Will Brian | I think the same idea can be used to show $\mathfrak{x} \leq \mathfrak{r}$. Roughly: instead of finding a convergent subsequence (given to you by $\mathfrak{r}_\sigma$), you just need to find a subsequence such that all subsequential limits are trapped within $\varepsilon/2$ of each other (which $\mathfrak{r}$ does). Then taking differences as in your previous comment, you get a subsequence where all subsequential limits are within $\varepsilon$ of $1$. This allows you to dodge any remote set. (Sorry for all the commenting -- if I have time today, I'll try to turn some of this into an answer.) | |
| Sep 18, 2021 at 13:34 | comment | added | Will Brian | Aha! That's a nice idea. | |
| Sep 18, 2021 at 13:28 | comment | added | Alex Ravsky | @WillBrian The key idea is for every $R\in\mathcal R$ pick a sequence $(r^R_n)_{n\in\omega}\in R^\omega$ such that $u^R_n=r^R_{n+1}-r^R_{n}<r^R_{n+2}-r^R_{n+1}$ for each $n\in\omega$. | |
| Sep 18, 2021 at 12:24 | comment | added | Will Brian | Perhaps I'm missing something, but I don't see why $\mathrm{cov}(\mathcal{A}(\mathbb T)) \leq \mathfrak{r}_\sigma$. The characterization of $\mathfrak{r}_\sigma$ you mention gives you an increasing sequence $(u_n)_{n \in \mathbb N}$ such that $(z^{u_n})_{n \in \mathbb N}$ converges, but you have no control over what this sequence converges to. Your definition of $\mathrm{cov}(\mathcal{A}(\mathbb T))$ requires that it converge to $1$. How do you do this? | |
| Sep 18, 2021 at 12:13 | comment | added | Alex Ravsky | @WillBrian Thanks for your suggestions. We added our proof of the bound $\mathfrak x\ge \mathrm{cov}(\mathcal M)$. | |
| Sep 18, 2021 at 12:12 | history | edited | Alex Ravsky | CC BY-SA 4.0 |
Added a proof of the bound $\mathfrak x\ge \mathrm cov(\mathcal M)$.
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| Sep 18, 2021 at 10:54 | comment | added | Will Brian | This shows (modulo my claim about Cohen-generic $z$'s) that $\mathfrak{x} \geq \mathrm{cov}(\mathcal M)$. Similarly, if you can show that a "random" $z$ has the same property, then this would show $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}$ also. | |
| Sep 18, 2021 at 10:53 | comment | added | Will Brian | My first thought is to consider what a "generic" $z$ does. For each $z \in \mathbb T$, define $R_z = \{n :\, |z^n-1| \geq 1/2 \}$. (Of course the "1/2" could be made smaller, but I doubt it matters.) I don't have a proof right now, but it seems like a Cohen-generic $z$ should have the property that $R_z \cap A$ is infinite for any infinite $A$ in the ground model. In other words: given $A$, there are only a meager set of $z$'s with $R_z \cap A$ finite. This means you need at least $\mathrm{cov}(\mathcal M)$ $A$'s to get a sufficiently large collection $\mathcal F$. . . . | |
| Sep 18, 2021 at 8:12 | history | edited | Alex Ravsky | CC BY-SA 4.0 |
deleted 5 characters in body
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| Sep 18, 2021 at 7:56 | history | asked | Alex Ravsky | CC BY-SA 4.0 |