Timeline for Preserve unbounded sets between different cofinality
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 29, 2020 at 11:43 | comment | added | Hanul Jeon | I get it, I confuse the topological closure and the $F$-closure under the given set. | |
| Apr 29, 2020 at 11:41 | comment | added | Holo | @HanulJeon $F(\xi)$ is a fixed value, but the closure of $F[\kappa\setminus\lambda]$ under $F$ will be $\bigcup\{F(\xi),F(F(\xi)),...\}$, which will be unbounded(where $\xi$ is some arbitrary value $>\lambda$) | |
| Apr 29, 2020 at 11:21 | comment | added | Holo | "fix some cofinal $(λ_i\mid i\in\omega)$, and for $k$ be the minimal $k$ such that $x∈λ_k$, map $x$ to some element of $\lambda_{k+1}\setminusλ_k$" If $\kappa=ω_{ω+1}$ and $λ=ω_ω$, send all elements in $\kappa\setminusλ$ to some $x$. If $x\inω$, then $F(x)∈ω_1$, if $x∈ω_1$ then $F(x)∈ω_2$ etc. Then the closure of every subset of $λ$ under $F$ is cofinal | |
| Apr 29, 2020 at 10:03 | comment | added | Holo | @HanulJeon but we are looking at the closure of the image, not just the image itself | |
| Apr 28, 2020 at 17:23 | comment | added | Holo | @JohannesSchürz doesn't similar thing works for $\kappa$ weakly compact? Regardless on $λ$ | |
| Apr 28, 2020 at 16:33 | comment | added | Johannes Schürz | If $\kappa$ is measurable and $\lambda$ regular such a function cannot exist. Assume the contrary. W.l.o.g. let $n=2$. Then we can define the function $G(\{\alpha,\beta\}):=\max(F((\alpha,\beta)), F((\beta,\alpha)))$. But there exists a measure 1 set $a$ such that $G\restriction [a]^2$ is constant. Therefore, $F$ cannot be preserving unbounded. | |
| Apr 28, 2020 at 16:21 | answer | added | Andreas Blass | timeline score: 4 | |
| Apr 28, 2020 at 16:07 | history | edited | Holo | CC BY-SA 4.0 |
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| Apr 28, 2020 at 14:58 | history | asked | Holo | CC BY-SA 4.0 |