Timeline for Induced subgraphs of the almost-disjointness graph
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2021 at 13:31 | comment | added | Will Brian | @bof: Hopefully my answer clears things up. It's a good question whether we can realize $K_{\aleph_1} \cup K_{\aleph_1}$ with a really "simple" family of sets (instead of using some kind of recursion). | |
| May 15, 2021 at 8:47 | comment | added | bof | @WillBrian I don't understand that "saturated" business, could you illustrate with a simple concrete example? How would you embed $K_{\aleph_1}\cup K_{\aleph_1}$, the union of two vertex-disjoint uncountable complete graphs? (Kicking myself in advance, as I'm sure I'm missing something obvious.) | |
| May 14, 2021 at 21:07 | comment | added | Will Brian | I think $|V| \leq \aleph_1$ should always be possible (building the graph recursively, using the fact that $\mathcal P(\omega) / \mathrm{fin}$ is countably saturated). So (if I'm right) a yes answer should at least be consistent. But I don't see how to get all graphs of size $\leq 2^{\aleph_0}$ when CH fails. | |
| May 14, 2021 at 19:33 | comment | added | Dominic van der Zypen | Thanks - this is already very helpful! I'll wait for a couple of days to see whether someone can answer for graphs with $|V|> \aleph_0$ and $|V|\leq 2^{\aleph_0}$ and then accept yours if it remains the only one. Hope that is ok with you. | |
| May 14, 2021 at 15:29 | history | answered | Louis D | CC BY-SA 4.0 |