Timeline for Injective choice function for infinite complete linear hypergraphs
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2020 at 14:10 | vote | accept | Dominic van der Zypen | ||
| Nov 6, 2020 at 23:12 | answer | added | Louis D | timeline score: 2 | |
| Nov 6, 2020 at 18:50 | comment | added | bof | @LouisD I guess you were thinking the same as I was when I posted the same "counterexample" in a now deleted comment. | |
| Nov 6, 2020 at 18:39 | comment | added | Louis D | @bof Of course. Not sure what I was thinking when I wrote that. I now see what you wrote in your first comment, so yes I believe that it does hold if some edge has order 1. | |
| Nov 6, 2020 at 18:23 | comment | added | bof | @LouisD $E=V\cup\{V\}$ doesn't satisfy the condition $|e_1\cap e_2|=1$. | |
| Nov 6, 2020 at 17:51 | comment | added | Louis D | Ok, here is a link to my more general question mathoverflow.net/questions/375823/… | |
| Nov 6, 2020 at 17:35 | comment | added | Louis D | As bof points out, a projective plane with $|V|=\kappa$ will have $|E|=\kappa$, $|e|=\kappa$ for all $e\in E$, and $|\{e\in E: v\in e\}|=\kappa$ for all $v\in V$. In such a case, an injection can be constructed as described here mathoverflow.net/questions/331656/… However, this leads me to wonder about a more general question which I will start a new thread for in a moment. | |
| Nov 6, 2020 at 15:50 | comment | added | Louis D | I was about to write an answer, but then realized that I don't immediately see how to find the injection in the last case. So I will post the first part of my answer as a comment here: If you allow for edges of order 1, then then answer is no as seen by letting $E=V\cup \{V\}$. If every edge has order at least 2 and some edge has order exactly 2, then $H$ is a "near-pencil." In this case, you can clearly find a bijection. So that leaves the case where every edge has order at least 3, in which case $H$ is a (non-degenerate) projective plane and thus every edge has the same order. | |
| Nov 6, 2020 at 14:41 | comment | added | bof | However, if we assume that $V\notin E$ and that $|e|\gt2$ for all $e\in E$, then we can prove that there is an infinite cardinal $\kappa$ such that $|V|=|E|=\kappa$, and $|e|=\kappa$ for all $e\in E$, and $|\{e\in E:v\in e\}|=\kappa$ f0r all $v\in V$. Unless I'm making another mistake. | |
| Nov 6, 2020 at 14:34 | comment | added | bof | Oops, I was wrong. Your definition allows a hypergraph with vertices $$w,v_1,v_2,v_3,\dots$$ and edges $$\{v_1,v_2, v_3,\dots\},\ \{w,v_1\},\ \{w,v_2\},\ \{w,v_3\},\dots$$ which has an injective choice function. | |
| Nov 6, 2020 at 12:51 | comment | added | bof | If $V\in E$ then either $E=\{V\}$ or else $E=\{V,\{v\}\}$ for some $v\in V$, and the existence of an injective choice function is trivial. Suppose $V\notin E$ and $V$ is infinite. Then there is an infinite cardinal $\kappa$ such that $|V|=|E|=\kappa$, and there are $\kappa$ points on each line and $\kappa$ lines through each point. Then $E$ has an injective choice function, since it's a collection of $\kappa$ sets, each of cardinality $\kappa$, where $\kappa$ is an infinite cardinal. | |
| Nov 6, 2020 at 9:30 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |