Timeline for Strongly minimal covers
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2022 at 5:46 | comment | added | domotorp | @Tri You are right, in this new version (0,1) will also be a clique! | |
| May 14, 2022 at 19:55 | comment | added | Tri | @domotorp Let $a_1$, $a_2$, $a_3$, ... be the terms of a convergent series of positive numbers that converges to, say, 1/1729. Let $X_1=(0,1)$. Let $X_2=(1-a_1,2-a_1)$. Let $X_3=(2-a_1-a_2,3-a_1-a_2)$. ... You get the idea. I think this is a strongly minimal cover because if you remove, say, 42 of the $X$'s, they overlap so little, you have to replace them with at least 42 sets. I have not proven this. | |
| May 14, 2022 at 19:46 | comment | added | domotorp | @Tri Isn't my example such? In the graph $G$ connect two vertices by an edge if and only if their distance is less than 1. The maximal edges will be exactly the open unit intervals. | |
| May 12, 2022 at 20:22 | comment | added | Tri | What about an example of a hypergraph without a strongly minimal cover when the hypergraph arises in the following way? Let $G=(V,E)$ be a graph with vertex set $V$ and edge set $E$. Let the hypergraph $H$ have vertex set $V$ and edge set the set of cliques of $G$. | |
| May 7, 2022 at 23:55 | comment | added | Tri | The family $\big\{\{r\}\mid r\in\mathbb R\big\}$ is a minimal cover. | |
| May 7, 2022 at 17:43 | vote | accept | Dominic van der Zypen | ||
| Jan 9, 2015 at 8:00 | vote | accept | Dominic van der Zypen | ||
| May 7, 2022 at 17:42 | |||||
| Jan 9, 2015 at 7:52 | comment | added | domotorp | @Noah: See the update. | |
| Jan 9, 2015 at 7:51 | history | edited | domotorp | CC BY-SA 3.0 |
added 474 characters in body
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| Jan 9, 2015 at 5:03 | comment | added | Włodzimierz Holsztyński | At least one can modify @domotorp example by considering rationals only: $\ V:=(0;\infty)\cap\mathbb Q);\ $ and it's enough to consider as maximal edges the intervals $\ (a;a+1)\cap\mathbb Q\ $ such that $\ a>0\ $ and $\ a\in \mathbb Q.\ $ Thus domotorp's example becomes countable with countable many maximal edges. | |
| Jan 9, 2015 at 3:34 | comment | added | Noah Schweber | What happens if we require every element of $E$ to be finite? Then nothing like this works, but it's not clear that there isn't some other example. | |
| Jan 9, 2015 at 2:37 | comment | added | Włodzimierz Holsztyński | Nice! (A set $\ V:=(0;\frac 32)\ $ would be a smaller example :-). | |
| Jan 8, 2015 at 23:08 | history | answered | domotorp | CC BY-SA 3.0 |