Timeline for Minimal number of edges for complete linear hypergraphs

10 events

when toggle format	what		by	license	comment
Nov 23, 2016 at 11:37	vote	accept	Dominic van der Zypen
Nov 22, 2016 at 20:54	history	edited	Pat Devlin	CC BY-SA 3.0	added 22 characters in body
Nov 22, 2016 at 20:52	comment	added	Pat Devlin		By the way, I agree that there cannot be ${n \choose 2}$ edges. But there can be ${L \choose 2}$ points.
Nov 22, 2016 at 20:48	history	edited	Pat Devlin	CC BY-SA 3.0	added 400 characters in body
Nov 22, 2016 at 20:42	comment	added	Pat Devlin		Here's another way to get the bound $n = {L \choose 2}$ that's more geometric. Namely, just take $L$ lines in the plane that are in general position (i.e., no three intersect in a point and no two are parallel). [This is what happens if you draw $L$ lines at random.] Only consider the $n = {L \choose 2}$ points where they intersect. Then each line has $L-1$ points on it, each point is on exactly $2$ lines, and any two points determine a line.
Nov 22, 2016 at 18:38	comment	added	Dominic van der Zypen		It is wrong that there can be $n\choose 2$ edges, see DeBruijn Erdos theorem. And it turns out I dont understand your lower bound either
Nov 22, 2016 at 17:51	comment	added	Pat Devlin		And it seems like you don't need the condition $\|e_i\| \geq 2$.
Nov 22, 2016 at 6:21	vote	accept	Dominic van der Zypen
Nov 22, 2016 at 18:36
Nov 22, 2016 at 1:45	comment	added	Pat Devlin		In fact, the proof also shows ${L \choose 2} \geq \sum_{i=1} ^{n} {\|V_i\| \choose 2} \geq n$.
Nov 22, 2016 at 0:24	history	answered	Pat Devlin	CC BY-SA 3.0