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Nov 23, 2016 at 11:37 vote accept Dominic van der Zypen
Nov 22, 2016 at 20:45 comment added Pat Devlin However, because any two edges intersect in $1$ point, we do in fact get $|E| \leq n$ by Fisher's inequality.
Nov 22, 2016 at 20:39 comment added Pat Devlin A few remarks. (1) The De Bruijn-Erdos theorem isn't quite applicable here because [a] that theorem is about geometric line configurations {not about hypergraphs} and [b] that theorem is a statement about the collection of ALL lines containing at least two points. (2) In either case, the theorem would actually imply $|E| \geq n$.
Nov 22, 2016 at 6:21 vote accept Dominic van der Zypen
Nov 22, 2016 at 18:36
Nov 22, 2016 at 0:24 answer added Pat Devlin timeline score: 3
Nov 21, 2016 at 19:14 answer added Aaron Meyerowitz timeline score: 1
Nov 21, 2016 at 14:30 answer added domotorp timeline score: 1
S Nov 21, 2016 at 14:11 history suggested Ali Taghavi
I add a tag
Nov 21, 2016 at 13:50 review Suggested edits
S Nov 21, 2016 at 14:11
Nov 21, 2016 at 13:46 history asked Dominic van der Zypen CC BY-SA 3.0