Timeline for A strengthening of Frankl's union-closed sets conjecture?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2016 at 3:23 | vote | accept | bof | ||
| Jan 14, 2016 at 17:50 | answer | added | Ilya Bogdanov | timeline score: 5 | |
| Jan 11, 2016 at 12:56 | comment | added | bof | @GordonRoyle If you allow the empty set as a member of the family, yes. For no particular reason, I stated Frankl's conjecture in the equivalent form with nonempty sets. | |
| Jan 11, 2016 at 12:51 | comment | added | Gordon Royle | Doesn't Frankl's conjecture just ask for an element in at least half the sets? (That is, not strictly more than half.) | |
| Jan 11, 2016 at 9:43 | comment | added | joro | This is my mistake, I misread. | |
| Jan 11, 2016 at 9:19 | comment | added | joro | I misunderstood, sorry. Indeed mine is not counterexample. | |
| Jan 11, 2016 at 8:57 | comment | added | joro | OK. $|\mathcal P(\{1,2\})|=2^2=4$. If you don't count the empty sets it makes it $3$. You divide by $2$. | |
| Jan 11, 2016 at 8:46 | comment | added | joro | What has $\mathcal P(\{1,2\})$ to do with my claim? | |
| Jan 11, 2016 at 8:41 | comment | added | joro | Maybe I misunderstood, but there 8 sets in the power set. $3$ is in half of the sets and not in the other half, so you have equality. If you don't count the empty set, this make $3$ not in $3$ sets and your inequality still doesn't hold. | |
| Jan 11, 2016 at 7:37 | comment | added | joro | Isn't the powerset of {1,2,3} counterexample to both? | |
| Jan 11, 2016 at 6:52 | history | edited | bof | CC BY-SA 3.0 |
added 7 characters in body
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| Jan 11, 2016 at 6:33 | history | asked | bof | CC BY-SA 3.0 |