Timeline for How to deconstruct a sum of intersecting upsets
Current License: CC BY-SA 3.0
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| Sep 4, 2012 at 7:26 | comment | added | Sean Eberhard | I can at least answer that side question negatively, by taking $f$ to be the function on $P([4])$ taking the value $0$ on the sets of size $\leq 1$, $1/2$ on $2$-sets and $3$-sets, and $1$ on the $4$-set. Now check that the total mass on the $2$-level is greater than the total mass on the $3$-level, which is not true of any intersecting upset. There are in fact many more inequalities satisfied by intersecting upsets, such as $f(A_1)+\cdots+f(A_n)\geq f(B_0)+\cdots+f(B_n)$ whenever $B_k\cup B_{k+1}\subset A_{k+1}$ for all $k=0,\ldots,n−1$ and $B_0\cap B_n=\emptyset$. | |
| Aug 29, 2012 at 6:13 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |