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Applying down-shifts until none changes the family terminates, and the result is closed under taking subsets
Statement
Starting from a finite family and repeatedly applying effective down-shifts eventually stops. The final family is closed under taking subsets.
Facts & Assumptions
Given: a finite family .
The weight is a natural number (The down-shift of a set family at a point ).
An effective shift strictly decreases the weight and preserves the number of sets (, and with equality only when ).
Proof
Every effective shift strictly decreases the natural number by [L1]. Therefore there cannot be an infinite sequence of effective shifts, so the process terminates.
Let be a family on which every down-shift is ineffective. If and , then the definition of an ineffective shift forces .
By repeatedly removing one element at a time and using step 1.2, every subset of every member of also lies in . So the terminal family is downward closed.
Remarks
- The proof spends no order on the points of beyond the ability to choose which shift to apply next; any effective shift decreases the same weight.
Depends on
- The down-shift $S_j$ of a set family at a point $j$
- $\lvert S_j(\mathcal{F})\rvert=\lvert\mathcal{F}\rvert$, and $w(S_j(\mathcal{F}))\le w(\mathcal{F})$ with equality only when $S_j(\mathcal{F})=\mathcal{F}$
- A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$
- The natural numbers $\mathbb{N}$ (von Neumann)
- Order on the natural numbers
Used by
Dependency tree · two levels
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Sources
- L. Babai and P. Frankl, Linear Algebra Methods in Combinatorics, §7.5 (standard reference, not scraped)