How to generate all order-relations consistent with an additive value-function? There is a finite set $X$. Each item $x\in X$ has a positive value, $v(x)$. Each value function $v$ induces an order-relation $\prec$ on the subsets of $X$:
$$ Y\prec Z \iff \sum_{y\in Y}v(y) < \sum_{z\in Z} v(z)$$
An order-relation on subsets of $X$ is called additive if there exists a positive value-function $v$ that induces it. For example, suppose $X := \{x,y\}$. Then the following order-relation is additive:
$$ \emptyset \prec \{x\} \prec \{y\} \prec \{x,y\} $$
bu the following order-relation is not:
$$ \emptyset \prec \{x\}\prec \{x,y\}  \prec \{y\} $$
MY QUESTION IS: given a set of items, what is a way to generate all the additive preference-relations over its subsets?
 A: I believe the following works . . .

First, some notation . . .

Let $X_0 = {\large{\varnothing}}$.

For each positive integer $n$,


*

*Let $X_n = \{1,...,n\}$.$\\[4pt]$

*Let $P_n$ denote the power set of $X_n$.$\\[4pt]$

*Let $Q_n = \{S \in P_n \mid n \in S\}$.


The goal is to find all additive orders on $X_n$ (i.e., a preference ordering of $P_n$).

We'll say an additive order on $X_n$ has "ascending singletons" if
$$1 < 2 < \cdots  < n$$
or more precisely, $i < j \implies \{j\}$ is preferred to $\{i\}$.

For simplicity (and without loss of generality), we only consider additive orders with ascending singletons (so in what follows, the term "additive order" means additive order with ascending singletons).

Determine the additive orders recursively.

To determine a unique additive order on $X_n$, first choose an additive order on $X_{n-1}$, retaining only the subset comparisons, not any particular positive valuation which might have induced the order.

The chosen additive order on $X_{n-1}$ forces a unique preference order on $Q_n$.

Thus, at this point, the preference order is already determined for pairs $(A,B)$ with 


*

*$A,B \in P_{n-1}$.$\\[4pt]$

*$A,B \in Q_n$.$\\[4pt]$

*$A \in P_{n-1}$, $B \in Q_n$, and $A \subset B$.$\\[4pt]$


We need the concept of "domination" (forced preference) . . .

It's recursive.

Given subsets $A,B$ of $X_n$, we say $B$ dominates $A$ if


*

*$B$ is already known to be preferred to $A$.

or, assuming there's not yet any known preference for $A$ vs $B$:


*if $A$ can be each be partitioned as $A = A_1 \cup A_2$, and $B$ can be partitioned as $B = B_1 \cup B_2$ where $B_1$ dominates $A_1$ and $B_2$ dominates $A_2$.


If $B$ dominates $A$, then we force $B$ to be preferred to $A$ (if that was not already the case).

Stated without proof, here's what I think is true . . .

Merge the orders of $P_{n-1}$ and $Q_n$, retaining the internal orders of $P_{n-1}$ and $Q_n$, while making sure that in the merge, preferences forced by the the principle of domination are respected. Moreover, as each element of $B$ is inserted, the domination relation may need to be updated.

Claim: Any choice of such a merge yields a unique additive order on $X_n$, and every additive order on $X_n$ which is consistent with the chosen additive order on $X_{n-1}$ can be obtained by this process.
