Theorem 4.5 — each $\bar z\in N$ uniquely minimizes the lexicographic weighted Tchebycheff program for $\bar\lambda$ of (4.3)

Let $Z\subseteq\mathbb R^k$ ($k\ge1$) be a compact set of criterion vectors with nondominated set $N$, let $z^*$ be an ideal criterion vector for $Z$, and let $\bar z\in N$. Let $\bar\lambda$ be the weights of eq. (4.3) for $\bar z$. Then $\bar\lambda\in\bar\Lambda=\{\lambda\in\mathbb R^k\mid \lambda_i\ge0,\ \sum_i\lambda_i=1\}$, and $\bar z$ uniquely minimizes the lexicographic weighted Tchebycheff program

$\bar z$ is a lexicographic minimizer, and every lexicographic minimizer equals $\bar z$.

So every nondominated criterion vector is computable, uniquely, by the lexicographic weighted Tchebycheff program, whatever the shape of the feasible region.

Formalization Note The paper states "there exists a $\bar\lambda\in\bar\Lambda$"; its proof ends "for $\bar\lambda$ as defined in (4.3)", and the statement here exhibits that witness, which implies the existential form. Compactness of $Z$ is an added hypothesis. The paper assumes only $S$ bounded; its proof, in the case $\bar z_j=z^*_j$, needs every first-stage minimizer to lie below some nondominated vector, which compactness provides. Without closedness the witness (4.3) can fail: $Z=\{(5,3,0),(0,10,0),(0,0,6)\}\cup\{(5,0,4+t):0\le t<1\}$, $z^*=(5,10,6)$, $\bar z=(5,3,0)$, $\bar\lambda=(1,0,0)$ has no lexicographic minimizer. Closedness is needed by the theorem itself, not only by this proof: adding the points $(5-s^2,\,3+s,\,0)$, $0<s\le1$, to that $Z$ (still bounded, $\varepsilon=0$ still admissible) leaves $\bar z=(5,3,0)\in N$ a lexicographic minimizer for no $\lambda\in\bar\Lambda$, so Theorems 4.5 and 4.6 are false for a bounded, non-closed $Z$. The priority "$P_1<<<P_2$" is printed reversed; the two-stage meaning of p. 336 is encoded (see the definition IsLexMin). "Uniquely" refers to the criterion vector; $x$ and $\alpha$ are eliminated.