Solution
= Solution
\b[Yes: this is a <monotone set function>.] Assume $V\ne\varnothing$, as required by its definition at the <empty set>. If $\varnothing\ne A\subseteq B$, every weight considered in the maximum over $A$ is considered in the maximum over $B$, hence $f(A)\leq f(B)$. If $A=\varnothing$ and $B\ne\varnothing$, the minimum over $V$ is at most every weight in $B$, hence again $f(A)\leq f(B)$. No sign condition on the weights is needed.