combinatorial and linear duality

Question

Let $S$ be a finite set, and let $W$ be a nonempty set of subsets of $S$; we will identify every subset of $S$ with its characteristic function, a 0-1 vector in $\mathbb R^S$. The combinatorial dual $\hat W$ of $W$ is the set of all subsets of $S$ that have non-empty intersection with every subset of $W$. The positive linear dual, for our purposes, is the set $W^\perp$ of all $v \in \mathbb R^S$ with non-negative entries such that $v \cdot w \ge 1$ for all $w \in W$. The convex production of a finite set $F \subset [0, \infty)^S$ is the set of vectors of the form $v + w$, where $v$ is in the convex hull of $F$, and $w \in [0, \infty)^S$. My first question is

Is $W^\perp$ always the convex production of $\hat W$?

It is clear that the latter is a subset of the former.

My second question is

Is the distance the origin of $W^\perp$ equal to the distance to the origin of the convex hull of $\hat W$?

Of course an affirmative answer to the first question would imply an affirmative answer to the second.

This question is motivated by a related question in the theory of combinatorial extremal length.

Answer 1

No, let $S=\{1,2,3\}$, let $W$ consist of the "large" subsets of $S$, i.e., the ones of cardinality 2 or 3. Then $\hat W=W$. Let $v=(1/3,2/3,2/3)$. Then $v\in W^\bot$. But any vector in the convex production of $\hat W$ must lie above the half plane $x+y+z\ge 2$, which $v$ does not.

Answer 2

Here is a counterexample to both: Let $W$ be the $2$-element subsets of a $4$-element set. The combinatorial dual is the collection of subsets of size $3$ or $4$. $W^\perp$ contains $(1/2, 1/2, 1/2, 1/2)$ which has coordinate sum $2$, while every element of $\hat W$ has coordinate sum at least $3$. The distance from the origin to the convex hull of $\hat W$ is $3/2$, while the distance to $(1/2,1/2,1/2,1/2)$ is $1$.