Let $V^n$ be $n$-dimensional vector space over $\mathbb{Q}$ with some fixed basis. If $a, b \in V^n$, $a = (a_1, \dots, a_n)$ and $b = (b_1, \dots, b_n)$, I'll say $a \leq b$ if $a_i \leq b_i$ for every $1 \leq i \leq n$.
Suppose $v \in V^n$ is some vector, $v > 0$ and $U \subseteq V^n$ is a subspace, generated by $u^1, \dots, u^m$, with every $u^k \geq 0$. Is it true that there is a finite set $P \subset U$, such that for every $p \in P$ we have $v \leq p$ and if for any $u \in U$ $v \leq u$ then there is some $p \in P$ such that $p \leq u$?
This is trivial if $v \in U$ of course, but if it's not I can't prove it no matter what I do. Is that even true?
Upd: edited to ensure $v$ and the basis vectors of $U$ are positive.
Here is a counterexample. Let $n=3$ and let $U$ be spanned by $(1,0,1)$ and $(0,1,1)$, and let $v=(0,0,1)$ (so $U$ consists of all $(a,b,c)$ such that $a+b=c$). Then for any $t\in[0,1]$, $(t,1-t,1)$ is a minimal element of $U$ greater than $v$: you can't decrease the third coordinate while staying above $v$, and if you decrease the first or second coordinates then you'll no longer be in $U$. Thus any $P$ would have to contain all of these points and so could not be finite.