Given a hyperplane $\alpha^T x = \beta$ in $\mathbb R^n$, with $\beta > 0, \alpha\_i > 0$ **for all** $i \in [n]$. Then for any $\{v^i\} \subseteq \{x \in \mathbb Z^n\_+ \mid \alpha^T x = \beta\}$, it's obvious to see that there must have: $\{v\_i\}$ forms an antichain with respect to the component-wise order. My question is, for a given set of less than $n$ positive integer vectors, to guarantee the existence of a hyperplane $\alpha^T x = \beta$ containing all of these integer points with $\alpha\_ i > 0$ for all $i \in [n]$, is the antichain condition also sufficient?


Formally speaking:


Given an antichain $\{v^i\}\_{i \in[d]} \subseteq \mathbb Z^n\_+$ with $d< n$. (Here antichain is with respect to the component-wise order: for any $i \neq j \in [d],$ there exists $t\_1, t\_2 \in [n],$ such that $ v^i\_{t\_1}>v^j\_{t\_1}, v^i\_{t\_2}<v^j\_{t\_2}$.) Then: does there always exist a hyperplane $\alpha^T x = \beta$ containing all these integer points, and $\alpha\_i > 0$ for any $i \in [n]$.



Existence of a hyperplane with strictly positive coefficients to contain an antichain in $\mathbb{Z}^n_+$
Here is a counterexample. Consider vectors $v^1 = (4,4,1,1)$, $v^2 = (1,1,4,4)$, and $v^3 =(3,3,3,3)$. They form an antichain, as required. Further, $d = 3 < n =4$. However, there is no hyperplane $\{x:\alpha x = \beta\}$ with all $\alpha\_i > 0$ that contains all vectors $v^i$.


**Discussion.** It's not hard to see that the required hyperplane exists if and only if the affine span $L$ of vectors $\{v^i\}$ is an antichain (that is, the difference between any two distinct vectors in $L$ is not in the orthant $Q\_- = \{x: x\_i \leq 0 \text{ for all } i\}$). In one direction the proof is immediate: If all $v^i$ lie in a hyperplane $H = \{x:\alpha x = \beta\}$ (with all $\alpha\_i > 0$), then $L \subset H$. Therefore, every distinct $u, v\in L$ must be incomparable. In the other direction, the criterion follows from duality.


In the counterexample above, points $(5/2,5/2,5/2,5/2) = \frac{(4,4,1,1)+ (1,1,4,4)}{2}$ and $(3,3,3,3)$ are in $L$. But the former is less than the latter in each coordinate.


