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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 $i \in [n]$.

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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.

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  • $\begingroup$ Thank you so much for your explanation! It's really a straightforward and elegant proof. $\endgroup$
    – HAORAN ZHU
    Commented Dec 31, 2020 at 1:52
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    $\begingroup$ I just realized that your "if and only if" proof does not use the "antichain" assumption? I mean for the second direction in your proof, the separating hyperplane might not necessarily have $\alpha_i>0$ (strictly positive), right? $\endgroup$
    – HAORAN ZHU
    Commented Dec 31, 2020 at 4:03
  • $\begingroup$ Corrected the issue with strict/non-strict inequalities. $\endgroup$
    – Yury
    Commented Dec 31, 2020 at 17:03

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