Let $Y$ be a closed and convex subset of $\mathbb{R}^n$ such that $ 0\in Y$ and $Y \cap \mathbb{R}^{n}_{+}=\{0\}$. If for $\bar{y}\in Y$ there does not exist $y \in Y$ such that $y>\bar{y}$, then there exist $p \in \mathbb{R}^n$, $p>0$ such that $p\bar{y}\geq py$ for all $y\in Y$.
I tried to prove as follows:
Note $\bar{y}\notin int(Y)$, otherwise $\exists\varepsilon>0$ such that $N_\varepsilon (\bar{y})\subset Y$ and this means that $\exists y'\in Y$ such that $y'>\bar{y}$, a contradiction. The Supporting Hyperplane Theorem assure that there exists $p\in \mathbb{R}^n$, $p\neq 0$ such that $p\bar{y}\geq py, \forall y\in Y$.
To finish the proof I just have to show that $p>0$. Is that even true? I have the intuition that if $p_i<0$ for some $i$, there would be another element of $y\in Y$ such that $y>\bar{y}$, but I can't formalize the argument.
