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Let us consider a polyhedron in $\mathbb{R}^n$ (in this context it must NOT be bounded) $\mathcal{P} = \{ x: A \cdot x \leq c\}$ for some matrix $A$. Let $\mathcal{I} = \{ x: A \cdot x < c\}$ be a non-empty set. I want to prove that the closure of $\mathcal{I}$ is $\mathcal{P}$. The inclusion $\overline{\mathcal{I}} \subset \mathcal{P}$ is trivial. Probably also the other one, but maybe there is some standart procedure to prove this, without any strange reasoning. My idea is to take $Q \in \mathcal{P}$ and any $R \in \mathcal{I}$. Then I would take the line from $R$ to $Q$ and try to prove that it is entirely contained in $\mathcal{I}$, save for the point $Q$. This would prove the statement.

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  • $\begingroup$ I'd rather call your set: a polyhedron. $\endgroup$ – dohmatob Sep 2 '15 at 17:53
  • $\begingroup$ you are right, I edited, though saying that it can be unbounded sort of clears the definition anyhow. Any suggestions for the proof? $\endgroup$ – Kore-N Sep 3 '15 at 7:59
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So here is the proof I managed to get. I´m not sure it´s the best there is. Take $Q$ and $R$ as above. Now let $T \in [QR]$ be a point in the segment of the line from $Q$ to $R$, with $T \neq Q,R$. Suppose $T$ is not in $\mathcal{I}$. Than some of the inequalities has to be an equality. Suppose $A_1 \cdot T -c_1= 0$, and let us call $r$ the line defined by $A_1 \cdot x -c_1= 0$, and $s$ the line passing through $Q$ and $R$. These lines have intersection point $T$, so either they are identical (which is impossible because $P$ does not lie on $r$), or they have only one intersection in $T$. This is impossible aswell, because we would find $P$ in the wrong halfspace.

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