How to prove that polyhedron is a closed set? [closed]

$\begingroup$

Given the definition of polyhedron, $\{x \in R^n : Ax \leq b\} $

$\endgroup$ 1

1 Answer

$\begingroup$

For $(a_i)_{i=1}^n \in \mathbb{R}^n$ and $b \in \mathbb{R}$

Is $\{(x_i)_{i=1}^n \in \mathbb{R}^n|\sum_{i=1}^na_ix_i \leq b \}$ a closed set ?

Remark : I have noted that some of the litterature on polyhedrons, and more generally operations research, has very complicated proofs for results that are very simple given a little topology. They often have good reasons to do it, like introducing some algorithm, but it has a "proof complexity" cost. Best be aware of this.

$\endgroup$