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Defenition of "Recession Direction":
Assume that $S \subseteq R^n$ and $p$ is a point in $S$. We say vector $d \neq 0$ is a Recession Direction for $p$ if :
$\forall \lambda \ge 0 \space\space \space \space p+\lambda d \in S$

Theorem: If $S$ is a closed convex set and $d \neq 0$ is a Recession Direction for some(at least one) $p \in S$, then $d$ is a Recession Direction for all of the points in $S$. (Assume that this Theorem is proved to us.)

Question:

Prove that if $S$ is not closed, the theorem above doesn't hold anymore.(Give a counterexample)

My problem:
I don't fully understand what a closed convex set is. The definition of a convex set is clear to me. But, the definition of a closed set is not. So, because of that, I can't find the counterexample that the question wants.

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    @M.Winter excuse me i forgot to write a small thing ... now its corrected – 2017-02-20

1 Answers 1

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Take the open upper half plane $\mathbb R^2_+:=\{(x,y)\in\mathbb R^2\mid y>0\}$ and add the origin to obtain $S$:

$$ S:=\mathbb R^2_+\cup\{(0,0)\}.$$

All points have the recession direction $(1,0)$, except the origin. $\mathbb R^2_+$ is not closed as its boundary $\mathbb R\times \{0\}$ is not part of it.