Why recession cone of a set $C$ is indeed a cone?

Question

The recession cone of a set $C$, i.e., $R_C$ is defined as the set of all vectors $y$ such that for each $x \in C$, $x-ty \in C$ for all $t\geq 0$.

On the other hand, a set $S$ is called a cone, if for every $z \in S$ and $\theta \geq 0$ we have $\theta z \in S$.

To prove the statement one can write:

Let $w \in R_C$, then there exists a vector $x \in C$ such that $x-tw \in C$ for all $t\geq 0$. Also, $R_C$ is a cone if for $\theta \geq 0$, $\theta w \in R_C$, that is $\theta(x-tw)=\theta x-\theta tw \in R_C$. Naming $\theta t$ az $\beta$, one has $\theta x-\beta w \in R_C$. To conclude the proof one needs to prove that $\theta x \in C$ if $x \in C$ which requires $C$ itself is a cone but definition of recession cone does not claim this?

Answer 1

You need to show that if $y \in R_C$ then $\lambda y \in R_C$ for all $\lambda >0$.

If $y \in R_C$, and $x \in C$, then $x+t y \in C$ for all $t \ge 0$.

Choose $\lambda >0$. Then $x + {t \over \lambda} \lambda y \in C$ for all $t \ge 0$ and so $x + s (\lambda y) \in C$ for all $s \ge 0$. Hence $\lambda y \in R_C$.

$C$ does not need to be a cone. For example, if $C= \mathbb{R} \times \{1\} \subset \mathbb{R}^2$, then $R_C = \mathbb{R} \times \{0\} $.