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recession cone
Let $C$ be a convex set in $\mathbb{R}^{n}$. If $C$ is bounded, then for any $x\in C$, any ray emanating from $x$ will eventually “exit” $C$ (that is, there is a point $z$ on the ray such that $z\notin C$). If $C$ is unbounded, however, then there exists a point $x\in C$, and a ray $\rho$ emanating from $x$ such that $\rho\subseteq C$. A direction $d$ in $C$ is a point in $\mathbb{R}^{n}$ such that for any $x\in C$, the ray $\{x+rd\mid r\geq 0\}$ is also in $C$ (a subset of $C$).
The recession cone of $C$ is the set of all directions in $C$, and is denoted by denoted by $0^{{+}}C$. In other words,
$0^{{+}}C=\{d\mid x+rd\in C,\ \forall x\in C,\ \forall r\geq 0\}.$ 
If a convex set $C$ is bounded, then the recession cone of $C$ is pretty useless; it is $\{0\}$. The converse is not true, as illustrated by the convex set
$C=\{(x,y)\mid 0\leq x<1,\ y\geq 1\}\cup\{(x,y)\mid 0\leq x\leq 1,\ 0\leq y\leq 1\}.$ 
Clearly, $C$ is not bounded but $0^{{+}}C=\{0\}$. However, if the additional condition that $C$ is closed is imposed, then we recover the converse.
Here are some other examples of recession cones of unbounded convex sets:

If $C=\{(x,y)\midx\leq y\}$, then $0^{{+}}C=C$.

If $C=\{(x,y)\midx<y\}$, then $0^{{+}}C=\overline{C}$, the closure of $C$.

If $C=\{(x,y)\midx^{{n}}\leq y,n>1\}$, then $0^{{+}}C=\{(0,y)\mid y\geq 0\}$.
Remark. The recession cone of a convex set is convex, and, if the convex set is closed, its recession cone is closed as well.
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