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 $\{x+rd\mid r\ge 0\}$ 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,
$${0}^{+}C=\{d\mid x+rd\in C,\forall x\in C,\forall r\ge 0\}.$$ 
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
$$ 
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:

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

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If $$, then ${0}^{+}C=\overline{C}$, the closure of $C$.

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

Canonical name  RecessionCone 
Date of creation  20130322 16:20:24 
Last modified on  20130322 16:20:24 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 52A20 
Classification  msc 52A07 
Defines  direction of a convex set 