Let $V$ be a vector space over a field $F$ equipped with a non-discrete valuation $\abs{\cdot}:F\to\mathbb{R}$ Let $A,B$ be two subsets of $V$ Then $A$ is said to absorb $B$ if there is a non-negative real number $r$ such that, for all $\lambda\in F$ with $\abs{\lambda}\geq r$ $B\subseteq\lambda A$ $A$ is said to be an absorbing set, or a radial subset of $V$ if $A$ absorbs all finite subsets of $V$

Equivalently, $A$ is absorbing if for any $x\in V$ there is a non-negative real number $r$ such that $x\in\lambda A$ for all $\lambda\in F$ with $\abs{\lambda}\geq r$ If a finite subset $B$ of $V$ consists of $x_1,\ldots,x_n$ then corresponding to each $x_i$ there is an $r_i\geq 0$ such that $x_i\in\lambda A$ such that $\vert\lambda\mid\geq r_i$ $\forall\lambda\in F$ So $x_i\in\lambda A$ with $\abs{\lambda}\geq r$ if we take $r=\max\lbrace r_1,\ldots,r_n\rbrace$ So $A$ absorbs $B$

Example. If $V=\mathbb{R}^n$ and $F=\mathbb{R}$ then any set containing an open ball centered at $0$ is absorbing. This implies that an absorbing set does not have to be connected, convex.

A closely related concept is that of a circled set, or a balanced set. Let $V$ and $F$ be defined as above. A subset $C$ of $V$ is said to be circled, or balanced, if $\lambda C\subseteq C$ for all $\abs{\lambda}\leq 1$ Clearly, $C$ absorbs itself ($C\subseteq\lambda^{-1}C$ $\abs{\lambda^{-1}}\geq 1$ , and $0\in C$ $C$ is also symmetric ($-C=C$ , for $-C\subseteq C$ and $C=-(-C)\subseteq -C$ As an example of a circled set that is neither absorbing nor convex, consider $V=\mathbb{R}^2$ and $F=\mathbb{R}$ and $C$ the union of $x$ and $y$ axes. For an example of an absorbing set that is not circled, take the union of a unit disk and an annulus centered at 0 that is large enough so it is disjoint from the disk.