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# absorbing set

Let $V$ be a vector space over a field $F$ equipped with a
non-discrete valuation $|{\cdot}|:F\to\mathbb{R}$. 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 $|{\lambda}|\geq r$, $B\subseteq\lambda A$. $A$ 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 $|{\lambda}|\geq r$. 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 $|\lambda\mid\geq r_{i}$, $\forall\lambda\in F$. So $x_{i}\in\lambda A$ with $|{\lambda}|\geq r$ if we take $r=\max\{r_{1},\ldots,r_{n}\}$. 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 $|{\lambda}|\leq 1$. Clearly, $C$ absorbs itself ($C\subseteq\lambda^{{-1}}C$,
$|{\lambda^{{-1}}}|\geq 1$), 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.

## Mathematics Subject Classification

46A08*no label found*15A03

*no label found*

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