Absorbing set

Remarks

An absorbing set $A$ contains $0$: for $x = 0$ there is $c \ne 0$ with $0 \in cA$, so $0 = c a$ for some $a \in A$, and then $a = 0$. If $A$ is balanced, then $A$ is absorbing if and only if for every $x \in V$ there is $t > 0$ with $t x \in A$. Example: the open unit ball $A = \set{x : \norm{x} < 1}$ of a normed space is absorbing; for $x \in V$ take $r = \norm{x} + 1$, and then $\norm{x / c} < 1$ whenever $\abs{c} \ge r$. Non-example: in $\R^2$, the axis $A = \R \times \set{0}$ is not absorbing, because $(0, 1) \notin c A$ for every $c$. In a topological vector space every neighbourhood of $0$ is absorbing, since $c \mapsto c^{-1} x$ is continuous and tends to $0$.