Absorbent Sets of Vectors

# Absorbent Sets of Vectors

Definition: Let $E$ be a vector space. A subset $A \subseteq E$ is Absorbent if for every vector $x \in E$ there exists a number $\lambda > 0$ such that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $x \in \mu A$. |

For example, let $A = \{ (x, 0) : x \in \mathbb{R} \} \subset \mathbb{R}^2$. Then $A$ is simply the $x$-axis. This set is convex, balanced (and hence absolutely convex), but it is not absorbent.

To see this, suppose otherwise. Suppose that there exists a $\lambda > 0$ such that if $\mu \in \mathbf{F}$ and $|\mu| \geq \lambda$ then $(0, 1) \in \mu A$. But then $(0, 1) = \mu(x, 0)$ for some $x \in \mathbb{R}$, which is never true. Thus $A$ is not absorbent.

On the other hand, the disk $B = \{ (x, y) : x^2 + y^2 \leq 1 \}$ is an absorbent subset of $\mathbb{R}^2$.

Proposition 1: Let $E$ be a vector space. If $\{ M_1, M_2, ..., M_n \}$ is a finite collection of absorbent subsets of $E$, then $\bigcap_{i=1}^{n} M_i$ is absorbent. |

**Proof:**Let $\{ M_1, M_2, ..., M_n \}$ be a finite collection of absorbent subsets of $E$ and let $M := \bigcap_{i=1}^{n} M_i$. Let $x \in E$. Since each $M_i$ is absorbent, there exists numbers $\lambda_1, \lambda_2, ..., \lambda_n > 0$ such that if $\mu \in \mathbf{F}$ and $|\mu| \geq \lambda_i$ then $x \in \mu M_i$, for each $1 \leq i \leq n$.

- So for each $x \in E$, take $\lambda := \max \{ \lambda_1, \lambda_2, ..., \lambda_n \}$. Then if $\mu \in \mathbf{F}$ and $|\mu| \geq \lambda$ then $x \in \mu M_i$ for each $1 \leq i \leq n$, and consequently, $x \in \mu M$. So $\bigcap_{i=1}^{n} M_i$ is absorbent. $\blacksquare$