Absorbent Sets

# Absorbent Sets

Definition: Let $X$ be a linear space and let $E \subseteq X$. Then $E$ is said to be Absorbent or an Absorbing Set if for every $x \in X$ there exists a $\lambda > 0$ such that $\lambda x \in E$. |

Proposition 1: Let $X$ be a seminormed linear space. Then the open unit ball $B(0, 1) = \{ x \in X : p(x) < 1 \}$ and the closed unit ball $\bar{B}(0, 1) = \{ x \in X : p(x) \leq 1 \}$ are both absorbent sets. |

**Proof:**Let $x \in X$. If $p(x) = 0$ then by definition, $x \in B(0, 1)$. So assume that $p(x) \neq 0$. Let $\lambda = \frac{1}{2p(x)} > 0$. Then:

\begin{align} \quad p(\lambda x) = p \left ( \frac{1}{2p(x)} x \right ) = \frac{1}{2p(x)} p(x) = \frac{1}{2} < 1 \end{align}

- This shows that $\lambda x \in B(0, 1)$. So for every $x \in X$ there exists a $\lambda > 0$ such that $\lambda x \in B(0, 1)$. Thus $B(0, 1)$ is absorbent. A similar argument shows that $\bar{B}(0, 1)$ is also absorbent. $\blacksquare$