Sum of Absorbing Sets is Absorbing

Theorem
Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $A, B \subseteq X$ be absorbing sets.

Then $A + B$ is absorbing.

Proof
Let $u \in A + B$.

Then there exists $a \in A$, $b \in B$ such that $u = a + b$.

Now, there exists $t_1 \in \R_{> 0}$ such that:
 * $a \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t_1$

There also exists $t_2 \in \R_{> 0}$ such that:
 * $b \in \alpha B$ for $\alpha \in \C$ with $\cmod \alpha \ge t_2$

Let $t = \max \set {t_1, t_2}$.

Then for $\alpha \in \C$ with $\cmod \alpha \ge t$, we have:
 * $u = a + b \in \alpha A + \alpha B = \alpha \paren {A + B}$

from Dilation of Subset of Vector Space Distributes over Sum.

So $A + B$ is absorbing.