Sum of Absorbing Sets is Absorbing
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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.
$\blacksquare$