Image of Absorbing Set under Surjective Linear Transformation is Absorbing

Theorem

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

Let $X$ and $Y$ be vector spaces over $\GF$.

Let $A \subseteq X$ be an absorbing set.

Let $T : X \to Y$ be a surjective linear transformation.

Then $T \sqbrk A$ is an absorbing set.

Proof

Let $y \in Y$.

Since $T$ is surjective, there exists $x \in X$ such that $y = T x$.

Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:

$x \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.

Then:

$y = T x \in T \sqbrk {\alpha A}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.

From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:

$y \in \alpha T \sqbrk A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.

So $T \sqbrk A$ is absorbing.

$\blacksquare$