Consider a vector space $X$ over the field $\mathbb{F}$ of real or complex numbers and a set $S\subset X$.
In this Wikipedia article about absorbing sets,
$S$ is called absorbing if for all $x\in X$ there exists a real number $r$ such that for all $\alpha\in\mathbb{F}$ with $\vert \alpha \vert \geq r$ we have $$ x\in \alpha S $$ where $\alpha S:=\{\alpha s\mid s\in S\}$.
According to this article about locally convex space, absorbing sets are defined slightly different:
$S$ is called absorbing if $$ \bigcup_{t>0}tS=X, $$ or equivalently for every $x\in X$, $tx\in S$ for some $t > 0$.
Here are my questions:
- Are these two definitions equivalent?
- Could anyone come up with a cited reference for each version of the definition?
- [Added:] Thanks to @ForgotALot, these two definition are not equivalent. Is there any extra assumption that one can have these two definitions equivalent?
[Some thoughts] The first one implies the second one when $\mathbb{F}=\mathbb{R}$. I don't see the case when $\mathbb{F}=\mathbb{C}$.
The set $S$ which is the union of the unit circle $T$ and $\{(0,0)\}$ in $\mathbb{R}^2$ is absorbing according to the second definition; in contrast, it is not absorbing according to the first since any $x$ is in $\alpha S$ only if $x=(0,0)$ or $|\alpha|=|x|$. Thus the two definitions are not equivalent.
When you say "I don't see the case when $\mathbb{R}=\mathbb{F}$" do you perchance mean $\mathbb{F}=\mathbb{C}$? In that case meeting the first definition implies that the second is met too since, per the first definition, we must have $x\in\alpha S$ for every $\alpha$ with $|\alpha|\geq r$, including the real ones.
Rudin's definition in Functional Analysis 1.33 is limited to convex sets; apologies for not noting that in my comment above. Perhaps if we limited ourselves to convex sets, which zaps my counterexample, we would get more sensible results.