Different definitions of absorbing sets from the Wikepedia
Emily Wong 
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 $\mathbb{F}=\mathbb{C}$.
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$Bourbaki-Topological Vector Spaces, Conway-A Course in Function Analysis and both references under the wiki page-absorbing set use the first definition. I didn't see your second definition before. Rudin uses an equivalent definition when the set is convex. I will give a proof below.
Proposition: If $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. $X$ is a vector space over $\mathbb{K}$. $C$ is a convex subset of $X$. Then the following statements are equivalent.
(1) $\forall x\in X$, $\exists r>0$ such that $x\in rC$.
(2) $\forall x\in X$, $\exists r>0$, $\forall \alpha\in\mathbb{K}$, $|\alpha|>r\Rightarrow x\in\alpha C$.
Proof: (2) $\Rightarrow$ (1) is obvious.
(1) $\Rightarrow$ (2): $0\in C$. If $\mathbb{K}=\mathbb{C}$. $\forall x\in X$, $\exists r_1,r_2,r_3,r_4\in\mathbb{R}^+$ such that $x\in r_1C$, $-x\in r_2C$, $ix\in r_3C$, $-ix\in r_4C$. Let $r=\max\{r_1,r_2,r_3,r_4\}$, then $r\in\mathbb{R}^+$. $\{\frac{1}{r}x,-\frac{1}{r}x,\frac{i}{r}x,\frac{-i}{r}x\}\subset C$. $\forall |\alpha|>\sqrt{2}r$. $\exists k_1,k_2\in\{-1,1\}$, $\alpha_1,\alpha_2\in [0,+\infty)$ such that $\frac{r}{\alpha}=k_1\alpha_1+ik_2\alpha_2$. Then $\alpha_1+\alpha_2<1$. Then $\frac{1}{\alpha}x=\frac{r}{\alpha}\frac{1}{r}x=(\alpha_1+\alpha_2)(\frac{\alpha_1}{\alpha_1+\alpha_2}\frac{k_1}{r}x+\frac{\alpha_2}{\alpha_1+\alpha_2}\frac{ik_2}{r}x)\in(\alpha_1+\alpha_2)C\subset C$. Then $x\in\alpha C$. If $\mathbb{K}=\mathbb{R}$, $\forall x\in X$, $\exists r>0$ such that $\{x,-x\}\subset rC$. $\forall \alpha\in\mathbb{K}$ satisfying $|\alpha|>r$, $\exists k\in\{-1,1\}$ such that $\alpha=k|\alpha|$. Then $\frac{1}{\alpha}x=\frac{r}{|\alpha|}\frac{k}{r}x\in\frac{r}{|\alpha|}C\subset C$.
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