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I'm studying functional analysis, currently a chapter about topological vector spaces. It is stated that every neighborhood of zero is an absorbing set. But I was wondering if the reversed statement is also true?

Isn't, as an counterexample, in $X =>R^1$ the disconnected set $A = [2,1) \cup \{0\} \cup (-1,-2]$ also absorbing and does not contain an open set, which contains zero (so isn't a neighborhood of zero)? For every $x\in X$ I can construct some $t>0, t=1/x \pm \epsilon$ such that $t \cdot x \in A$. For $x=0$ I can choose any $t>0$.

Edit: We defined absorbing as follows. Let $X$ be a vector space, $A$ a subset of $X$. Then $A$ is called absorbing, if for every $x \in X$ there exists a $t > 0$ such that $tx \in A$.

FeWa
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    Do you mean $[-2,-1) \cup {0} \cup (1,2]$? – Arthur Sep 13 '18 at 12:16
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    Is this your definition of an absorbing set? Please edit your question to include a definition. In any event, if that is your definition, the set which you describe doesn't look like an absorbing set to me. – Xander Henderson Sep 13 '18 at 12:17
  • The definition you gave in your question is not the definition Wikipedia gives. For your definition, the set ${0,\pm 1}$ is an absorbing set in $\mathbb{R}$. – Xander Henderson Sep 13 '18 at 12:22
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    I noticed that, I got the definition straight from my printed lecture notes though. Thanks anyway, I'm going to double check them! – FeWa Sep 13 '18 at 12:31

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(Based on the definition given by wikipedia)

That doesn't look absorbing to me: given any non-zero $x$ as you increase the scaling of $\alpha([-2,-1) \cup \{0\} \cup (1,2])$ you'll eventually find that $x$ falls into the gap between $0$ and the interval.

Instead, I'd look for a counterexample by using an infinite dimensional space. For example, take the space of polynomials and the 'box' around $0$ whose width in degree $n$ is $\frac{1}{n}$. For each polynomial $p$ you can just look at the restriction to the finite-dimensional subspace up to its degree, on which this set is a neighbourhood and therefore absorbing. Therefore the original box is absorbing for all $p$ and is thus an absorbing set.

However, it is not a neighbourhood of $0$, because the ball of radius $\varepsilon$ will not fit in the box in all degrees greater than $\frac{1}{\varepsilon}$.

(A good exercise might be to show that your hypothesis holds for finite-dimensional spaces.)

Chessanator
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