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From planetmath

Let $V$ be a vector space over a field $F$ equipped with a non-discrete valuation $|\cdot|:F\to \mathbb{R}$ . Let $A$ and $B$ be two subsets of $V$. Then $A$ is said to absorb $B$ if there is a non-negative real number $r$ such that, for all $\lambda \in F$ with $|\lambda| \geq r$ , $B \subseteq \lambda A$.

I am thinking about those subsets that can absorb themselves.

  1. A subset consisting of a single nonzero vector cannot absorb itself, but a subspace can. I was wondering if it is possible to characterize those subsets that can absorb themselves? Must such a subset contain the zero vector? Must it be a subspace?
  2. Same questions for open subsets in a topological vector space.

Thanks and regards!

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2  
It certainly need not be a subspace, take a ball around zero for example. As for characterizing the self-absorbing subsets, there's no hope. Just try to draw a few pictures in $\mathbb{R}^2$... For example, every bounded set containing a neighborhood of zero is self-absorbent. Or the union of any number of rays starting from the origin (those rays need not contain zero). –  t.b. Jan 5 '12 at 13:24
    
There was a now deleted comment asking for an open subset not containing $0$ which is self-absorbing: try $V \smallsetminus \{0\}$, for example. –  t.b. Jan 5 '12 at 13:31
    
@t.b.: Thanks! Sorry, I deleted my comment because I saw your modification to your first comment. In a topological vector space, is a neighborhood of zero always self-absorbing? –  Tim Jan 5 '12 at 13:37
1  
No: $(-1,1) \cup (2, \infty)$ in $\mathbb{R}$. Yes if it is bounded. –  t.b. Jan 5 '12 at 13:41
    
@t.b.: Thanks! How about a connected neighborhood of zero? –  Tim Jan 5 '12 at 13:45

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