Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(V,\,+,\,\cdot\,)$ be a vectorspace and $D\subset V$ a set with the following properties

For $\;\lambda D:=\{\lambda d\mid d\in D\}\;\textrm{ and}\;\;\lambda,\, \mu\ge0$: $$0\in D$$ $$\bigcup_{\lambda\in\mathbb{R}}\lambda D=V$$ $$\lambda<\mu\Rightarrow\lambda D\subset\mu D$$ In other words, the set can be bloated up to fill the entire space. It would be practical to give this property a name for the paper I am working on, because it is referenced rather often.

Is there a naming convention? If not, what would you suggest?

share|cite|improve this question
You could describe such a set as $V\backslash \mathbb{R}$, but that is of course a description of the set and not the property... IT also seems to be similar to the concept om hypercyclic operators, but with a set instead of an operator, but I guess that doesn't help either.. :-( – malin Jan 2 '13 at 13:59
@malin indeed it doesnt help ;) Maybe I will just have to invent a name for it, but I want to check if there is a name for it already – CBenni Jan 2 '13 at 14:02
Related to absorbing set as seen here – GEdgar Jan 2 '13 at 14:23
@GEdgar yup, thats what I was looking for. You can post that as an answer ;) – CBenni Jan 2 '13 at 14:25
up vote 2 down vote accepted

Related to absorbing set found in wikipedia

share|cite|improve this answer

Do you know that $\lambda D\cap\mu D=(0)$ for all $\lambda\neq\mu$? If so, then the name fundamental domain is the most common, since it is almost the fundamental domain of $V$ under the $\mathbb{R}$ action.

share|cite|improve this answer
oh, I totally forgot about 1 property >_< sorry. I edited my OP. The sets are subsets of one another – CBenni Jan 2 '13 at 13:54

The trivial subspace $D=\{0\}$ is the name of your set.

For all negative numbers $\lambda$ you have $\lambda D\subseteq 0D=\{0\}$, so $D=\{0\}$

Or did you mean your properties to hold only for non-negative reals?

In case you restrict to non-negative $\lambda$, wouldn't your condition then just mean that $D$ is a neighborhood of the origin? Assuming $V$ is finite dimensional...

share|cite|improve this answer
Agreed ... this is not a good quesiton if negative $\lambda$ are allowed. – GEdgar Jan 2 '13 at 14:21
Yes, indeed, I meant only non-negative Integers. – CBenni Jan 2 '13 at 14:24
@ufr, about D being a neighboorhood of the origin, not quite. If D contains a nonempty, open neighborhood of the origin, the properties are fulfilled. GEdgar found that exactly these sets are called "absorbing" – CBenni Jan 2 '13 at 14:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.