Take the 2-minute tour ×
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.

From Wikipedia:

Given a topological vector space $(X,τ)$ over a field $F$, $S$ is called bounded if for every neighborhood $N$ of the zero vector there exists a scalar $α$ so that $$ S \subseteq \alpha N $$ with $$ \alpha N := \{ \alpha x \mid x \in N\}. $$

I was wondering if the concept is still the same when "for every neighborhood $N$ of the zero vector" is replaced by "there exists a neighborhood $N$ of the zero vector"? Is it true that every neighborhood of the zero vector can be scaled to contain any other neighborhood of the zero vector?

Thanks and regards!

share|improve this question
3  
The whole space is a neighborhood of the zero vector. –  Mariano Suárez-Alvarez Jan 3 '12 at 4:35
    
@MarianoSuárez-Alvarez: Thanks! Is it true that not any neighborhood of the zero vector can be scaled to the whole space, because infinity is not a scalar in the field $F$? –  Tim Jan 3 '12 at 4:42
4  
That "argument" is not a good one. Show rather that if a neighborhood $U$ of zero contains a scaled copy of the whole space, then it is in fact the whole space. –  Mariano Suárez-Alvarez Jan 3 '12 at 4:44
    
@MarianoSuárez-Alvarez: Thanks! (1) In a topological vector space, is "a neighborhood U of zero contains a scaled copy of the whole space" the same as "a scaled copy of a neighborhood U of zero is the whole space"? (2) In a vector space instead of a topological vector space, is it true that if a subset U of zero contains a scaled copy of the whole space, then it is in fact the whole space? –  Tim Jan 3 '12 at 14:17
add comment

1 Answer 1

up vote 1 down vote accepted

No. For example, as Mariano commented, $X$ is a neigbhorhood of the zero vector.

If $N$ is a neighborhood of the zero vector such that every neighborhood of the zero vector can be scaled to contain $N$, then $N$ is bounded. It is true that $S$ is bounded if there exists a bounded neighborhood $N$ of the zero vector and a scalar $\alpha$ such that $S\subseteq \alpha N$, but this is not equivalent to boundedness in general (See Matt E's comment below).

E.g., think of $\mathbb R^2$, where boundedness is the same as being contained in a big enough circle. A set like $\{(x,y):x^2+y^2<1\}$ cannot be scaled to contain a set like $\{(x,y):x>-1\}$. More generally, in a normed space boundedness of $S$ is equivalent to $\sup\{\|x\|:x\in S\}<\infty$, and the idea is that $S$ can be scaled to be contained in an open neighborhood $N$ of the zero vector, no matter how small $N$ is. (Of course in general $X$ need not even be metrizable, so this intuition isn't perfect.)

share|improve this answer
1  
Dear Jonas, What are your ambient assumptions in the statement "It is true that $S$ is bounded if and only if there exists a bounded neighbourhood ..."? I was under the impression that if a topological vector space (at least a locally convex one, over say $\mathbb R$ of $\mathbb C$) contains any bounded neighbourhood at all, then it is normable. (And not all topological vector spaces are normable.) Regards, –  Matt E Jan 3 '12 at 4:53
    
Matt: Oops, I was really off there. Thanks for catching that. When writing that I was thinking of normed spaces, not heeding my own warning in the final sentence. –  Jonas Meyer Jan 3 '12 at 4:57
1  
@MattE: I think I am already very polite. You, however, are much much more polite than I am. :-) –  Tim Jan 3 '12 at 4:58
1  
(1) If it is also locally convex and Hausdorff, yes. This is known as Kolmogorov's Normability Criterion (q.v.). (2) Yes. For example, it is straightforward to show that a subset of a bounded set is bounded and a scalar multiple of a bounded set is bounded. (3) This would be true in the locally convex Hausdorff case by the above, but I am not sure what can be said in general. (4) There are always bounded sets regardless, for example finite sets. An example of a TVS with no bounded neighborhood of the zero vector is the weak topology on an infinite dimensional Hilbert space $H$. This is..... –  Jonas Meyer Jan 4 '12 at 5:54
1  
@Tim: ...the coarsest topology on $H$ that makes all of the functionals $x\mapsto\langle x,y\rangle$ continuous. It is coarser than the norm topology, and therefore all norm bounded subsets of $H$ are also weakly bounded. But the weak topology has no open nonempty weakly bounded subsets. That the weak topology is not normable follows among other ways from the open mapping theorem, because it is strictly coarser than the norm topology, and comparable Banach space norms are equivalent. It is not even metrizable; one way to see that is here: math.stackexchange.com/q/42337. –  Jonas Meyer Jan 4 '12 at 5:59
show 2 more comments

Your Answer

 
discard

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.