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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!

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    @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?2012-01-03

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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.)

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    @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: http://math.stackexchange.com/q/42337.2012-01-04