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!
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.)