# The closed unit ball to generate a linear normed space.

Let $X$ be a linear normed space and let B denote the closed unit ball of $X$. Then we can stretch the unit ball to get every vector in $X$: $$X=\bigcup_{n=1}^\infty nB.$$ Is this true? Why?

Given $x\in X$, there is $n\in\Bbb N$ with $\|x\|\le n$. Then $x\in nB$.
Indeed, in a normed space every neighbourhood of $0$ is absorbing, as this property is also called. For, if $x \in X$, take $n = 1 + \lceil\|x\|\rceil \in \mathbb{N}$ so that $\|x\| < n$. Then $\|\frac{1}{n}\cdot x\| = \frac{1}{n}\|x\| <1$ So $x = n \cdot (\frac{1}{n}\cdot x) \in nB$, as required.
IIRC, a locally convex topological vector space is normable exactly when this property holds for all (open) neighbourhoods of $0$.