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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?

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Given $x\in X$, there is $n\in\Bbb N$ with $\|x\|\le n$. Then $x\in nB$.

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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 $0$.

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  • $\begingroup$ Thanks for your reply. Because it is also a linear space is X=span(B)? $\endgroup$ – Answer Lee Aug 25 '17 at 4:38
  • $\begingroup$ Sure, this then follows. @AnswerLee $\endgroup$ – Henno Brandsma Aug 25 '17 at 5:02

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