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If not, what would a counterexample be?

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A bounded set is exactly the one which is contained in some ball $B(x,r)$. By definition. –  Ilya Jan 28 '13 at 7:31
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According to the standard definition of boundedness in topological vector spaces, a bounded set is one which is absorbed by every neighbourhood of 0. I'm not sure how for every neighbourhood $U$ of 0 there is $t$ such that $B_1(0) \subset sU$ for all $s > t$. –  user59922 Jan 28 '13 at 7:35
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up vote 3 down vote accepted

Assuming the "unit ball" to mean $\{y: d(y,0) < 1\}$, where $d$ is any metric compatible with the topology, then the answer is no. For example, take $\mathbb R$ with the metric $d(x,y) = \dfrac{|x-y|}{1+|x-y|}$. Then the unit ball consists of all of $\mathbb R$, and this is not bounded (in the topological vector space sense).

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And what about normable spaces? –  user59922 Jan 28 '13 at 8:01
    
The unit ball for a norm is bounded, because any neighbourhood of $0$ must contain some $B_\epsilon(0)$, and $B_\epsilon(0) = \epsilon B_1(0)$. –  Robert Israel Jan 28 '13 at 8:09
    
Why exactly is that? –  user59922 Jan 28 '13 at 8:11
    
the answer is still no. my understanding here is that you are taking just any old compatible metric, compatible with a normable topology, and asking if the unit ball is bounded? I think that you've confused everyone with this question. Can you edit it please. –  Rabee Tourky Jan 28 '13 at 8:13
    
Why does any neighbourhood of $0$ contain some $B_\epsilon(0)$? From the definition of the topology corresponding to a metric. Why is $B_\epsilon(0) = \epsilon B_1(0)$? Because $\|\epsilon x\| = \epsilon \|x\|$. –  Robert Israel Jan 28 '13 at 8:46
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if it's a unit ball, it contains all points within a distance of $1$ from an origin $z$. A not bounded set would include points $x,y$ arbitrary far away from each other, but using the triangle inequality this cannot be true, since: $$d(x,y) \le d(x,z) + d(z,y) < 2$$

Am I missing something?

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