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Just to be sure: In an infinite dimensional Banach space the open unit ball cannot be totally bounded, right? The context is that I need this in order to find a lack in here...

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C-star-W-star
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2 Answers2

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You are right: it is not totally bounded. Riesz's lemma directly leads to an infinite uniformly separated subset of unit ball, as the Wikipedia article shows.

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    But don't we have to be careful about the open ball? Sure, the closed ball is not totally bounded but subsets of it (as the open ball) still can be... – C-star-W-star Aug 11 '14 at 23:58
  • @Freeze_S Sure, the open ball of radius $1$ contains closed ball of radius $1/2$. –  Aug 12 '14 at 00:13
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If the open unit ball were totally bounded, then so would its closure, which is the closed unit ball. The closed unit ball, in turn, is complete (as a closed subset of a Banach space). Hence, the closed unit ball would be compact, which it can be shown it is not, given that the space is infinite-dimensional.

triple_sec
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    I doubt that you can carry over arguments of totally boundedness on closures without additional arguments... – C-star-W-star Aug 11 '14 at 23:28
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    @Freeze_S Let $E$ be a totally bounded subset of a metric space $(X,d)$. Fix $\varepsilon>0$ and let ${x_j}{j=1}^N\subseteq X$ and $N\in\mathbb N$ be such that $E\subseteq\bigcup{j=1}^N B(\varepsilon/2,x_j)$, which is possible by the definition of total boundedness. It is not difficult to see that $\overline E\subseteq\bigcup_{j=1}^N B(\varepsilon,x_j)$. This shows that if $E$ is totally bounded, then so is $\overline E$. – triple_sec Aug 11 '14 at 23:35
  • Your 'it is not difficult to see that' is wrong (take $E=B(\epsilon,x_0)$ itself). – C-star-W-star Aug 11 '14 at 23:50
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    @Freeze_S If $E=B(\varepsilon,x_0)$ and $x\in\overline E$, then there exists some $y\in E$ such that $d(x,y)<\varepsilon$ (by the definition of closure). Since $y\in E$, it follows that $d(y,x_0)<\varepsilon$. By the triangle inequality, $d(x,x_0)<2\varepsilon$. Conclusion: $\overline E\subseteq B(2\varepsilon,x_0)$. I don't really see what's the glitch here. – triple_sec Aug 12 '14 at 00:35
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    Ah I missed the $\frac12$ that makes sense of course - thx ;) – C-star-W-star Aug 12 '14 at 01:55
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    @Freeze_S My pleasure, I'm glad I could help. :-) – triple_sec Aug 12 '14 at 03:30