# Banach Space: Open Unit Ball Totally Bounded?

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...

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.

• 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... Aug 11, 2014 at 23:58
• @Freeze_S Sure, the open ball of radius $1$ contains closed ball of radius $1/2$.
– user147263
Aug 12, 2014 at 0:13

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.

• I doubt that you can carry over arguments of totally boundedness on closures without additional arguments... Aug 11, 2014 at 23:28
• @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 $\overline E$. Aug 11, 2014 at 23:35
• @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. Aug 12, 2014 at 0:35
• Ah I missed the $\frac12$ that makes sense of course - thx ;) Aug 12, 2014 at 1:55
• @Freeze_S My pleasure, I'm glad I could help. :-) Aug 12, 2014 at 3:30