Exhibiting open covers with no finite subcovers.

How do I exhibit an open cover of the closed unit ball of the following:

(a) $X = \ell^2$
(b) $X=C[0,1]$
(c) $X= L^2[0,1]$

that has no finite subcover?

• Hint: First find a sequence without convergent subsequence. – t.b. Apr 15 '12 at 20:09
• I was going to expand on this ^ comment when his answer appeared but since I use different words, I'll post what I was going to write anyway: For a), if you take $(0, \dots, 1, 0 , \dots )$ the sequences that are $1$ at position $k$ and zero otherwise then they're all in the closed unit ball in $\ell^2$. Take your open cover to be the $\varepsilon$-balls around these sequences and $\varepsilon < \sqrt{2}$. (to be continued) – Rudy the Reindeer Apr 15 '12 at 20:26
• Then pick any finite subcover of this. Then there will be an $N$ such that all the sequences in your cover are zero after $N$. Pick a point $(0, \dots, 0, 1, 0, \dots )$ where $1$ appears somewhere above $N$. Then this point is not covered by your finite subcover since its distance to any of the sequences in your basis is $\sqrt{2}$. – Rudy the Reindeer Apr 15 '12 at 20:29
• @Matt: the collection of balls you start with doesn't cover the unit ball. – t.b. Apr 15 '12 at 20:30
• @t.b. I'm trying to see why they don't... – Rudy the Reindeer Apr 15 '12 at 20:35

Suppose $\{y_n\}_{n=1}^\infty$ is a sequence in the closed unit ball $B$ which has no accumulation point. Then $U_{N} = B \smallsetminus \{y_n\,:\,n \geq N\}$ is open and $B \subset \bigcup_{N=1}^\infty U_N$. However, $\{U_{N}\}_{N=1}^\infty$ has no finite subcover.
For a) take the standard basis, for b) take $y_{n}(t) = t^n$ and for c) try $y_n(t) = \exp{(2\pi i \,n \,t)}$.
• It was brought to my attention that it may be confusing that $U_N$ is open in $B$ but not in the surrounding space $X$. If this is the case just replace $U_N$ by $V_N = X \smallsetminus \{y_n\,:\,n \geq N\}$. – t.b. Apr 15 '12 at 20:51
• @MattN. In these specific instances ($\ell^2$, $C[0,1]$ and $L^2[0,1]$) you can exhibit a sequence without accumulation points directly. In an abstract infinite dimensional Banach space you need to argue that such a sequence exists. Riesz's lemma allows you to do that. – t.b. Apr 20 '12 at 18:55