# Covering theorem for Banach spaces

Are there any Banach space $X$ with $\operatorname{dim}(X)=\infty$ satisfying $S_X=\lbrace x\in X| |x|=1\rbrace$ is covered by for some $B_1,B_2,\ldots,B_N$, where $B_N$ are balls in $X$ with $0\notin B_i$ for $i=1,\ldots, N$?

I can find some literatures concerning countably many ball cases (e.g. T. W. Koerner, J. Lond. Math. Soc. (1970) 643-646). Also, I can prove that there are no such Hilbert space $X$ (using, for example, orthonormal bases?)

But, I couldn't find any literature about finite ball case (which is my question). Is it trivial?

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I don't understand, if $0 \notin B_i$ for all $i$, how can $S_X$ be covered by $B_1,\ldots,B_N$? Do you mean $S_X - \{0\}$? – HYL Jun 18 '11 at 17:16
Oh, I meant $S_X$ is the set of norm=1 elements in X. I fixed it. – Greg morse Jun 18 '11 at 17:44
"Ball-covering property of Banach spaces" by L. Cheng looks helpful. – Jonas Meyer Jun 18 '11 at 19:17

If $\mathrm{dim}(X)=+\infty$ and $S_X\subset\bigcup_{j=1}^NB_j$ where $B_j$ are closed balls (with $0\notin B_j$ for all $j$) then since the weak closure of $S_X$ is $\left\{x\in X, \lVert x\rVert \leq 1\right\}$. Since each $B_j$ is closed for the topology given by the norm and convex we know that each $B_j$ is closed for the weak topology. Hence we get $\left\{x\in X, \lVert x\rVert \leq 1\right\} \subset\bigcup_{j=1}^N B_j$. We get a contradiction since we assumed that $0\notin \bigcup_{j=1}^NB_j$.
So perhaps Greg means $B_n$ are open balls... – GEdgar Jun 18 '11 at 20:03
Assuming you mean open balls $B_j$. By Hahn-Banach, there is a linear functional $f_j$ that is positive on $B_j$. The set $\{x\in S_X : f_j(x)=0, 1\le j\le n\}$ is disjoint from the union, and it not empty if $X$ has dimension $> n$.