# The norm-closed unit ball of $c_0$ is not weakly compact

Show that the norm-closed unit ball of $c_0$ is not weakly compact; recall that $c_0^*=\ell_1$.

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Did you mean $C_0$ instead of $c_0$? – Zev Chonoles May 19 '12 at 8:11
@Zev: What is $C_0$? I'm pretty sure $c_0$ here means the space of sequences converging to $0$, and that is standard notation. On the other hand, "$C_0$" is often followed by a topological space to denote the space of continuous complex-valued functions on that space that vanish at infinity. E.g. $C_0(\mathbb N)=c_0$ (when $\mathbb N$ is given the discrete topology). – Jonas Meyer May 19 '12 at 21:15

Hint: Let $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$. Suppose $z\in c_0$ is a weak cluster point of $(x_n)$. By considering the action of the standard unit vectors of $\ell_1$ on the $x_n$, obtain a contradiction by showing that we must have $z=(1,1,\ldots)$.
In general, $B(X)$ is weakly compact if and only if $X$ is reflexive. Additionally, Eberlein-Smulian Theorem tells us that in this case $B(X)$ is weakly sequentially compact. – Theo May 18 '12 at 23:35