Define $e_k : \mathbb{N} \rightarrow \mathbb{R}$ by $$ e_k (n) = \left\{ \begin{array}{ll} 1, & \quad k = n \\ 0, & \quad k \neq n \end{array} \right. $$
Is $E:= \{e_k : k \in \mathbb{N} \}$ a weakly compact subset of $l^{\infty} (\mathbb{N})$? Prove.
My proof is as follows:
$E$ is a closed subset of $\text{Ball } l^{\infty}$, thus, we only need to check if if $\text{Ball } l^{\infty}$ were weakly compact.
According to Kakutani's Theorem, $\text{Ball } l^{\infty}$ is weakly compact if and only if $l^{\infty}$ were reflexive. But since $l^{\infty}$ is not reflexive, then $E$ is not weakly compact.