Let X be an infinite dimensional normed $\mathbb K$-linear space .Prove that $\partial B(0,1)=\{x\in X: ||x||= 1\}$ is not weakly closed in $X$.
We know that a convex subset of $X$ is weakly closed iff it is closed w.r.t. the normed topology. But here $\partial B(0,1)$ is not convex in X.
(definition of weak topology): Let $X$ be a normed $\mathbb K$ -linear space . The weak topology of $X$ is the smallest topology $J_{\omega}$(which is a subset of the normed topology of X) such that $f:(X,J_{\omega})\to\mathbb K$ is continuous for all $f \in X^*.$
Please someone give some hints.
Thank you.
