Calculate the weak closure of $S_{c_0} = \{x \in c_0 : \|x\|_\infty = 1 \}$ I need to find the weak closure of this set. First I tried using McKay theorem, but found out that $S_{c_0}$ is not convex. Now I'm totally lost. I've seen proofs online that the weak closure of the unit sphere in a normed space is the unit ball, and I kind of understand the proof, but I wanted to find it specifically for this set. Are there any tips on how to find the weak closure of any given tips that could help me in the future?
 A: Suppose $\|x\|_{\infty} \leq 1$ so that $|x_n| \leq 1$ for all $n$. Consider $x^{n}=(x_1,x_2,...,x_n,1,0,0,....)$. Clearly, $\|x^{n}\|_{\infty} =1$. I claim that $x^{n} \to x$ in the weak topology. For this I have to show that $\sum x_iy_i =\lim_{n \to \infty} \sum\limits_{i=1}^{n} x_iy_i+y_{n+1}$ for any $(y_i) \in c_0^{*}=\ell^{1}$. This is easy and I will let you prove this. Finally, conclude that the weak closure is the closed unit ball of $c_0$.
A: I am assuming that the ambient space is $l_\infty$. It can be shown that $c_0$ is norm closed and since it is convex it is weak closed. Similarly the norm closed unit ball is convex and hence is weak closed. Let $B$ denote the intersection of $c_0$ and the norm closed unit ball. This is weak closed and so
$\overline{S}^w_{x_0} \subset B$.
Now suppose $x \in B$ and $\|x\|_\infty <1$. Suppose $W$ is a weak open set containing $x$. Then there is some weak open neighbourhood of $x$ of the form $\{ y | | \phi(x-y)| < \epsilon \}$ for some $\epsilon>0$ and a finite collection $\phi_1,...\phi_n \in l_\infty^*$. Since $\bigcap_k \ker \phi_k$ has codimension $\le n$, it contains some non zero $d \in c_0$ and we can find a $t$ such that
$\|x+td\|_\infty = 1$. Hence $x \in \overline{S}^w_{x_0} $.
In particular, $\overline{S}^w_{x_0}  = B$.
