Closed $\iff$ weakly closed subspace On this link
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2004;task=show_msg;msg=1414.0001
is the argument that a linear subspace in a normed space is closed w.r.t. norm iff it is weakly closed.
On the other hand, $c_0$ (sequences convergent to $0$) is a norm-closed linear subspace of $l_\infty$ (bounded sequences), but it is not weakly closed, since the base vectors $e_i$ are weakly dense in $l_\infty$.
Since I studied func.an. quite a while ago, my question is - what am I missing?
 A: Edit: the previous answer was wrong. Below is a new version. 
Weak convergence of a net $\{x_n\}$ to $1\in\ell^\infty$ means that, for any functional $\varphi$ in the dual of $\ell^\infty$, $\varphi(x_n-1)\to0$. 
We can see $\ell^\infty$ as $C(\beta\mathbb N)$, the continuous functions on the Stone-Čech compactification of $\mathbb N$. Let $\omega\in\beta\mathbb N\setminus\mathbb N$ (in other words, $\omega$ is a free ultrafilter). Then $1(\omega)=1$ and $x(\omega)=0$ for all $x\in c_0$. So we have, letting $\varphi_\omega$ be the point-evaluation at $\omega$,
$$
\varphi_\omega(x-1)=\varphi_\omega(x)-\varphi_\omega(1)=0-1=-1,
$$
and so no net in $c_0$ will make the limit go to zero. This means that $c_0$ is not weakly dense in $\ell^\infty$. Weak-star density works because one has to deal with less functionals. 
As was mentioned, the Hahn-Banach theorem guarantees that $c_0$ (being convex) is both norm and weakly closed.  
A: $c_0$ is not weakly dense in $\ell_\infty$. Indeed, let $u:=\sum_{k=0}^{+\infty}e^{2k}$, where $e^k_j:=\delta_{kj}$. Take a Banach limit $L$. Then 
$$U:=\{x\in\ell_\infty, |L(x-u)|<1/3\}$$
is a neighborhood of $u$ in the weak topology. If $x\in U\cap c_0$, then $L(x)=0$, hence $|L(u)|<1/3$. It's a contradiction, since $L(u)=1/2$.
A: There is a fundamental fact about convex sets in locally convex spaces (and, in particular normed linear spaces) that a convex set is closed if and only if it is closed in the weak topology (the topology of convergence on the (continuous)
dual.  WARNING: convergence is of nets, NOT sequences.  Also, every space has a weak topology since every space has  continuous dual (albeit possibly only the zero vector).  There is SOMETIMES a weak*-topology if the space in question is itself a dual but only then.
