# Closed extensions in the weak* topology

Let $\ell^\infty$ be the Banach space of bounded sequences with the usual norm. and let $\ell_0(x) = \lim_{n \rightarrow \infty} x_n$, for convergent sequences. Show that the sett L consisting of all continuous extensions of $\ell_0$ to $\ell^\infty$ is closed in the weak* topology on $(\ell^\infty)'$.

The extension is done by Hahn-Banach, but how do I show that something is closed in the weak*? Can I use sequentially closed here? are the topology metriceble? I can show that this is not a Hilbert space. Is it reflexive? Im a little bit unsure about all this weak* stuff, Please help me out and merry christmas!

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To show $L$ is weak-* closed, you want to show that its complement is weak-* open, i.e. any $\phi \in (\ell^\infty)' \backslash L$ has a weak-* neighborhood disjoint from $L$. In fact, if $\phi \in (\ell^\infty)' \backslash L$ there is some sequence $s$ that converges to a limit $m$ that is not $\phi(s)$. Consider $\{\psi \in (\ell^\infty)': |\psi(s) - \phi(s)|<\epsilon\}$.
Great, I did not know this was the way to solve it, but how do I know that the set is open? Since the $\psi$ do not converge to m either? –  Johan Dec 24 '12 at 22:49
The topology that makes all the linear functionals on the form $x(u) = u(x)$ continuous? –  Johan Dec 25 '12 at 9:27
Not quite: it's the weakest topology that does that. But $\psi \to \psi(s)$ being continuous is all you need to show that $\{\psi: |\psi(s) - \phi(s)|<\epsilon\}$ is open. –  Robert Israel Dec 25 '12 at 20:35
The continuity of $\psi$ is not an issue here. The map $\psi \to \psi(s)$ is continuous (because that's what the weak-* topology says is continuous), so we are done. –  Robert Israel Dec 25 '12 at 21:22