= A norm-closed dual subspace need not be weak-star closed
{title2=$c_0\subsetneq\ell^\infty=(\ell^1)',\qquad(c_0)_\circ=0$}
The <continuous dual space> of $\ell^1$ is $\ell^\infty$ under the complex-linear coordinate pairing. Its subspace $c_0$ is <norm> closed: uniform approximation by a sequence with a small tail leaves a small tail. Nevertheless its preannihilator in $\ell^1$ is zero, because the coordinate <unit vectors> belong to $c_0$ and test every coordinate. Its double continuous annihilator is therefore all of $\ell^\infty$. It is weak-star dense directly as well: the finite truncations of any bounded sequence lie in $c_0$ and converge against every $\ell^1$ test by absolute summability. This demonstrates why <norm> <closure> cannot replace <closure> in the <weak-star topology> in the dual-side annihilator theorem.
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