# Understanding weakly dense.

Let $$c_0 = \{x \in l^\infty \ | \ x_i \ \to 0 \}$$. Show that $$c_0$$ is not weakly dense in $$l^\infty$$.

This means that the closure of $$c_0$$ w.r.t the weak topology is not $$l^\infty$$, right?

Does that mean that $$\exists x\in l^\infty$$ such that $$\not\exists (x_1, x_2, ....), x_i \in c_0: x_i \to^w x$$?

• Weak topology is not metrizable so you cannot use sequences to characterize denseness. Jan 9 at 11:31
• Hint: The (geometric) Hahn-Banach theorem might come handy. Jan 9 at 11:41
• This may help. Jan 9 at 12:13

Claim: If $$C$$ is a convex subset of a normed space, then the weak closure and the norm closure of $$C$$ are equal.
Proof: Apply the Hahn-Banach separation theorem. Details are left for you. $$\quad \square$$
Corollary: The weak closure of $$c_0$$ is equal to the norm closure of $$c_0$$ in $$\ell^\infty$$. Since $$c_0$$ is norm-closed in $$\ell^\infty$$, we deduce that $$c_0$$ is weakly closed in $$\ell^\infty$$ as well. So basically you are asked to show that the inclusion $$c_0 \subseteq \ell^\infty$$ is strict, which is trivial (consider the function $$n \mapsto 1$$).