Annihilator of Subspace of Banach Space is Weak-* Closed/Proof 1

Proof
From Set is Closed iff Equals Topological Closure, we aim to show:
 * $M^\bot = \map {\cl_{w^\ast} } {M^\bot}$

From Set is Subset of its Topological Closure, we have:
 * $M^\bot \subseteq \map {\cl_{w^\ast} } {M^\bot}$

Now let:
 * $f \in \map {\cl_{w^\ast} } {M^\bot}$

From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {f_\lambda}_{\lambda \in \Lambda}$ in $M^\bot$ converging to $f$ in $\struct {X^\ast, w^\ast}$.

That is, from Characterization of Convergent Net in Weak-* Topology:
 * for each $x \in X$ the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ in $\GF$.

Since $f_\lambda \in M^\bot$ for each $\lambda \in \Lambda$, we have:
 * $\map {f_\lambda} x = 0$ for each $x \in M$ and $\lambda \in \Lambda$.

From Constant Net is Convergent, we obtain:
 * the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ and $0$ in $\GF$.

From Metric Space is Hausdorff, $\GF$ is Hausdorff.

Hence from Characterization of Hausdorff Property in terms Nets, we obtain $\map f x = 0$ for each $x \in M$.

So:
 * $f \in M^\bot$

So we obtain:
 * $\map {\cl_{w^\ast} } {M^\bot} \subseteq M^\bot$

and hence:
 * $M^\bot = \map {\cl_{w^\ast} } {M^\bot}$

From Set is Closed iff Equals Topological Closure, we have that $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$.