Banach Space is F-Space

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$.

From Normed Vector Space is Topological Vector Space, we may consider $\struct {X, \norm {\, \cdot \,} }$ as a topological vector space.

With this identification, $\struct {X, \norm {\, \cdot \,} }$ is an $F$-Space.

Proof
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.

From Norm Topology Induced by Metric Induced by Norm, $d$ induces the topology on $\struct {X, \norm {\, \cdot \,} }$.

Since $\struct {X, \norm {\, \cdot \,} }$ is a Banach space, $\struct {X, d}$ is a complete metric space.

From Metric Induced by Norm is Translation-Invariant, $d$ is translation-invariant.

Hence $\struct {X, \norm {\, \cdot \,} }$ is an $F$-Space.