Combination Theorem for Continuous Mappings/Normed Division Ring/Product Rule

Theorem
Let $T = \struct{S, \tau}$ be a topological space.

Let $\struct{R, +, *, \norm{\,\cdot\,}}$ be a normed division ring.

Let $f: T \to R$ and $g: T \to R$ be continuous mappings.

Then:
 * $f * g : T \to R$ is continuous.

where $f * g : T \to R$ is the mapping defined by:
 * $\forall x \in T: \map {\paren{fg}} x = \map f x * \map g x$

Proof
Let $\tau_R$ be the topology induced by the norm $\norm{\,\cdot\,}$.

Let $\tau_{R \times R}$ be the product topology on $R \times R$.

Let $f \times g : T \to \struct{R \times R, \tau_{R \times R}}$ be the mapping defined by:
 * $\forall x \in U : \map {\paren{f \times g}} x = \tuple{\map f x, \map g x}$

Let $* : \struct{R \times R, \tau_{R \times R}} \to \struct{R, \tau_R}$ be the mapping defined by:
 * $\forall x, y \in R : \map * {x,y} = x * y$

Lemma 2
From Addition on Normed Division Ring is Continuous, the mapping $* : \struct{R \times R, \tau_{R \times R}} \to \struct{R, \tau_R}$ is continuous.

From Composite of Continuous Mappings is Continuous, the composition $* \circ \paren{f \times g}$ is continuous.

Hence $fg$ is continuous.