Combination Theorem for Continuous Mappings/Metric Space/Absolute Value Rule

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $\R$ denote the real numbers.

Let $f: M \to \R$ be a real-valued function from $M$ to $\R$

Let $f$ be continuous at $a \in M$.

Then:
 * $\size f$ is continuous at $a$

where:
 * $\map {\size f} x$ is defined as $\size {\map f x}$.

Proof
Let $\epsilon \in \R_{>0}$ be a positive real number.

Because $f$ is continuous at $a$:
 * $\exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\map f x - \map f a} < \epsilon$

From Reverse Triangle Inequality:
 * $\size {\size {\map f x} - \size {\map f a} } \le \size {\map f x - \map f a}$

and so:


 * $\exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$

$\epsilon$ is arbitrary, so:


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$

That is, by definition:
 * $\size f$ is continuous at $a$.