Combination Theorem for Continuous Mappings/Metric Space/Minimum Rule

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

Let $\R$ denote the real numbers.

Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.

Let $\min \set {f, g}: M \to \R$ denote the pointwise maximum of $f$ and $g$.

Then:
 * $\min \set {f, g}$ is ‎continuous on $M$.

Proof
Let $a \in M$ be arbitrary.

From Min Operation Representation on Real Numbers
 * $\min \set {x, y} = \dfrac 1 2 \paren {x + y - \size {x - y} }$

Hence:
 * $\min \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$

From Difference Rule for Continuous Mappings on Metric Space:
 * $\map f x - \map g x$ is continuous at $a$.

From Absolute Value Rule for Continuous Mappings on Metric Space:
 * $\size {\map f x - \map g x}$ is continuous at $a$.

From Sum Rule for Continuous Mappings on Metric Space:
 * $\map f x + \map g x$ is continuous at $a$

From Difference Rule for Continuous Mappings on Metric Space:
 * $\map f x + \map g x - \size {\map f x - \map g x}$ is continuous at $a$

From Multiple Rule for Continuous Mappings on Metric Space:
 * $\dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$ is continuous at $a$.

As $a$ is arbitrary:


 * $\min \set {f, g}$ is ‎continuous on $M$.