Uniformly Convergent iff Difference Under Supremum Metric Vanishes

Theorem
Let $X$ and $Y$ be metric spaces

Let $\sequence {f_n}$ be a sequence of mappings defined on $X$.

Let $f: X \to Y$ be a mapping.

Let $d_S: S \times S \to Y$ denote the supremum metric on $S \subseteq X$.

Then:
 * $\sequence {f_n}$ converges uniformly to $f$ on $S$


 * $\map {d_S} {f_n, f} \to 0$ as $n \to \infty$.
 * $\map {d_S} {f_n, f} \to 0$ as $n \to \infty$.

Proof
We have the following string of equivalences:

Start by :
 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N, \forall x \in S: \map d {\map {f_n} x, \map f x} < \epsilon$

By, this is equivalent to:
 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \epsilon$ is an upper bound of $\set {\map d {\map {f_n} x, \map f x} : x \in X}$.

So, by, this is equivalent to:
 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \sup_{x \mathop \in X} \map d {\map {f_n} x, \map f x} \le \epsilon$

By, this is equivalent to:
 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \map {d_S} {f_n, f} \le \epsilon$

By, this is equivalent to:
 * $\map {d_S} {f_n, f} \to 0$ as $n \to \infty$