Uniformly Convergent iff Difference Under Supremum Metric Vanishes

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Theorem

Let $X$ be a set.

Let $\struct {Y, d}$ be a metric space.

Let $S$ be the set of all bounded mappings from $X$ to $Y$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $S$.

Let $f \in S$.

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


Then:

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

if and only if:

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


Proof

\(\ds \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N_{>N}: \forall x \in X: \, \) \(\ds \map d {\map {f_n} x, \map f x}\) \(<\) \(\ds \epsilon\) Definition of Uniform Convergence
\(\ds \leadstoandfrom \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N_{>N}: \, \) \(\ds \sup_{x \mathop \in X} \map d {\map {f_n} x, \map f x}\) \(\le\) \(\ds \epsilon\) Definition of Supremum of Set
\(\ds \leadstoandfrom \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N_{>N}: \, \) \(\ds \map {d_S} {f_n, f}\) \(\le\) \(\ds \epsilon\) Definition of Supremum Metric
\(\ds \leadstoandfrom \ \ \) \(\ds \map {d_S} {f_n, f}\) \(\to\) \(\ds 0\) \(\ds \text{as } n \to \infty\) Definition of Limit of Sequence in Metric Space

$\blacksquare$