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 $S$
- $\map {d_S} {f_n, f} \to 0$ 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 |
$\blacksquare$