Supremum Metric on Continuous Real Functions/Examples/Closure of Open 1-Ball of 0 on Unit Interval

Example of Supremum Metric on Continuous Real Functions

Let $\closedint 0 1 \subseteq \R$ be the closed unit interval.

Let $\mathscr C \closedint 0 1$ be the supremum space of continuous functions $f: \closedint 0 1 \to \R$.

Then:

$\map \cl {\map {B_1} \bszero} = \set {f \in \mathscr C \closedint 0 1: \map {d_\infty} {f, \bszero} \le 1}$

where:

$\map {B_1} \bszero$ denotes the open $1$-ball of $\bszero$

$\d_\infty$ denotes the Chebyshev distance

$\bszero$ denotes the constant function $f_0$.

Proof

Let $S = \set {f \in \mathscr C \closedint 0 1: \map {d_\infty} {f, \bszero} \le 1}$.

We have that:

$\map \cl {\map {B_1} \bszero} \subseteq S$

Suppose $f \in S$.

Let $\epsilon \in \R_{>0}$ be given.

Then:

$\paren {1 - \dfrac \epsilon 2} \in \map {B_1} \bszero \cap \map {b_\epsilon} f$

and so:

$f \in \map \cl {\map {B_1} \bszero}$

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 36$