Open Balls of Supremum Metric on Continuous Real Functions on Closed Interval

== Theorem ==
Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.
Let $d: \mathscr C^2 \to \R$ be the supremum metric on $\mathscr C \closedint a b$ defined as:
:$\ds \forall f, g \in \mathscr C \closedint a b: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$
where $\sup$ denotes the supremum.
Let $f, g \in \mathscr C \closedint a b$ be such that:
:$\forall x \in \closedint a b: \map f x < \map g x$
Consider the set $S$, defined as:
:$S = \set {h \in \mathscr C \closedint a b: \forall x \in \closedint a b: \map f x < \map h x < \map g x}$
Then:
:$S$ is an open ball of $\struct {\mathscr C \closedint a b, d}$
:
:$\exists c \in \R_{>0}: \map g x := \map f x + c$
That is, $f$ and $g$ differ by a constant.
== Proof ==
Recall the definition of open ball:
The open $\epsilon$-ball of $a$ in $M = \struct {A, d}$ is defined as:
:$\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$
In this context, the open $\epsilon$-ball of $\phi$ in $\mathscr C \closedint a b$ is defined as:
:$\ds \map {B_\epsilon} \phi := \set {\rho \in \mathscr C \closedint a b: \sup_{x \mathop \in \closedint a b} \size {\map \rho x - \map \phi x} < \epsilon}$
=== Necessary Condition ===
Suppose $g \in \mathscr C \closedint a b$ is defined as:
:$\forall x \in \closedint a b: \map g x := \map f x + c$
Let $\epsilon = \dfrac c 2$.
Let $\phi \in \mathscr C \closedint a b$ be defined as:
:$\forall x \in \closedint a b: \map \phi x = \map f x + \epsilon$
Then we have that:
and it is seen by definition of open $\epsilon$-ball of $\phi$:
:$S = \map {B_\epsilon} \phi$
=== Sufficient Condition ===
Let $S$, defined as:
:$S = \set {h \in \mathscr C \closedint a b: \forall x \in \closedint a b: \map f x < \map h x < \map g x}$
be an open ball in $\mathscr C \closedint a b$.
Thus there exists $\epsilon \in \R_{>0}$ and $\phi \in \mathscr C \closedint a b$ such that:
:$S = \map {B_\epsilon} \phi$
That is:
:$\ds \exists \epsilon \in \R_{>0}: \forall \rho \in S: \sup_{x \mathop \in \closedint a b} \size {\map \rho x - \map \phi x} < \epsilon$
 it is not the case $f$ and $g$ are such that:
:$\forall x \in \closedint a b: \map g x - \map f x = c$
for some constant $c \in \R$.
Take $h \in S$ and $r \in \R_{>0}$ such that $\map {B_r} h \subseteq S$.
It is to be shown that $\map {B_r} h \subsetneq S$.
Take $N \in \N$ such that $\dfrac 1 N < r$.
For each $n \ge N$, let $g_n = h + r - \dfrac 1 n $.
Then:
:$g_n \in \map {B_r} h$
and therefore:
:$g_n \in S$
which implies that:
:$\forall x \in \closedint a b: \map {g_n} x < \map g x$
But then:
:$\ds \forall x \in \closedint a b: \map h x + r = \lim_{n \mathop \to \infty} \map h x + r - \dfrac 1 n = \lim_{n \mathop \to \infty} \map {g_n} x \le \map g x$
By a similar argument:
:$\forall x \in \closedint a b: \map h x - r \ge \map f x$
and therefore:
:$\forall x \in \closedint a b: \map g x - \map f x \ge 2 r$
But $g - f$ cannot be the constant function $2 r$.
Hence there is some $x_0 \in \closedint a b$ such that $\map g {x_0} - \map f {x_0} > 2 r$.
So it must be the case that:
:$\map g {x_0} > \map h {x_0} + r$
or:
:$\map f {x_0} < \map h {x_0} - r$
or both.
If:
:$(1): \quad \map g {x_0} > \map h {x_0} + r$
then:
:$g - h$ is continuous
:$g - h$ is always greater than or equal to $r$
and $(1)$ holds.
Take $k < \map g {x_0} - \map h {x_0} - r$
and such that $g - k >f$.
Then $g - k \in S$.
But $g - k \notin \map {B_r} h$, because $\map g {x_0} - k - \map h {x_0} > r$.
It follows that $S$ cannot be an open $\epsilon$-ball of $\phi$ in $\mathscr C \closedint a b$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 25$
* 
Category:Supremum Metric
Category:Open Balls