Open Ball in Supremum Space of Continuous Functions on Interval I'm having trouble with an exercise in W.A. Sutherland's "Introduction to Metric and Topological Spaces" 1st Edition (Oxford Science Publications, 1975), Chapter $2$, Exercises $2.6: 25$.
Let $\mathscr C [a, b]$ be the set of all continuous real functions $f: [a, b] \to \mathbb R$, where $[a, b]$ denotes the closed real interval from $a$ to $b$: $[a, b] = \{x \in \mathbb R: a \le x \le b\}$.
Let $d: \mathscr C^2 \to \mathbb R$ be the supremum metric on $\mathscr C [a, b]$ defined as:
$\displaystyle \forall f, g \in \mathscr C: d (f, g) := \sup_{x \mathop \in [a, b]} |f(x) - g (x)|$
where $\sup$ denotes the supremum function.
Let $f, g \in \mathscr C [a, b]$ be such that $\forall x \in [a, b]: f(x) < g(x)$
Consider the set $S$, defined as:
$S = \{h \in \mathscr C [a, b]: \forall x \in [a, b]: f(x) < h(x) < g(x)\}$
Is $S$ an open ball of $(\mathscr C [a, b], d)$?
The answer given is that $S$ is an open ball of $(\mathscr C [a, b], d)$ if and only if $g(x) - f(x)$ is constant.
Proving the necessary condition is straightforward: if $g(x) - f(x)$ is constant, then $S$ is the open ball whose center is $\rho$ defined such that $\rho(x) = f(x) + \dfrac {g(x) - f(x)} 2$. Proving this is an open ball is easy.
Going the other way is a little more difficult.
This is what I've done so far.
Let it be assumed that $S$ is an open ball.
Then there exists $\phi \in \mathscr C [a, b]$ and $\epsilon \in \mathbb R_{>0}$ such that $B_\epsilon(\phi) = S$.
That is:
$$\forall \rho \in S: \sup_{x \mathop \in [a, b]} | \rho (x) - \phi (x) | < \epsilon$$
Aiming for a contradiction, suppose it is not the case $f$ and $g$ are such that $\forall x \in [a, b]: g(x) - f(x) = c$ for some constant $c \in \mathbb R$.
Then
$$\exists \xi, \zeta \in [a, b]: g (\xi) - f (\xi) > g (\zeta) - f (\zeta)$$
Then:
$$\sup_{x \mathop \in [a, b]} |g (x) - f (x)| \ge g (\xi) - f (\xi) > g (\zeta) - f (\zeta)$$
As $f$ and $g$ are continuous:
$$\exists (p, q) \subset [a, b]: \forall x \in (p, q): |g (x) - f (x)| > g (\zeta) - f (\zeta)$$
Let $h \in S$ such that:
$$\exists r, s \in (p, q): |h(r) - h(s)| > g (\zeta) - f (\zeta)$$
That is, in the interval $(p, q)$, the oscillation on $h$ is greater than the minimum distance of $g - f$.
The intention is to prove that $h$ cannot be in the open ball that is $S$, but I've lost the train of my thinking and I can't see my way through to the end.
Can someone guide me out of the woods?
 A: Suppose that $g-f$ is not constant. Take $h\in S$ and $r>0$ such that $B_r(h)\subset S$; I will prove that $B_r(h)\varsubsetneq S$. Take $N\in\Bbb N$ such that $\frac1N<r$. For each $n\geqslant N$, let $g_n=h+r-\frac1n$. Then $g_n\in B_r(h)$ and therefore $g_n\in S$, which implies that $(\forall x\in[a,b]):g_n(x)<g(x)$. But then, for each $x\in[a,b]$,\begin{align}h(x)+r&=\lim_{n\to\infty}h(x)+r-\frac1n\\&=\lim_{n\to\infty}g_n(x)\\&\leqslant g(x).\end{align}By a similar argument,$$(\forall x\in[a,b]):h(x)-r\geqslant f(x)$$and therefore$$(\forall x\in[a,b]):g(x)-f(x)\geqslant2r.$$But $g-f$ cannot be the constant function $2r$ and so there is some $x_0\in[a,b]$ such that $g(x_0)-f(x_0)>2r$. So, you must have $g(x_0)>h(x_0)+r$ or $f(x_0)<h(x_0)-r$ (or both). If$$g(x_0)>h(x_0)+r,\tag1$$then the function $g-h$ is continuous, it is always greater than or equal to $r$, and $(1)$ holds. Take $k<g(x_0)-h(x_0)-r$ and such that $g-k>f$. Then $g-k\in S$. But $g-k\notin B_r(h)$, since $g(x_0)-k-h(x_0)>r$.
