Showing a set is nowhere dense in $C([0,1])$ Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that
\begin{align*}
\lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert,
\end{align*}
for all $x \in [0,1]$. Why is $E_n$ nowhere dense in $C\big([0,1]\big)$.
I have been able to show that $E_n$ is closed in $C\big([0,1]\big)$, but haven't been able to show why the interior is empty. Any advice?...
 A: I assume you mean $C[0,1]$ with sup-norm.

Hint: Try to show that for each $n$ and arbitrary $\varepsilon>0$ there exists $g\notin E_n$ such that $\|g\|_\infty<\varepsilon$.
Try to show that $f\in E_n$ and $g\notin E_{2n}$ implies $f+g\notin E_n$.
Using these two facts you should be able that if $f\in E_n$ then in any ball $B(f,r)$ there is a function which does not belong to $E_n$.
A: Let $g(x)=\lvert x\rvert$ in $[-1,1]$, and extend $g$ to be periodic in $\mathbb R$, with period $2$.
Set
$$
g_{k,\ell}(x)=\frac{1}{k} g(\ell x).
$$
It is not hard to see that 
$$
g_{k,\ell}\in E_n \quad\text{iff}\quad n\ge \frac{\ell}{k}. 
$$
Fix now $n\in\mathbb N$. We shall show that $E_n$, which is a closed subset of $C[0,1]$, has empty interior.
Let $f\in E_n$ and $\varepsilon>0$. We shall show that $B(f,\varepsilon)\not\subset E_n$.
Let $k,\ell\in\mathbb N$, so that
$$
\frac{1}{\ell}<\varepsilon\quad\text{and}\quad \frac{k}{\ell}>2n.
$$
Then
$$
f+g_{k,\ell}\in B(f,\varepsilon) \quad\text{and}\quad f+g_{k,\ell}\not\in E_n.
$$ 
