Showing interior of a set is empty Consider the metric space $(C[0,1],d_{\infty})$. For $x_{0} \in [0,1]$ and $M > 0$ define the set $A \subset C[0,1]$ by
$$    A = \{f \in C[0,1] \phantom{.}|\phantom{.} |f(x) - f(x_{0})| \leqslant M|x - x_{0}| \text{ for all } x \in [0,1] \}.
$$
Show that $A$ is closed and that Int$(A) = \emptyset$. 
I have shown that $A$ is closed. I know the definition of $Int(A)$ but am not sure how to apply it here. How may I show the empty character of $Int(A)$?
 A: Let $f$ be a function in $A$. $f$ belonging to the interior of $A$ means that there is a "small ball" in $A$ wrapping it. So to disprove it you need to find a way to show that this kind of "small ball"s does not exist. The simplest and straightest way is to find a sequence $(f_n)_{n\in\mathbb{N}} \notin A$ that converges to $f$ in this metric. For example you can try some functions that are close to $f$ but change fast near $x_0$.
A: Another approach. Let 
$$U=\{f\in C[0,1]:\mbox{ there is $K$ such that}|f(x)-f(x_0)|\le K|x-x_0|\,\,\forall x\in[0,1]\}$$
Note that $A\subset U$.
On the other hand, if $f,g\in U$ and $a$ is scalar:
$\begin{eqnarray}|(f+ag)(x)-(f+ag)(x_0)|&=&|f(x)+ag(x)-f(x_0)-ag(x_0)|\\
&\le&|f(x)-f(x_0)|+|ag(x)-ag(x_0)|\\
&\le&K_1|x-x_0|+|a|K_2|x-x_0|\\
&=&(K_1+|a|K_2)|x-x_0|\,\,\,\forall x\in[0,1]\end{eqnarray}$
Thus $U$ is a vector subspace of $C[0,1]$. Since $U\neq C[0,1]$, therefore $\mathrm{int}(U)=\emptyset$. But $A\subseteq U$. Then $\mathrm{int}(A)\subseteq\mathrm{int}(U)=\emptyset$.
Thus $\mathrm{int}(A)=\emptyset$
