Let $(I_n)_{n \in \mathbb N}$ be the sequence of intervals of $[0,1]$ with rational endpoints, and for every $n \in \mathbb N~$ let $E_n=\{f \in C[0,1] : f \:\text{is monotone in}\: I_n\}$. Prove that for every $n \in \mathbb N$, $E_n$ is closed and nowhere dense in $(C[0,1],d_\infty)$. Deduce that there are continuous functions in the interval $[0,1]$ which aren't monotone in any subinterval.
For a given $n$, $E_n$ can be expressed as $E_n=E_{n\nearrow} \cup E_{n\swarrow}$ where $E_{n\nearrow}$ and $E_{n\swarrow}$ are the sets of monotonically increasing functions and monotonically decreasing functions in $E_n$ respectively. I am having problems trying to prove that these two sets are closed. I mean, take $f \in (C[0,1],d_\infty)$ such that there is $\{f_k\}_{k \in \mathbb N} \subset E_{n\nearrow}$ with $f_k \to f$. How can I prove $f \in E_{n\nearrow}$?. Suppose I could prove this, then I have to show that $\overline {E_n}^\circ=E_n^\circ=\emptyset$. This means that for every $f \in E_n$ and every $r>0$, there is $g \in B(f,r)$ such that g is not monotone. Again, I am stuck. If I could solve this two points, it's not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.
Best Answer
It is easy to show that $E_n$ is close. Just like what you did, let $E_n^1$ be the subset of $E_n$ containing all functions increasing on $I_n$. Let $\{ f_k\}$ be a sequence in $E_n$ converging to $f\in C[0,1]$. Pick $x, y\in I_n$, $x<y$. Then $f_n(x) \leq f_n(y)$ for all $n$. Take $n\to \infty$ gives $f(x) \leq f(y)$. Thus $E_n^1$ is closed. Now $E_n$ is the union of two close set, so it is close.
To show that $E_n$ has empty interior, let $f\in E_n$. Assume that $f$ is inceasing (the case when $f$ is decreasing can be done similarly). We perturb $f$ a little bit to get a function $h$ which is not monotone. Let Write $I_n = [a, b]$. Let $\epsilon >0$ be arbitrary. Let $c_1, c_2\in (a, b)$ such that
$$ c_1 < c_2, \text{ and }\ f(b) - \epsilon < f(c_2) \leq f(b).$$
$c_1, c_2$ can be found as $f$ is continuous. Let $g$ be a continuous function on $[0,1]$ with the properties:
$$|g|\leq \epsilon,\ g(a) = g(b) = -\epsilon,\ g(c_1)= g(c_2)=0,\ .$$ Then $h=f-g$ satisfies
$$d(f, h)_\infty \leq \epsilon$$
and $h(a) = f(a) - \epsilon < f(a)\leq f(c_1)= h(c_1)$ and $h(c_2) = f(c_2) > f(b) - \epsilon = h(b)$. Thus $h$ is not monotone. As $f$, $\epsilon>0$ are arbitrary, $E_n$ contains no open set. Hence $E_n$ are nowhere dense. (The conclusion is quite interesting).