 # Proving the existence of a non-monotone continuous function defined on [0,1] wstecznyg5 2022-07-22 Answered
Proving the existence of a non-monotone continuous function defined on [0,1]
Let $\left({I}_{n}{\right)}_{n\in \mathbb{N}}$ be the sequence of intervals of [0,1] with rational endpoints, and for every let ${E}_{n}=\left\{f\in C\left[0,1\right]:f\phantom{\rule{mediummathspace}{0ex}}\text{is monotone in}\phantom{\rule{mediummathspace}{0ex}}{I}_{n}\right\}$. Prove that for every $n\in \mathbb{N}$, ${E}_{n}$ is closed and nowhere dense in $\left(C\left[0,1\right],{d}_{\mathrm{\infty }}\right)$. 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 ${E}_{n}={E}_{n↗}\cup {E}_{n↙}$ where ${E}_{n↗}$ and ${E}_{n↙}$ 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 $f\in \left(C\left[0,1\right],{d}_{\mathrm{\infty }}\right)$ such that there is $\left\{{f}_{k}{\right\}}_{k\in \mathbb{N}}\subset {E}_{n↗}$ with ${f}_{k}\to f$. How can I prove $f\in {E}_{n↗}$?. Suppose I could prove this, then I have to show that ${\overline{{E}_{n}}}^{\circ }={E}_{n}^{\circ }=\mathrm{\varnothing }$. This means that for every $f\in {E}_{n}$ and every $r>0$, there is $g\in B\left(f,r\right)$ 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.
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Step 1
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 $\left\{{f}_{k}\right\}$ be a sequence in ${E}_{n}$ converging to $f\in C\left[0,1\right]$. Pick $x,y\in {I}_{n}$, $x. Then ${f}_{n}\left(x\right)\le {f}_{n}\left(y\right)$ for all n. Take $n\to \mathrm{\infty }$ gives $f\left(x\right)\le f\left(y\right)$. 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 ${I}_{n}=\left[a,b\right]$. Let $ϵ>0$ be arbitrary. Let ${c}_{1},{c}_{2}\in \left(a,b\right)$ such that
${c}_{1},{c}_{2}$ can be found as f is continuous. Let g be a continuous function on [0,1] with the properties:

Step 2
Then $h=f-g$ satisfies $d\left(f,h{\right)}_{\mathrm{\infty }}\le ϵ$ and $h\left(a\right)=f\left(a\right)-ϵ and $h\left({c}_{2}\right)=f\left({c}_{2}\right)>f\left(b\right)-ϵ=h\left(b\right)$. Thus h is not monotone. As f, $ϵ>0$ are arbitrary, ${E}_{n}$ contains no open set. Hence ${E}_{n}$ are nowhere dense. (The conclusion is quite interesting).

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