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To prove the existence of nowhere monotone continuous functions, one step is to show that for any interval $I \subset [0,1]$ of positive length, the set $A_I= \{f \in C([0,1]): f|_I$ is increasing$\}$ is closed (i.e. see http://www.apronus.com/math/nomonotonic.htm). I am not sure how to do prove this.

My idea is that if $(f_n) \in A_I$ converges to $f$, then for all $x <y \in I$, consider $f(x)-f(y) = (f(x)-f_n(x))+(f_n(x)-f_n(y))+(f_n(y)-f(y))$. The first and third brackets can be made arbitrarily small and the middle bracket is negative, so the left side intuitively should also be negative, which would show that $f \in A_I$. I am not sure how to actually prove this.

Thank you.

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2 Answers 2

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It is much easier, and directly follows from the result that if $a_n \le b_n$ for all $n$, and if $a_n \to a$ and $b_n \to b$, then $a \le b$.

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Your approach perhaps isn’t quite the simplest one, but it can be made to work:

$$\begin{align*} f(x)-f(y) &= \big(f(x)-f_n(x)\big)+\big(f_n(x)-f_n(y)\big)+\big(f_n(y)-f(y)\big)\\ &\le\big(f(x)-f_n(x)\big)+\big(f_n(y)-f(y)\big) \end{align*}$$

for all $n\in\Bbb N$, so

$$\begin{align*} f(x)-f(y)&\le\lim_{n\to\infty}\Big(\big(f(x)-f_n(x)\big)+\big(f_n(y)-f(y)\big)\Big)\\ &=\lim_{n\to\infty}\big(f(x)-f_n(x)\big)+\lim_{n\to\infty}\big(f_n(y)-f(y)\big)\\ &=0+0\\ &=0\;, \end{align*}$$

which is exactly what you want.

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