Set of Increasing Functions on an Interval is Closed 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.

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 $a \le b$.
