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I need proof this: The set $F=\left\{f\in[0,1]^{[0,1]}: \ f\ es\ monotonic\right\}$ is compact respect to product Topology. I think that $F$ is closed and then is compact because $F\subset[0,1]^{[0,1]}$ and $[0,1]^{[0,1]}$ is compact by Tychonoff theorem. But I don´t know as proof that $F$ is closed.

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Show that the set of non-monotonic functions is open : Let $f$ be a non-monotonic member of $[0,1]^{[0,1]}.$ Let $$0\leq a_1<a_2<a_3\leq 1$$ $$\text {where } f(a_2)\not \in [\min (f(a_1),f(a_3)),\max (f(a_1),f(a_3))].$$ Choose $d>0$ where $$(-d+f(a_2),d+f(a_2))\cap [\min (f(a_1),f(a_3)),\max (f(a_1),f(a_3))]=\phi.$$ For $j\in \{1,2,3\}$ let $$I_{a_j}=[0,1]\cap (-d/2+f(a_j),(d/2+f(a_j)).$$ For $x\in [0,1]\backslash \{a_1,a_2,a_3\}$, let $I_x=[0,1].$ Then $U=\prod_{x\in [0,1]}I_x$ is a nbhd of $f$ in the Tychonoff product topology on $[0,1]^{[0,1]}.$ And no $g\in U$ is monotonic because $$g\in U\implies g(a_2)\not \in [\min (g(a_1),g(a_3)),\max (g(a_1),g(a_3))].$$

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  • $\begingroup$ Note also that the set of strictly monotonic functions is not closed. For if $f_n(x)=1-(1-x)/n$ for $n\in N$ and $g(x)=1$ (for all $x\in [0,1]$ ) then $g$ is in the closure of $\{f_n :n \in N\}.$ $\endgroup$ – DanielWainfleet Dec 11 '15 at 3:46
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Assume $f$ is not in $F$. Then there exists $x>y$ such that $f(x)<f(y)$. Then $$G=\{g:[0,1]\rightarrow[0,1]| g(x)<[f(x)+f(y)]/2\,\,\mathrm{and}\,\, g(y)>[f(x)+f(y)]/2\}$$ is an open set containing $f$. Hence $F$ is closed.

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  • $\begingroup$ not only that, but $G$ is missing $F$ :) $\endgroup$ – Mirko Dec 11 '15 at 3:28

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