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Let X be a non-empty complete metric space and let $\{f_n:X\to \Bbb R\}^\infty_{n=1}$ be a sequence of continuous functions with the following property: for each $c\in X$, there exists an integer $N_x$ so that $\{f_n(x)\}_{n\ge N_x}$ is either a monotone increasing or decreasing sequence. Prove that there is a non-empty open subset $U\subseteq X$ and an integer $N$ so that the sequence $\{f_n(x)\}_{n\ge N}$ is monotone for all $x\in U$.

I am sure that I have to use Baire Category theorem, $X$ is complete metric space and I have to produce closed set $A_n$ with the given condition so that $X=\bigcup^\infty_1 A_n$ so that one must be no where dense. But my query is how $A_n$ can be written so that $A_n$ is closed with given property. Please write me only how $A_n$ looks like. Thank you

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Let for $n \in \mathbb N$ $$ A_n = \Bigl\{x \in X \mid \bigl(f_k(x)\bigr)_{k \ge N} \text{ is monotone}\Bigr\} $$ To see that $A_n$ is closed, write it is $$ A_n = \bigcap_{k\ge n}\{f_k \le f_{k+1}\} \cup \bigcap_{k\ge n} \{f_k \ge f_{k+1}\} $$

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    $\begingroup$ I'm sorry, why it is closed? Could u explain me little bit, please.. $\endgroup$
    – Toeplitz
    Jun 23 '14 at 17:17
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    $\begingroup$ For two continuous functions $f,g$, the set $\{f \ge g\} = (f-g)^{-1}([0,\infty))$ is closed. The intersection of closed sets is closed. $\endgroup$
    – martini
    Jun 23 '14 at 17:32

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