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


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 $$ A_n = \bigcap_{k\ge n}\{f_k \le f_{k+1}\} \cup \bigcap_{k\ge n} \{f_k \ge f_{k+1}\} $$

  • 1
    $\begingroup$ I'm sorry, why it is closed? Could u explain me little bit, please.. $\endgroup$
    – Toeplitz
    Jun 23 '14 at 17:17
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.