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I want to prove the following.

Let $f : \mathbb R \mapsto \mathbb R $. Show that there exists an interval $(a,b) \in \mathbb R $ and $c >0 $: such that for any $x \in (a,b) $ there is a sequence $\{x _n \} $ with $x _n \to x $ and $|f(x _n )| \le c $.

(no assumption on $f $ being continuous )

I have no idea how to go about to prove this...

Thanks in advance!

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    $\begingroup$ A wild guess, can Baire category theorem be used? $\endgroup$
    – Alexander
    Nov 1, 2014 at 11:37
  • $\begingroup$ Consider the sets $A_n=\bigl\{\, x\mid |f(x)|\le n\,\bigr\}$. The closure of some $A_n$ contains an open interval. Show that this gives what you want. $\endgroup$ Nov 1, 2014 at 11:43

1 Answer 1

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This can be can dealt with rather easily if you now Baire Category Theorem.

As $f[\mathbb R]\subset \mathbb R$, then $\mathbb R=\bigcup_{n\in\mathbb N}f^{-1}(-n,n)$. Let $F_n=\overline{f^{-1}(-n,n)}$, and observe now that $\mathbb R$ is a countable union of closed sets. Baire's Theorem provides that at least one of the has nonempty interior, i.e., there are $a<b$ and $n\in\mathbb N$, such that $$ (a,b)\subset \overline{f^{-1}(-n,n)}. $$

Clearly, this $(a,b)$ does our job for $c=n$.

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