For $f :\mathbb R \to \mathbb R $, there exists an $(a,b)$, such that $f$ is bounded on a sequence with limit $x$, for all $x\in(a,b)$

Question

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!

Answer 1

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 $c=n$.