# 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 $x\in(a,b)$

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...

• 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. Nov 1, 2014 at 11:43
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 $c=n$.