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!

  • 1
    $\begingroup$ 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


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 $c=n$.


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.