Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f: \mathbb{R} \to \mathbb{R}$. For every $x \in \mathbb{R}$, there exists $\delta$, for every $y \in N(x, \delta)$ ($N$ stands for neighborhood) $f(y) \geq f(x)$.

Show that the range of $f$ is countable.

share|cite|improve this question
Please do not give orders. I'm sure you will find people are more willing to help if you just ask! – JavaMan Jun 13 '11 at 1:53
thank you for your better fashion! – Leitingok Jun 13 '11 at 2:02
No problem. The grammar to your post is still not quite right. If there is anything you would like to change, please do so. – JavaMan Jun 13 '11 at 2:08
up vote 12 down vote accepted

Suppose $a$ is in the range of $f(x)$. Then if $x_0$ is any number such that $f(x_0) = a$, one can find an interval $[b,c]$ containing $x_0$ with $b$ and $c$ rational such that $f(x) \geq a$ on $[b,c]$. Thus $a \rightarrow (b,c)$ gives a mapping from the range of $f(x)$ to ${\mathbb Q} \times {\mathbb Q}$, and this mapping is clearly injective. Hence the range of $f$ is countable.

share|cite|improve this answer
Simple and nice. – leo Jun 13 '11 at 5:54
How come the mapping is injective? You said "clearly" but I am not quite convinced... though if I would the proof would be pretty! EDIT: duh... I worked it out. Though you could've added one more line... It is not that much obvious. – Patrick Da Silva Jun 13 '11 at 6:40
@Patrick: It looks like it could be an assignment so maybe I shouldn't write too much. – Zarrax Jun 13 '11 at 11:06
Okay, cool then. – Patrick Da Silva Jun 13 '11 at 20:03

In other words, $f$ has a local minimum everywhere. More generally, the same result holds if $f$ has a local extremum everywhere; according to this (Problem 2010-4/B), the result can be found in several places in the literature (a proof is also provided). Interestingly, as a corollary in the case where $f$ is further continuous, we get an answer to this question (note my answer given there).

EDIT (elaborating on the second part): Suppose that $f:\mathbb{R} \to \mathbb{R}$ is continuous and has a local extremum everywhere. From the latter property, it follows that the range of $f$ is countable. Hence we conclude from the intermediate value theorem that $f$ must be constant (for otherwise the range would be uncountable, a contradiction).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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