0
$\begingroup$

Question: If every point is a local maximum point of $f$, prove that $f$ takes on only a countable set of values.

Attempt: I tried attributing to each point $x$, $2$ rational numbers $a_x,b_x$ such that $f(x)$ is the maximum value on $(a_x,b_x)$, and since there are is a countable set of rational numbers, there's only a countable set of such intervals and so a countable set of values of $f(x)$. However, I can't prove that I can make it so $a_x\neq a_y$ or $b_x \neq b_y$ for $x \neq y$, so I can't remove multiplicities and assume the set of all the intervals is countable. This is where I'm stuck and would love some help.

$\endgroup$
1
$\begingroup$

So for each $x\in\Bbb R$, there are $a_x<x<b_x$ such that $f(y)\ge f(x)$ for all $y\in (a_x,b_x)$. By density of $\Bbb Q$, wlog $a_x,b_x\in\Bbb Q$. If $a_x=a_\xi$ and $b_x=b_\xi$, then $f(y)\ge f(x)$ and $f(y)\ge f(\xi)$ for all $y\in (a_x,b_x)$, so by considereing $y=x$ and $y=\xi$, we find $f(x)\ge f(\xi)$ and $f(\xi)\ge f(x)$, i.e., $f(x)=f(\xi)$. But if $f(x)$ depends only on $\langle a_x,b_x\rangle\in\Bbb Q^2$, $f$ can take only $|\Bbb Q^2|$-many values.

$\endgroup$

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.