0
$\begingroup$

I'm reading some notes, and on page 11 the author uses the function f(x):[0,2], f(x) = 1/(x^2 - 2) over the rational numbers to show that the extreme value theorem doesn't hold(over the rationals).

I'm wondering if it's correct to even apply the extreme value theorem for this function - isn't one of the requirements that the function be bounded(which this one isn't)?

Apologies for the lack of formatting, I have no knowledge about that (yet).

$\endgroup$
  • $\begingroup$ A much easier example would be $f:[0,\sqrt{2}]\rightarrow\mathbb{R}$, $f(x)=x^2$... $\endgroup$ – JP McCarthy May 29 '14 at 8:33
  • $\begingroup$ On page 111 of the notes you linked, the only condition is that $f$ be continuous on a closed finite interval. That $f$ is bounded is part of the conclusion of the theorem, not a requirement. $\endgroup$ – David May 29 '14 at 8:43
0
$\begingroup$

As pointed out in the comments, that example does work. But if it makes you feel better, think about $\sin(x)$ on, say, $[0,2]$. The maximum of $\sin$ occurs at $x= \pi/2, 5\pi/2$, etc., none of which are rational. You can show that $\sin(x)$ has no maximum on the rational numbers in $[0,2]$, since you can plug in rational values of $x$ arbitrarily close to $\pi/2$ and the $y$-values approach 1, but never get there.

$\endgroup$
  • $\begingroup$ Ok - I think I understand. But the example function makes a jump, although at an irrational number. I guess the limit and derivative exist at any given point between 0 and 2, but it just seems intuitively wrong that a function with a jump is continuous..? Thanks a lot in any case! $\endgroup$ – Zak Laberg May 29 '14 at 9:34
  • $\begingroup$ Well, that's exactly the point the notes are trying to make about why we don't do calculus on the rational numbers. On the real numbers, continuous functions look like nice, non-jumpy curves. But if you restrict to the rational numbers, all of a sudden, even continuous functions make little jumps all over the place. Those jumps are why theorems that should be true about continuous functions don't really work on the rationals. $\endgroup$ – coolpapa May 29 '14 at 9:38

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.