3
$\begingroup$

If there is, what's an example. If not, how do I prove none exists?

$\endgroup$
  • 2
    $\begingroup$ Hint: A continuous function $f: [0,1]\to (0,1)$ must attains it's maximum as $[0,1]$ is closed and bounded. $\endgroup$ – user99914 May 11 '15 at 5:16
  • 2
    $\begingroup$ The image of $f$ is doomed to be compact. $\endgroup$ – Michael Hoppe May 11 '15 at 5:32
  • 1
    $\begingroup$ Without compactness you can argue that the inverse image of an open set is open. $\endgroup$ – hjhjhj57 May 11 '15 at 5:40
  • 2
    $\begingroup$ @hjhjhj57 That doesn't work. The closed interval $[0,1]$ is open in the subspace topology. $\endgroup$ – Mike Haskel May 11 '15 at 5:48
  • $\begingroup$ @MikeHaskel You're right, thanks! $\endgroup$ – hjhjhj57 May 11 '15 at 5:49
16
$\begingroup$

No. The continuous image of a compact set is compact.

$\endgroup$

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.