0
$\begingroup$

If $ f $ is a continuous mapping from a compact, connected metric space M to the real numbers and there exists a real number s such that f(m) never equals s, then there exists a constant $ c>0 $ such that either $ f(m)>s+c $ or $ f(m)<s-c.$ Intuitively this makes sense but I can't come up with a proof.

$\endgroup$
  • 2
    $\begingroup$ This is related to your previous question, asked from a different account. (If you are going to continue using this site, you should register your account: see the profile link on top of the screen). In addition to what was done there, use the fact that a continuous function function attains extreme values on a compact set. $\endgroup$ – user147263 Feb 2 '15 at 2:15
  • $\begingroup$ sorry, I couldn't figure out how to ask it with the same one $\endgroup$ – Gigi Feb 2 '15 at 2:18
  • $\begingroup$ does the metric space necessarily have to be compact for this to be true? $\endgroup$ – Gigi Feb 2 '15 at 2:24
  • $\begingroup$ Think of a continuous function on $(0,1)$ that does not attain its supremum and infimum. $\endgroup$ – user147263 Feb 2 '15 at 2:25
  • 1
    $\begingroup$ You need compactness to be sure that the image is compact, and connectedness to be sure that the image is connected. You need both properties to conclude your result. $\endgroup$ – Ian Feb 2 '15 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.