2
$\begingroup$

So, I've started studying Calculus less than a month ago but now I've been stuck for two days trying to prove the extreme value theorem and it's really stressing me out. I've searched on other Calculus books but none seems to provide a proof for it, and trying to look for a proof on the internet has only gotten me to study yet other theorems which seem hardly within the scope of Calculus.

So far I've managed to prove that if $f$ is continuous in $[a,b]$, then it must also be bounded in that interval. I also know how to prove that $f$ attains its least upper bound in $[a,b]$, provided that it exists. Basically, all I want to prove now is that if $f$ is continuous and bounded in $[a,b]$, then it must also have a least upper bound. But however obvious it may seem to me, I can't seem to come up with a proof. Is there a way to prove it?

Thank you very much for any help.

  • 0
    Have you seen Weierstrass' theorem already? – 2017-02-07
  • 0
    If you're referring to Bolzano-Weierstrass theorem, yes. I used it to prove that f is bounded in [a,b]. – 2017-02-08
  • 0
    I am referring to "every function $f(x)$ that is continuous on a compact interval $[a,b]$ attains its maximum and minimum in $[a,b]$". – 2017-02-08
  • 0
    Thats known as the extreme value thereom – 2017-02-08

2 Answers 2

1

A basic property of the real numbers is that every non-empty set with an upper bound has a least upper bound. Your desired result follows immediately from this (given what you've already shown).

1

You are almost done: just observe that the real line has the so-called supremum property: any nonempty bounded subset of it has a supremum. You can very well assume that the real line has this property as an axiom, even though this is a consequence of the way it is built from the rationals.