Theorem. Let $I:=[a,b]$ be a closed bounded interval and let $f:I\to \mathbb{R}$ be continuous on $I$. Then $f$ has an absolute maximum and absolute minimum on I.

Note that the absolute maximum portion is all I need to answer my question in this proof.


By hypothesis the set $f(I) := \{f(x): x\in I\}$ is bounded in $\mathbb{R}$ [proven in previous thm]. Let $s = \sup f(I)$. I claim that that there exists an $x_s$ in $I$ such that $s = f(x_s)$.

Since $ s = \sup f(I)$, if $n \in \mathbb{N}$, then the number $s - 1/n$ is not an upperbound of the set $f(I)$. Consequently there exists a number $x_n$ s.t.

$$s - 1/n < f(x_n) \leq s \, , \, \, \forall n \in \mathbb{N} \tag{$\star$}$$

Since $I$ is bounded, the sequence $X = (x_n)$ is bounded. Therefore, by the Bolzano-Weierstrass Theorem, there exists a subsequence $X' = (x_{n_k})$ of $X$ that converges to some number $x_s$. Since the elements of$X'$ belong to $I$ it follows that $x_s\in I$. Therefore $f$ is continuous at $x_s$ so that $\lim(f(x_{n_k})) = f(x_s)$. Applying this fact to ($\star$) and using the squeeze theorem we conclude that $\lim(f(x_{n_k})) = s$ and we have

$$f(x_s) = \lim(f(x_{n_k})) = s = \sup f(I)$$

which means $x_s$ is an absolute maximum point of $f$ on $I$.

The part I'm having difficulty with is the the "$\leq$" in inequality $(\star)$. Why can we assume that $f(x_n)$ might be equal to $\sup f(I)$? It seems as though we're assuming that which is to be proved might be true.

And furthermore, I can not recall using the squeeze theorem on strict inequalities (the left side of $(\star)$.

  • $\begingroup$ Consider a constant function $f$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 5 '18 at 20:09
  • $\begingroup$ Well, some member of $A$ may be equal to $\sup A$. After all the definition of supremum does not exclude the possibility that supremum might belong to the set. However I admit that the idea of supremum is non-trivial only when it does not belong to the set, otherwise the supremum is more commonly known as maximum element of the set. $\endgroup$ – Paramanand Singh Feb 6 '18 at 14:11
  • 1
    $\begingroup$ Also squeeze theorem can apply to strict inequalities for the simple reason that the weak inequality includes the case of strong inequality and is more general. $\endgroup$ – Paramanand Singh Feb 6 '18 at 14:12
  • $\begingroup$ Looks like your text has made the Bolzano-Weierstrass as the central theorem. You should try to prove this result (and other key results) using other formulations like Heine Borel or Nested interval principle. In this case you can skip these completeness theorems entirely and prove the theorem just using the fact that continuous functions are bounded on closed intervals. If $f(x) \neq s$ for all $x\in I$ then $g(x) =1/(f(x)-s)$ is continuous on $I$ and hence bounded. But by definition of supremum we can have $f(x) $ as close to $s$ as we want and this makes $g$ unbounded. Contradiction!! $\endgroup$ – Paramanand Singh Feb 6 '18 at 14:20
  • $\begingroup$ Yes, the text has one section concerning the Nested Interval Principle, and rarely refers to it. But it seems like the Bolzano-Weierstrass Thm is used in every other theorem for limits and continuity. Heine Borel is new to me, my text doesn't mention it. For my curiosity, are the three approaches equivalent or do you consider one more powerful? $\endgroup$ – Zduff Feb 7 '18 at 2:07

We are considering the possibility that $f(x_n)=s$, we are not assuming this is the case. It would be wrong to say $f(x_n)<s$ (the theorem shows this is not always the case!).

To put this another way, we do not know that the equality is obtained (you are correct, this is what we are trying to prove). However, by definition of the supremum, we can say with certainty that for any $x\in I$ that $f(x)\le s$.


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.