0
$\begingroup$

I have a question about proof of intermediate value theorem. I will post the entire proof here.

Let $I\subset \Bbb{R}$ be interval $I=(a,b)$, and $f:I\rightarrow\Bbb{R}$ be a continous function. then if some $u$ is between $f(a),f(b)$, then there exists some $c\in (a,b)$ such that $f(c)=u$.

WLOG assume $f(a)<f(b)$. For $f(a)>f(b)$ the proof is very simillar.

Let $S=\{x\in(a,b);f(x)<u\}$. S is non-empty, because $a\in S$ and $S$ is bounded above by $b$. By completeness, $\sup{S}$ exists, call it $c$. We claim, that $f(c)=u$.

Choose some $\epsilon>0$. $f$ is continuous, thus $$f(x)-\epsilon<f(c)<f(x)+\epsilon$$ for all $x\in (c-\delta,c+\delta)$.

By the properties of the supremum, we can chose some $a^*$ from $(c-\delta,c)$ such that $a^*\in S$. We have $$f(c)<f(a^*)+\epsilon\leq u+\epsilon$$

Similarly we can chose $a^{**}$ from $(c,c+\delta)$ such that $a^{**}\notin S$. We have $$f(c)>f(a^{**})-\epsilon\geq u-\epsilon$$

which together gives $$u-\epsilon<f(c)<u+\epsilon$$ which holds for all $\epsilon >0$, thus $f(c)=u$ is the only possible value, as claimed.

My question is: Why can't we conclude the proof after saying that the supremum exist? I don't understand in which sense we used the fact that $f$ is continuous. As I know, if we don't use all assumptions in a proof, it is somehow wrong. Would anyone please explain me, what could happen if $f$ wasn't continuous? Why can't we say directly from definition of $S$ that $\sup{S}=u=f(c)$, I must be missing something.

$\endgroup$
2
$\begingroup$

The continuity of the function $f$ is used to choose some $a^*$ and $a^{**}$ that are within $\epsilon$ of the value $f(c)$. You cannot assume that any choice of $a^*\in(c-\delta,c)$ will satisfy $f(c)<f(a^*)+\epsilon$ without the continuity of $f$ (the continuity of $f$ implies $|f(x)-f(c)|< \epsilon$ for all $x\in(c-\delta,c+\delta)$). Similarly with the choice of $a^{**}$.

$\endgroup$
4
  • $\begingroup$ Yeah, that makes sense. But why do i have to work with the continuity part at all? By the definition of $S$ itself, i have that $\sup S=u=f(c)$. What am i missing? $\endgroup$ – Michal Dvořák Jul 4 '18 at 15:53
  • $\begingroup$ This is not quite correct. You only know that $c\ge x$ for all $x\in S$. This is the definition of $\sup S$. In other words, $c$ is the least upper bound on all of the values $x$ for which $f(x)<u$. This doesn't allow you to immediately conclude anything about the value $f(u)$ itself. In short, the supremum of a set allows you to conclude things about $\textit{that set}$. You cannot jump to facts about a function's behavior on the set, etc. $\endgroup$ – Beckham Myers Jul 4 '18 at 16:03
  • $\begingroup$ Ah, that's it. So basically we can have that anywhere close to $f(u)$, the function is well behaving, and if $f$ was discontinuous at $u$, then i would be wrong, is that correct? $\endgroup$ – Michal Dvořák Jul 4 '18 at 16:07
  • 2
    $\begingroup$ Yes, the continuous behavior of $f$ around $c$ allows you to "shrink" the interval of possible values around $u$ arbitrarily small so that you can conclude that $f(c)=u$ as desired. $\endgroup$ – Beckham Myers Jul 4 '18 at 16:10
2
$\begingroup$

Suppose that $a=-1$, $b=1$, and $f\colon[-1,1]\longrightarrow\mathbb R$ is defined by$$f(x)=\begin{cases}-1&\text{ if }x<0\\1&\text{ otherwise.}\end{cases}$$Take $u=0$. Then $S=[-1,0)$ and therefore $c=0$. However, $f(c)=f(0)=1\neq0$.

The problem here is that $f$ is discontinuous at $0$. Otherwise, we would necessarily have $f(c)=0$.

$\endgroup$

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.