Theorem: Let $f$ be continuous on $[a,\,b]$ and assume $f(a)<f(b)$. Then for every $k$ such that $f(a)<k<f(b)$, there exists a $c\in[a,\,b]$ such that $f(c)=k$.

proof: $f$ continuous at $a\implies$ for $\varepsilon=k-f(a)>0$, $\exists\delta>0$ s.t. $$|f(x)-f(a)|<\varepsilon=k-f(a)\quad\forall x\colon |x-a|<\delta.$$ Consider the set $H=\{x\in[a,\,b]\colon f(x)<k\}\not=\emptyset \implies c=\sup{(H)}$. Show $f(c)=k$, suppose $f(c)<k\iff k-f(c)>0$. We know $f$ is continuous at $c$ so $\forall\varepsilon>k-f(c)>0$ $\exists\delta>0:|f(x)-f(c)|<\varepsilon=k-f(c)$ when $|x-c|<\delta$. $\implies f(x)-f(c)<k-f(c)$. Say $x=c+\delta/2\implies f(x)<k\implies c+\delta/2\in H$ which contradicts the fact $c=\sup{(H)}$, since $\delta>0$. Same proof works if $f(c)>k$, thus $f(c)=k$.

This is a proof for the intermediate value theorem given by my lecturer, I was wondering if someone could explain a few things:

  • What is the set $H$, what does it define?
  • Why does contradicting the fact that $c=\sup{(H)}$ prove that $f(c)\not<k$?
  • How would I continue to actually finish this proof, ie. show $f(c)\not>k$?
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    $\begingroup$ And the question is ... ? $\endgroup$
    – enzotib
    Jan 5, 2015 at 22:23
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    $\begingroup$ the first line of th proof has no sense... i.e $\forall \varepsilon=k-f(a)>0$... $\endgroup$
    – idm
    Jan 5, 2015 at 22:25
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    $\begingroup$ @user2850514 But if the $\;\epsilon\;$ is "specific" then that quantifier there makes no sense at all. Even worse, it confuses. $\endgroup$
    – Timbuc
    Jan 5, 2015 at 22:34
  • $\begingroup$ The "Therefore $\;f\;$ is continuous (where? at all $\;x\in [a,b]\;$?) iff (??) $\;f(x)<k\;$" is very strange, and if true then you've proved the IVT isn't true as $\;f(x)\neq k\;\;\forall\,x\in [a,b]\;$ ...! Unless, of course, you meant something else. There are other incorrect, or at least messy, things: $\;f\;$ is understood to be continuous at $\;a\;$ only from the right, and thus a neighborhood $\;|x-a|<\delta\;$ doesn't make much sense, etc. $\endgroup$
    – Timbuc
    Jan 5, 2015 at 22:38
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    $\begingroup$ I deleted that question. I have changed the quantifier. $\endgroup$ Jan 5, 2015 at 23:08

2 Answers 2


The set $H$ is the set of elements $x\in [a,b]$ for which $f(x)<k$.

Since this set is bounded, it has a supremum $c\in[a,b]$. There are three cases:

  1. $f(c)=k$
  2. $f(c)<k$
  3. $f(c)>k$

In the first case, you're done.

Suppose $f(c)<k$. In particular $c\in H$ and, moreover $a<c<b$, because by assumption, $f(a)<k$ and $f(b)>k$. By continuity of $f$ at $c$, there is $\delta>0$ such that $a<c-\delta<c+\delta<b$ and $$ |x-c|<\delta \implies f(x)<k $$ (by considering the continuous function $k-f(x)$ which is positive at $c$). But then $c+\delta/2\in H$, because $f(c+\delta/2)<k$. A contradiction to the fact that $c=\sup H$.

Suppose $f(c)>k$. By the same argument as before, there exists $\delta>0$ such that $a<c-\delta<c+\delta<b$ and $$ |x-c|<\delta \implies f(x)>k $$ But, by definition of supremum, there is $x\in H$ such that $c-x<\delta$; for this $x$ we have both $f(x)<k$ and $f(x)>k$, a contradiction.

  • $\begingroup$ Thanks, I have a few questions here: Why does $f(a)<k<f(b)\implies a<c<b$? Secondly, did you derive $|x-c|<\delta\implies f(x)<k$ the same way my lecturer did? (ie. same was as in the above post) $\endgroup$ Jan 5, 2015 at 23:50
  • 1
    $\begingroup$ @user2850514 In the case $f(c)<k$ it's clear that $c<b$. It's also $c>a$ or $H=\{a\}$ (then argue why this can't happen). If $f(c)>k$, then $c>a$; if $c=b$, then $H=[a,b)$ (argue why this can't happen). Yes, I did the same as your lecturer: just consider $k-f(x)>0$. $\endgroup$
    – egreg
    Jan 5, 2015 at 23:58
  • $\begingroup$ @egreg One question. For the case that f(x) < k, would be correct to say, instead of your argument, that in that case, delta + c would be the supremum, and as "If a set has a supremum, then it has only one supremum" then there are two supremums, a contradiction. $\endgroup$
    – Beginner
    Oct 19, 2016 at 20:31
  • $\begingroup$ @Beginner Why should $\delta+c$ be a supremum? $\endgroup$
    – egreg
    Oct 19, 2016 at 20:59
  • $\begingroup$ @ because the absolute value of x-c is less than delta because f is continuous, right? $\endgroup$
    – Beginner
    Oct 19, 2016 at 21:02

Ideas for a possibly clearer, easier proof:

1) Prove that if $\;f\;$ is continuous at $\;[a,b]\;$ and $\;f(a)f(b)<0\;$ then there exists $\;c\in (a,b)\;\;s.t.\;\;f(c)=0\;$

2) prove the complete IVT by choosing $\;g(x):=f(x)-k\;$ and applying (1) to $\;g\;$ .

Highlights of proof of (1) : WLOG assume $\;f(a)<0\;,\;\;f(b)>0\;$ . Take the point $\;w_1:=\frac{a+b}2\;$ . If $\;f(w_1)=0\;$ we're done, otherwise choose the interval

$$\begin{cases}\left[\,a\,,\,w_1:=\frac{a+b}2\,\right]&,\;\;\text{if}\;\;f(w_1)>0\\{}\\or\\{}\\ \left[\,w_1:=\frac{a+b}2\,,\,b\,\right]&,\;\;\text{if}\;\;f(w_1)<0\end{cases}$$

Take now $\;w_2=\frac{a+w_1}2\;\;or\;\;w_2=\frac{w_1+b}2\;$ , depending on what interval we chose above, resp. If $\;f(w_2)=0\;$ we're done, otherwise do as above.

Continue the process above inductively

Apply now Cantor's Theorem (of embedded closed interval which lengths tend to zero) to deduce there's one single point $\;c\;$ in the intersection all these interval, use continuity of $\;f\;$ to show that $\;f(\text{left endpoints of intervals})\to f(c)\;$ , and likewise for right end points, and by construction get

$$0\le f(c)\le 0\implies f(c)=0$$

Fill in details.

  • $\begingroup$ Hi, thanks for the answer. I haven't however met cantor's theorem and am looking for a much more rigorous proof (by the definition of continuity and such) rather than using numerical methods to approximately find the root. $\endgroup$ Jan 5, 2015 at 23:05
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    $\begingroup$ Cantor's Theorem is usually used to prove basic theorems, like Bolzano-Weierstrass in sequences. Haven't you messed with this? And please observe that the above proof is all the rigurous you can expect and there is no approximation at all of roots (?) or whatever! $\endgroup$
    – Timbuc
    Jan 5, 2015 at 23:15

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