My final for my introductory analysis course is tomorrow and my teacher gave us a list of possible theorems to prove. If anyone could please show me a proof for The Intermediate Value Theorem that is short and easy to follow, so even if I still cannot understand it I can at least memorize it. Also, I have looked through numerous texts and the internet, but they all seem to confuse me. I know that itis an insult to all you math experts to memorize proofs, but I am desperate at this point. Thank you

  • $\begingroup$ Have I committed a math exchange faux pas? Is this why I am being down voted? I can remove the question if this is the case. $\endgroup$
    – Dick
    Apr 10 '13 at 11:55
  • 3
    $\begingroup$ I don't know what you have done wrong, and i don't know why you got 3 downvotes $\endgroup$ Apr 10 '13 at 12:00
  • $\begingroup$ I don't know exactly why this post is so heavily downvoted, especially with no comments given (I generally think that if I don't know why it was downvoted, then many other people don't know either so bad behavior is not warned against). But I think it may be because you devote much of your question to self-hate comments. These really don't belong here. It would probably serve you better to just focus on the math. Secondly, in this course, you probably have a text/have proven the IVT already. Perhaps they are downvoting because of 'no research effort?' I don't know. But I wish you luck. $\endgroup$
    – davidlowryduda
    Apr 10 '13 at 12:00
  • $\begingroup$ I want to thank you very much for your proof, even if do not know as of yet what the nested interval theorem is. As far as my research, most of the proofs I come across have lemmas I do not understand or that do a lot of referencing to theorems that if where not proven in class and I can't use them with out proving them as well. Which as you can see is my problem. But thank you again. $\endgroup$
    – Dick
    Apr 10 '13 at 12:09
  • $\begingroup$ @Dick: Have a look at Nested intervals theorem on Wikipedia. Maybe you had this theorem in your course, but under a different name? BTW putting this comment under Dominic's post wold have been better, since it relates to his proof. $\endgroup$ Apr 10 '13 at 12:41

The indermediate value theorem says:

Let $f:[a,b]\to \mathbb{R}$ be continuous and $f(a)<0$ and $f(b)>0$, then there exists a $\xi \in (a,b)$ such that $f(\xi)=0$.

You can prove it by using nested intervals:
You look at $f(\frac{a+b}{2})$, when it is bigger than null you look at $f$ on the interval $[a,\frac{a+b}{2}]$, if it is smaller than 0 we look at $[\frac{a+b}{2},b]$, when it is $0$ we are done. Lets denote the left endpoints with $a_n$ and the right endpoints with $b_n$.

As the diameter of our nested intervals is $(b-a)\cdot 2^{-n}$ which clearly converges to zero we have $$\lim_{n \to \infty} a_n =\lim_{n\to \infty} b_n=\xi$$ As $f$ is continuous we get $$\lim_{n\to \infty} f(a_n)=\lim_{n\to \infty} f(b_n)=f(\xi)$$ On the other hand we know $$f(a_n) < 0 \quad \forall n$$ and $$ f(b_n)>0 \quad \forall n $$ Hence we know $$\lim_{n\to \infty} f(a_n)\leq 0$$ and $$\lim_{n\to \infty} f(b_n)\geq 0$$ Hence $$0\leq f(\xi) \leq 0$$ Hence $f(\xi)=0$

You use that when $C_i$ is closed, bounded and non empty for all $i$ and $C_{i+1} \subset C_i$ for all $i$ then $$\bigcap_{i \in \mathbb{N}} C_i \neq \varnothing$$

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    $\begingroup$ As a brief note, the "nested intervals" theorem is equilivant to the "least upper bound" axiom and various other things (en.wikipedia.org/wiki/Fundamental_axiom_of_analysis) - so if in your course you don't use the nested interval theorem, you can still adapt this proof to use whatever axiom your course requires you to use. $\endgroup$
    – Andrew D
    Apr 10 '13 at 13:34

The important part is to apply the definition of continuity:

$f(x)$ is continuous at point $c$ if $\forall \epsilon>0, \exists \delta>0 \quad S.T \quad |x-c|<\delta \Rightarrow |f(x)-f(c)|<0$ (Alternatively speaking, there always exist $\delta>0$ such that for any $\epsilon>0,|f(x)-f(c)|<0$).


Define $c=sup\{x:f(x)\leq u\}$ ($\star$), and claim $f(c)=u$

Assum $x\in (-\delta+c,c+\delta),\delta>0$ ($x$ is in the $\epsilon$ neighborhood of $x$)

By definition of continuity $\forall \epsilon>0, |f(x)-f(c)|<\epsilon \Rightarrow -\epsilon+f(c)<f(x)<f(c)+\epsilon$

In the following, we will manipulate both side of this inequality and prove the intermediate value theorem by contradiction:

  1. If $f(c)>u$, then $f(c)-u>0$, so set $\epsilon=f(c)-u \Rightarrow f(x)>f(c)-\epsilon=f(c)-(f(c)-u)=u$(use the left side of the inequality ) $\Rightarrow \forall x\in (-\delta+c,c+\delta),f(x)>u$, so it means that $(c-\delta)$ is the least upper bound of the set $\{x:f(x)\leq u\}$, which contradicts with the definition of $c$(also a least upper bound and it's not possible to have 2 least upper bounds at the same time.)
  2. If $f(c)<u$, then $u-f(c)>0$,so set $\epsilon=u-f(c) \Rightarrow f(x)<f(c)+\epsilon=f(c)+(u-f(c))=u$(use the right side of the inequality)$\Rightarrow \forall x\in (-\delta+c,c+\delta),f(x)<u$. This means that there exist $x>c$ such that $f(x)<u$, which contradicts with the definition of $c$ again. (because $c$ is the sup of the set $\{x:f(x)\leq u\}$)

If you set $u=0$, then it's the answer of your question. Hope this can help.

  • $\begingroup$ Nice and neat, but I didn't understand how did you make the contradiction. $\endgroup$
    – SomeOne
    Jan 9 '14 at 8:23

This theorem is a case when the most intuitive proof requires relatively advanced technique. here's a proof using general topology.

Consider a continuous function $f: \mathbb{R} \to \mathbb{R}$ which takes values in $a, b$ but not $c \in (a, b)$. Then $f$ factors through the embedding $i: \mathbb{R} \setminus \{c\} \to \mathbb{R}$: $f = i \circ \hat{f}$, where $\hat{f}: \mathbb{R} \to \mathbb{R} \setminus \{c\}$. $\hat{f}$ is continuous: any open set of $\mathbb{R} \setminus \{c\}$ is obtained by removing $c$ from an open subset of the whole $\mathbb{R}$. But since $f$ is nowhere equal $c$, removing this point doesn't affect the preimage, which is open. $\mathbb{R}$ is path-connected (and thus connected), but $\mathbb{R} \setminus \{c\}$ has two connected components. Therefore, the image of $\hat{f}$ must lie to the one side of $c$ - contradiction.


Let $f : [a,b]\to \mathbb R$ be continuous, $f(a)< 0, f(b) > 0$. Because $[a,b]$ is connected, and $f$ continuous, $f([a,b]) \subset \mathbb R$ is connected. But the only connected sets in $\mathbb R$ are intervals, so $I=f([a,b])$ is an interval with $I \ni f(a) < 0$ and $I \ni f(b) > 0$. Thus $0\in I = f([a,b])$, so there exists a $\xi \in [a,b]$, $f(\xi ) = 0$.

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    $\begingroup$ Ok try to prove what you use without the intermediate value theorem ... $\endgroup$ Apr 10 '13 at 12:08
  • $\begingroup$ Could you please tell me what connected means. $\endgroup$
    – Dick
    Apr 10 '13 at 12:11
  • $\begingroup$ @Dick If you haven't had connectedness yet, discard this proof. Also as Dominic pointed out, it is kind of a circular proof. $\endgroup$
    – Stefan
    Apr 10 '13 at 12:15
  • $\begingroup$ @Dick A space $X$ isconnected when there are no nonempty open sets $O_1$, $O_2$ with a trivial intersection ($O_1\cap O_2=\varnothing$) such that $X=O_1 \cup O_2$. $\endgroup$ Apr 10 '13 at 12:16

Here is a short proof via Heine Borel Theorem. We need to establish the following:

Intermediate Value Theorem: Let $f$ be continuous on $[a, b]$ and let $f(a)f(b) < 0$. Then there is a point $c \in (a, b)$ for which $f(c) = 0$.

Let's extend definition of $f$ beyond $[a, b]$ by setting $f(x) = f(a)$ for $x < a$ and $f(x) = f(b)$ for $x > b$. Then $f$ is continuous everywhere. Let's now assume on the contrary that $f$ does not vanish anywhere in $(a, b)$.

By continuity of $f$ on each point $x \in [a, b]$ there is a neighborhood $I_{x}$ of $x$ on which $f$ maintains a constant sign. The collection of such intervals $I_{x}$ forms an open cover for $[a, b]$ and hence by Heine-Borel Theorem a finite number of such intervals cover $[a, b]$.

The end points of these chosen finite intervals $I_{x}$ which lie in $[a, b]$ form a partition of $[a, b]$ say $$P = \{a = x_{0}, x_{1}, x_{2}, \ldots, x_{n} = b\}$$ and let $\delta = \max_{i = 1}^{n}(x_{i} - x_{i - 1})$ be the norm of partition $P$.

If we have another partition $P' = \{x_{0}', x_{1}', \ldots, x_{m}'\}$ of $[a, b]$ such that norm of $P'$ is less than $\delta$ then each sub-interval $[x_{j - 1}', x_{j}']$ generated by $P'$ lies completely inside one of the chosen intervals $I_{x}$. Hence $f$ maintains constant sign on each of the sub-intervals $[x_{j - 1}', x_{j}']$ and therefore is of constant sign on $[a, b]$. This contradiction shows that $f$ must vanish at some point $c \in (a, b)$.


Let $ f : [a, b] \longrightarrow \mathbb{R} $ be continuous, with $ f(a) < f(b) $. And say $ \eta $ is a real number such that $ f(a) < \eta < f(b) $.

We should now exhibit an $ x \in [a, b] $ at which $ f(x) = \eta $.

Let's call a point $ p $ in $ [a, b] $ "low" if $ f(p) < \eta $, and "high" if $ f(p) > \eta $. Let $ L $ be the set of all low points.
$ L $ contains $ a $ and is bounded above by $ b $, so $ l := \sup(L) $ exists.

Note $ a < l < b $ (Since $ a $ is low, by continuity at $ a $ there is a $ \delta > 0 $ such that all points of $ [a, a + \delta] $ are low. So $ a < a + \delta \leq l $, giving $ a < l $. Similarly $ l < b $ too).

Is $ l $ low ? If it were, by continuity at $ l $ there would be a $ \delta > 0 $ such that all points of $ [l - \delta, l + \delta] $ are low. But by definition of $ l $, no point $ > l $ can be low, a contradiction.
Is $ l $ high ? If it were, by continuity at $ l $ there would be a $ \delta > 0 $ such that all points of $ [l-\delta, l+\delta] $ are high. But by definition of $ l $, there must be a point in $ [l-\delta, l] $ which is low, a contradiction.

So $ l $ is neither low nor high, i.e. $ f(l) = \eta $, as needed.


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