I've read the following article:


It states that there are two versions of the theorem namely

Let $I=[a,b] \subset \mathbb R$ and $f: I \rightarrow \mathbb R$ then:

Version 1:

If $u \in \mathbb R: f(a) < u < f(b) \lor f(a) > u > f(b)$ then $\exists c \in (a,b): f(c) = u$

Version 2:

The image set $f(I) = \{f(x) : x \in I\}$ is also an interval, and either it contains $[f(a), f(b)]$, or it contains $[f(b), f(a)]$; that is $f(I)$ is an interval and $[f(a), f(b)] \lor [f(b), f(a)] \subseteq f(I)$.

Are these versions equivalent ?? For me they seem equivalent if we leave out $f(I)$ is an interval.

Also why does version 1 only apply to $u$ between $f(a)$ and $f(b)$ ??

If $f(I)$ is an interval then version 1 should apply to $u$ not between $f(a)$ and $f(b)$ ?

Thanks for your input.

  • 2
    $\begingroup$ The whole point of the IVT is that the image of an interval under a continuous function is an interval. That Wikipedia article is atypically obscure. $\endgroup$ – lhf Dec 16 '13 at 15:39
  • $\begingroup$ What if $u$ not between $f(a)$ and $f(b)$ ? Why don't the theorem imply that we can find a $c$ then such that $f(c) = u$ ? $\endgroup$ – Shuzheng Dec 16 '13 at 16:06

Yes, the two versions in the Wikipedia article are equivalent. Versions 1 says that given any $u$ between $f(a)$ and $f(b)$ there is a $c$ between $a$ and $b$ such that $f(c) = u$. What this is saying is exactly that the interval $[f(a), f(b)]$ (or $[f(b), f(a)]$) is contained in $f([a,b])$.

You could also say that given any $u$ in the closed interval $[f(a), f(b)]$ (or $[f(b), f(a)]$), you can find a $c$ in the closed interval $[a,b]$ (or $[b,a]$) such that $f(c) = u$. Note that, for example, if $u = f(a)$, then $c = a$. So there isn't really much of a point in allowing $u$ to be $f(a)$ or $f(b)$. That is, version 1 also "applies" to $u$ in the closed interval.

More general, if $u$ is between $\min\{f(x) : x\in I\}$ and $\max\{f(x): x\in I\}$, then there is a $c$ between $a$ and $b$ such that $f(c) = u$. Note that the minimum and maximum exist because you have a continuous function on a closed interval.

  • $\begingroup$ What about $u$ not between $f(a)$ and $f(b)$ ? Suppose we have a graph where the values decrease from $a$ and rise to $b$ ? $\endgroup$ – Shuzheng Dec 16 '13 at 16:03
  • 1
    $\begingroup$ @NicolasLykkeIversen: I updated my answer. $\endgroup$ – Thomas Dec 16 '13 at 16:25
  • $\begingroup$ "The intermediate value theorem states that for each value between the least upper bound and greatest lower bound of the image of a continuous function" $$ $$ But a continuous function is bounded on a closed interval right - thus having maximum and minimum. Why does the article mention greatest lower bound and least upper bound ? Can we consider $f: (0,\sqrt {2}) \rightarrow \mathbb R$ for example ? And thus for $$\lim_{n \rightarrow \infty} [0+\frac 1 n,\sqrt{2} -\frac 1 n]$$ we have an interval of the image of $f$. Does endpoints also apply ? I mean do we have $f(0)$.... $\endgroup$ – Shuzheng Dec 16 '13 at 19:42
  • $\begingroup$ $f(\sqrt 2) \in f((0,\sqrt{2}))$ ? What I ask is, when it says "between" LUB and GLB, does that include these points ? Is there points such that these values lies in the image of $f$ ? $\endgroup$ – Shuzheng Dec 16 '13 at 19:50
  • 1
    $\begingroup$ If one allows open intervalls and picks some value $\inf f(I) < u < \sup f(I)$, then because of the properties of infimum and supremum there are points $c$, $d$ in the interval whose function values lie in between $$\inf f(I)<f(c)<u<f(d)<\sup f(I)$$ and then version I applies to the sub-interval $[c,d]$ (or $[d,c]$). $\endgroup$ – LutzL Dec 17 '13 at 9:12

I believe I was the last to edit the section of the wikipedia article stating those two versions (the state before that was even more confusing, the first version was repeated as some kind of version 2 1/2). Please edit the article or leave proposals and criticism at the bottom of the talk page.

As to the interval version: We just do not know what real numbers outside of $[f(a), f(b)]$ are actually values of the function. But what we do know is that if $c<d$ are elements of $f([a,b])$, then by the first version of the theorem, all intermediate values $u\in [c,d]$ are also function values, i.e., $[c,d]\subset f([a,b])$. Thus there are no gaps in $f([a,b])$, which makes it an interval. From the Bolzano-Weierstraß theorem we then know that this interval is closed for $a,b$ finite.

  • $\begingroup$ "In mathematical analysis, the intermediate value theorem states that for each value between the least upper bound and greatest lower bound of the image of a continuous function there is at least one point in its domain that the function maps to that value." $$$$ Suppose we have $f: (0, \sqrt{2}) \rightarrow \mathbb R$. Then clearly 2 is a least upper bound for the image of $f$. But does there exists a $c$ such that $f(c) = 2$ ? The text says between GLB and LUB, does this include these two values ? In general a continuous function on bounded interval contains its max and min, so why GLB/LUB? $\endgroup$ – Shuzheng Dec 16 '13 at 20:54
  • $\begingroup$ Since the theorem is based on the completeness of the real numbers, supremum and infimum of the set of function values exist. They may be infinite. -- However, You have a valid concern, this formulaton is rather heavy for the leading summary, and furthermore, it is not reflected in the following text, so that it is not really a summary. I'll think about it, but feel free to open a section on this topic on the talk page, I'm sure there are many math editors that have this article on their watchlist. $\endgroup$ – LutzL Dec 17 '13 at 9:06

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