Intermediate value theorem: are these two versions equivalent ? I've read the following article:
http://en.wikipedia.org/wiki/Intermediate_value_theorem
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.
 A: 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.
A: 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.
