IVP implies EVP? I define Intermediate Value Property (IVP) and Extreme Value Property (EVP) as follows:


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*IVP: If $I$ is an interval, and $f:I\rightarrow\mathbb{R}$, we say that 
$f$ has the intermediate value property iff whenever $a<b$ are point's in $I$ and $f(a)\leq c\leq f(b)$, there is a $d\in$  $(a,b)$ such that $f(d)=c$.

*EVP: If $I$ is an interval, and $f:I\rightarrow\mathbb{R}$, we say that $f$ has the extreme value property iff $f$ has maximum and minimum value, each at least once. That is, $\exists a,b\in I$ such that $f(a)\leq f(x) \leq f(b)$ for all $x\in I$.


My question is, does IVP imply EVP? I don't think so, but I still can't find the counter example. Cheers!
 A: Non-explicit example: Let $f$ be any differentiable function on an interval $I$ with unbounded derivative. Then $f'$ has the IVP by Darboux's theorem, but misses either a maximum or minimum.
A: On the interval $0 < x < 1$, consider $f(x) = x$. It has the intermediate value property, but not the extreme value property.
A: No, not at all.
Even if you stricten up your definition to avoid the most simple counter example, by borrowing from the IVT and EVT and replacing the requirement of continuity by IVP and EVP respectively. For example:

A function $f$ is said to have the IVP in $E$ if for every closed interval $[a,b]\subseteq E$ and for every $c$ between $f(a)$ and $f(b)$ there's a $\xi\in I$ such that $f(\xi) = c$.

This one avoids counterexamples where you just insert a discontinuity as you would then normally have a gap and can find an interval where the IVP is not fulfilled. Also since $E$ is a subset of the domain of $f$ you need the function to be defined everywhere in the interval making it harder to use unbounded functions.
However there are functions that would avoids such attempts at covering the loopholes. One way is to use an everywhere surjective function. That's functions whose range on every non-trivial interval is $\mathbb R$. One such function is:
$$\phi(x) = \begin{cases}
\lim_{n\to\infty} \tan(n!x) & \text{ if the limit exists} \\
0 & \text{ otherwise}
\end{cases}$$
This means trivially that $\phi$ would have the IVP (in fact it would take every other value on the interval too), but since it's unbounded there it would mean the function does not have the EVP (anywhere). 
We can even make a bounded counter example by composing such a function to produce an arbitrary range. For example we have that $f(x) = 1/(1+\phi(x)^2)$ will have the range $(0,1]$ on every non-trivial interval. It will have the IVP, be bounded yet not having the EVP since $\inf_I f(x)=0$, but $f(x)\ne 0$.
