I always thought that a function $f:\mathbb{R} \to \mathbb{R}$ has the intermediate value property (IVP) iff it maps every interval to an interval (Darboux property):


Let $f$ have the Darboux property and let $a<b$ and $f(a) < f(b)$. Then $f([a,b])$ is an interval and so contains $[f(a),f(b)]$. If $u > \in [f(a),f(b)]$ then of course $u \in f([a,b])$ and thus there exists some $k \in [a,b]$ such that $f(k) = u$, i.e. $f$ has IVP.

Now let $f$ have the IVP and let $a < b$, $x,y \in f([a,b])$ and $z > \in \mathbb{R}$ with $x<z<y$. We claim $z \in f([a,b])$. Indeed we have $x = f(x')$ and $y = f(y')$ for some $x',y' \in [a,b]$. W.L.O.G assume that $x' < y'$. Then by the IVP there is some $c\in [x', y']$ such that $f(c) = z$, i.e. $z \in f([a,b])$ and therefore $f([a,b])$ is an interval.

But on this blog in problem 5. the author says that they are not equivalent:

"This [Darboux property] is slightly different from continuity and intermediate value property. Cotinuity implies Darboux and Darboux implies Intermediate value property."

Have I missed something and if yes, where does the proof given above break?

  • $\begingroup$ Looks good to me. I commented at that blog post linking here. (It is "awaiting moderation." The author is also a Math.SE user.) $\endgroup$ – Jonas Meyer Jul 6 '16 at 21:32
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    $\begingroup$ I think that we consider slightly different definitions for the intermediate value property... As I was taught in my first university year, a function $f: I \to J$ has the intermediate value property if $f(I)$ is an interval, i.e. the image of the whole domain of definition is an interval. This generalizes the Darboux property which says that the image of ANY interval contained in the domain of definition is an interval. I am aware that this definition of the intermediate value property may not be universal... I should edit the problem to make it clear... $\endgroup$ – Beni Bogosel Jul 8 '16 at 9:11

That is correct according to the definition of intermediate value property saying that for all $a<b$ in the domain, for all $u$ between $f(a)$ and $f(b)$, there exists $k\in(a,b)$ such that $f(k) = u$. The two properties are equivalent, and your proof of that is correct.

The blog's author Beni Bogoşel clarified in a comment that he was using a different definition of intermediate value property, meaning that the entire image is an interval. The ambiguity is understandable given that the Intermediate Value Theorem is often stated in terms of a particular interval: $f:[a,b]\to \mathbb R$ continuous implies $f([a,b])$ contains the interval $\{\min\{f(a),f(b)\},\max\{f(a),f(b)\}\}$. (In this case, because the restriction of a continuous function is also continuous, this theorem automatically implies the stronger intermediate value property for continuous functions on intervals.)

The author acknowledged that the convention is not universal, and he may edit to clarify.

  • $\begingroup$ So to understand it right, the function $f:[0,1] \to [0,1]$ with $f(x) = x$ for $x \in (0,1)$, $f(1) = 0$ and $f(0) = 1$ would have the (weak) IVP since $f([0,1]) = [0,1]$? $\endgroup$ – Andrei Kh Jul 9 '16 at 11:15
  • $\begingroup$ @AndreiKh: Yes. $\endgroup$ – Jonas Meyer Jul 9 '16 at 13:00

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