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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):

Proof:

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?

S.C.
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Andrei Kh
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  • Looks good to me. I commented at that blog post linking here. (It is "awaiting moderation." The author is also a Math.SE user.) – Jonas Meyer Jul 06 '16 at 21:32
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    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... – Beni Bogosel Jul 08 '16 at 09:11
  • For the second direction, note that if $f([a,b])$ is a singleton (meaning that $f|_{[a,b]}$ takes any $x \in [a,b]$ and maps it to some constant $c$), then $f([a,b])$ is a trivial interval...namely a singleton (because $[c,c] = {c}=f([a,b])$). Therefore, we suppose that $f([a,b])$ is not a singleton, which implies that there are at least two elements $\in f([a,b])$ such that $x \lt y$. Consider any arbitrary two of them. Also, note that the definitions being used in this proof are summarized nicely here: https://mathproblems123.files.wordpress.com/2010/07/strange-functions.pdf – S.C. Sep 22 '22 at 03:31
  • For the first direction, note that if $f(a)=f(b)$, then we would have that for these particular $a \lt b$, the IVP definition is vacuously true because the antecedent requires that $f(a) \lt z \text{ and } z \lt f(b)$...but if $f(a)=f(b)$, then no such $z$ will make the antecedent evaluate to true (because we would need a $z$ that satisfies "$z \lt f(a) \land z \gt f(a)$", which is impossible). Therefore, the implication always evaluates to true for these particular $a$ and $b$. – S.C. Sep 22 '22 at 03:59

1 Answers1

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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.

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Jonas Meyer
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