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The way I understand the intermediate value theorem is this: if you have a function f that is continuous over a domain $[a,b]$ then there is a value $f(c)$, where $f(a)≤f(c)≤f(b)$, such that $a≤c≤b$.

This seems self-evident. If $f$ is continuous, then there exists an $f(c)$ such that $a≤c≤b$. But isn't this just a restatement of the fact that $f$ is continuous?

Isn't the intermediate value theorem self-evident for continuous functions?

  • 2
    What do you mean by Self-Evident? You probably may trace your ancestry to Von Neumann, Princeton! :-)2012-02-22
  • 5
    No you need the fact that the real line as a topological space is connected (plus the fact that the image of a connected set under a continuous function is connected).2012-02-22
  • 1
    Well, continuous functions have IVP but those that have IVP need not be continuous. Look up Darboux functions. And, therefore, having IVP is weaker than continuity and hence not a equivalent restatement.2012-02-22
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    "$f$ is continuous on $[a,b]$" means "for every $x$ in $[a,b]$, for every $\epsilon >0$, there exists $\delta >0$ such that, for every $y$ in $[a,b]$, if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$". Can you please explain how the intermediate value theorem is an obvious restatement of this condition?2012-02-22
  • 8
    The intermediate value theorem is intimately connected with the fact that the reals are *complete* (every nonempty set bounded above has a supremum; equivalently, every Cauchy sequence coverges). In fact, it is *equivalent* to the supremum property/every Cauchy sequence converges, so it is not entirely "obvious" (in that the proof relies on these rather deep properties of the real numbers that are not easy to prove from first principles).2012-02-22
  • 23
    You have a misunderstanding of the intermediate value theorem. The intermediate value theorem says that $f(x)$ must take every possible value between $f(a)$ and $f(b)$. It might "seem" self-evident, but it isn't true for example, if you look at continuous functions on the rational numbers, for example.2012-02-22
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    My favorite example of a continuous function from rationals to rationals that fails the intermediate value test is simply $f(x)=1/(2-x^2)$: rational function, gotta be continuous, but it blows up around the “missing” point $\sqrt{2}$.2012-02-22
  • 7
    Somebody should probably mention that the statements written in the question, in whatever way one interprets them, are NOT what is called the intermediate value theorem. The OP might want, first and foremost, to reach a correct statement.2012-02-22
  • 3
    What you wrote is trivially true even for non-continuous functions. Just take `c=b` and you're done.2012-02-22
  • 2
    If one were to reject classical logic, there are [intuitionist models of real numbers](http://math.andrej.com/2008/08/13/intuitionistic-mathematics-for-physics/) (see the section on Smooth Infinitesimal Analysis for a quick intro to what the theory is all about) in which *all functions are continuous* but the usual exact statement of intermediate value theorem is false. (I believe a construction was given in Moerdijk and Reyes’s _Models of smooth infinitesimals analysis_, but I don't have the book to check now.)2012-02-22

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