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How can someone prove that continuity implies the intermediate value property?

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

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  • $\begingroup$ Thats the intermediate value theorem, $\endgroup$ – José Carlos Santos Aug 1 '17 at 1:40
  • $\begingroup$ Functions, not necessarily continuous, that have this property are also known as Darboux functions. Examples include the derivatives of functions that are differentiable everywhere (even if they're not continuously differentiable). $\endgroup$ – Theo Bendit Aug 1 '17 at 2:05
  • $\begingroup$ I have given proofs of intermediate value theorem for continuous function in this post paramanands.blogspot.com/2011/06/… You can see that each proof uses a different version of completeness property of real numbers. $\endgroup$ – Paramanand Singh Aug 1 '17 at 2:29
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Hint: This theorem can be proven using an epsilon/delta approach, given the completeness property of the real numbers.

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Yes. This follows from the fact that the image of a connected space under a continuous map is again connected.

Suppose there were some point $a \in f(I)$ so that that there were no $x$ sub that $f(x)=a.$ by the extreme value theorem, $a$ is on the interior of $f(I)$ and hence $f(I)$ would be the disjoint union of two open sets $\{x \in I \mid f(x)<a\}$ and $\{x \in I \mid f(x)>a\}$ in the sub space topology, a contradiction.

This specialization of the aforementioned fact is sometimes called the intermediate value theorem for calculus.

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  • $\begingroup$ No need to appeal to the extreme value theorem here. All we need to do is show that the only connected subsets of $\mathbb{R}$ are intervals, and then $f(I)$ must be an interval and must contain every value between $f(a)$ and $f(b)$. $\endgroup$ – Michael Lee Aug 1 '17 at 1:46
  • $\begingroup$ That's true as well. I'm not convinced that this is easier to prove than the extreme value theorem, but you are fully correct. $\endgroup$ – Andres Mejia Aug 1 '17 at 1:48
  • $\begingroup$ See the Necessary Condition part of the proof here. $\endgroup$ – Michael Lee Aug 1 '17 at 1:49
  • $\begingroup$ Oh, I see your point. I'm on my phone, so I cannot edit right now, but yes. I agree $\endgroup$ – Andres Mejia Aug 1 '17 at 1:50
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Note that if $f(a) = f(b)$, then $c = f(a) = f(b)$, so $c$ can be chosen as $a$ or $b$. Otherwise $f(a) \neq f(b)$, and without loss of generality, $f(a) < f(b)$ (otherwise consider $-f$).

Let $S = \lbrace x \in [a, b] : f(x) \le c\rbrace$. Note that $a \in S$ and $S$ is bounded (above), so it has a supremum $p$. For all points $x \in (p, b]$, we have $x \notin S$, so $f(x) > c$. Therefore, using the continuity of $f$,

$$f(p) = \lim_{x\rightarrow p} f(x) = \lim_{x\rightarrow p^+} f(x) \ge c.$$

On the other hand, since $p$ is the supremum of $S$, there must be a sequence $x_n \in S$ (i.e. such that $f(x_n) \le c$) such that $x_n \rightarrow p$. Therefore, by the continuity of $f$,

$$f(p) = f\left(\lim_{n\rightarrow \infty} x_n\right) = \lim_{n\rightarrow \infty} f(x_n) \le c.$$

Putting together, $f(p) = c$, as required.

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I guess it depends on the logical framework you're using.

In smooth infinitesimal analysis (which uses intuitionistic logic rather than classical logic), every function is continuous, and the intermediate value theorem fails!

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  • $\begingroup$ Why the IVT fails there? $\endgroup$ – asn32 Feb 6 '18 at 0:40
  • $\begingroup$ @AngghaNugraha I don't know! $\endgroup$ – étale-cohomology Feb 6 '18 at 13:51

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