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Assume $f: [a,b] \to \mathbb{R}$ is a monotone function that satisfies the Intermediate Value Theorem. Prove that $f$ is continuous.

It is sort of confusing how they say it satisfies IVT. Don't only continuous functions satisfy IVT? If instead they mean the condition that if it takes any value between $[a,b]$ then it also take any value between $f(a)$ and $f(b)$, then we will need to use the definition of the monotonicity. Suppose that $x<y$ and $f(x) < f(y)$ for any $x,y \in [a,b]$. How can we use this with IVT here to prove that $f$ is continuous?

user19405892
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2 Answers2

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It's not restrictive to assume that $f$ is increasing (otherwise use $-f$). Prove that, for $c\in(a,b)$, $$ \lim_{x\to c^-}f(x)=\sup\{f(x):a\le x<c\}, \qquad \lim_{x\to c^+}f(x)=\inf\{f(x):c<x\le b\}, $$ and that $$ \lim_{x\to a^+}f(x)=\inf\{f(x):a<x\le b\}, \qquad \lim_{x\to b^-}f(x)=\sup\{f(x):a\le x< b\}. $$ Suppose that for some $c\in(a,b)$ the limits from the left and from the right are different and apply the hypothesis about the IVT. Finish up with the values at the extremes $a$ and $b$.

egreg
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  • What do you mean by "Prove that"? Also, isn't the the part after the "and that" obvious by definition of it being monotonic? – user19405892 Mar 02 '16 at 01:19
  • @user19405892 I mean that you should prove those statements. The part after “and that" is proved in the same way, it just didn't fit in the format. – egreg Mar 02 '16 at 07:05
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An $\epsilon,\delta$-proof:

Let $x_0 \in (a,b)$ be arbitrary and let $\epsilon > 0$. Let $s_1 = \min(f(x_0)+\epsilon/2, f(b))$. Then the number $f(x_0)+s_1$ satisfies $f(x_0) < f(x_0)+s_1 < f(x_0)+\epsilon$. By the intermediate value property, there exists then some $\delta_1 > 0$ such that $f(x_0 + \delta_1) = f(x_0)+s_1$ (by the choice of $s_1 = \min(f(x_0)+\epsilon/2, f(b))$ we are guaranteed to stay within the domain of $f$). Similarly, let $s_2 = \max(f(x_0)-\epsilon/2, f(a))$ and find a $\delta_2>0$ such that $f(x_0-\delta_2) = f(x_0)-s_2$. Take $\delta = \min(\delta_1,\delta_2) > 0$.

Now consider the punctured neighbourhood $U_\delta = \{x \in (a,b) : 0 < |x - x_0| < \delta\}$. Without loss of generality, we may assume that $f$ is increasing. Let $x \in U_\delta$ be arbitrary. If $x > x_0$ we have, as $f$ is increasing, $f(x_0) \leq f(x) \leq f(x_0+\delta_1) = f(x_0)+s_1 < f(x_0)+\epsilon$, whence $|f(x)-f(x_0)| < \epsilon$. If $x < x_0$, we have, as $f$ is increasing, $f(x_0)-\epsilon < f(x_0)-s_2 = f(x_0-\delta_2) \leq f(x) \leq f(x_0)$, whence $|f(x)-f(x_0)| < \epsilon$.

You can work out the proof for the endpoints as an exercise!

Étienne Bézout
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  • Do we need to prove that $f$ is continuous everywhere or just on the interval? – user19405892 Mar 02 '16 at 01:45
  • The interval is the domain of $f$. We certainly can't prove that it's continuous outside of its domain. We prove that it is continuous at every point in its domain. That's what it means for a function to be continuous. – Étienne Bézout Mar 02 '16 at 01:46
  • I am confused in egreg's solution. Why is he talking about the endpoints of the interval $a,b$ if we already know the function exists there? – user19405892 Mar 02 '16 at 01:52
  • @user19405892 Both he an I are talking about proofs of continuity for points in the interior of the domain, i.e. $(a,b)$. However, you must establish continuity at the whole domain and thus also prove continuity at the end points $a$ and $b$. – Étienne Bézout Mar 02 '16 at 09:29
  • The upper bound you've established at the beginning is not necessarily true. $f(x_0)+s_1=2f(x_0)+\frac{\epsilon}{2}\overset{??}{<}f(x_0)+\epsilon$. While your idea is correct, you must construct the $s_1$, $s_2$ in a slightly different way. – Philipp Jul 11 '21 at 13:58