Proof of Rolle's Theorem on an Open Interval

Let $|\cdot|$ be the absolute value, and regard $\mathbb{R}$ as a metric space through the metric $d_{\mathbb{R}}$ of The Absolute Value Metric on the Real Line, and let $\mathcal{T}_{d_{\mathbb{R}}}$ be the collection of subsets of $\mathbb{R}$ that are open in $(\mathbb{R},d_{\mathbb{R}})$, which is a topology on $\mathbb{R}$ by Metric Open Sets Form a Topology. By An Open Interval is an Interval All of Whose Points Are Interior the set $(p,q)$ is an interval all of whose points are interior points of it. Let $[a,b]$ be the closed interval of Interval in the Real Line. Claim numbers refer to Elementary Order Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field as indicated.

Step 1: $[a,b]\subseteq(p,q)$. If $x\in[a,b]$ then $a\le x\le b$ with $a,b\in(p,q)$, so $x\in(p,q)$ because $(p,q)$ is an interval.

Step 2: $[a,b]$ is nonempty and compact. It contains $a$. Since $a<b$, the strict order gives $a\le b$, so Closed Interval $[a,b]$ is Compact in $\mathbb{R}$ applies and shows that $[a,b]$ is compact in $(\mathbb{R},\mathcal{T}_{d_{\mathbb{R}}})$.

Step 3: the restriction of $g$ to $[a,b]$ is continuous. Let $x\in[a,b]$. By Step 1, $x\in(p,q)$, and $g$ is differentiable at $x$, so by Differentiability at a Point Implies Continuity There $g$ is continuous at $x$ relative to $(p,q)$. The defining condition quantifies over points of the domain, and $[a,b]\subseteq(p,q)$, so the same $\delta$ witnesses continuity at $x$ relative to $[a,b]$ of the restriction $g_{0}$ of $g$ to $[a,b]$. Hence $g_{0}$ is continuous on $[a,b]$.

Step 4: extrema. By claim 2 of Semicontinuity Under Negation and Characterization of Continuity applied at each point, $g_{0}$ is both upper semicontinuous and lower semicontinuous on $[a,b]$. By Semicontinuous Functions Attain Their Extrema on a Compact Set, applied to the metric space $(\mathbb{R},d_{\mathbb{R}})$ and the nonempty compact subset $[a,b]$, there are $x_{\max},x_{\min}\in[a,b]$ with $g(x)\le g(x_{\max})$ and $g(x_{\min})\le g(x)$ for every $x\in[a,b]$.

Step 5: a local extremum strictly between $a$ and $b$. We produce $c$ with $a<c<b$ at which $g$ has a local maximum or a local minimum relative to $(p,q)$.

First a remark used twice. Suppose $a<c<b$ and let $\delta$ be the least of $c-a$ and $b-c$, which exists by claim 9 and is positive by claim 1. If $y\in(p,q)$ satisfies $|y-c|<\delta$, then $-\delta<y-c<\delta$ by claim 9 of Properties of the Absolute Value in an Ordered Field, so by claim 1 we get $c-\delta<y<c+\delta$; since $\delta\le c-a$ gives $a\le c-\delta$ and $\delta\le b-c$ gives $c+\delta\le b$, claim 2 yields $a\le y\le b$, that is $y\in[a,b]$.

Case 1: $g(x_{\max})=g(a)$ and $g(x_{\min})=g(a)$. Then every $x\in[a,b]$ satisfies $g(a)=g(x_{\min})\le g(x)\le g(x_{\max})=g(a)$, so $g(x)=g(a)$ by antisymmetry. Let $c=(a+b)\cdot2^{-1}$. From $a<b$: claim 1 gives $a+a<a+b$ and $a+b<b+b$, and claim 10 with the multiplier $2^{-1}$, positive by claim 8, gives $a<c$ and $c<b$. With $\delta$ as in the remark, every $y\in(p,q)$ with $|y-c|<\delta$ lies in $[a,b]$, so $g(y)=g(a)=g(c)$ and in particular $g(y)\le g(c)$. So $g$ has a local maximum at $c$ relative to $(p,q)$.

Case 2: $g(x_{\max})\ne g(a)$. Since $a\in[a,b]$ we have $g(a)\le g(x_{\max})$, hence $g(a)<g(x_{\max})$. Also $g(b)=g(a)<g(x_{\max})$. So $x_{\max}\ne a$ and $x_{\max}\ne b$, while $a\le x_{\max}\le b$; hence $a<x_{\max}<b$. Put $c=x_{\max}$ and take $\delta$ as in the remark: every $y\in(p,q)$ with $|y-c|<\delta$ lies in $[a,b]$, so $g(y)\le g(x_{\max})=g(c)$. So $g$ has a local maximum at $c$ relative to $(p,q)$.

Case 3: $g(x_{\min})\ne g(a)$. Symmetrically $g(x_{\min})<g(a)=g(b)$, so $a<x_{\min}<b$; putting $c=x_{\min}$ and taking $\delta$ as in the remark gives $g(c)\le g(y)$ for every $y\in(p,q)$ with $|y-c|<\delta$, so $g$ has a local minimum at $c$ relative to $(p,q)$.

The three cases are exhaustive, since if Case 1 fails then $g(x_{\max})\ne g(a)$ or $g(x_{\min})\ne g(a)$.

Step 6. In every case $c$ satisfies $a<c<b$, so $c\in(a,b)$, and $g$ is differentiable at $c$. By Vanishing of the Derivative at an Interior Local Extremum, $g'(c)=0$.