Proof of Slice Function and the Partial Derivative

We use the elementary order arithmetic of the ordered field $\mathbb{R}$ and the properties of the absolute value.

Step 0 (a strict two-sided bound). For $p,c\in\mathbb{R}$, the inequality $|p|<c$ holds if and only if both $-c<p$ and $p<c$ hold. Indeed, if $|p|<c$, then $p\le|p|$ by claim 3 of Properties of the Absolute Value in an Ordered Field gives $p<c$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, while $-|p|\le p$ together with $-c<-|p|$, the latter by claim 4 of Elementary Order Arithmetic in an Ordered Field, gives $-c<p$. Conversely, if $-c<p$ and $p<c$, then $|p|$ equals $p$ or $-p$ by claim 1 of Properties of the Absolute Value in an Ordered Field, and $p<c$ in the first case, while claim 4 of Elementary Order Arithmetic in an Ordered Field gives $-p<c$ in the second.

In particular, by claim 1 of Elementary Order Arithmetic in an Ordered Field (adding $a_i$), the condition $|s-a_i|<\rho$ on a real number $s$ is equivalent to the conjunction of $a_i-\rho<s$ and $s<a_i+\rho$. Hence $I$ is exactly the set of $s\in\mathbb{R}$ with $|s-a_i|<\rho$.

Step 1 (claim 1). Since $a\in U$ and $U$ is open in $\mathbb{R}^n$, there is $\rho\in\mathbb{R}$ with $0<\rho$ such that every point of $\mathbb{R}^n$ whose Euclidean distance to $a$ is less than $\rho$ lies in $U$.

Let $s\in\mathbb{R}$ satisfy $|s-a_i|<\rho$. The points $a[s]$ and $a$ have equal $k$th coordinates for every $k\ne i$, and their $i$th coordinates differ by $s-a_i$, so by the definition of the Euclidean distance the distance from $a[s]$ to $a$ is the nonnegative real number whose square is $(s-a_i)^2$. Now $|s-a_i|$ is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field and satisfies $|s-a_i|^2=(s-a_i)^2$ by claim 4 of the same lemma, applied with both arguments equal to $s-a_i$. Moreover distinct nonnegative reals have distinct squares: if $0\le p$, $0\le q$ and $p<q$, then $0<q$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, so $q\,p<q\,q$ by claim 10 of that lemma, while $p\,p\le q\,p$ (an equality if $p=0$, and claim 10 with multiplier $p$ if $0<p$), whence $p^2<q^2$ by claim 2. Therefore the distance from $a[s]$ to $a$ equals $|s-a_i|$, which is less than $\rho$, and hence $a[s]\in U$.

Step 2 ($I$ is an interval with $a_i$ as an interior point, and $g$ is defined). Fix $\rho$ as in claim 1.

First, $I$ is an interval: if $x,z\in I$ and $y\in\mathbb{R}$ satisfies $x\le y\le z$, then $a_i-\rho<x$ and $x\le y$ give $a_i-\rho<y$, and $y\le z$ with $z<a_i+\rho$ gives $y<a_i+\rho$, both by claim 2 of Elementary Order Arithmetic in an Ordered Field; hence $y\in I$.

Second, $a_i$ is an interior point of $I$. By claim 8 of Elementary Order Arithmetic in an Ordered Field the element $\eta=\rho\cdot 2^{-1}$ satisfies $0<\eta$ and $\eta<\rho$. Put $u=a_i-\eta$ and $v=a_i+\eta$. Claim 1 of Elementary Order Arithmetic in an Ordered Field applied to $0<\eta$ gives $u<a_i<v$, and applied to $\eta<\rho$ it gives $a_i-\rho<u$ and $v<a_i+\rho$; with claim 2 of that lemma this places $u$ and $v$ in $I$. So $u,v\in I$ with $u<a_i<v$, which is the defining condition for $a_i$ to be an interior point of $I$. In particular $a_i\in I$.

Finally, by Step 0 every $s\in I$ satisfies $|s-a_i|<\rho$, so $a[s]\in U$ by Step 1 and $g(s)=f(a[s])$ is defined for every $s\in I$.

Step 3 (the two difference quotients agree). Let $h\in\mathbb{R}$ satisfy $0<|h|<\rho$. Then $|(a_i+h)-a_i|=|h|<\rho$, so $a_i+h\in I$ by Step 0. The point $a[a_i+h]$ is by definition the point whose $i$th coordinate is $a_i+h$ and whose $k$th coordinate is $a_k$ for $k\ne i$, that is the point $(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n)$ appearing in the definition of the partial derivative, and $a[a_i]=a$. Hence

both sides being defined because $a[a_i+h]$ and $a$ lie in $U$.

Step 4 (claim 2). Let $L\in\mathbb{R}$. We show that $L$ satisfies the defining condition of the partial derivative of $f$ with respect to the $i$th variable at $a$ if and only if $L$ satisfies the defining condition of differentiability of $g$ at $a_i$.

In either direction, suppose the relevant condition holds for $L$, and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ be given, furnishing some $\delta_0$ with $0<\delta_0$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $\delta$ with $\delta\le\delta_0$, $\delta\le\rho$, and $\delta$ equal to $\delta_0$ or to $\rho$; in either case $0<\delta$. Every $h$ with $0<|h|<\delta$ satisfies $0<|h|<\delta_0$ and $0<|h|<\rho$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, so the assumed condition applies to it, and moreover $a_i+h\in I$ by Step 3. By Step 3 the two difference quotients at such $h$ are equal, so the inequality asserted by one condition at $h$ is the inequality asserted by the other. Hence the condition in the other definition holds for this $\varepsilon$ with this $\delta$. Note that each definition carries a membership clause: differentiability of $g$ at $a_i$ requires $a_i+h\in I$, and the partial derivative condition requires $a[a_i+h]\in U$. Both are automatic once $\delta\le\rho$, the first by Step 0 and the second by Step 1, so restricting attention to $\delta\le\rho$ loses nothing in either direction.

Therefore the real numbers $L$ satisfying the two defining conditions are the same. In particular one of the two derivatives exists if and only if the other does, and when they exist the values they denote, being characterised by these identical conditions, coincide: