Proof of Continuity of the Reciprocal of a Nonvanishing Real-Valued Function on a Metric Space

All references to numbered claims below are to Elementary Order Arithmetic in an Ordered Field unless another item is named. Unfolding $d_{\mathbb{R}}$, continuity of a map $h:A\to\mathbb{R}$ at $x$ relative to $A$ says: for every $\varepsilon$ with $0<\varepsilon$ there is $\delta$ with $0<\delta$ such that every $z\in A$ with $d(x,z)<\delta$ satisfies $|h(z)-h(x)|<\varepsilon$.

Claim 1. Assume $f$ is continuous at $x$ relative to $A$, and write $u=f(x)$, so $u\ne 0$ by hypothesis. By claim 1 of Properties of the Absolute Value in an Ordered Field we have $0\le|u|$, and $|u|=0$ would force $u=0$; hence $|u|\ne 0$ and therefore $0<|u|$. By claim 8 the element $|u|\cdot 2^{-1}$ satisfies $0<|u|\cdot 2^{-1}$ and $|u|\cdot 2^{-1}+|u|\cdot 2^{-1}=|u|$.

Step 1: a lower bound for $|f(z)|$ near $x$. By continuity with tolerance $|u|\cdot 2^{-1}$ there is $\delta_1$ with $0<\delta_1$ such that every $z\in A$ with $d(x,z)<\delta_1$ satisfies $|f(z)-u|<|u|\cdot 2^{-1}$. Fix such a $z$ and write $c=|f(z)-u|$. By claim 7 of Properties of the Absolute Value in an Ordered Field we have $\bigl||f(z)|-|u|\bigr|\le c$, and claim 6 of that lemma gives $-c\le|f(z)|-|u|$; adding $|u|$, which preserves $\le$ by compatibility of the order with addition in an ordered field, yields $|u|-c\le|f(z)|$. From $c<|u|\cdot 2^{-1}$ and claim 4 we get $-|u|\cdot 2^{-1}<-c$, and claim 1 gives $|u|-|u|\cdot 2^{-1}<|u|-c$. Since $|u|\cdot 2^{-1}+|u|\cdot 2^{-1}=|u|$ we have $|u|-|u|\cdot 2^{-1}=|u|\cdot 2^{-1}$, so claim 2 yields

$$|u|\cdot 2^{-1}<|f(z)|.$$

In particular $0<|f(z)|$ by claim 2.

Step 2: inverting the bound. Both $|u|\cdot 2^{-1}$ and $|f(z)|$ are positive, so by claim 7 their multiplicative inverses exist and are positive, and by claim 5 the product $(|u|\cdot 2^{-1})^{-1}\cdot|f(z)|^{-1}$ is positive. Multiplying the inequality of Step 1 by that positive element and using claim 10, then simplifying with the defining property of the multiplicative inverse, gives

$$|f(z)|^{-1}<(|u|\cdot 2^{-1})^{-1}.$$

Step 3: the estimate. Let $0<\varepsilon$ and put $\eta=\varepsilon\cdot|u|\cdot|u|\cdot 2^{-1}$, which is positive by repeated use of claim 5. By continuity there is $\delta_2$ with $0<\delta_2$ such that every $z\in A$ with $d(x,z)<\delta_2$ satisfies $|f(z)-u|<\eta$. By claim 9 there is $\delta$ with $\delta\le\delta_1$, $\delta\le\delta_2$ and $\delta$ equal to $\delta_1$ or to $\delta_2$; in either case $0<\delta$.

Let $z\in A$ with $d(x,z)<\delta$. By claim 2 we have $d(x,z)<\delta_1$ and $d(x,z)<\delta_2$, so Steps 1 and 2 apply to this $z$ and $|f(z)-u|<\eta$.

Since $f(z)$ and $u$ are nonzero, the field identities give

$$r(z)-r(x)=f(z)^{-1}-u^{-1}=\bigl(u-f(z)\bigr)\cdot f(z)^{-1}\cdot u^{-1}.$$

For a nonzero $w$ we have $|w|\cdot|w^{-1}|=|w\cdot w^{-1}|=|1|=1$ by claim 4 of Properties of the Absolute Value in an Ordered Field together with the definition of the absolute value and $0\le 1$ from claim 1 of Elementary Arithmetic in an Ordered Field; hence $|w^{-1}|=|w|^{-1}$. Applying claim 4 of Properties of the Absolute Value in an Ordered Field to the displayed product therefore gives

$$|r(z)-r(x)|=|u-f(z)|\cdot|f(z)|^{-1}\cdot|u|^{-1},$$

and $|u-f(z)|=|f(z)-u|$ by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field together with claim 2 of Properties of the Absolute Value in an Ordered Field.

Put $Q=|f(z)|^{-1}\cdot|u|^{-1}$, which is positive by claim 5. From $|f(z)-u|<\eta$ and claim 10 we get $|f(z)-u|\cdot Q<\eta\cdot Q$. From Step 2 and $0\le\eta\cdot|u|^{-1}$, claim 5 of Elementary Arithmetic in an Ordered Field gives

$$\eta\cdot Q=\eta\cdot|u|^{-1}\cdot|f(z)|^{-1}\le\eta\cdot|u|^{-1}\cdot(|u|\cdot 2^{-1})^{-1}.$$

By associativity and commutativity of multiplication together with the defining properties of the multiplicative inverse and of the multiplicative identity, the right-hand side equals $\varepsilon$. Hence $|r(z)-r(x)|<\eta\cdot Q\le\varepsilon$, and claim 2 gives $|r(z)-r(x)|<\varepsilon$. As $\varepsilon$ was arbitrary, $r$ is continuous at $x$ relative to $A$.

Claim 2. If $f$ is continuous on $A$, it is continuous at every point of $A$ relative to $A$, so claim 1 applies at every such point; hence $r$ is continuous on $A$.