Proof of Continuity of the Absolute Value, Maximum and Minimum of Real-Valued Functions 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. Let $0<\varepsilon$, and let $\delta$ be as provided by continuity of $f$ at $x$ relative to $A$ for this $\varepsilon$. For $z\in A$ with $d(x,z)<\delta$, claim 7 of Properties of the Absolute Value in an Ordered Field gives

and $|f(z)-f(x)|<\varepsilon$, so claim 2 gives $\bigl||f|(z)-|f|(x)\bigr|<\varepsilon$.

A common choice of $\delta$ for claims 2 and 3. Let $0<\varepsilon$. Continuity of $f$ and of $g$ at $x$ relative to $A$ gives $\delta_1$ and $\delta_2$, both positive, such that $d(x,z)<\delta_1$ implies $|f(z)-f(x)|<\varepsilon$ and $d(x,z)<\delta_2$ implies $|g(z)-g(x)|<\varepsilon$, for $z\in A$. 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 $0<\delta$. Fix $z\in A$ with $d(x,z)<\delta$. By claim 2 both $|f(z)-f(x)|<\varepsilon$ and $|g(z)-g(x)|<\varepsilon$ hold, so claim 9 of Properties of the Absolute Value in an Ordered Field gives

Adding $f(x)$, respectively $g(x)$, and using claim 1 together with the field identities for sums and differences, this says

Claim 2. Write $m=\max\{f(z),g(z)\}$ and $m'=\max\{f(x),g(x)\}$. By claim 1 of Elementary Properties of the Maximum of Two Elements we have $f(x)\le m'$ and $g(x)\le m'$, so adding $\varepsilon$ gives $f(x)+\varepsilon\le m'+\varepsilon$ and $g(x)+\varepsilon\le m'+\varepsilon$. With the bounds above and claim 2 we get $f(z)<m'+\varepsilon$ and $g(z)<m'+\varepsilon$. By claim 2 of Elementary Properties of the Maximum of Two Elements the element $m$ equals $f(z)$ or $g(z)$, so in either case $m<m'+\varepsilon$. Exchanging the roles of $z$ and $x$ in this argument, and using the bounds $f(z)-\varepsilon<f(x)$ and $g(z)-\varepsilon<g(x)$, which follow from the displayed inequalities by claim 1, gives $m'<m+\varepsilon$.

Adding $-m'$ to $m<m'+\varepsilon$ and simplifying by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field gives $m-m'<\varepsilon$. Adding $-m$ to $m'<m+\varepsilon$ likewise gives $m'-m<\varepsilon$; by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field we have $m'-m=-(m-m')$, so claim 4 turns this into $-\varepsilon<m-m'$. Hence $-\varepsilon<m-m'<\varepsilon$, and claim 9 of Properties of the Absolute Value in an Ordered Field gives $|m-m'|<\varepsilon$, that is, $\bigl|\max\{f,g\}(z)-\max\{f,g\}(x)\bigr|<\varepsilon$.

Claim 3. Write $n=\min\{f(z),g(z)\}$ and $n'=\min\{f(x),g(x)\}$. By claim 1 of Elementary Properties of the Minimum of Two Elements we have $n'\le f(x)$ and $n'\le g(x)$, so adding $-\varepsilon$ gives $n'-\varepsilon\le f(x)-\varepsilon$ and $n'-\varepsilon\le g(x)-\varepsilon$. With the bounds $f(x)-\varepsilon<f(z)$ and $g(x)-\varepsilon<g(z)$ and claim 2 we get $n'-\varepsilon<f(z)$ and $n'-\varepsilon<g(z)$. By claim 2 of Elementary Properties of the Minimum of Two Elements the element $n$ equals $f(z)$ or $g(z)$, so $n'-\varepsilon<n$ in either case. Exchanging the roles of $z$ and $x$ gives $n-\varepsilon<n'$.

Adding $\varepsilon$ and then $-n'$ to $n'-\varepsilon<n$, and simplifying as in claim 2, gives $n'-n<\varepsilon$; likewise $n-n'<\varepsilon$. By claim 6 of Additive Cancellation and Elementary Additive Identities in a Field and claim 4, the first of these reads $-\varepsilon<n-n'$. Hence claim 9 of Properties of the Absolute Value in an Ordered Field gives $|n-n'|<\varepsilon$, that is, $\bigl|\min\{f,g\}(z)-\min\{f,g\}(x)\bigr|<\varepsilon$.

Claim 4. If $f$ and $g$ are continuous on $A$, they are continuous at every point of $A$ relative to $A$, so claims 1, 2 and 3 apply at every such point; hence $|f|$, $\max\{f,g\}$ and $\min\{f,g\}$ are continuous on $A$.