Proof of Continuity of Sums and Products of Real-Valued Functions on a Metric Space

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

Claim 1. Let $b\in\mathbb{R}$, let $h$ be the constant map with value $b$, and let $0<\varepsilon$. Take $\delta=1$, which satisfies $0<1$ by claim 6. For $z\in A$ we have $h(z)-h(x)=b-b=0$ by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field, and $|0|=0$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so $|h(z)-h(x)|=0<\varepsilon$.

Claim 2. Let $0<\varepsilon$. By claim 8 the element $\eta=\varepsilon\cdot 2^{-1}$ satisfies $0<\eta$ and $\eta+\eta=\varepsilon$. 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 $|f(z)-f(x)|<\eta$ and $d(x,z)<\delta_2$ implies $|g(z)-g(x)|<\eta$, 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 $0<\delta$.

Let $z\in A$ with $d(x,z)<\delta$. By claim 2 we get $d(x,z)<\delta_1$ and $d(x,z)<\delta_2$, hence $|f(z)-f(x)|<\eta$ and $|g(z)-g(x)|<\eta$. By commutativity and associativity of addition together with claim 6 of Additive Cancellation and Elementary Additive Identities in a Field,

so claim 5 of Properties of the Absolute Value in an Ordered Field gives

By claim 3 the right-hand side is smaller than $\eta+\eta=\varepsilon$, so claim 2 gives $|(f+g)(z)-(f+g)(x)|<\varepsilon$.

Claim 3. Let $0<\varepsilon$, and set $M=|g(x)|+1$ and $N=|f(x)|+1$. By claim 1 of Properties of the Absolute Value in an Ordered Field we have $0\le|g(x)|$, so $0+1\le|g(x)|+1$ by compatibility of the order with addition in an ordered field; since $0<1$ by claim 6 and $0+1=1$, claim 2 gives $0<M$, and likewise $0<N$. By claim 7 the inverses $M^{-1}$ and $N^{-1}$ exist and are positive. By claim 8 the element $\eta=\varepsilon\cdot 2^{-1}$ is positive with $\eta+\eta=\varepsilon$, and by claim 5 the elements $\eta\cdot M^{-1}$ and $\eta\cdot N^{-1}$ are positive.

Continuity of $f$ and of $g$ at $x$ relative to $A$ gives positive $\delta_0,\delta_1,\delta_2$ such that, for $z\in A$: $d(x,z)<\delta_0$ implies $|g(z)-g(x)|<1$; $d(x,z)<\delta_1$ implies $|f(z)-f(x)|<\eta\cdot M^{-1}$; and $d(x,z)<\delta_2$ implies $|g(z)-g(x)|<\eta\cdot N^{-1}$. Applying claim 9 twice produces $\delta$ with $\delta\le\delta_0$, $\delta\le\delta_1$, $\delta\le\delta_2$ and $\delta$ equal to one of them, so $0<\delta$.

Let $z\in A$ with $d(x,z)<\delta$, and put $u=|f(z)-f(x)|$ and $v=|g(z)-g(x)|$; by claim 2 all three implications above apply, and $0\le u$ and $0\le v$ by claim 1 of Properties of the Absolute Value in an Ordered Field.

First, $|g(z)|=|(g(z)-g(x))+g(x)|\le v+|g(x)|$ by claim 5 of Properties of the Absolute Value in an Ordered Field, and $v<1$ gives $v+|g(x)|<1+|g(x)|=M$ by claim 1; so $|g(z)|<M$ by claim 2, and in particular $|g(z)|\le M$. Also $|f(x)|+0\le|f(x)|+1=N$ by compatibility of the order with addition applied to $0\le 1$, that is $|f(x)|\le N$.

By distributivity, commutativity and associativity in the field,

so claims 5 and 4 of Properties of the Absolute Value in an Ordered Field give

Since $0\le u$ and $|g(z)|\le M$, claim 5 of Elementary Arithmetic in an Ordered Field gives $u\,|g(z)|\le u\,M$. Since $u<\eta\cdot M^{-1}$ and $0<M$, claim 10 gives $M\,u<M\cdot(\eta\cdot M^{-1})$, and the right-hand side equals $\eta$ by associativity and commutativity of multiplication together with the defining property of the multiplicative inverse and of the multiplicative identity. As $u\,M=M\,u$, claim 2 yields $u\,|g(z)|<\eta$. The same argument with $v$, $|f(x)|$ and $N$ in place of $u$, $|g(z)|$ and $M$ yields $|f(x)|\,v<\eta$.

By claim 3, $u\,|g(z)|+|f(x)|\,v<\eta+\eta=\varepsilon$, and claim 2 gives $|f(z)g(z)-f(x)g(x)|<\varepsilon$.

Claim 4. By claim 1 the constant map with value $c$ is continuous at $x$ relative to $A$, and $cf$ is its pointwise product with $f$. Applying claim 3 with that constant map in place of $f$ and with $f$ in place of $g$ gives the assertion.

Claim 5. If $f$ and $g$ are continuous on $A$, they are continuous at every point of $A$ relative to $A$, so claims 2, 3 and 4 apply at every such point; hence $f+g$, $fg$ and $cf$ are continuous at every point of $A$ relative to $A$, that is, continuous on $A$.