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

Since $d_{\mathbb{R}}(s,t)=|s-t|$ by The Distance on the Real Numbers §distance, Continuous Map Between Metric Spaces §continuous-at says that a map $h:A\to\mathbb{R}$ is continuous at $x$ relative to $A$ if and only if for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $|h(y)-h(x)|<\varepsilon$. We verify this form. $\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, so Rules of Arithmetic and Order in an Ordered Field and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares apply to it, and The Maximum and Minimum of Two Elements of a Total Order applies to its total order.

Minima. For real $\delta_{1},\delta_{2}>0$, let $\delta=\min\{\delta_{1},\delta_{2}\}$, which exists by The Maximum and Minimum of Two Elements of a Total Order §exists. As $\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, its order is total by Ordered Fields §ordered-field, so $\delta_{1}\le\delta_{2}$ or $\delta_{2}\le\delta_{1}$, so $\delta\in\{\delta_{1},\delta_{2}\}$ by The Maximum and Minimum of Two Elements of a Total Order §values and $0<\delta$; and $\delta\le\delta_{1}$, $\delta\le\delta_{2}$ by The Maximum and Minimum of Two Elements of a Total Order §bounds. If $y\in A$ and $d(x,y)<\delta$, then $d(x,y)<\delta_{1}$ and $d(x,y)<\delta_{2}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed.

Constants. Let $b\in\mathbb{R}$, let $h:A\to\mathbb{R}$ be the map with constant value $b$, and let $\varepsilon>0$ be real. Take $\delta=1$, positive by Rules of Arithmetic and Order in an Ordered Field §squares. For $y\in A$, $h(y)-h(x)=b+(-b)=0$ by Negatives, Differences, Reciprocals and Quotients §negative, and $|0|=0$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value; so $|h(y)-h(x)|=0<\varepsilon$.

Sums. Let $\varepsilon>0$ be real, and let $\eta=\varepsilon/2$; then $0<\eta$ and $\eta+\eta=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. Choose, by continuity of $f$ and of $g$ at $x$ relative to $A$ applied with $\eta$, real $\delta_{1},\delta_{2}>0$ such that every $y\in A$ with $d(x,y)<\delta_{1}$ satisfies $|f(y)-f(x)|<\eta$ and every $y\in A$ with $d(x,y)<\delta_{2}$ satisfies $|g(y)-g(x)|<\eta$; let $\delta=\min\{\delta_{1},\delta_{2}\}>0$ as in the paragraph Minima. Let $y\in A$ with $d(x,y)<\delta$. By the field laws and $-(u+v)=(-u)+(-v)$ (Rules of Arithmetic and Order in an Ordered Field §signs),

$$(f+g)(y)-(f+g)(x)=\bigl(f(y)-f(x)\bigr)+\bigl(g(y)-g(x)\bigr),$$

so Rules of Arithmetic and Order in an Ordered Field §triangle gives $|(f+g)(y)-(f+g)(x)|\le|f(y)-f(x)|+|g(y)-g(x)|$, and the right-hand side is $<\eta+\eta=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum (a strict inequality gives $\le$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict). Hence $|(f+g)(y)-(f+g)(x)|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed.

Products. Let $\varepsilon>0$ be real. Put $M=|g(x)|+1$ and $K=|f(x)|+1$. As $0<1$ (Rules of Arithmetic and Order in an Ordered Field §squares) and $0\le|g(x)|$, $0\le|f(x)|$ (Rules of Arithmetic and Order in an Ordered Field §absolute-value), Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum and $0=0+0$ give $0<M$ and $0<K$; and $|f(x)|=|f(x)|+0\le K$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Let $\eta=\varepsilon/2$, so $0<\eta$ and $\eta+\eta=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving; then $\eta/M$ and $\eta/K$ are positive by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product.

In this order, choose by continuity at $x$ relative to $A$: a real $\delta_{0}>0$ for $g$ and the tolerance $1$, a real $\delta_{1}>0$ for $f$ and the tolerance $\eta/M$, and a real $\delta_{2}>0$ for $g$ and the tolerance $\eta/K$. Let $\delta=\min\{\delta_{0},\min\{\delta_{1},\delta_{2}\}\}$, positive and, by the paragraph Minima applied twice, such that every $y\in A$ with $d(x,y)<\delta$ satisfies $d(x,y)<\delta_{j}$ for $j=0,1,2$.

Let $y\in A$ with $d(x,y)<\delta$, and put $u=|f(y)-f(x)|$ and $v=|g(y)-g(x)|$, so $u<\eta/M$, $v<\eta/K$ and $v<1$, and $0\le u$, $0\le v$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value.

First, $g(y)=(g(y)-g(x))+g(x)$ by the field laws, so $|g(y)|\le v+|g(x)|$ by Rules of Arithmetic and Order in an Ordered Field §triangle, and $v+|g(x)|<1+|g(x)|=M$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum; hence $|g(y)|\le M$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed.

Next, by the field laws and $(s-t)w=sw-tw$ (Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs),

$$f(y)g(y)-f(x)g(x)=\bigl(f(y)-f(x)\bigr)g(y)+f(x)\bigl(g(y)-g(x)\bigr),$$

so Rules of Arithmetic and Order in an Ordered Field §triangle and the product rule $|st|=|s|\,|t|$ of Rules of Arithmetic and Order in an Ordered Field §absolute-value give $|(fg)(y)-(fg)(x)|\le u\,|g(y)|+|f(x)|\,v$.

Since $0\le u$ and $|g(y)|\le M$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling gives $u\,|g(y)|\le u\,M$; since $0<M$ and $u<\eta/M$, Rules of Arithmetic and Order in an Ordered Field §order-product gives $u\,M<(\eta/M)\,M=\eta$, the last equality by Negatives, Differences, Reciprocals and Quotients §reciprocal and associativity. So $u\,|g(y)|<\eta$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. In the same way, from $0\le v$, $|f(x)|\le K$, $0<K$ and $v<\eta/K$ we get $|f(x)|\,v\le K\,v<\eta$, so $|f(x)|\,v<\eta$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum (a strict inequality gives $\le$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict), $u\,|g(y)|+|f(x)|\,v<\eta+\eta=\varepsilon$, and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $|(fg)(y)-(fg)(x)|<\varepsilon$.

Scalar. Let $h_{c}:A\to\mathbb{R}$ be the map with constant value $c$; it is continuous at $x$ relative to $A$ by the clause constants. Its product with $f$, formed as in the statement, has value $c\,f(z)=(cf)(z)$ at each $z\in A$, so it is $cf$. The argument of the clause products used only that both factors are continuous at $x$ relative to $A$; applied to the pair $h_{c},f$ in place of $f,g$, it shows that $cf$ is continuous at $x$ relative to $A$.

On-set. If $f$ and $g$ are continuous on $A$, then by Continuous Map Between Metric Spaces §continuous they are continuous at every $x'\in A$ relative to $A$, so the clauses sums, products and scalar, applied at $x'$ in place of $x$, show that $f+g$, $fg$ and $cf$ are continuous at every $x'\in A$ relative to $A$, that is, continuous on $A$ by Continuous Map Between Metric Spaces §continuous.