Proof of Continuity and Uniform Continuity of a Uniform Limit of Real-Valued Maps

Preliminaries. For $s,t,u\in\mathbb{R}$, $|s-t|\le|s-u|+|u-t|$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (xi); and $|s-t|=|t-s|$. Since $d_{\mathbb{R}}(g(y),g(z))=|g(y)-g(z)|$ for a map $g:A\to\mathbb{R}$ and $y,z\in A$ (The Distance on the Real Numbers Is a Map with Values the Absolute Differences (i)), the condition of Continuous Map Between Metric Spaces (i) for $g$ to be continuous at $x\in A$ relative to $A$ reads: for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $y\in A$ with $d_{X}(x,y)<\delta$ satisfies $|g(y)-g(x)|<\varepsilon$; and that of Uniformly Continuous Map Between Metric Spaces (i) reads likewise with $|g(y)-g(z)|<\varepsilon$ for all $y,z\in A$ with $d_{X}(y,z)<\delta$.

A third of $\varepsilon$. For a real $\varepsilon>0$ let $\eta=(\varepsilon/2)/2$. Two applications of Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii) give $0<\eta<\varepsilon/2<\varepsilon$, $\eta+\eta=\varepsilon/2$ and $\varepsilon/2+\varepsilon/2=\varepsilon$. If $\alpha,\beta,\gamma\in\mathbb{R}$ satisfy $\alpha<\eta$, $\beta<\eta$ and $\gamma<\eta$, then $\alpha+\beta<\eta+\eta=\varepsilon/2$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii), $\gamma<\varepsilon/2$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i), and so $\alpha+\beta+\gamma<\varepsilon/2+\varepsilon/2=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii).

Proof of (i). Let $T\subseteq S\subseteq A$, and suppose that $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f$ on $S$. Let $\varepsilon>0$ be real. By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) there is $K\in\mathbb{N}$ with $|f_{k}(x)-f(x)|<\varepsilon$ for every $k\in\mathbb{N}$ with $k\ge K$ and every $x\in S$. Every $x\in T$ lies in $S$, so the same $K$ serves for $T$; as $\varepsilon$ was arbitrary, $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f$ on $T$ by Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii). For a fixed $x\in S$ the same $K$ gives $|f_{k}(x)-f(x)|<\varepsilon$ for every $k\ge K$; as $\varepsilon$ was arbitrary, $(f_{k}(x))_{k\in\mathbb{N}}$ converges to $f(x)$ by Convergent Sequences of Real Numbers (i), and as $x\in S$ was arbitrary, $(f_{k})_{k\in\mathbb{N}}$ converges pointwise to $f$ on $S$ by Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (i).

Proof of (ii). Let $x$, $r$ and $S$ be as there. First, $x\in S$: indeed $d_{X}(x,x)=0<r$ by Metric Space (iii), so $x\in B_{d_{X}}(x,r)$ by Open Ball in a Metric Space (ii), and $x\in A$, so $x\in S$ by the hypothesis on $S$.

The choices are made in this order: a real $\varepsilon>0$ is given; then $\eta=(\varepsilon/2)/2$; then $K$; then $\delta_{0}$; then $\delta$. By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) applied to $\eta$, there is $K\in\mathbb{N}$ with $|f_{k}(y)-f(y)|<\eta$ for every $k\in\mathbb{N}$ with $k\ge K$ and every $y\in S$; we use it for $k=K$ only. Since $f_{K}$ is continuous at $x$ relative to $A$, there is a real $\delta_{0}>0$ such that every $y\in A$ with $d_{X}(x,y)<\delta_{0}$ satisfies $|f_{K}(y)-f_{K}(x)|<\eta$. By Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (viii) there is a real $\delta$ with $0<\delta$, $\delta\le\delta_{0}$ and $\delta\le r$.

Let $y\in A$ with $d_{X}(x,y)<\delta$. Then $d_{X}(x,y)<r$ and $d_{X}(x,y)<\delta_{0}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i). Hence $y\in B_{d_{X}}(x,r)$ by Open Ball in a Metric Space (ii), and as $y\in A$, $y\in S$; so $|f(y)-f_{K}(y)|=|f_{K}(y)-f(y)|<\eta$. As $x\in S$, also $|f_{K}(x)-f(x)|<\eta$, and $|f_{K}(y)-f_{K}(x)|<\eta$ by the choice of $\delta_{0}$. Two applications of the triangle inequality of the preliminaries give

$$|f(y)-f(x)|\le|f(y)-f_{K}(y)|+|f_{K}(y)-f(x)|\le|f(y)-f_{K}(y)|+|f_{K}(y)-f_{K}(x)|+|f_{K}(x)-f(x)|,$$

and the right-hand side is smaller than $\varepsilon$ by the third of $\varepsilon$. As $\varepsilon>0$ was arbitrary, $f$ is continuous at $x$ relative to $A$.

Proof of (iii). Let $x\in A$. The real number $1$ is positive by Rules of Arithmetic and Order in an Ordered Field (vii), and $A$ contains every point of $B_{d_{X}}(x,1)$ that lies in $A$. Each $f_{k}$, being continuous on $A$, is continuous at $x$ relative to $A$ by Continuous Map Between Metric Spaces (ii), and $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f$ on $A$. So (ii), applied with $r=1$ and $S=A$, shows that $f$ is continuous at $x$ relative to $A$. As $x\in A$ was arbitrary, $f$ is continuous on $A$ by Continuous Map Between Metric Spaces (ii).

Proof of (iv). The choices are made in this order: a real $\varepsilon>0$ is given; then $\eta=(\varepsilon/2)/2$; then $K$; then $\delta$. By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) applied to $\eta$, there is $K\in\mathbb{N}$ with $|f_{k}(y)-f(y)|<\eta$ for every $k\in\mathbb{N}$ with $k\ge K$ and every $y\in A$; we use it for $k=K$ only. Since $f_{K}$ is uniformly continuous on $A$, there is a real $\delta>0$ such that all $u,v\in A$ with $d_{X}(u,v)<\delta$ satisfy $|f_{K}(u)-f_{K}(v)|<\eta$.

Let $u,v\in A$ with $d_{X}(u,v)<\delta$. Then $|f(u)-f_{K}(u)|=|f_{K}(u)-f(u)|<\eta$, $|f_{K}(u)-f_{K}(v)|<\eta$ and $|f_{K}(v)-f(v)|<\eta$, and two applications of the triangle inequality of the preliminaries give

$$|f(u)-f(v)|\le|f(u)-f_{K}(u)|+|f_{K}(u)-f_{K}(v)|+|f_{K}(v)-f(v)|<\varepsilon ,$$

the last step by the third of $\varepsilon$. As $\varepsilon>0$ was arbitrary, $f$ is uniformly continuous on $A$.