Proof of The Uniform Cauchy Criterion for Sequences of Real-Valued Maps

Two facts on $\mathbb{R}$ are used repeatedly. For real $\varepsilon>0$, the number $\varepsilon/2$ satisfies $0<\varepsilon/2<\varepsilon$ and $\varepsilon/2+\varepsilon/2=\varepsilon$, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii). 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 $|s-t|=|t-s|$.

Step 1: a uniform limit is a pointwise limit. Suppose that $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to a map $g:S\to\mathbb{R}$ on $S$. Let $x\in S$ and let $\varepsilon>0$ be real. By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii), read with $E=S$ and $g$ in place of $f$, there is $K\in\mathbb{N}$ with $|f_{k}(y)-g(y)|<\varepsilon$ for every $k\in\mathbb{N}$ with $k\ge K$ and every $y\in S$; in particular $|f_{k}(x)-g(x)|<\varepsilon$ for every $k\ge K$. As $\varepsilon$ was arbitrary, $(f_{k}(x))_{k\in\mathbb{N}}$ converges to $g(x)$ by Convergent Sequences of Real Numbers (i); as $x\in S$ was arbitrary, $(f_{k})_{k\in\mathbb{N}}$ converges pointwise to $g$ on $S$ by Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (i).

Step 2: the map of the limits. Suppose that $(f_{k}(x))_{k\in\mathbb{N}}$ is convergent for every $x\in S$. Then Sequences of Real-Valued Maps: the Map of Pointwise Limits on a Subset, and the Sequence of Pointwise Partial Sums (i), applied to the set $D$, the subset $S$ and the sequence $(f_{k})_{k\in\mathbb{N}}$ in $\mathbb{R}^{D}$, gives exactly one map $L:S\to\mathbb{R}$ with $L(x)=\lim_{k\to\infty}f_{k}(x)$ for every $x\in S$; this is the map $x\mapsto\lim_{k\to\infty}f_{k}(x)$, and $(f_{k}(x))_{k\in\mathbb{N}}$ converges to $L(x)$ for every $x\in S$ by The Limit of a Convergent Sequence (i). If $g:S\to\mathbb{R}$ and $(f_{k})_{k\in\mathbb{N}}$ converges pointwise to $g$ on $S$, then for every $x\in S$ the sequence $(f_{k}(x))_{k\in\mathbb{N}}$ converges to $g(x)$ by Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (i) and to $L(x)$, so $g(x)=L(x)$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii); hence $g=L$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)).

Proof of (ii). Suppose that $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to a map $h:S\to\mathbb{R}$ on $S$. By Step 1 it converges pointwise to $h$ on $S$, so $(f_{k}(x))_{k\in\mathbb{N}}$ converges to $h(x)$, and is convergent, for every $x\in S$. Step 2 now provides the map $L=(x\mapsto\lim_{k\to\infty}f_{k}(x))$ and shows that every $g:S\to\mathbb{R}$ to which $(f_{k})_{k\in\mathbb{N}}$ converges pointwise on $S$ equals $L$; in particular $h=L$, so $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $L$ on $S$. Finally, if $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $g:S\to\mathbb{R}$ on $S$, it converges pointwise to $g$ on $S$ by Step 1, so $g=L$.

Proof of (i), from left to right. Suppose that $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f:S\to\mathbb{R}$ on $S$, and let $\varepsilon>0$ be real. Apply Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii), read with $E=S$, to the real number $\varepsilon/2>0$: there is $K\in\mathbb{N}$ with $|f_{k}(x)-f(x)|<\varepsilon/2$ for every $k\in\mathbb{N}$ with $k\ge K$ and every $x\in S$. Let $m,k\in\mathbb{N}$ with $m\ge K$ and $k\ge K$, and let $x\in S$. Then $|f_{m}(x)-f(x)|<\varepsilon/2$ and $|f(x)-f_{k}(x)|=|f_{k}(x)-f(x)|<\varepsilon/2$, so

$$|f_{m}(x)-f_{k}(x)|\le|f_{m}(x)-f(x)|+|f(x)-f_{k}(x)|<\varepsilon/2+\varepsilon/2=\varepsilon ,$$

the strict inequality by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii) and the two together by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i). So $K$ is as required.

Proof of (i), from right to left. Suppose that for every real $\varepsilon>0$ there is $K\in\mathbb{N}$ with $|f_{m}(x)-f_{k}(x)|<\varepsilon$ for all $m,k\in\mathbb{N}$ with $m\ge K$ and $k\ge K$ and every $x\in S$.

First, for every $x\in S$ the sequence $(f_{k}(x))_{k\in\mathbb{N}}$ is convergent. Indeed, for every real $\varepsilon>0$ the $K$ of the hypothesis gives $|f_{m}(x)-f_{k}(x)|<\varepsilon$ for all $m,k\in\mathbb{N}$ with $m,k\ge K$, so $(f_{k}(x))_{k\in\mathbb{N}}$ is a Cauchy sequence by Cauchy Sequences of Real Numbers (i), hence convergent by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences (v). Let $L:S\to\mathbb{R}$, $L(x)=\lim_{k\to\infty}f_{k}(x)$, be the map of Step 2; $(f_{k}(x))_{k\in\mathbb{N}}$ converges to $L(x)$ for every $x\in S$.

We show that $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $L$ on $S$. The choices are made in this order: first a real $\varepsilon>0$ is given; then $K\in\mathbb{N}$ is taken from the hypothesis applied to $\varepsilon/2$, once for all points of $S$; then $k\in\mathbb{N}$ with $k\ge K$ and $x\in S$ are arbitrary. So $|f_{m}(x)-f_{k}(x)|<\varepsilon/2$ for every $m\in\mathbb{N}$ with $m\ge K$, that is, $-\varepsilon/2\le f_{m}(x)-f_{k}(x)\le\varepsilon/2$ by Rules of Arithmetic and Order in an Ordered Field (x), and so $f_{k}(x)-\varepsilon/2\le f_{m}(x)\le f_{k}(x)+\varepsilon/2$ by Rules of Arithmetic and Order in an Ordered Field (iv). Since $(f_{m}(x))_{m\in\mathbb{N}}$ converges to $L(x)$, Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x), applied with $f_{k}(x)+\varepsilon/2$ and $f_{k}(x)-\varepsilon/2$ in place of $\lambda$, gives $f_{k}(x)-\varepsilon/2\le L(x)\le f_{k}(x)+\varepsilon/2$. Hence $-\varepsilon/2\le f_{k}(x)-L(x)\le\varepsilon/2$ by Rules of Arithmetic and Order in an Ordered Field (iv), so $|f_{k}(x)-L(x)|\le\varepsilon/2$ by Rules of Arithmetic and Order in an Ordered Field (x), and therefore $|f_{k}(x)-L(x)|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i). This is the condition of Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii), read with $E=S$ and $L$ in place of $f$; so $(f_{k})_{k\in\mathbb{N}}$ converges uniformly on $S$ to the map $L:S\to\mathbb{R}$.