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

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named in the step where it is cited. Since $d_{\mathbb{R}}(s,t)=|s-t|$ by The Absolute Value Metric on the Real Numbers, the defining condition of Continuous Map Between Metric Spaces for a map $g$ from $A$ into $\mathbb{R}$, continuous at $x$ 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 $|g(y)-g(x)|<\varepsilon$; the condition of Uniformly Continuous Map Between Metric Spaces is read in the same way.

A quarter of $\varepsilon$. Let $\varepsilon$ be a real number with $0<\varepsilon$. Applying claim 8 of Elementary Order Arithmetic in an Ordered Field first to $\varepsilon$ and then to the number it produces, we obtain a real $\eta$ with $0<\eta$ and $\eta+\eta+\eta+\eta=\varepsilon$. Claim 3 of Elementary Order Arithmetic in an Ordered Field, applied to the strict inequality $0<\eta$ and the non-strict inequality $\eta+\eta+\eta\le\eta+\eta+\eta$, gives

$$\eta+\eta+\eta=0+(\eta+\eta+\eta)<\eta+(\eta+\eta+\eta)=\varepsilon .$$

We refer to such an $\eta$ below as a quarter of $\varepsilon$. We also use twice, in Claims 2 and 4, the following consequence of claim 3 of Elementary Order Arithmetic in an Ordered Field: if $\alpha<\eta$, $\beta<\eta$ and $\gamma<\eta$, then $\alpha+\beta+\gamma<\eta+\eta+\eta$, by applying that claim first to $\alpha<\eta$ together with $\beta\le\eta$ and then to the resulting strict inequality together with $\gamma\le\eta$.

Claim 1. Let $T\subseteq S\subseteq A$ and suppose that $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f$ on $S$.

For uniform convergence on $T$, let $\varepsilon>0$. By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) there is $K\in\mathbb{N}$ such that $|f_{k}(x)-f(x)|<\varepsilon$ for every $k\in\mathbb{N}$ with $K\le k$ and every $x\in S$. Every $x\in T$ lies in $S$, so the same $K$ witnesses that condition with $T$ in place of $S$; hence $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f$ on $T$.

For pointwise convergence on $S$, fix $x\in S$ and let $\varepsilon>0$. With $K$ as above, $|f_{k}(x)-f(x)|<\varepsilon$ for every $k\in\mathbb{N}$ with $K\le k$, which is the condition of Limit of a Sequence of Real Numbers for the sequence of real numbers $(f_{k}(x))_{k\in\mathbb{N}}$ and the limit $f(x)$. As $x\in S$ was arbitrary, Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (i) gives pointwise convergence to $f$ on $S$.

Claim 2. Let $x$, $r$ and $S$ be as in the statement and let $\varepsilon>0$. The choices below are made in this order: $\varepsilon$ is given, then a quarter $\eta$ of $\varepsilon$, then $K$, then $\delta_{0}$, then $\delta$.

First, $x\in S$. Indeed $d_{X}(x,x)=0<r$ by the axioms of Metric Space, so $x$ lies in the open ball $B_{d_{X}}(x,r)$ by Open Ball in a Metric Space; as also $x\in A$, the hypothesis on $S$ places $x$ in $S$.

By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) there is $K\in\mathbb{N}$ such that $|f_{k}(y)-f(y)|<\eta$ for every $k\in\mathbb{N}$ with $K\le k$ and every $y\in S$; we use this 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 $|f_{K}(y)-f_{K}(x)|<\eta$. Let $\delta$ be the least of the two numbers $\delta_{0}$ and $r$, which is positive and satisfies $\delta\le\delta_{0}$ and $\delta\le r$, by claim 9 of Elementary Order Arithmetic in an Ordered Field.

Now let $y\in A$ satisfy $d_{X}(x,y)<\delta$. Then $d_{X}(x,y)<r$, so $y\in B_{d_{X}}(x,r)$, and $y\in A$, so $y\in S$; hence $|f(y)-f_{K}(y)|<\eta$, using claim 2 of Properties of the Absolute Value in an Ordered Field to exchange the two arguments. Since $x\in S$, likewise $|f_{K}(x)-f(x)|<\eta$. Also $d_{X}(x,y)<\delta_{0}$, so $|f_{K}(y)-f_{K}(x)|<\eta$. Two applications of the triangle inequality, claim 5 of Properties of the Absolute Value in an Ordered Field, give

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

and the right-hand side is smaller than $\eta+\eta+\eta$ by the consequence recorded above, hence smaller than $\varepsilon$. As $\varepsilon>0$ was arbitrary, $f$ is continuous at $x$ relative to $A$.

Claim 3. Let $x\in A$. The number $1$ is positive by claim 6 of Elementary Order Arithmetic in an Ordered Field, and the set $A$ 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, and $(f_{k})_{k\in\mathbb{N}}$ converges uniformly to $f$ on $A$. So Claim 2, 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 $A$.

Claim 4. Let $\varepsilon>0$ and let $\eta$ be a quarter of $\varepsilon$. The choices are made in this order: $\varepsilon$, then $\eta$, then $K$, then $\delta$. By Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) there is $K\in\mathbb{N}$ with $|f_{k}(y)-f(y)|<\eta$ for every $k\in\mathbb{N}$ with $K\le k$ and every $y\in A$; we use this for $k=K$ only. Since $f_{K}$ is uniformly continuous on $A$, Uniformly Continuous Map Between Metric Spaces provides 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 $u,v\in A$ satisfy $d_{X}(u,v)<\delta$. Then, by two applications of the triangle inequality, claim 5 of Properties of the Absolute Value in an Ordered Field, together with claim 2 of that lemma,

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

and each of the three terms on the right is smaller than $\eta$, so the right-hand side is smaller than $\eta+\eta+\eta$, hence smaller than $\varepsilon$. As $\varepsilon>0$ was arbitrary, $f$ is uniformly continuous on $A$.