Proof of Extension of a Uniformly Continuous Real Function from a Dense Subset

Each result cited is universally quantified over the data in its own statement.

Step 0 (two facts used repeatedly). First, for every $x\in X$ there is a sequence in $S$ converging to $x$ in $(X,d)$: since $S$ is dense, the closure of $S$ in $(X,\mathcal{T}_{d})$ is $X$, so $x$ belongs to that closure, and Sequential Characterization of the Closure in a Metric Space supplies the sequence. Secondly, if a sequence $(a_{m})_{m\in\mathbb{N}}$ of real numbers converges to $a$ and $c\in\mathbb{R}$ satisfies $a_{m}\le c$ for every $m\in\mathbb{N}$, then $a\le c$: otherwise $\eta=a-c$ is positive, and convergence gives an $m$ with $|a_{m}-a|<\eta$, hence $a_{m}>a-\eta=c$ by claim 9 of Properties of the Absolute Value in an Ordered Field, contradicting $a_{m}\le c$.

Claim 2. Let $g_{1},g_{2}$ be as stated and let $x\in X$. By Step 0 choose a sequence $(x_{m})_{m\in\mathbb{N}}$ in $S$ converging to $x$. By Continuity Between Metric Spaces is Equivalent to Sequential Continuity the sequences $(g_{1}(x_{m}))_{m\in\mathbb{N}}$ and $(g_{2}(x_{m}))_{m\in\mathbb{N}}$ converge to $g_{1}(x)$ and to $g_{2}(x)$ respectively. These are the same sequence of real numbers, because $g_{1}$ and $g_{2}$ agree on $S$ and every $x_{m}$ lies in $S$; since limits are unique by The Real Numbers: Standing Notation and Background (iii), $g_{1}(x)=g_{2}(x)$.

Claim 1. Let $f:S\to\mathbb{R}$ be uniformly continuous on $S$, let $x\in X$ and let $(x_{m})_{m\in\mathbb{N}}$ be a sequence in $S$ converging to $x$.

(i) The sequence $(f(x_{m}))_{m\in\mathbb{N}}$ converges. Let $\varepsilon\in\mathbb{R}$ be positive. By uniform continuity of $f$ on $S$ there is a positive $\rho$ such that all $a,b\in S$ with $d(a,b)<\rho$ satisfy $|f(a)-f(b)|<\varepsilon$. By convergence there is $N\in\mathbb{N}$ with $d(x_{m},x)<\tfrac{\rho}{2}$ for every $m$ with $N\le m$, and then $d(x_{m},x_{n})\le d(x_{m},x)+d(x,x_{n})<\rho$ for all such $m,n$, whence $|f(x_{m})-f(x_{n})|<\varepsilon$. Thus $(f(x_{m}))_{m\in\mathbb{N}}$ is a Cauchy sequence, and it converges by The Real Numbers: Standing Notation and Background (iii).

(ii) The limit does not depend on the sequence. Let $(y_{m})_{m\in\mathbb{N}}$ be a sequence in $S$ converging to $x$, let $a$ be the limit of $(f(x_{m}))_{m\in\mathbb{N}}$ and $b$ that of $(f(y_{m}))_{m\in\mathbb{N}}$, and let $\varepsilon$ be positive. With $\rho$ as in (i) there is $N$ such that $d(x_{m},x)<\tfrac{\rho}{2}$ and $d(y_{m},x)<\tfrac{\rho}{2}$, hence $d(x_{m},y_{m})<\rho$ and $|f(x_{m})-f(y_{m})|<\varepsilon$, for every $m$ with $N\le m$; enlarging $N$ we may also assume $|f(x_{N})-a|<\varepsilon$ and $|f(y_{N})-b|<\varepsilon$. Then, by the triangle inequality of claim 5 of Properties of the Absolute Value in an Ordered Field,

$$|a-b|\le|a-f(x_{N})|+|f(x_{N})-f(y_{N})|+|f(y_{N})-b|<3\varepsilon .$$

As $\varepsilon$ was an arbitrary positive real and $3\varepsilon$ ranges over all positive reals with it, Comparison of Real Numbers with Arbitrary Positive Slack gives $|a-b|\le0$, and since $0\le|a-b|$ we get $a=b$.

By (i) and (ii), for each $x\in X$ there is exactly one real number $L$ such that $(f(x_{m}))_{m\in\mathbb{N}}$ converges to $L$ for every sequence $(x_{m})_{m\in\mathbb{N}}$ in $S$ converging to $x$: such an $L$ exists because at least one such sequence exists by Step 0 and (ii) identifies the limits of all of them, and it is unique for the same reason. Let $g:X\to\mathbb{R}$ be the function whose value at $x$ is that number.

(iii) $g$ agrees with $f$ on $S$. For $x\in S$ the sequence all of whose terms are $x$ lies in $S$ and converges to $x$ in $(X,d)$, because $d(x,x)=0$ is smaller than every positive real, and the sequence of its values, all equal to $f(x)$, converges to $f(x)$ by Constant Sequences and Index-Shifted Sequences of Real Numbers (i); hence $g(x)=f(x)$ by the definition of $g$.

(iv) $g$ is uniformly continuous, hence continuous, on $X$. Let $\varepsilon$ be positive and choose a positive $\rho$ such that all $a,b\in S$ with $d(a,b)<\rho$ satisfy $|f(a)-f(b)|<\tfrac{\varepsilon}{2}$. Let $x,y\in X$ satisfy $d(x,y)<\tfrac{\rho}{3}$ and choose sequences $(x_{m})_{m\in\mathbb{N}}$ and $(y_{m})_{m\in\mathbb{N}}$ in $S$ converging to $x$ and to $y$. There is $N$ with $d(x_{m},x)<\tfrac{\rho}{3}$ and $d(y_{m},y)<\tfrac{\rho}{3}$ for every $m$ with $N\le m$, so that $d(x_{m},y_{m})\le d(x_{m},x)+d(x,y)+d(y,y_{m})<\rho$ and hence $|f(x_{m})-f(y_{m})|\le\tfrac{\varepsilon}{2}$ for all such $m$. Replacing the two sequences by the subsequences with terms $x_{m+N}$ and $y_{m+N}$, which converge to the same limits by A Subsequence of a Convergent Sequence Has the Same Limit, we may assume that this bound holds for every $m\in\mathbb{N}$. The sequences $(f(x_{m}))_{m\in\mathbb{N}}$ and $(f(y_{m}))_{m\in\mathbb{N}}$ converge to $g(x)$ and $g(y)$ by the definition of $g$, so by claim 3 of Arithmetic of Limits of Real Sequences and claim 4 of Order Properties of Limits of Real Sequences the sequence with terms $|f(x_{m})-f(y_{m})|$ converges to $|g(x)-g(y)|$. By Step 0, $|g(x)-g(y)|\le\tfrac{\varepsilon}{2}<\varepsilon$. Thus the condition of Uniformly Continuous Map Between Metric Spaces holds with $\tfrac{\rho}{3}$, so $g$ is uniformly continuous on $X$; taking, for a given point and a given $\varepsilon$, the same number $\tfrac{\rho}{3}$ shows that $g$ is continuous on $X$.

Claim 3. Let $g$, $\varepsilon$ and $\theta$ be as stated and let $x,y\in X$ satisfy $d(x,y)<\theta$; put $\mu=\theta-d(x,y)$, a positive real. Choose sequences $(x_{m})_{m\in\mathbb{N}}$ and $(y_{m})_{m\in\mathbb{N}}$ in $S$ converging to $x$ and to $y$, and $N$ with $d(x_{m},x)<\tfrac{\mu}{2}$ and $d(y_{m},y)<\tfrac{\mu}{2}$ for $N\le m$. For such $m$,

$$d(x_{m},y_{m})\le d(x_{m},x)+d(x,y)+d(y,y_{m})<d(x,y)+\mu=\theta ,$$

so $d(x_{m},y_{m})\le\theta$ and the hypothesis gives $|g(x_{m})-g(y_{m})|\le\varepsilon$, both points lying in $S$. Passing to the subsequences with terms $x_{m+N}$, $y_{m+N}$ as in (iv), this holds for every index. By Continuity Between Metric Spaces is Equivalent to Sequential Continuity the sequences $(g(x_{m}))_{m\in\mathbb{N}}$ and $(g(y_{m}))_{m\in\mathbb{N}}$ converge to $g(x)$ and $g(y)$, so by claim 3 of Arithmetic of Limits of Real Sequences and claim 4 of Order Properties of Limits of Real Sequences the sequence with terms $|g(x_{m})-g(y_{m})|$ converges to $|g(x)-g(y)|$; Step 0 gives $|g(x)-g(y)|\le\varepsilon$.

Claim 4. Let $g$ and $C$ be as stated and let $x\in X$. Choose a sequence $(x_{m})_{m\in\mathbb{N}}$ in $S$ converging to $x$. Then $|g(x_{m})|\le C$ for every $m$, the points $x_{m}$ lying in $S$, and by Continuity Between Metric Spaces is Equivalent to Sequential Continuity and claim 4 of Order Properties of Limits of Real Sequences the sequence with terms $|g(x_{m})|$ converges to $|g(x)|$. By Step 0, $|g(x)|\le C$.