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Let $(S,d_S)$ and $(T,d_T)$ be two metric spaces with $T$ complete and suppose that a mapping $f:A\subseteq S\to T$ is uniformly continuous on its domain. I want to show that $f$ can be uniquely extended to the closure $\overline{A}$ with the extension uniformly continuous on $\overline{A}$.

Actually, this is a homework exercise from my introductory-analysis class, and before this exercise, several subsidiary questions had been deployed so I kind of cracked this question. For example, with every $x\in\overline{A}$, there comes a sequence $\{x_n\}_{n=1}^\infty$ in $A$ s.t. $x_n\to x$. This sequence is mapped to a Cauchy sequence $\{f(x_n)\}_{n=1}^\infty$ since $f$ is uniformly continuous on $A$. Then, thanks to completeness of $T$, we are assured of convergence of this Cauchy sequence, and hence able to designate the limit to be the value of a newborn $g:\overline{A}\to T$ at $x$. This $g$ will be the desired extension once we justify the following claims.

First, we have to show that $g$ is well-defined. By this, I mean another sequence $\{x_n\}_{n=1}^\infty\subseteq A$ converging to $x$ does not yield a different limit of $\{f(x_n)\}_{n=1}^\infty$. After establishing this claim, we need to show that $g$ is uniformly continuous on its domain, which requires employing uniform continuity of $f$. Last but not least, we have to show that $g$ is the restriction of $f$ to $A$. This can be done by recalling that $f$ is a continuous mapping.

So far, I've been able to resolve all the doubts about the claims mentioned above, but as indicated in the first paragraph, I fail to explain why a uniformly continuous extension of $f$ to $\overline{A}$ is unique. Does anyone have an idea? Thank you.

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    $\begingroup$ Assume we have such an extension function $f$. If $a$ is a limit point of $A$ then choose any sequence $\{a_i\}_{i=1}^{\infty}$ of points in $A$ that converges to $a$. Then $f(a)$ must have the value $\lim_{n\rightarrow\infty} f(a_i)$ (and the limit must exist) else $f$ is not continuous. It is not possible to choose $f(a)$ to have any other value. There were no choices about $f(a_i)$ since the function was already defined at those points. $\endgroup$
    – Michael
    Jul 15 at 0:41
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    $\begingroup$ At most one continuous extension to the closure: requires only that $f$ be continuous. Exists at least one continuous extension: this follows from uniform continuity. $\endgroup$
    – GEdgar
    Jul 15 at 1:27

3 Answers 3

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Uniqueness (Alternative)

Suppose $f_1, f_2$ be two uniform extension of $f$ on $\overline{A}$

Then $f_1=f=f_2$ on $A$

Let $D=\{x\in \overline{A} : f_1(x) =f_2(x) \}$

Then clearly $D$ is a closed set i.e $\overline{D}=D$ (the set of all points where two continuous map agrees is a closed set, provided the target space is Hausdorff.)

$ A\subset D\subset \overline{A}$

Then $\overline{A}\subset\overline{ D}\subset \overline{A}$

Implies $\overline{A}\subset{ D}\subset \overline{A}$

Hence $D=\overline{A}$ , implies $f_1=f_2$ on $\overline{A}$.

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  • $\begingroup$ Thank you. Since $\overline{A}$ is the smallest closed set that contains $A$, that $\overline{A}\subseteq S$ should be obvious. $\endgroup$
    – Wombat
    Jul 15 at 2:46
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    $\begingroup$ Exactly. You are correct. $\endgroup$ Jul 15 at 2:51
  • $\begingroup$ Your argument that $\mathcal{S}$ (to be distinguished from the metric space $(S,d_S)$) is closed in $S$ is new to me. Does that take a large amount of knowledge of topology? $\endgroup$
    – Wombat
    Jul 15 at 3:18
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    $\begingroup$ I used two facts : 1) set of all points where two continuous functions (between two metric spaces) agree is closed set . 2) If $D \subset S$ is closed then $D\cap \overline{A}$ is closed in $(\overline{A}, d_{\overline{A}}) $ $\endgroup$ Jul 15 at 3:55
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    $\begingroup$ Since $\overline{A}$ is closed, for any $D\subset \overline{A}$ is closed in the subspace iff colsed in the whole space. Hence no need to worry about closed (doesn't matter where) set $D$ $\endgroup$ Jul 15 at 4:02
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A brief note in the contra style.

Let $(x_n)_n$ and $(y_n)_n$ be any sequences in $A,$ both converging to $a\in\overline A.$ Let $c_{2n}=x_n$ and $c_{2n+1}=y_n.$ Then $(c_n)_n$ converges to $a.$

Suppose by contradiction that $(f(c_n))_n$ is not a Cauchy sequence in $T.$ Then for some $r>0$ we have $\lim_{m\to\infty}\sup_{m<n<n'}d_T(f(c_n),f(c_{n'}))>r.$ But $\lim_{m\to\infty}\sup_{m<n<n'}d(c_n,c_{n'})=0.$ So for every $e>0$ there exist $n,n'$ with $d(c_n,c_{n'})<e$ and $d_T(f(c_n),f(c_{n'}))>r,$ contrary to the uniform continuity of $f$.

Therefore $(f(c_n))_n$ is Cauchy in $T.$ By completeness of $d_T,$ therefore $(f(c_n))_n$ converges to some $b\in T$. For $f$ to be continuous at $a$ it is necessary that $b=\lim_n f(c_n)=f(\lim_n c_n)=f(a).$ And $(f(x_n))_n ,\, (f(y_n))_n$ are subsequences of $(f(c_n))_n$ so they both converge to $b.$

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Owing to the suggestions of @Michael and @GEdgar, some thoughts came into my mind. I hope I didn't misunderstand them. Suppose there is another such extension $h$. We need to show that $\forall x\in\overline{A}$, $g(x)=h(x)$. Grab a sequence $\{x_n\}_{n=1}^\infty$ in $A$ that converges to $x$. Because uniform continuity implies continuity, we have $$\text{$g(x_n)\to g(x)$ and $h(x_n)\to h(x)$}.$$ Now that both $g$ and $h$ restrict to $f$ on $A$, each pair of $g(x_n)$ and $h(x_n)$ agree. In fact, they are all equal to $f(x_n)$, which makes those two limits $g(x)$ and $h(x)$ identical. The work is done!

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