Contractibility is Not Equivalent to One-Point Homotopy Type (9.58.8)

James Munkres Topology, chapter 9.58, exercise 8:

Find a space $X$ and a point $x_0∈X$ such that the inclusion $\{x_0\}→X$ is a homotopy equivalence, but $\{x_0\}$ is not a deformation retract of $X$.

Proof: Let $X$ be the subset of $ℝ^2$ consisting of those lines $(1/n)×I$ for $n∈ℕ$, as well as $0×I$ and $I×0$, and let $x_0=(0,1)$. Then the map $$F : X×I→X$$ $$F((x,y),t)=\left\{ \begin{array} \{ (x,(1-3)y) & t∈[0,1/3] \\ ((2-3t)x,0) & t∈[1/3,2/3] \\ (0,3t-2) & t∈[2/3,1] \end{array} \right.$$ is a homotopy between the identity on $X$ and the constant map onto $x_0$, so that $X$ is contractible. But suppose $\{x_0\}$ is a deformation retract of $X$ via the map $F : X×I→X$, i.e. a homotopy between the two mentioned above such that $F(x_0×I)=\{x_0\}$. For each $n$ let $x_n=(1/n)×1$; then since each $ρ_n : I→X$ given by $ρ_n(t)=F(x_n,t)$ is a path from $x_n$ to $x_0$, let $t_n∈I$ be such that $π_2(F(x_n,t_n))=0$. Since $I$ is compact, let $t_{n_i}→α$ be a convergent subsequence. Then $(x_{n_i},t_{n_i})→(x_0,α)$ yet $F(x_{n_i},t_{n_i}) \not → x_0$ since every term of this sequence has second coordinate $0$.$~\square$

Contractible Spaces and Homotopy Classes (9.51.3)

James Munkres Topology, chapter 9.51, exercise 3:

A space $X$ is said to be contractible if the identity map $i : X→X$ is nulhomotopic.
(a) Show that $I$ and $ℝ$ are contractible.
(b) Show that contractible spaces are path connected.
(c) Show that if $Y$ is contractible, then for any $X$, the set $[X,Y]$ has a single element.
(d) Show that if $X$ is contractible and $Y$ is path connected, then $[X,Y]$ has a single element.

Proof: (a) Since $I$ and $ℝ$ are both convex, the linear homotopies suffice, for $z : I→\mathcal{C}(ℝ,ℝ)$ (say) given by $(1-t)·f(x)+t·g(x)$ is a path in $\mathcal{C}(ℝ,ℝ)$ in the compact-open topology between arbitrary $f,g∈\mathcal{C}(ℝ,ℝ)$.

(b) Let $X$ be path connected, and let $x,y∈X$. If $G : X×I→X$ is a nulhomotopy to the constant map onto $e$, then $p_1 : I→X$ given by $p_1(t)=G(x,t)$ is a path from $p_1(0)=x$ to $p_1(1)=e$. Similarly there is a path from $y$ to $e$, so that $x$ and $y$ are connected by a path.

(c) First, let $Y$ simply be path connected, let $y,e∈Y$ be arbitrary, and let $p$ be a path from $y$ to $e$. Then $P : Y×I→Y$ given by $P(x,t)=p(t)$ is a homotopy from the constant map onto $y$ to the constant map onto $e$, so that all constant maps into a path connected space are homotopic. Hence, when $Y$ is contractible, it suffices to show that an arbitrary map $f : X→Y$ is homotopic to a constant map; and indeed, if $G$ is a nulhomotopy in $Y$ onto a constant map $x↦e$, then $F : X×I→Y$ given by $F(x,t)=G(f(x),t)$ is continuous such that $F(x,0)=G(f(x),0)=f(x)$ and $F(x,1)=G(f(x),1)=e$.

(d) As we saw, all constant maps $X→Y$ are homotopic, so it suffices to show an arbitrary map $f : X→Y$ is homotopic to a constant map. Let $G$ be a nulhomotopy in $X$ onto a constant map $x↦e$; then $F : X×I→Y$ given by $F(x,t)=f(G(x,t))$ is continuous such that $F(x,0)=f(G(x,0))=f(x)$ and $F(x,1)=f(G(x,1))=f(e)$ is constant.$~\square$

Hierarchy of Conditions on Locally Euclidean Spaces (8.50.Supp 2-6)

James Munkres Topology, chapter 8.50, supplementary exercises 2-6:

Let $X$ be locally $m$-euclidean.

2. Consider the following conditions on $X$:
(i) $X$ is compact Hausdorff (ii) $X$ is an $m$-manifold (iii) $X$ is metrizable (iv) $X$ is normal (v) $X$ is Hausdorff

Show (i) $⇒$ (ii) $⇒$ (iii) $⇒$ (iv) $⇒$ (v).

3. Show $ℝ$ is locally $1$-euclidean and satisfies (ii) but not (i).

4. Show that $ℝ×ℝ$ in the dictionary order topology is locally $1$-euclidean and satisfies (iii) but not (ii).

5. Show that the long line is locally $1$-euclidean and satisfies (iv) but not (iii).

Proof: 2. [(i) $⇒$ (ii)] Since $X$ is locally metrizable, Hausdorff compactness of $X$ implies by the Smirnov metrization theorem that $X$ is metrizable. Since compact metric spaces are second countable, it follows $X$ is an $m$-manifold. [(ii) $⇒$ (iii)] This follows by the Urysohn metrization theorem. [(iii) $⇒$ (iv)] Metric spaces are necessarily normal. [(iv) $⇒$ (v)] Normal spaces are necessarily Hausdorff.

3. $ℝ$ is clearly Hausdorff locally $1$-euclidean with a countable basis, but is not compact.

4. It is clear $ℝ×ℝ$ in the dictionary order topology is locally $1$-euclidean and does not have a countable basis, seeing as $r×(0,1)$ for $r∈ℝ$ is an uncountable collection of disjoint nonempty open sets in $ℝ×ℝ$. However, by paracompactness of $ℝ$, it is seen that $ℝ×ℝ$ is paracompact, so that by the Smirnov metrization theorem the space is metrizable.

5. Every order topology is normal, and by the results of the exercises of chapter 26 we know that the long line is locally $1$-euclidean. However, the long line is limit point compact but not compact, so that it cannot be metrizable.$~\square$

Imbedding Theorem on m-Manifolds (8.50.6-7)

James Munkres Topology, chapter 8.50, exercises 6-7:

6. Prove the following theorem: Let $X$ be a locally compact, second-countable Hausdorff space such that every compact subspace of $X$ has topological dimension at most $m$. Then $X$ can be imbedded as a closed subspace into $ℝ^{2m+1}$.
(a) Given $f : X→ℝ^N$, we say $f(x)→∞$ (as $x→∞$) if for all $n∈ℕ$ there exists a compact subspace $C⊆X$ such that $|f(x)| > n$ whenever $x∈X \setminus C$. When $ρ$ is the bounded metric on $\mathcal{C}(X,ℝ^N)$, show that if $ρ(f,g) < 1$ and $f(x)→∞$, then $g(x)→∞$.
(b) Show that if $f(x)→∞$, then $f$ extends to a continuous mapping of one-point compactifications. Conclude that if $f$ is injective, then $X$ can be imbedded as a closed subspace into $ℝ^N$.
(c) When $C⊆X$ is compact and given $ε > 0$, define $$U_ε(C)=\{f~|~Δ(f|_C) < ε\}$$ Show $U_ε(C)$ is compact.
(d) Show that if $N=2m+1$, then $U_ε(C)$ is dense in $\mathcal{C}(X,ℝ^N)$.
(e) Show there exists a continuous map $F:X→ℝ^N$ such that $F(x)→∞$.
(f) Complete the proof.

7. Show that every $m$-manifold can be imbedded as a closed subspace into $ℝ^{2m+1}$.

Proof: (a) Given $n$, let $C⊆X$ be compact such that $|f(x)| > n+1$ for all $x∈X \setminus C$. Then $|g(x)| > n$ for all $x∈X \setminus C$.

(b) If $f(x)→∞$, then define $F: X^*→(ℝ^N)^*$ by $F(Ω_X)=Ω_{ℝ^N}$ and $F(x)=f(x)$ otherwise. Since $X$ is first-countable, it suffices to show $f(x_n)→f(x)$ whenever $x_n→x$. This is evident by continuity of $f$ when $x≠Ω_X$, and it follows from the definition of $f(x)→∞$ and of one-point compactifications when $x=Ω_X$. And when $f$ is injective, we see $F$ is a homeomorphism whose image is closed in $(ℝ^N)^*$, so that $f$ is a homeomorphism onto a closed subspace of $ℝ^N$.

(c) Note that $X$ is metrizable by the Urysohn metrization theorem, so that for each compact $C⊆X$ we see the image of the restriction of $U_ε(C)$ (technically, it requires specifying it is relative to $C$ rather than $X$, though by the Tietze extension theorem the point is moot) is open in $\mathcal{C}(C,ℝ^N)$ by the result proved in Theorem 50.5, which by nature of the bounded metric $ρ$ implies $U_ε(C)$ is open in $\mathcal{C}(X,ℝ^N)$.

(d) Let $f : X→ℝ^N$ and $δ > 0$ be given. By the result in Theorem 50.5, let $g : C→ℝ^N$ be such that $|f(x)-g(x)| < δ$ for all $x∈C$ and $Δ(g) < ε$. Extend $g-f|_C$ to a continuous map $h : X→[-δ,δ]^N$ by the Tietze extension theorem; then $k = h+f$ is such that $ρ(f,k) < δ$, and since $k|_C=g-f|_C+f|_C=g$, we have $Δ(k) < ε$.

(e) Let $\{U_i\}$ be a countable basis for $X$. First, define a sequence $D_n$ of compact subsets of $X$ such that $∪D_n=X$, such as by letting $D_n$ be the union of those basis elements $U_i$ for $i < n$ with compact closure. Let $C_0=ø$. Given compact $C_n$, by local compactness of $X$ cover $C_n$ by finitely many sets open in $X$ of compact closure, and let $C_{n+1}$ be the union of these closures together with $D_n$; then $C_n ⊆ \text{Int }C_{n+1}$ for each $n$, and $∪C_n=X$.

For all $n$, let $S_n=C_n-\text{Int }C_n$. Let $f_0 : C_n→ℝ$ be void. Given a function $$f_n : C_n→ℝ$$ such that $f_n(x)=n$ for all $x∈S_n$, let $$g_{n+1} : C_{n+1} \setminus \text{Int }C_n → [n,n+1]$$ be such that $g_{n+1}(S_n)=\{n\}$ and $g_{n+1}(S_{n+1})=\{n+1\}$, and define $$f_{n+1} : C_{n+1}→ℝ$$ by $f_{n+1}(x)=f_n(x)$ if $x∈C_n$ and $f_{n+1}(x)=g_{n+1}(x)$ otherwise. Then $f_{n+1}$ is continuous by the pasting lemma, and is $n+1$ on $S_{n+1}$.

Since we see $f_n(x)=f_m(x)$ for all $x∈C_n$ whenever $n≤m$, define $f : X→ℝ$ by $f(x)=f_n(x)$ when $x∈C_n$. Since every compact subset of $X$ must be contained in $C_n$ for some $n$ (lest $\{\text{Int }C_n\}$ be a cover with no finite subcover), and since $X$ is compactly generated, we see $f$ is continuous. Further, $f(x)→∞$ because $f(x) ≥ n$ whenever $x∈X \setminus C_n$. Finally, define $F : X→ℝ^N$ by $π_i(F(x))=f(x)$ for all $i$. It follows that $F$ is continuous and $F(x)→∞$.

(f) By the Baire property and previous arguments, $∩U_{1/n}(C_n)$ is dense in $\mathcal{C}(X,ℝ^N)$. Hence, let $λ : X→ℝ^N$ be such that $λ∈∩U_{1/n}(C_n)$ and $ρ(λ,f) < 1$. It follows that $Δ(λ)=0$ so that $λ$ is injective, hence by (b) $λ$ is an imbedding of $X$ onto a closed subspace of $ℝ^N$.

7. Being regular and locally Euclidean, $m$-manifolds are locally compact, and being Hausdorff and second countable the other qualities necessary to apply the previous theorem follow together with an application of Theorem 50.1.$~\square$

Hausdorffness and Regularity of Compact-Open C(X,Y) (7.46.6)

James Munkres Topology, chapter 7.46, exercise 6:

Let $\mathcal{C}(X,Y)$ be under the compact-open topology. Show $\mathcal{C}(X,Y)$ is Hausdorff if $Y$ is Hausdorff, and regular if $Y$ is regular.

Proof: Hausdorffness is simple, as $\mathcal{C}(X,Y)$ inherits a topology at least as fine as that of a subspace under the product topology of $Y^X$, which is Hausdorff when $Y$ is. So suppose $Y$ is regular, let $f∈\mathcal{C}(X,Y)$, and let $K⊆\mathcal{C}(X,Y)$ be a closed subset not containing $f$. Then there exists compact $C_1,...,C_n⊆X$ and open $U_1,...,U_n⊆Y$ such that $f(C_i)⊆U_i$ and for all $g∈K$ there exists $i_g$ such that $g(C_{i_g})⊈U_{i_g}$. Since $f(C_i)$ is compact for each $i$, by regularity of $Y$ choose neighborhoods $V_i$ of these sets such that $\overline{V_i}⊆U_i$. Then $∩B(C_i,V_i)$ is a hood of $f$, and since $$\overline{B(C_i,V_i)}⊆B(C_i,\overline{V_i})$$ we observe $$\overline{∩B(C_i,V_i)}⊆∩\overline{B(C_i,V_i)}⊆∩B(C_i,U_i)$$ is disjoint from $K$.$~\square$

Compact Convergence of a Power Series (7.46.5)

James Munkres Topology, chapter 7.46, exercise 5:

Consider the sequence of functions $f_n : (-1,1)→ℝ$ defined by $$f_n(x)=\sum_{k=1}^n kx^k$$ (a) Show $(f_n)$ converges in the topology of compact convergence; conclude that the limit function is continuous.
(b) Show $(f_n)$ does not converge uniformly.

Proof: (a) First, note that $f(x)=\sum kx^k$ converges for all $|x| < 1$ by the ratio test. Since each compact subset of $(-1,1)$ is contained in some interval $[-x,x]⊆(-1,1)$, it will suffice to show $f_n$ converges uniformly on $[-x,x]$ for all $x∈(0,1)$. Therefore let $|y| ≤ |x|$ and observe $$|f(y)-f_n(y)| = |\sum^∞_{k=1}ky^k-\sum^n_{k=1}ky^k| = |\sum^∞_{k=n+1} ky^k| ≤$$ $$\sum^∞_{k=n+1} k|x|^k →0$$ as $n→∞$.

(b) Suppose some $n$ such that $|f(x)-f_n(x)| < 1/2$ for all $x∈(-1,1)$. Simply choose $x∈(0,1)$ so that $(n+1)x^{n+1} ≥ 1/2$, and observe $$|f(x)-f_n(x)| = |\sum_{k=n+1}^∞ kx^k| ≥ 1/2~~\square$$

Grayscale Functions

Let $X$ be a topological space. For $A⊆X$, let $ι_A(x)=1$ if $x∈A$, and $ι_A(x)=0$ otherwise. Let $\mathcal{U}(X)=\{U~|~U⊆X \text{ is open}\}$. When $(U_n)$ is a sequence in $\mathcal{U}(X)$, we write $U_n→U$ for $U∈\mathcal{U}(X)$ if $$\overline{U}=\overline{\bigcup_{n∈ℕ}(\bigcap_{N≥n} U_N)}$$ (Convergence need not be unique) If $φ : X→\mathcal{U}(X)$ is such that $x∈φ(x)$ when $φ(x)≠ø$, and $φ(x_n)→φ(x)$ whenever $x_n→x$, we say $φ$ is a gradated open assignment on $X$. Given a sequence $σ=(x_i)$ in $X$ and a subset $A⊆X$, let $$λ_σ(A)=\lim_{N→∞} \sum^N \dfrac{ι_A(x_i)}{N}$$ wherever the convergence exists. If $φ$ is a gradated open assignment such that $f = λ_σ ∘ φ : X→[0,1]$ is a well-defined continuous mapping, then the pair $(φ,σ)$ is called a grayscale pair and $f$ the corresponding grayscale function. If every continuous function $f∈\mathcal{C}(X,[0,1])$ is a grayscale function, then $X$ is said to be a grayscale domain.
~~~~~
Every infinite discrete space is grayscale, and a finite discrete space is never grayscale. In fact, if $X=\{x_1,...,x_n\}$ is a finite discrete space, then the function $f : X→[0,1]$ is grayscale if and only if $f(x_i)=0,1$ for some $i$, or there exists some $n×n$ binary matrix $M$ with $1$s on the main diagonal and a non-negative-valued $n×1$ column vector $A$ whose entries sum to $1$ such that $$MA=\begin{bmatrix} f(x_1) \\ f(x_2) \\ \cdots \\ f(x_n) \end{bmatrix}$$ ~~~~~ We prove $[0,1]$ is itself grayscale. If $f : [0,1]→[0,1]$ is continuous, then let $$φ(x) = \left\{ \begin{array} \{ [0,f(x)) & x=0 \\ (1-f(x),1] & x=1 \\ (x(1-f(x)),~f(x)+x(1-f(x)) & x∈(0,1) \end{array} \right.$$ Note that the length of the interval $φ(x)$ is $f(x)$, and $x∈φ(x)$ if $φ(x)≠ø$. To prove gradation, note that when $x_n→x$, we have $x_n(1-f(x_n))→x(1-f(x))$ and $f(x_n)+x_n(1-f(x_n))→f(x)+x(1-x)$. When $μ(n)→0$ denotes the maximum of the difference between the two pairs of an $n^\text{th}$ term and its limit, we see that if $B(y,ε)⊆φ(x)$ for $x∈(0,1)$, we have $y∈φ(x_n)$ for all $n≥N$ when $N$ is such that $μ(n) < ε$ for all $n≥N$. Exclusion of those elements $y∉\overline{φ(x)}$ from the final intersection is similar. The cases for $x=0,1$ are straightforward edge cases.

Let $σ=(x_n)$ be the sequence of rationals in the order $\dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{4},\dfrac{2}{4},\dfrac{3}{4},...$ (we shall refer to the subcollection of those with $n$ in the denominator as the $n$-strip elements).

It now remains to show $λ_σ(x)=f(x)$. To observe this, first fix $x$ and let $α=f(x)$. Note that each $n$-strip collection refers to a subset of $[0,1]$ containing $n-1$ elements spaced $1/n$ apart from each other. Therefore, when $g(n)$ denotes the number of $n$-strip elements contained in $φ(x)$, we see $g(n)∈[α(n-1),~α(n-1)-2]$, and when $$G(n)=\sum^N_{i=1} g(i)$$ we see $G(n)∈[\dfrac{1}{2}αn(n-1)-2n,~\dfrac{1}{2}αn(n-1)]$. Since the number of elements up to and including the $n$-strip elements is $β(n)=\dfrac{1}{2}n(n-1)$, we see $G(n)/\sum β(n)∈[α-\dfrac{4}{n-1},α]$. This shows that a subsequence of the limit involved in $λ_σ(x)$ converges to $α=f(x)$, so it suffices to show that the limit inferior and limit superior are the same. To wit, note that the minimum value between the partial sums $G(n)/\sum β(n)$ and $G(n+1)/\sum β(n+1)$ is bounded below by $$G(n)/\sum β(n+1)∈[α\dfrac{n-5}{n+1},α\dfrac{n-1}{n+1}]$$ The case for limit superior is parallel. Thus $λ_σ(x)$ is well defined and equals $f(x)$.

By a similar argument, $(0,1)≅ℝ$ is also grayscale.

Completeness, Total Boundedness, and Compactness of the Hausdorff Metric (7.45.7b-d)

James Munkres Topology, chapter 7.45, exercise 7:

Let $(X,d)$ be a metric space, and let $(\mathcal{H},D)$ be its associated Hausdorff metric. Show completeness, total boundedness, and compactness are equivalent conditions in both spaces.

Proof: Note that there is a natural isometry of $(X,d)$ with a closed subspace of $(\mathcal{H},D)$, so that completeness, total boundedness, and compactness of $\mathcal{H}$ implies the corresponding quality in $X$ (to cover a subset $A$ of a totally bounded space $B$ with finitely many $ε$ balls, cover $B$ with $ε/2$ balls, remove those disjoint from $A$, and place one $ε$ ball centered in $A$ per $ε/2$ ball from $B$). Since completeness and total boundedness imply compactness, it will suffice to prove (a) completeness of $(X,d)$ implies completeness of $(\mathcal{H},D)$, and (b) total boundedness of $(X,d)$ implies total boundedness of $(\mathcal{H},D)$.

(a) Let $(A_n)$ be a Cauchy sequence in $\mathcal{H}$. If necessary, take a subsequence so that $D(A_n,A_{n+1}) < 1/2^n$ for all $n$. Now, let $A$ be the set of all limit points of subsequences of $(a_n)$ of $X$ such that $a_n∈A_n$ for each $n$. Since $D(A_1,A_n) < 1$ for each $n$, it is clear $A$ is bounded, nonempty, and (by diagonalization of limits) closed. We show $A_n → A$. Let $ε > 0$. Since the size of the neighborhood of $A_n$ required to contain $A$ approaches $0$, it suffices to show $A_n ⊈ B_D(A,ε)$ for only finitely many $n$. To wit, let $N$ be such that $\sum_{i=N}^∞ 1/2^i < ε/2$. Then if $a_n∈A_n$ for $n≥N$, set $b_0=a_n$ and given $b_i$, choose $b_{i+1}$ so that $d(b_i,b_{i+1}) < 1/2^n$. Then $(b_i)$ is a Cauchy sequence, and appending cursory points in each of $A_1,...,A_{n-1}$ we can find a point of $A$—namely, $b$ when $b_i→b$—such that $d(a_n,b) < ε$ and now $a_n∈B_D(A,ε)$.

(b) Let $ε > 0$. Cover $X$ by finitely many $ε$ balls centered about the points $a_1,...,a_n$. Let $J=\mathcal{P}(\{a_1,...,a_n\}) \setminus \{ø\}$, and center around each point $j∈J⊆\mathcal{H}$ an $ε$ ball. This is seen to be a finite covering of $\mathcal{H}$ by $ε$ balls, with an arbitrary element $A∈\mathcal{H}$ being within distance $ε$ of the element of $J$ which minimally (with regard to set containment) covers $A$ considered as a subset of $X$.$~\square$

R^ω Under the l^2 Metric is Complete (7.43.7)

James Munkres Topology, chapter 7.43, exercise 7:

Show that the subspace of $ℝ^ω$ of those sequences $(x_n)$ such that $\sum x_n^2$ converges is complete under the $\ell^2$ metric.

Proof: Let $(f_n)$ be a Cauchy sequence under this metric. Since the $\ell^2$ distance between any two points is at least as large as the uniform distance in $ℝ^ω$, and since the metric under the latter is complete is complete, let $f_n→f$ in the uniform topology. It suffices to show $f_n→f$ in the $\ell^2$ metric. Let $ε > 0$; let $N$ be such that $d_{\ell^2}(f_n,f_m) < ε/2$ for $n,m≥N$. We shall proceed by showing $d_{\ell^2}(f,f_N) ≤ ε/2$ so that $d_{\ell^2}(f,f_n) < ε$ for all $n≥N$, and this former will be demonstrated by showing $d_{\ell^2}(f,f_N) > ε/2$ implies a neighborhood about $f$ in the uniform topology that does not intersect any $f_n$ for $n≥N$, a contradiction.

Therefore, assume $d_{\ell^2}(f,f_N) > ε/2$, and let $$\sqrt{(f(1)-f_N(1))^2+...+(f(n)-f_N(n))^2} = ε/2+δ$$ for some $n$ and $δ > 0$. Consider the uniform $δ/\sqrt{n}$ neighborhood $U$ about $f$; for any $g∈U$, it is evident that $d_{\ell^2}(f_N,g) ≥ ε/2$ (otherwise, consider the first $n$ coordinates of $f$, $f_N$, and $g$ in $ℝ^n$ and apply the triangle inequality), so that $g≠f_m$ for any $m≥N$.$~\square$

Uniform Extensions into Complete Metric Spaces (7.43.2)

James Munkres Topology, chapter 7.43, exercise 2:

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, with $Y$ complete. Show that if $A⊆X$ and $f : A→Y$ is uniformly continuous, there exists a uniformly continuous extension of $f$ to $\overline{A}$.

Proof: Let $(x_n)$ be a Cauchy sequence in $X$; we show $(f(x_n))$ is a Cauchy sequence in $Y$. Let $ε > 0$. Then if $δ > 0$ is such that $d_Y(f(a),f(b)) < ε$ for all $a,b∈A$ such that $d_X(a,b) < δ$, and $N$ is such that $d_X(x_n,x_m) < δ$ for all $n,m≥N$, we see $d_Y(f(x_n),f(x_m)) < ε$ for all $n,m≥N$, so that $f(x_n)$ is Cauchy.

For all $x∈\overline{A}$, choose some Cauchy sequence $(x_n)$ in $A$ converging to $x$. Then $f(x_n)→y_x$ since $Y$ is complete. Define $g: \overline{A}→Y$ by $g(x)=y_x$. Since $f$ is continuous, $y_x=f(x)$ for all $x∈A$. Now it suffices to show $g$ is uniformly continuous. Let $ε > 0$; let $δ > 0$ be such that $d_Y(f(a),f(b)) < ε/3$ whenever $a,b∈A$ are such that $d_X(a,b) < δ$. Let $x,y∈\overline{A}$ be such that $d_X(x,y) < δ/3$; let $(x_n)→x$ and $(y_n)→y$ be the chosen sequences as before. Choose $n$ such that $$d_Y(g(x),g(x_n)),d_Y(g(y),g(y_n)) < \text{min }\{ε/3,δ/3\}$$ We see $d_X(g(x_n),g(y_n)) = d_X(f(x_n),f(y_n)) < ε/3$ since $d_X(x_n,y_n) ≤ d_X(x_n,x)+d_X(x,y)+d_X(y,y_n)$. Finally, we observe $$d_Y(g(x),g(y)) ≤ d_Y(g(x),g(x_n)) + d_Y(g(x_n),g(y_n)) + d_Y(g(y_n),g(y)) < ε$$