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 $$d_Y(g(x),g(y)) ≤ d_Y(g(x),g(x_n)) + d_Y(g(x_n),g(y_n)) + d_Y(g(y_n),g(y)) < ε$$