Degree of Continuous Maps on S^n (9.58.9-10)

James Munkres Topology, chapter 9.58, exercise 9-10:

9. Let $h : S^1→S^1$ be a continuous map. Let $b_0=(1,0)$, and let $γ$ generate $π_1(S^1,b_0)$. For any given point $x∈S^1$, define $γ(x)=\hat{α}(γ)$ where $α$ is a path from $b_0$ to $x$; note that the choice of $α$ is immaterial since $π_1(S^1,b_0)$ is abelian. Since $\hat{α}$ is an isomorphism, $γ(x)$ will generate $π_1(S^1,x)$. As well, we see $h_*$ is a homomorphism between $π_1(S^1,x)$ and $π_1(S^1,h(x))$, so $$h_*(γ(x))=d_x·γ(h(x))$$ for some integer $d_x$, if the fundamental groups are considered additively. This $d_x$ is called the degree of $h$ (relative to $x$), and is independent of choice of $γ$ since choosing the other generator of $π_1(S^1,b_0)$ will change signs accordingly to result in the same $d_x$.

(a) Show that $d_x$ is independent of the choice of $x∈S^1$, so we may denote it more generally by $d$.
(b) Show that if $h, k : S^1→S^1$ are homotopic, then their degrees are the same.
(c) Show that $\text{deg}(h∘k)=(\text{deg }h)·(\text{deg }k)$.
(d) Compute the degrees of the constant map, the identity map, the reflection map $(x,y)↦(x,-y)$, and the map $z↦z^n$ when $S^1$ is considered as a subgroup of $ℂ$.
(e) Show that if $h,k : S^1→S^1$ have the same degree, they are homotopic.

10. Suppose that to every map $h : S^n→S^n$ we have assigned an integer, denoted by $\text{deg }h$ and called the degree of $h$, such that:

1. Homotopic maps have the same degree
2. $\text{deg }(h∘k) = (\text{deg }h)·(\text{deg }k)$
3. The identity map has degree $1$, any constant map degree $0$, and the reflection maps $(x_1,...,x_i,...,x_{n+1})↦(x_1,...,-x_i,...,x_{n+1})$ degree $-1$.

Prove the following:
(a) There is no retraction $r : B^{n+1}→S^n$
(b) If $h : S^n→S^n$ has degree different than $(-1)^{n+1}$, then $h$ has a fixed point.
(c) If $h : S^n→S^n$ has degree different than $(-1)^n$, then $h(x)=-x$ for some $x$.
(d) If $S^n$ has a nonvanishing tangent vector field $v$, then $n$ is odd.

Proof: Notation shall be mildly abused in the explanations that follow, in that $γ$ will refer to an element of $π_1(S^1,b_0)$ when outside brackets, and a loop about $b_0$ whose homotopy class is such $γ$ when inside brackets.

9. (a) We shall show $d_x=d_{b_0}$ for all $x∈S^1$. We have $$h_*(γ)=d_{b_0}·\hat{α}(γ)$$ where $α$ is a path from $b_0$ to $h(b_0)$. Let $β$ be a path from $b_0$ to $x$, and let $δ$ be a path from $b_0$ to $h(x)$, and observe $$h_*(γ(x))=h_*(\hat{β}(γ))=[h∘\overline{β}]*[h∘γ]*[h∘β]=\widehat{h∘β}(h_*(γ))=$$$$\widehat{h∘β}(d_{b_0}·\hat{α}(γ))=d_{b_0}·\widehat{h∘β}∘\hat{α}(γ)=d_{b_0}·\widehat{α*(h∘β)}(γ)=$$$$d_{b_0}·\hat{δ}(γ)=d_{b_0}·γ(h(x))$$ (b) Choose $x∈S^1$. Since $h$ and $k$ are homotopic, let $α$ be a path from $h(x)$ to $k(x)$ such that $k_*=\hat{α}∘h_*$. Then if $d$ is the degree of $h$ and $β$ is a path between $b_0$ and $h(x)$, we have $$k_*(γ(x))=\hat{α}∘h_*(γ(x))=d·\hat{α}(γ(h(x))=$$$$d·\hat{α}∘\hat{β}(γ)=d·\widehat{β*α}(γ)=d·γ(k(x))$$ (c) We simply observe $$(h∘k)_*(γ(x))=h_*∘k_*(γ(x))=(\text{deg }k)·h_*(γ(k(x))=$$$$[(\text{deg }k)·(\text{deg }h)]·γ((h∘k)(x))$$ (d) The constant map induces trivial homomorphisms, so $d=0$ in this case. The identity map induces identity homomorphisms, so $d=1$ in this case. Let $f$ be the reflection map; when $p : ℝ→S^1$ is the standard covering map $x↦(\text{cos }2πx,\text{sin }2πx)$ and $\tilde{γ} : I→ℝ$ is given by $x↦x$, we see $γ=p∘\tilde{γ}$ generates $π_1(S^1,b_0)$. As well, $p(-\tilde{γ}(x))=(\text{cos }(-2πx),\text{sin}(-2πx))=(\text{cos }2πx,-\text{sin}(2πx))=f∘γ(x)$ so $\widetilde{f∘y}=-\tilde{γ}$. Since $-\tilde{γ}$ is a path from $0$ to $-1$ in $ℝ$ we may observe $f_*(γ)=[f∘γ]=-γ$ so that $d=-1$. A similar appeal to covering maps beyond when $n=-1$ shows that generally $z↦z^n$ is degree $n$.

(e) Lemma: If $G : I×I→S^1$ is a homotopy between $h∘γ$ and $k∘γ$, then $h, k$ are homotopic. Proof: Consider $S^1$ as the circle group in $ℂ$. We see $p_1(t)=G(0,t)$ and $p_2(t)=G(1,t)$ are paths from $h(b_0)$ to $k(b_0)$, and let $q(x,t)=(\dfrac{p_1(t)}{p_2(t)})^x$. First we show $J: I×I→S^1$ defined by $J = G·q$ is another homotopy, but such that $J(0,t)=J(1,t)$ for all $t∈I$. As such, it is clear $$J(x,0)=G(x,0)·1^x=h∘γ(x)$$ and $$J(x,1)=G(x,1)·1^x=k∘γ(x)$$ and also $J(0,t)=G(0,t)·1=p_1(t)=p_2(t)·\dfrac{p_1(t)}{p_2(t)}=J(1,t)$.

Thus, since $γ$ is also a quotient map (if $γ$ weren't surjective, then $γ$ would map into $S^1-p≅ℝ$ for some $p∈S^1$ and thus be nulhomotopic), we may factor $J$ through $S^1×I$ via the quotient map $γ×i$ to obtain $K$. For $x∈S^1$, let $γ(v)=x$, and we see $$K(x,0)=K∘(γ×i)(v,0)=J(v,0)=h∘γ(v)=h(x)$$ $$K(x,1)=K∘(γ×i)(v,1)=J(v,1)=k∘γ(v)=k(x)$$ so that $K$ is a homotopy between $h$ and $k$.$~\square$

Let $α$ be a path from $b_0$ to $h(b_0)$, and let $β$ be a path from $h(b_0)$ to $k(b_0)$. Then $$\hat{β}∘h_*(γ)=\hat{β}(d·\hat{α}(γ))=d·\widehat{α*β}(γ)=k_*(γ)$$ so that $\hat{β}∘h_*=k_*$. As such, let $F : I×I→S^1$ be a path homotopy between $\overline{β}*(h∘γ)*β$ and $k∘γ$ (however, by path homotopy equivalence, let this former piecewise function be split such that $(\overline{β}*(h∘γ)*β)(1/3+t/3)=h∘γ(t)$). Also, let $H : I×I→I$ be a homotopy between the functions $f,g : I→I$ given by $f(t)=t$ and $g(t)=1/3+t/3$, specifically a homotopy from $g$ to $f$. Define $K : I×I→S^1$ by $K(x,y)=F(H(x,y),y)$. We claim $K$ is a homotopy between $h∘γ$ and $k∘γ$, so that the statement follows by the proceeding lemma. To wit, observe $$K(x,0)=F(H(x,0),0)=(\overline{β}*(h∘γ)*β)(1/3+x/3)=h∘γ(x)$$ $$K(x,1)=F(H(x,1),1)=k∘γ(x)$$ 10. (a) If $j : S^n→B^{n+1}$ is the inclusion map, then we see if $r$ were a retraction, then since $B^{n+1}$ is convex that $i_{S^n} \simeq r∘j \simeq 0∘j$, and so the identity map is homotopic to the constant map on $S^n$, contradicting (1) and (3).

(b) Suppose $h$ does not have a fixed point; then extend $h$ to $ℝ^{n+1}-0$ via $H(x)=h(\dfrac{x}{||x||})$. Now define a homotopy $G$ from $H$ to the circular antipodal map $λ(x)=-\dfrac{x}{||x||}$ on $ℝ^{n+1}-0$ by $G(x,t)=(1-t)·H(x)+t·λ(x)$. Note that by construction $G(x,t) \not = 0$ for any $x,t$ so the range is indeed within $ℝ^{n+1}-0$. Now when $j : S^n→ℝ^{n+1}-0$ is inclusion and $r : ℝ^{n+1}-0→S^n$ the standard retraction, we see $F : S^n×I→S^n$ given by $F(x,t)=r(G(j(x),t))$ is a homotopy between $H_{|S^n}=h$ and the antipodal map $x↦-x$ on $S^n$, and since this latter is seen to have degree $(-1)^{n+1}$ by (2) and (3), it follows by (1) that $h$ has the same degree.

(c) This is merely part (b) applied to the composite of the antipodal map with $h$.

(d) If a nonvanishing vector $v(x)$ is tangent to a point $x$ of $S^n$, then particularly $t·v(x) ≠ x$ for $t∈ℝ$. Now, when $j : S^n→ℝ^{n+1}-0$ is the inclusion map, consider the straight-line homotopy between $j$ and $v$ on $ℝ^{n+1}-0$. This homotopy is well defined by the nonvanishing and non-orthogonality of the vector field. As well, the straight-line homotopy between $v$ and $-j$ is also defined. Therefore $j$ is homotopic to $-j$ in $ℝ^{n+1}-0$, and since $S^n$ is a retract of this latter, we find the identity map is homotopic to the antipodal map in $S^n$. Since the antipodal map has degree $-1$ when $n$ is even, it follows $n$ must be odd.$~\square$

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

Paracompactness and Perfect Maps (6.41.8)

James Munkres Topology, chapter 6.41, exercise 8:

Let $p:X→Y$ be a closed surjective continuous map such that $p^{-1}\{y\}$ is compact for each $y∈Y$. If $X$ is Hausdorff, show $X$ is paracompact if and only if $Y$ is.

Proof: ($⇐$) Let $Y$ be paracompact, and let $\mathcal{B}=\{U_α\}$ be an open cover of $X$. For each $y∈Y$, let $\mathcal{B}_y⊆\mathcal{B}$ be a finite subcover of $p^{-1}\{y\}$, and obtain open $V_y⊆X$ such that $p^{-1}\{y\}⊆V_y⊆∪\mathcal{B}_y$ and $p(V_y)⊆Y$ is a neighborhood of $y$. Then $\{p(V_y)\}$ is an open cover for $Y$, so let $\mathcal{A}$ be a locally finite open refinement covering $Y$. For each $y∈Y$, let $y∈A_y∈\mathcal{A}$. We claim $\mathcal{C}=\{p^{-1}(A_y)∩B~|~y∈Y, B∈\mathcal{B}_y\}$ is a locally finite open refinement of $\mathcal{B}$ covering $X$. It is clear $\mathcal{C}$ is an open refinement, and given $x∈X$, we have $x∈p^{-1}(y)∩B⊆p^{-1}(A_y)∩B$ for some $y∈Y$ and $B∈\mathcal{B}_y$, therefore $\mathcal{C}$ covers $X$. As well, let $U$ be a neighborhood of $p(x)$ intersecting $A_y$ for only finitely many $y∈Y$. Then $p^{-1}(U)$ is a neighborhood of $x$ intersecting $p^{-1}(A_y)$ for only finitely many $y∈Y$, hence intersecting $p^{-1}(A_y)∩B$ for only finitely many pairs $(y,B)$ when $B∈\mathcal{B}_y$. Therefore $\mathcal{C}$ is locally finite and $X$ is paracompact.

($⇒$) Let $X$ be paracompact, and let $\mathcal{B}$ be an open cover of $Y$. Then let $\mathcal{A}$ be a locally finite open refinement of $\{p^{-1}(B)~|~B∈\mathcal{B}\}$ covering $X$. We claim $\mathcal{C}=\{p(A)~|~A∈\mathcal{A}\}$ is a locally finite refinement of $\mathcal{B}$ covering $Y$, so that since $Y$ is regular by normality of $X$, $Y$ will be paracompact by Lemma 41.3. The only nontrivial quality to check is local finiteness; given $y∈Y$, for each $x∈p^{-1}(y)$ let $U$ be a neighborhood of $x$ intersecting only finitely many elements of $\mathcal{C}$. Since $p^{-1}(y)$ is compact, there exists a neighborhood about it intersecting only finitely many elements of $\mathcal{C}$, and furthermore a saturated sub-neighborhood $V_y$ of this one. Being saturated, $V_y∩A=ø$ implies $p(V_y)∩p(A)=ø$ for all $A∈\mathcal{A}$, so that $p(V_y)$ is a neighborhood of $y$ intersecting only finitely many elements of $\mathcal{C}$.$~\square$

Unions of Paracompact Spaces (6.41.7)

James Munkres Topology, chapter 6.41, exercise 7:

Let $X$ be a regular space. Show that (a) if $X$ is a finite union of closed paracompact subspaces, or (b) if $X$ is covered by the interiors of countably many closed paracompact subspaces, then $X$ is paracompact.

Proof: (a) Suppose $X=K_1∪K_2$ where $K_1,K_2⊆X$ are closed and paracompact. By induction the general case will follow from this one. Suppose $\{U_α\}$ is an open cover of $X$. Then $\{U_α∩K_1\}$ and $\{U_α∩K_2\}$ are open covers of $K_1$ and $K_2$ respectively, so let $\mathcal{B}_1 = \{B_β\}$ and $\mathcal{B}_2=\{B_γ\}$ be locally finite refinements of the two. We claim $\mathcal{C}=\mathcal{B}_1∪\mathcal{B}_2$ is a locally finite refinement of $\{U_α\}$ covering $X$. The refinement and covering conditions are evident, so we proceed to demonstrate local finiteness; assume $x∉K_1$. If $U$ is a neighborhood of $x$ such that $K_2∩U$ intersects only finitely many members of $\mathcal{B}_2$, then $U∩(X-K_1)$ is a neighborhood of $x$ intersecting among the same finite family from $\mathcal{B}_2$, and is disjoint from the members of $\mathcal{B}_1$. So we may assume $x∈K_1$, and similarly $x∈K_2$. As such let $U$ and $V$ be neighborhoods of $x$ such that $K_1∩U$ and $K_2∩V$ intersect only finitely many members of $\mathcal{B}_1$ and $\mathcal{B}_2$ respectively. If $U∩V$ intersects an element of $\mathcal{B}_1$, then since that element is within $K_1$ so too does $U∩V∩K_1⊆U∩K_1$, so that only finitely many elements of $\mathcal{B}_1$ are intersected. Similarly too for $\mathcal{B}_2$, and now $U∩V$ is a neighborhood of $X$ intersecting only finitely many elements of $\mathcal{B}$.

(b) Let $X=∪\text{int }K_i$ where each $K_i$ is paracompact. Again let $\{U_α\}$ be an open cover of $X$. For each $i$, let $\mathcal{B}_i$ be a locally finite refinement of $\{U_α∩K_i\}$ covering $K_i$, and further let $\mathcal{A}_i=\{B∩\text{int }K_i~|~B∈\mathcal{B}_i\}$. Then $\mathcal{A}_i$ is a locally finite open cover of $\text{int }K_i$ for each $i$, so that $\mathcal{A}=∪\mathcal{A}_i$ is a countably locally finite open refinement of $\{U_α\}$ covering $X$, so that $X$ is paracompact by Lemma 41.3.$~\square$

Product of a Compact and Paracompact Space (6.41.2a)

James Munkres Topology, chapter 6.41, exercise 2a:

Show that the product of a paracompact space $X$ and a compact space $Y$ is paracompact.

Proof: Let $\{U_α\}$ be an open cover of $X×Y$. For each $x∈X$, the space $\{x\}×Y≅Y$ is compact, so let it be covered by finitely many $U_{x_1},...,U_{x_n}$ with nontrivial intersection with $\{x\}×Y$, and let $W_x=∩π_1(U_{x_i})$. Then $\{W_x\}_{x∈X}$ is an open cover of $X$, so let $\mathcal{A}$ be a locally finite open refinement covering $X$. For each $A∈\mathcal{A}$, finitely many $U_{x_1},...,U_{x_n}$ cover $A×Y⊆W_x×Y$ (for some $x$), so let $C_A=\{(A×Y)∩U_{x_i}\}$. We claim $\mathcal{C}=∪_{A∈\mathcal{A}}C_A$ is a locally finite open refinement of $\{U_α\}$ covering $X×Y$.

Every element of each $C_A$ is contained in some $U_α$, so $\mathcal{C}$ is clearly a refinement. As well, given $z=x×y∈X×Y$, let $x∈A∈\mathcal{A}$ so that $z∈A×Y=∪C_A$ implying $z$ is contained in an element of $C_A$ and now $\mathcal{C}$ covers $X$. Finally, choose a neighborhood $U$ of $x$ intersecting only finitely many members $A_i∈\mathcal{A}$. Then $U×Y$ is a neighborhood of $z$ that can intersect only among the members of $C_{A_i}$, all of which are finite. Therefore $\mathcal{C}$ is locally finite and $X×Y$ is paracompact.$~\square$

Countable Local Finiteness (6.39.5-6)

James Munkres Topology, chapter 6.39, exercises 5-6:

5. If $X$ is second-countable, show a collection $\mathcal{A}$ of subsets of $X$ is countably locally finite if and only if it is countable.

6. Let $ℝ^ω$ have the uniform topology. Given $n$, let $\mathcal{B}_n$ be the collection of all subsets of the form $\prod A_i$ where $A_i=ℝ$ for $i≤n$ and $A_i$ equals $\{0\}$ or $\{1\}$ otherwise. Show $\mathcal{B}=∪\mathcal{B}_n$ is countably locally finite, but is neither countable nor locally finite.

Proof: (5) It suffices to consider when $\mathcal{A}$ is an uncountable, countably locally finite collection of subsets of $X$. Since countable unions of countable sets are countable, there exists an uncountable locally finite collection $\mathcal{B}$. Assume $ø∉\mathcal{B}$, and construct a choice function $f : \mathcal{B}→∪\mathcal{B}$ such that $f(B)∈B$ for each $B∈\mathcal{B}$. If $\{U_α\}$ is a countable basis for $X$ and $\mathcal{B}$ is locally finite, then the countable set $$\{V_n\}=\{U_α~|~f^{-1}(U_α)\text{ finite}\}$$ covers $X$. But now $$\mathcal{B}=f^{-1}(X)=f^{-1}(∪V_n)=∪f^{-1}(V_n)$$ is a countable union of finite sets, so $\mathcal{B}$ is countable, a contradiction.

(6) It's clear that $\mathcal{B}_n$ is uncountable for any $n$, so $\mathcal{B}$ is not countable. As well, $1^ω$ is contained in every subset of the form $\prod A_i$ where, for some $N$, $A_i=ℝ$ for every $i≤N$, and $A_i=\{1\}$ otherwise, so that $\mathcal{B}$ is not even point-finite, let alone locally finite. But the $1/2$-neighborhood of any element in $ℝ^ω$ intersects at most one element of $\mathcal{B}_n$, so $\mathcal{B}_n$ is evidently locally finite, hence $\mathcal{B}$ is countably locally finite.$~\square$

Stone-Cech Compactification of Discrete Spaces (5.38.7-8)

James Munkres Topology, chapter 5.38, exercises 7-8:

7. Let $X$ be a discrete space.
(a) Show that if $A⊆X$, then $\overline{A}∩\overline{X-A}=ø$ where their closures are taken in $β(X)$.
(b) Show that if $U⊆β(X)$ is open, then $\overline{U}$ is open.
(c) Show that $X$ is totally disconnected.

8. Show the cardinality of $β(ℕ)$ is at least as great as $I^I$ where $I=[0,1]$.

Proof: 7. (a) Define a function $f : X→ℝ$ by $f(A)=1$ and $f(X-A)=0$. Letting $F : β(X)→ℝ$ extend $f$, we see $F^{-1}(1)$ and $F^{-1}(0)$ are disjoint closed sets in $β(X)$ containing $A$ and $X-A$ respectively, so these latters' closures are disjoint.

(b) Note $\overline{U∩X}∪\overline{X-U∩X}$ is the whole space $β(X)$ since its complement is an open set not intersecting $X$, so by part (a) $\overline{U∩X}$ is open. Now evidently $\overline{U∩X}⊆\overline{U}$, but also $\overline{U}⊆\overline{U∩X}$ since if there exists $x∈\overline{U}$ with a neighborhood $V$ disjoint from $U∩X$, let $y∈V∩U$ and now $V∩U$ is a neighborhood of $y$ not intersecting $X$ hence $y∉\overline{X}$, a contradiction.

(c) Let $x,y∈β(X)$ be distinct. Since $β(X)$ is Hausdorff let $U$ be open such that $x∈U$ and $v∉\overline{U}$. Then by part (b) $\overline{U}∪β(X)-\overline{U}$ is a separation of $β(X)$ disconnecting $x$ from $y$.

8. As we've seen (cf. 5.31.16a), $ℝ^I≅(0,1)^I$ has a countable dense subset, so $\overline{(0,1)^I}=[0,1]^I$ does as well, call it $S$. Letting $f : ℕ→S⊆[0,1]^I$ be a surjection that is automatically continuous, we obtain a map of $β(ℕ)$ into $I^I$ containing $S$, and since images of compact sets are compact hence closed in a Hausdorff space, the map is a surjection and the claim is proven.$~\square$

Munkres Review Chapters 1-4

(1) The ordered square is connected. Proof: Linear continua are connected.

(2) $ℝ^ω$ in the uniform topology is disconnected. Proof: The sets of bounded and unbounded sequences in $ℝ^ω$ are both open in this metric, and form a separation.

(3) The ordered square is not path connected. Proof: Suppose there exists a path from $0×0$ to $1×1$. Since $I_0^2$ is a linear continuum, this implies the path is surjective and $I_0^2$ is an image of the separable $[0,1]$. But $I_0^2$ itself is not separable, for $r×(1/4,3/4)$ for $r∈I$ is a collection of uncountably many disjoint open subsets.

(4) $ℝ_K$ is not path connected. Proof: Suppose $f : [0,1]→ℝ_K$ is a path from $0$ to $1$. Then $f^{-1}(0)$ is a closed set not containing $1$, so $r = \text{sup }f^{-1}(0) < 1$. Since $[r,1]≅[0,1]$ we may assume $f(x) > 0$ for $x > 0$.

Now, let $a_n = \text{inf } f^{-1}(1/n)$ for $n∈ℕ^+$. We see $a_{n+1} < a_n$ by connectivity of continuous images, but also $a_n \nrightarrow 0$ in $[0,1]$ since $1/n \nrightarrow 0$ in $ℝ_K$. Hence let $a > 0$ be such that $a < a_n$ for all $n$. But then $f(a) > 0$ so $1/N < f(a)$ for some $N$, implying $a_N < a$ again by connectivity of continuous images, a contradiction.

(5) The ordered square is not locally path connected. Proof: The proof of (3) extends to show that the path components of $I_0^2$ are precisely $r×[0,1]$ for $r∈[0,1]$, which are not open.

(6) $ℝ_l$ is not locally compact at any of its points. Proof: Let $x∈ℝ_l$ and $U$ be a hood of $x$. Suppose $V$ is a hood of $x$ such that $\overline{V}⊆U$ is compact. Then $x∈[a,b)⊆V$ for some $a,b∈ℝ$, and since $[a,b)$ is also closed in $\overline{V}$ this implies $[a,b)$ is compact. However, $[a,b)$ is not compact even in $ℝ$.

(7) $ℝ^ω$ in the uniform topology is not locally compact. Proof: Suppose $C$ is a compact subset of $ℝ^ω$ containing a hood of $i=(0,0,...)$. Then $B[i,ε]=[-ε,ε]^ω$ is compact for some $ε∈(0,1)$. But $\{e_n\}_{n∈ℕ^+}$ (when $e_n$ is the point with zeros in every coordinate except the $n\text{th}$ in which it is $ε$) is an infinite subset containing no limit point (as $d(e_n,e_m) = ε$ for every $n≠m$), so $[-ε,ε]^ω$ cannot even be limit point compact.

(8) $ℝ^ω$ in the uniform topology is not second countable, separable, or Lindelof. Proof: We see each pair of distinct $x,y∈\{0,1\}^ω$ are of distance $1$ in $ℝ^ω$, so that $ℝ^ω$ cannot be separable and hence not second countable. As well, $\{B(x,3/4)~|~x∈\{0,1\}^ω\}$ is an uncountable open cover of the closed subset $[0,1]^ω$, yet there are not even any proper subcovers, so $ℝ^ω$ cannot be Lindelof.

(9) $ℝ^I$ is not locally metrizable. Proof: Suppose some basis element $U=\prod_{i∈I} U_i$ of $ℝ^I$ were metrizable. Then since $U_i=ℝ$ for all but finitely many $i∈I$, and since $I-F$ for any finite subset $F⊆I$ is still uncountably infinite, we see $ℝ^I$ can be imbedded in $U$. But $ℝ^I$ itself is not metrizable, as it is not normal.

(10) $ℝ^I$ is not Lindelof. Proof: Regular Lindelof spaces are normal, which $ℝ^I$ is not.

Countability Axioms and Topological Groups (4.30.18)

James Munkres Topology, chapter 4.30, exercise 18:

Show in a first-countable topological group $G$ that second countability, separability, and the Lindelof condition are equivalent.

Proof: Let $\{B_n\}$ be a countable basis about $e$. We may presume $B_n⊇B_{n+1}$ for all $n$, and if necessary by setting $B_n'=B_n∩B_n^{-1}$ we may presume $B_n^{-1}=B_n$.

Suppose $G$ has a countable dense subset $D$, and that $U$ is a hood about $g$. Then since $g×e∈m^{-1}(U)$ and $\{gB_n\}$ is a basis about $g$, there exists $n$ such that $gB_n×B_n⊆m^{-1}(U)$. Choose $d∈D∩gB_n$ and we see $d=gb$ for some $b∈B_n$, hence $g=db^{-1}∈dB_n$, so $dB_n$ is a hood about $g$. Furthermore, given $c∈B_n$ we see $dc=gbc∈m(gB_n×B_n)⊆U$, so that $dB_n⊆U$. Therefore $\{dB_n~|~d∈D,n∈ℕ\}$ is a countable basis for $G$.

Suppose $G$ is Lindelof. Then for each $n∈ℕ$, there is a countable subset $D_n⊆G$ such that $∪_{d∈D_n}dB_n=G$. We show $D=∪D_n$ is dense in $G$; let $U⊆G$ be a hood about $g$. Then $gB_n⊆U$ for some $n$, and we also see $g∈∪_{d∈D_n}dB_n$ so write $g=db$ for some $d∈D_n⊆D$ and $b∈B_n$. But now $d=gb^{-1}∈gB_n⊆U$ so $d∈U$ and $D$ is a countable dense subset of $G$.$~\square$

Separability of Large Euclidean Product Spaces (3.30.16)

James Munkres Topology, chapter 3.30, exercise 16:

(a) Show that the product space $ℝ^I$ contains a countable dense subset.
(b) Show that if $|J| > |\mathcal{P}(ℕ)|$, then $ℝ^J$ does not contain a countable dense subset.

Proof: (a) For every oddly finite sequence $q_1,...,q_{2n+1}$ of rationals such that $0 ≤ q_2 < q_4 < ... < q_{2n} ≤ 1$, let there be an associated function $I → ℝ$ that takes value $q_{2k-1}$ on $(q_{2k-2},q_{2k})$ for each $k = 1, ..., n$ (where $q_0=0$), takes value $q_{2n+1}$ on $(q_{2n},1]$, and is zero on each of $q_2,...,q_2n$. We see the collection of such functions is countable, and we claim it forms a countable dense subset of $ℝ^I$: For a given basis element $∏_{i∈I}U_i$ of open sets $U_i⊆ℝ$ such that $U_i \not = ℝ$ for only finitely many $i∈I$, let $i_1 < ... < i_n$ be exactly all such that $U_{i_j} \not = ℝ$. We may choose rationals $q_2,...,q_{2(n-1)}$ such that $i_1 < q_2 < i_2 < q_4 < ... < q_{2(n-1)} < i_n$ (unless $n=1$ where one chooses $i_1 < q_2$ and say $q_3=0$) and for each $q_1,q_3,...,q_{2n-1}$ choose rationals such that $q_{2k-1}∈U_{i_k}$ for each $k=1,...,n$. Then the associated function on $q_1,...,q_{2n-1}$ takes a value within $U_i$ on each $i$, and hence is contained in the basis element.

(b) Suppose $D$ is a countable subset of $ℝ^J$. Fix some nonempty interval $(a,b)$, and choose some disjoint nonempty interval $(c,d)$. Then since the function $f : J → \mathcal{D}$ given by $f(α) = D ∩ π_α^{-1}(a,b)$ cannot be injective, we find some distinct $α,β∈ℝ$ such that for each $d∈D$ we have $d(α)∈(a,b)$ iff $d(β)∈(a,b)$. But now the basis element $∏U_j$ where $U_α=(a,b)$ and $U_β=(c,d)$ and $U_j=ℝ$ otherwise cannot contain a point of $D$.$~\square$

G-Delta Sets and First Countability (3.30.1)

James Munkres Topology, chapter 3.30, exercise 1:

(a) A $G_δ$ set in a space $X$ is a countable intersection of open sets of $X$. Show that in a first-countable $T_1$ space, every one-point set is a $G_δ$ set.
(b) There is a familiar space wherein every one-point set is a $G_δ$ set, which is nevertheless not first countable. What is it?

Proof: (a) Let $x∈X$, and let $\{B_i\}$ be a countable basis at $x$. Suppose $y∈∩B_i$ while $y \neq x$. Then since $X$ is $T_1$, let $U$ be a neighborhood of $x$ not containing $y$. We see $U$ is open and contains $x$, yet contains no element of the basis $\{B_i\}$ seeing as $y \not \in U$, a contradiction.

(b) Let $ℝ^\omega$ be under the box topology. Then given $x∈ℝ^\omega$, when we let $U_n=∏(x_n-1/n,x_n+1/n)$ we see $∩U_n=\{x\}$, so that every one-point set in $ℝ^\omega$ is $G_δ$. To show that $ℝ^\omega$ is not first countable, suppose $\{B_n\}$ is a countable basis at any particular point $x$. Then for each $n∈ℕ$, we may choose an interval $(a_n,b_n)$ such that $x_n∈(a_n,b_n) \subset π_n(B_n)$. Hence, let $U=∏(a_n,b_n)$; we see $x∈U$ yet $B_n \not \subseteq U$ for all $n$ since $π_n(B_n) \not \subseteq π_n(U)$, so that $\{B_n\}$ is not a countable basis at $x$, a contradiction.$~\square$