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.$~\square$

Fixed Point Analysis and Partial Differential Equations

Suppose $f : ℝ^2→ℝ$ is continous and globally Lipschitz in its second coordinate, i.e. there exists some $m∈ℝ$ such that for all $x,y_1,y_2∈ℝ$ $$|f(x,y_1)-f(x,y_2)|≤m·|y_1-y_2|$$ Further suppose $h > 0$ such that for the aforementioned $m$ we observe $mh < 1$. Then when $c[0,h]$ is the space of continuous functions $[0,h]→ℝ$ under the maximum absolute difference metric $d$, and $y_0∈ℝ$ is fixed, show that the operator $$T : c[0,h]→c[0,h]$$ $$T(φ)(x)=y_0+∫_0^x f(t,φ(t)) \mathrm{d}t$$ is well defined and has a unique fixed point.

Proof: Note that by the fundamental theorem of calculus, $T(φ)$ is continuous on $[0,h]$ so that $T$ is indeed well defined. Since $c[0,h]$ is a complete metric space, by Banach's theorem it suffices to show that $T$ is a contracting map to ensure existence and uniqueness of a fixed point. Observe the inequalities given $φ_1,φ_2∈c[0,h]$: $$d(T(φ_1),T(φ_2)) = \max_{x∈[0,h]} |(y_0+∫_0^x \! f(t,φ_1(t))~\mathrm{d}t) - (y_0+∫_0^x \! f(t,φ_2(t))~\mathrm{d}t)| =$$ $$\max_{x∈[0,h]} |∫_0^x \! f(t,φ_1(t)) - f(t,φ_2(t))~\mathrm{d}t| ≤ \max_{x∈[0,h]} ∫_0^x \! |f(t,φ_1(t)) - f(t,φ_2(t))|~\mathrm{d}t≤$$ $$\max_{x∈[0,h]} m·∫_0^x \! |φ_1(t)-φ_2(t)|~\mathrm{d}t ≤ mh·\max_{x∈[0,h]} |φ_1(x)-φ_2(x)| = mh·d(φ_1,φ_2)$$ so that $T$ is a contracting map of factor no larger than $mh < 1$.

Banach Fixed Point Theorem

(Banach) Let $X$ be a nonempty complete metric space, and let $T : X→X$ be a contracting map, i.e. $d(T(x),T(y)) ≤ θ·d(x,y)$ for all $x,y∈X$ for some fixed $θ < 1$. Then $T$ has a unique fixed point.

Proof: (Banach) Uniqueness of fixed points is clear, since if $T$ fixes $x$ and $y$, then $θ·d(x,y) ≥ d(T(x),T(y))=d(x,y)$ and $θ ≥ 1$ unless $d(x,y)=0$ and $x=y$. To exhibit the fixed point, let $x_0∈X$ be any point, and define $x_{n+1}=T(x_n)$. Then when $α=d(x_0,x_1)$ it follows by induction that $d(x_n,x_{n+1}) ≤ θ^nα$. As well, by the triangle inequality it follows that $d(x_n,x_m) ≤ \sum_{k=n}^{m-1} θ^kα$ when $n≤m$. Since $θ < 1$ this forms part of a convergent geometric series, showing $(x_n)$ forms a Cauchy sequence in $X$. Let $x$ be the point of convergence of this sequence. $T$ is continuous since contraction maps are generally continuous, so $T(x)$ is the point of convergence of the sequence $(T(x_n))=(x_{n+1})$, showing $T(x)=x$.

Function Spaces from Compact Sets into Complete Metric Spaces

Let $X$ be a compact space, and let $(Y,d)$ be a complete metric space. We show that the set $F$ of continuous functions from $X$ to $Y$ under the metric $$h(f,g)=\max_{x∈X}~d(f(x),g(x))~~~~~\text{for }f,g∈F$$ induces a complete metric space.

Note that when $f,g : X→Y$ are continuous, then since $d : Y×Y→ℝ$ is continuous, we observe $d∘(f×g) : X→ℝ$ is continuous and due to compactness of $X$ indeed attains a maximum value, so that the function $h : F×F→ℝ$ is well defined. To verify that it is a metric, we see $h(f,g)=0$ iff $d(f(x),g(x))=0$ iff $f(x)=g(x)$ for all $x∈X$, i.e. $f=g$. As well, $$h(f,g)=\max_{x∈X}~d(f(x),g(x))=\max_{x∈X}~d(g(x),f(x))=h(g,f)$$ Finally, let $a,b,c∈F$. Then $$h(a,c) = \max_{x∈X}~d(a(x),c(x)) ≤ \max_{x∈X}~[d(a(x),b(x))+d(b(x),c(x))] ≤$$$$\max_{x∈X}~[d(a(x),b(x))]+\max_{x∈X}~[d(b(x),c(x))] = h(a,b)+h(b,c)$$ so that $h(a,c) ≤ h(a,b)+h(b,c)$ and $(F,h)$ is a metric space.

Now we show completeness. Let $(f_n)$ be a Cauchy sequence in $F$. We show that for each $x∈X$, the sequence $(f_n(x))$ is Cauchy in $Y$. To wit, given $ε > 0$, choose $N$ such that $n,m > N$ implies $h(f_n,f_m) < ε$. Then $d(f_n(x),f_m(x)) ≤ \max_{x∈X}~d(f_n(x),f_m(x)) = h(f_n,f_m) < ε$. Thus since $Y$ is complete write $f_n(x)→f(x)$ for each $x∈X$. It now suffices to show $f : X→Y$ is continuous and $f_n→f$. We shall prove the latter convergence is uniform and the former will follow by the uniform limit theorem.

Let $ε > 0$ be given. Since $(f_n)$ is Cauchy, let $N$ be such that $n,m > N$ imply $h(f_n,f_m) < ε/2$. Then $d(f_n(x),f_m(x)) < ε/2$ for all $x∈X$. Now let $n > N$ be given. Since $f_n(x)→f(x)$ choose $m > N$ such that $d(f_m(x),f(x)) < ε/2$. Then $d(f_n(x),f(x)) ≤ d(f_n(x),f_m(x))+d(f_m(x),f(x)) < ε$ so the convergence is uniform.

Local Compactness and Products of Quotient Maps (3.29.11)

James Munkres Topology, chapter 3.29, exercise 11:

(a) Let $Z$ be a locally compact Hausdorff space, and let $p : X→Y$ be a quotient map. Show that the map $$π=p×1 : X×Z→Y×Z$$ is a quotient map.

(b) Let $p : A→B$ and $q : C→D$ be quotient maps, and let $B$ and $C$ be locally compact Hausdorff. Show $p×q$ is a quotient map.

Proof: (a) It is clear that $π$ is continuous, so it suffices to show $A⊆Y×Z$ is open when $π^{-1}(A)⊆X×Z$ is open. For any given $y×z∈A$, when $x×z$ is such that $π(x×z)=y×z$ (i.e. $x∈p^{-1}(y)$) we find a neighborhood $U$ of $x$ saturated with respect to $p$ and a neighborhood $V$ of $z$ such that $U×V⊆π^{-1}(A)$, so that $π(U×V)=p(U)×V$ will be a neighborhood of $y×z$ contained in $A$. In this fashion, let $U_1×V$ be a basis element in $π^{-1}(A)$ containing $x×z$ since $π^{-1}(A)$ is open. Since $Z$ is locally compact Hausdorff, we may assume $V$ is such that $\overline{V}$ is compact and $U×\overline{V}⊆π^{-1}(A)$.

Given open $U_i$ such that $x×z∈U_i×\overline{V}⊆π^{-1}(A)$, we construct $U_{i+1}$ such that $x×z∈U_{i+1}×\overline{V}⊆π^{-1}(A)$ and $p^{-1}(p(U_i))⊆U_{i+1}$, and we shall set $U=∪U_i$. To see this construction, note that for each $u∈p^{-1}(p(U_i))$ we see $u×\overline{V}$ is compact, so that by the tube lemma we may choose a neighborhood $V_u$ of $u$ such that $V_u×\overline{V}⊆π^{-1}(A)$. Letting $U_{i+1}=∪V_u$, we see each of the two desired properties are fulfilled.

We see $p^{-1}(p(U))=U$ by the second desired property of each $U_i$, so that $U$ is saturated. Since also $U×V⊆U×\overline{V}⊆π^{-1}(A)$ we have demonstrated the proper $U$ and $V$, so that $π$ is a quotient map.

(b) This is evident by considering the composition of quotient maps $p×q=(1×q)∘(p×1)$.$~\square$