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.