$ \newcommand{\func}{\operatorname} \newcommand{\lcm}{\func{lcm}} \newcommand{\hl}[1]{\textbf{#1}} \newcommand{\lf}{\lfloor} \newcommand{\rf}{\rfloor} \newcommand{\f}[1]{\left\lf #1 \right\rf} \newcommand{\bb}[1]{\left( #1 \right)} \newcommand{\Sb}[1]{\left\lbrack #1 \right\rbrack} \renewcommand{\ss}[1]{\left\lbrace #1 \right\rbrace} \renewcommand{\mod}[1]{\left\lvert #1 \rightrvert} \newcommand{\R}{\boldsymbol{R}} \newcommand{\C}{\boldsymbol{C}} \newcommand{\Z}{\boldsymbol{Z}} \newcommand{\Q}{\boldsymbol{Q}} \renewcommand{\P}{\mathbb{P}} \newcommand{\Po}{\mathcal{P}} \newcommand{\F}{\mathcal{F}} \newcommand{\Zz}{\Z_{\geq 0}} \renewcommand{\c}[1]{\left\langle #1 \right\rangle} \renewcommand{\mod}[1]{\left\lvert #1 \right\rvert} \newcommand{\norm}[1]{\left\lVert #1 \right\rVert} \renewcommand{\bar}[1]{\overline{#1}} \newcommand{\eqbreak}{\phantom{{} = {}}} \newcommand{\impliesbreak}{\phantom{{} \implies {}}} \newcommand{\iffbreak}{\phantom{{} \iff {}}} \newcommand{\del}{\partial} \renewcommand{\d}[1]{\, \mathrm{d} #1} \newcommand{\dydt}{\frac{\d{y}}{\d{t}}} \newcommand{\dydx}{\frac{\d{y}}{\d{x}}} \newcommand{\pd}[2]{\frac{\del #1}{\del #2}} \newcommand{\pp}[3]{\frac{\del^2 #1}{\del #2 \del #3}} \newcommand{\dd}[2]{\frac{\mathrm{d} #1}{\mathrm{d} #2}} \newcommand{\Inn}{\func{Inn}} \newcommand{\Aut}{\func{Aut}} \newcommand{\diag}{\func{diag}} \newcommand{\id}{\func{id}} \newcommand{\GL}{\func{GL}} \newcommand{\SL}{\func{SL}} \newcommand{\End}{\func{End}} \newcommand{\Hom}{\func{Hom}} \newcommand{\Tr}{\func{Tr}} \newcommand{\rank}{\func{rank}} \newcommand{\ORD}{\mathbf{ORD}} \newcommand{\OD}{\mathbf{OD}} \newcommand{\HOD}{\mathbf{HOD}} \newcommand{\ZF}{\mathsf{ZF}} \newcommand{\ZFC}{\mathsf{ZFC}} \newcommand{\ZC}{\mathsf{ZC}} \renewcommand{\AC}{\mathsf{AC}} \newcommand{\DC}{\mathsf{DC}} \newcommand{\CC}{\mathsf{CC}} \newcommand{\CH}{\mathsf{CH}} \newcommand{\SH}{\mathsf{SH}} \newcommand{\KH}{\mathsf{KH}} \newcommand{\GCH}{\mathsf{GCH}} \newcommand{\PA}{\mathsf{PA}} \newcommand{\BST}{\mathsf{BST}} \newcommand{\MA}{\mathsf{MA}} \newcommand{\Con}{\func{Con}} \newcommand{\ON}{\mathbf{ON}} \newcommand{\dom}{\func{dom}} \newcommand{\ran}{\func{ran}} \newcommand{\pred}{\func{pred}} \newcommand{\mos}{\func{mos}} \newcommand{\WF}{\mathbf{WF}} \newcommand{\type}{\func{type}} \newcommand{\V}{\mathbf{V}} \renewcommand{\L}{\mathcal{L}} \newcommand{\cl}{\func{cl}} \newcommand{\trcl}{\func{tc}} \newcommand{\TC}{\func{TC}} \newcommand{\A}{\mathfrak{A}} \newcommand{\B}{\mathfrak{B}} \newcommand{\add}{\func{add}} \newcommand{\cov}{\func{cov}} \newcommand{\non}{\func{non}} \newcommand{\cf}{\func{cf}} \newcommand{\forces}{\Vdash} \newcommand{\one}{\mathbbm{1}} \newcommand{\restrictedto}{\mathord{\upharpoonright}} \newcommand{\Fn}{\func{Fn}} \newcommand{\Col}{\func{Col}} \newcommand{\bigtriangle}{\triangle} \newcommand{\U}{\mathcal{U}} \newcommand{\W}{\mathcal{W}} \newcommand{\N}{\mathcal{N}} \newcommand{\Seq}{\mathit{Seq}} \newcommand{\lh}{\func{lh}} \newcommand{\Ult}{\func{Ult}} \newcommand{\In}{\func{In}} \newcommand{\otp}{\func{otp}} \newcommand{\symdiff}{\, \triangle \,} \newcommand{\crit}{\func{crit}} \newcommand{\sat}{\func{sat}} \renewcommand{\int}{\func{int}} \newcommand{\length}{\func{length}} \newcommand{\Lim}{\func{Lim}} \newcommand{\Succ}{\func{Succ}} \newcommand{\G}{\mathcal{G}} \newcommand{\T}{\mathcal{T}} \newcommand{\ext}{\func{ext}} \newcommand{\height}{\func{height}} \renewcommand{\diamond}{\diamondsuit} \newcommand{\D}{\mathcal{D}} \newcommand{\NS}{\mathrm{NS}} $
Exercise 4.1.
Solution. Let $C(\R)$ be the set of all continuous functions $\R \to \R$. Following the hint, by identifying the continuous functions $f : \R \to \R$ by their definition on $\Q$, there is an injection $F : C(\R) \to {}^{\R}\Q$ given by $F(f) = f\restrictedto\Q$. But $\vert {}^{\R}\Q\vert = \vert \R\vert ^{\vert \Q\vert } = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0} = \mathfrak{c}$.$\square$
Exercise 4.2.
Solution. We shall write:$$ \begin{align*}
\tau_a = a_0 + \xi_0 + a_1 + \xi_1 + a_2 + \xi_2 + \cdots
\end{align*}$$ Suppose $\tau_a = \tau_b$, so there exists an isomorphism $f : \tau_a \to \tau_b$. Suppose for a contradiction that there is some $n$ such that $a_n \neq b_n$. Let $n$ be the least natural number such that $a_n \neq b_n$, and assume WLOG that $a_n > b_n$. Let $y = \max b_n$, and suppose $f(x) = y$. Then $y < f(x + 1) \in \xi_n$. Since $\xi_n$ is the order-type of the integers, there exists a $y' \in \tau_b$ such that $f(x) < y' < f(x + 1)$. By order-preserving property, there is no $x'$ such that $f(x') = y'$, contradicting bijectivity.
$\square$
Exercise 4.3.
Lemma 4.3.A. Let $\Z[x]$ denote the the set of all polynomials of integer coefficients. Then $\Z[x]$ is countably infinite.
Proof. Note that $\Z$ is countable, as the map $f : \boldsymbol{N} \to \Z$ defined by:
$$ \begin{align*}
f(n) =
\begin{cases}
k, &\text{if $n = 2k$ for some $k \in \boldsymbol{N}$} \\
-k, &\text{if $n = 2k + 1$ for some $k \in \boldsymbol{N}$}
\end{cases}
\end{align*}$$ is a bijection. We may now define a map $g : \Z[x] \to \boldsymbol{N}$ defined by:
$$ \begin{align*}
g(a_nx^n + \cdots + a_1x + a_0) = 2^{f^{-1}(a_0)}3^{f^{-1}(a_1)}\cdots p_{n+1}^{f^{-1}(a_n)}
\end{align*}$$ where $p_n$ is the $n^\text{th}$ smallest prime number. Fundamental Theorem of Arithmetic then tells us that $g$ is injective (in fact it's bijective, but we do not need it here). Thus, $\Z[x]$ is countable, and clearly it is infinite.
$\blacksquare$
Lemma 4.3.B. $\Z[x]$ is well-orderable.
Proof. We define $\prec \; \subseteq \Z[x] \times \Z[x]$ as follows: Given $a_nx^n + \cdots + a_1x + a_0,b_mx^m + \cdots + b_1x + b_0 \in \Z[x]$, we define:
$$ \begin{gather*}
a_nx^n + \cdots + a_1x + a_0 \prec b_mx^m + \cdots + b_1x + b_0 \\
\updownarrow \\
n < m \text{ or } (n = m \wedge a_n < b_m)
\end{gather*}$$ It's easy to check that $\prec$ defines a well-order, with the zero polynomial being the minimal element.
$\blacksquare$
Solution. Let $X$ be the set of all algebraic reals. Define $h : X \to \boldsymbol{N} \times \Z[x]$ by $h(a) = (k,p(x))$, where $p(x)$ is the $\prec$-least polynomial (as in Lemma 4.3.B), and $a$ is the $(k + 1)^\text{th}$ smallest root of $p(x)$. Then $h$ is clearly injective, and since $\Z[x]$ is countable by Lemma 4.3.A, $\boldsymbol{N} \times \Z[x]$ is also countable, so $X$ must be countable.
$\square$
Remark.. If we assume $\AC$, then this exercise becomes far easier as countable union of countable sets is countable (which makes proving Lemma 4.3.A slightly easier), and every set is well-orderable (trivialising Lemma 4.3.B).
Exercise 4.4.
Solution. Using a choiceless bijection we may work with the Baire space $\N = \omega^\omega$ in place of $\R$, and $S \subseteq \N$. Let $S = \lbrace f_i : i \in \omega\rbrace$. Define:$$ \begin{align*}
X = \prod_{i \in \omega} \lbrace f_i(i)+1,f_i(i)+2\rbrace
\end{align*}$$ Clearly $X \cap S = \emptyset$, for if $f \in X$ then for any $i \in \omega$, $f(i) > f_i(i)$. Thus, $X \subseteq \N - S$. Clearly $\vert X\vert = 2^{\aleph_0} = \mathfrak{c}$, so $\vert \N - S\vert \geq \mathfrak{c}$. On the other hand, $\vert \N - S\vert \leq \vert \N\vert = \mathfrak{c}$.
$\square$
Exercise 4.5.
Exercise 4.5(i).
Solution. This follows from Exercise 4.4 and that $\Q$ is countable.$\square$
Exercise 4.5(ii).
Solution. This follows from Exercise 4.4 and that by Exercise 4.3, the set of all algebraic numbers is countable.$\square$
Exercise 4.6.
Lemma 4.6.A. Every open set of reals is a countable union of disjoint open intervals.
Note.. For this exercise, we use $(a,b)$ to denote the open interval, and $\c{a,b}$ to denote ordered pair.
Proof. Let $U \subseteq \R$ be open. First observe that for $x \in U$, there exists a maximal open interval $U_x \subseteq U$ containing $x$ - this is because all open sets are generated by open intervals as basic sets, and if $U_x,U_x'$ are open intervals containing $x$, then $U_x \cup U_x'$ is also an open interval containing $x$.
For each $x \in U$, let $U_x \subseteq U$ be the maximal open interval that contains $x$. Let $\U := \lbrace U_x : x \in U\rbrace$ be the set of all maximal open intervals in $U$. Then they must be disjoint, as if $U_x \cap U_y \neq \emptyset$ then $U_x \cup U_y$ is a larger open interval. Clearly $\bigcup \U = U$, so it remains to show that $\U$ is countable.
Write $\U = \lbrace U_i : i \in I\rbrace$, where $I$ is an index set and $U_i \cap U_j = \emptyset$ for $i \neq j$. For each $i$, there exists a rational number $q_i \in U_i$. Since the $U_i$'s are disjoint, the map $U_i \mapsto q_i$ is one-to-one. Thus $I$ is countable as desired. Note that the map $U_i \mapsto q_i$ does not require the Axiom of Choice, as there is a choiceless algorithm to obtain a rational number $q_i \in U_i$.
$\blacksquare$
Solution. Let $X := \lbrace \c{a,b} \in \R \times \R : a < b\rbrace$, and let $\mathcal{T}$ denote the set of all open subsets of $\R$. let $\mathcal{B}$ be the set of all open intervals of $\R$. Consider the one-to-one map $f$ of $X$ onto $\mathcal{B}$ defined by $f(\c{a,b}) = (a,b)$. Clearly $\vert X\vert = \mathfrak{c}$, so $\vert \mathcal{B}\vert = \mathfrak{c}$.
Now define $g : [\mathcal{B}]^\omega \to \mathcal{T}$ by:
$$ \begin{align*}
g(\lbrace B_i : i \in \omega\rbrace) = \bigcup_{i \in \omega} B_i
\end{align*}$$ By Lemma 4.6.A, $g$ is onto. Thus:
$$ \begin{align*}
\vert \mathcal{T}\vert \leq \vert [\mathcal{B}]^\omega\vert = \vert \mathcal{B}\vert ^{\aleph_0} = \mathfrak{c}^{\aleph_0} = \mathfrak{c}
\end{align*}$$ On the other hand, we have$\vert \mathcal{T}\vert \geq \vert \mathcal{B}\vert = \mathfrak{c}$, so equality holds.
$\square$
Exercise 4.7.
Solution. We first show that the Cantor set $\C$ is closed. Let $x \in \R$, and write $x = \sum_{n=1}^\infty \frac{a_n}{3^n}$ where $a_n \in \lbrace 0,1,2\rbrace$. Suppose on the contrary that $a_N = 1$ for some $N$, and it's not the case that $a_n = 0$ for all $n > N$ or $a_n = 2$ for all $n > N$. We write:$$ \begin{align*}
x = \sum_{n=0}^{N-1} \frac{a_n}{3^n} + \frac{1}{3^N} + s = \sum_{n=0}^{N-1} \frac{a_n}{3^n} + \frac{2}{3^N} - t
\end{align*}$$ where $0 < s,t < \frac{1}{3^N}$. Now let $y = \sum_{n=1}^\infty \frac{b_n}{3^n} \in \C$. If $y \leq x$, then it can't be that $y > \sum_{n=1}^{N-1} \frac{a_n}{3^n} + \frac{1}{3^N}$, for otherwise we must have $b_N = 1$. Thus $y \leq \sum_{n=1}^{N-1} \frac{a_n}{3^n} + \frac{1}{3^N}$, hence $x - y \geq s$. By similar reasoning, if $y \geq x$ then $y - x \geq t$. Thus, $x$ is not a limit point of $\C$, so all points not in $\C$ are not limit points.
We now show that $\C$ has no isolated points. Fix $y = \sum_{n=1}^\infty \frac{b_n}{3^n} \in \C$. For any $\epsilon > 0$, let $N$ be large enough so that $\frac{2}{3^N} < \epsilon$. If $b_n = 2$ for some $n \geq N$, then $z := y - \frac{2}{3^n} \in \C$ and $\vert y - z\vert < \epsilon$. Otherwise, $z := y + \frac{2}{3^N} \in \C$ and $\vert y - z\vert < \epsilon$. Thus $y$ is not isolated.
$\square$
Exercise 4.8.
Solution. The proof of Theorem 4.5. may be repeated. We provide a similar proof that can also serve as an alternative proof to Theorem 4.5.We shall define a one-to-one function $f$ with $\dom(f) = 2^{<\omega}$ and $f(s)$ is an open set for all $s \in \omega^{<\omega}$, with $f(s) \cap P \neq \emptyset$. We define this inductively.
Let $f(\emptyset) = (a,b)$. Suppose for all $s \in \omega^{<\omega}$ with $\length(s) = n$, $f(s)$ are defined such that $f(s) \cap P \neq \emptyset$, and $f(s_1) \cap f(s_2) = \emptyset$ for $s_1 \neq s_2$ of length $n$. Say $x \in f(s) \cap P$, and let $(c,d) \subseteq f(s)$ such that $x \in (c,d)$. Since $P$ is perfect, $x$ is a limit point so there exists $x_0,x_1 \in P \cap (c,d)$ such that $x_0 \neq x_1$ and $\vert x_0 - x\vert > 0$, $\vert x_1 - x\vert > 0$. We define $f(s^\frown(0)),f(s^\frown(1))$ as the open sets containing $x_0$ and $x_1$ respectively such that $x,x_1 \notin f(s^\frown(0))$ and $x,x_0 \notin f(s^\frown(1))$ (clearly such open sets exist). Note that if $s' \neq s$ and $\length(s') = n$, then since $f(s^\frown(i)) \subseteq f(s)$ we have $f(s^\frown(i)) \cap f(s'^\frown(i')) = \emptyset$ for $i,i' \in \lbrace 0,1\rbrace$.
Now for each $l \in 2^\omega$, let $X_l := \bigcap_{n=0}^\infty f(l\restrictedto n)$. $X_l$ is a nested intersection of bounded open sets, so it contains a single point $x_l \in X_l$. Then $\lbrace x_l : l \in 2^\omega\rbrace \subseteq P \cap (a,b)$, so $\vert P \cap (a,b)\vert = \mathfrak{c}$.
$\square$
Exercise 4.9.
Solution. Let $x \in P_2 - P_1$. Since $x \notin P_1$ and $P_1$ is perfect, $x$ is isolated from $P_1$, i.e. there exists an open neighbourhood $(a,b) \ni x$ such that $(a,b) \cap P_1 = \emptyset$. Then $\emptyset \neq P_2 \cap (a,b) \subseteq P_2 - P_1$, and by Exercise 4.8 $\vert P_2 \cap (a,b)\vert = \mathfrak{c}$. Thus $\vert P_2 - P_1\vert = \mathfrak{c}$.$\square$
Exercise 4.10.
Solution. Since $P$ is closed, $P^\ast \subseteq P$. On the other hand, for every $x \in P$ is a limit point, and hence a condensation point by Exercise 4.8, so $P \subseteq P^\ast $.$\square$
Exercise 4.11.
Solution. Clearly if $P \subseteq F$ then $P^\ast \subseteq F^\ast $, and $P = P^\ast $.$\square$
Exercise 4.12.
Solution. As in the hint, for each $a \in F^\ast $, since every neighbourhood of $a$ has uncountably many points and $\vert F - P\vert \leq \aleph_0$, must have that $a$ is a limit point of $P$. Since $P$ is closed, $a \in P$.$\square$
Exercise 4.13.
Solution. This is an immediate consequence of Exercise 4.12.$\square$
Exercise 4.14.
Solution. Note that we require Axiom of Choice to prove the Baire Category Theorem.Suppose $\Q = \bigcap_{n=0}^\infty U_n$, where each $U_n$ is open. Since $\Q$ is dense, so is each $U_n$. Thus $\R - \Q = \bigcup_{n=0}^\infty (\R - U_n)$, where each $\R - U_n$ is closed nowhere dense. Clearly $\Q = \bigcup_{q \in \Q} \lbrace q\rbrace$ is also a countable union of closed nowhere dense sets, therefore $\R$ is a countable union of closed nowhere dense sets. This contradicts the Baire Category Theorem, which implies that $\R$ is not a countable union of closed nowhere dense sets.
$\square$
Exercise 4.15.
Solution. Since $f$ is continuous, we have that:- If $B$ is open, then $f_{-1}(B)$ is open.
- $f_{-1}(\R - B) = \R - f_{-1}(B)$.
- $f_{-1}\bb{\bigcup_{n=0}^\infty B_n} = \bigcup_{n=0}^\infty f_{-1}(B_n)$.
$\square$
Exercise 4.16.
Solution. For $x \in \R$ and $r > 0$, let $B_r(x)$ denote the open ball of radius $r$ around $x$. We have:$$ \begin{align*}
&\iffbreak \text{$f$ is continuous at $x$} \\
&\iff (\forall \epsilon > 0)(\exists \delta > 0)(\forall y \in \R)(\vert x - y\vert \leq \delta \to \vert f(x) - f(y)\vert < \epsilon) \\
&\iff (\forall \epsilon > 0)(\exists \delta > 0)(\forall y \in \R)(y \in \bar{B_\delta(x)} \to f(y) \in B_\epsilon(f(x))) \\
&\iff (\forall \epsilon > 0)(\exists \delta > 0)(B_\delta(x) \subseteq f_{-1}(B_\epsilon(f(x)))) \\
&\iff x \in \bigcap_{n=1}^\infty \bigcup_{\substack{U \text{ open} \\ U \subseteq f_{-1}\bb{B_{\frac{1}{n}}(f(x)))}}} U
\end{align*}$$ which is a $G_\delta$ set.
$\square$
Exercise 4.17.
Exercise 4.17(i).
Solution. Define $F : \N \times \N \to \N$ by stipulating that for all $f \in \N$, we have $F(f) = (g,h)$, where $g(n) = f(2n)$ and $h(n) = f(2n+1)$ for all $n$. Clearly $F$ is a one-to-one and onto function. We shall show that $F$ is in fact a homeomorphism.We first show that $F$ is continuous. Fix $s,t$ be two finite sequences, and we wish to show that $F_{-1}(O(s) \times O(t))$ is open. Fix $f \in F_{-1}(O(s) \times O(t))$. We then have for $n > \max\lbrace 2 \cdot \length(s), 2 \cdot \length(t) + 1\rbrace$, if $g \in \N$ and $f\restrictedto n = g\restrictedto n$, then $g \in F_{-1}(O(s) \times O(t))$. In other words, $O(f\restrictedto n) \subseteq F_{-1}(O(s) \times O(t))$, so $F_{-1}(O(s) \times O(t))$ is open.
A similar reasoning can be used to show that $F$ is an open map. Fix a finite sequence $s$. Then for $(f,g) \in F"(O(s))$, only the first $\leq\left\lfloor{\frac{\length(s)}{2}}\right\rfloor + 1$ values of $f$ and $g$ are restricted, so $O(f\restrictedto k) \times O(g\restrictedto l) \subseteq F"(O(s))$ for some $k,l$.
$\square$
Exercise 4.17(ii).
Solution. Let $\Gamma : \omega \times \omega \to \omega$ be Cantor's pairing function. Define $G : \N \to \N^\omega$ be stipulating that for all $f \in \N$, if $G(f) = \c{f_n : n \in \omega}$, then $f_n(k) = f(\Gamma(n,k))$ for all $k$. Then $G$ is one-to-one and onto, and one may repeat a proof similar to that of Exercise 4.17(i) that $G$ is a homeomorphism. Note that $\N^\omega$ is equipped with the product topology, so basic open sets are of the form $\prod_{i \in \omega} U_i$, where $U_i = \N$ for all but finitely many $i$, and if $U_i \neq \N$ then $U_i = O(s_i)$ for some finite sequence $i$.$\square$
Exercise 4.18.
Solution. If $s \in T_F$, then $s \subseteq f$ for some $f \in F$, so we may extend $s$ in the direction of $f$ (i.e. take $t \subseteq f$ such that $\length(t) > \length(s)$, and we have that $s \subseteq t$).We show that $F \mapsto T_F$ is a one-to-one correspondence. Let $F,F'$ be two closed subsets of $\N$, and suppose $T_F = T_{F'}$. Let $f \in F$, so $f\restrictedto n \in T_F = T_{F'}$ for all $n$. Thus, for each $n$ there exists an $f_n \in T_{F'}$ such that $f\restrictedto n = f_n\restrictedto n$. Then $f_n \to f$ under the metrisable topology of $\N$, so $f \in T_{F'}$. Thus $F \subseteq F'$, and by symmetry of argument we have $F = F'$. This gives that the map is one-to-one.
To see that the map is onto, let $T$ be a sequential tree without maximal node. We shall show that $T_{[T]} = T$. $T_{[T]} \subseteq T$ is clear. Conversely, if $s \in T$ then since $T$ has no maximal node, there exists an infinite sequence $s \subseteq s_1 \subseteq s_2 \subseteq \dots$ in $T$. Then $f := \bigcup_{i=1}^\infty s_i$ is an infinite branch in $T$, so $f \in [T]$. Then $s \subseteq f$ so $s \in T_{[T]}$.
$\square$
Exercise 4.19.
Solution. It suffices to show that the perfect set $P$ constructed in Theorem 4.5 is homeomorphic to the Cantor set. For each $f \in \lbrace 0,1\rbrace^\omega$, define $f' \in \lbrace 0,2\rbrace^\omega$ by $f'(n) = 0$ if $f(n) = 0$, and $f'(n) = 2$ if $f(n) = 1$. We also let $a_f$ be the unique element in $P \cap \bigcap_{n=0}^\infty I_{f\restrictedto n}$. Define a function $F$ from $P$ to $\C$ by:$$ \begin{align*}
F(a_f) = \sum_{n=0}^\infty \frac{f'(n)}{3^{n+1}}
\end{align*}$$ We now show that $F$ is a homeomorphism (under the subspace topologies of $P$ and $\C$ respectively). Note the subspace topology of $\C$ is generated by $O_k(x)$ for $x \in \C$ and $k \in \omega$, where:
$$ \begin{align*}
O_k(x) := \ss{y \in \C : y = \sum_{n=1}^\infty \frac{b_n}{3^n} \wedge (\forall n \leq k) \, a_n = b_n}
\end{align*}$$ The subspace topology of $P$ is generated by $\lbrace a_f : f \supseteq s\rbrace$ for all $s \in S$ (or $\lbrace a_f : f \in O(s)\rbrace$).
Fix $U \subseteq \C$ open, and let $a_f \in F_{-1}(U)$ for some $f \in \lbrace 0,1\rbrace^\omega$. Then $F(f) \in U$, and since $U$ is open we have that $O_k(F(f)) \subseteq U$ for some $k$ large enough. But this implies that for all $s \in O(f\restrictedto k)$ and $g \supseteq s$, we have that $a_g \in F_{-1}(U)$. Thus $\lbrace a_g : g \in O(f\restrictedto k)\rbrace \subseteq F_{-1}(U)$, so $F_{-1}(U)$ is open. The same approach can be used to show that $F$ is an open map, so it is a homeomorphism.
$\square$
Exercise 4.20.
Solution. Let $X$ be a Polish space, and let $D$ denote the dense subset of $X$. We shall show that the function $F : X \to [0,1]^\omega$ in the hint (w.r.t. $D$) is one-to-one, and a homeomorphism to its image (under the subspace topology).To see that $F$ is one-one, let $x,y \in X$ and suppose $x \neq y$. Let $d(x,y) = \delta$. Let $x_n \in B_{\frac{\delta}{3}}(x)$ (possible by denseness). Then $d(x,x_n) < \frac{\delta}{3}$ but $d(y,x_n) \geq d(y,x) - d(x,x_n) > \frac{2\delta}{3}$, so $F(x) \neq F(y)$.
Let $U \subseteq \lbrace 0,1\rbrace^\omega$ be open, and we wish to show that $F_{-1}(U)$ is open. Let $x \in F_{-1}(U)$, and let $\epsilon_0,\dots,\epsilon_n > 0$ be such that $\prod_{i<n} B_{\epsilon_i}(F(x)(i)) \times [0,1]^\omega \subseteq U$. Let $V := \bigcap_{i<n} B_{\epsilon_i}(x)$, which is open. If $y \in V$, then for $i < n$ we have:
$$ \begin{align*}
\vert d(y,x_i) - d(x,x_i)\vert \leq d(y,x) < \epsilon_i \implies F(y)(i) \in B_{\epsilon_i}(F(x)(i))
\end{align*}$$ Thus $F(y) \in U$, so $x \in V \subseteq F_{-1}(U)$. Hence $F_{-1}(U)$ is open.
Now let $V \subseteq X$ be open, and we wish to show that $F"(V) \subseteq F"(X)$ is open under the subspace topology. Let $f \in F"(V)$, and let $x \in V$ such that $F(x) = f$. Let $\epsilon > 0$ such that $B_\epsilon(x) \subseteq V$. WLOG suppose $x_0 \in B_{\frac{\epsilon}{3}}(x)$. Then we see that $f \in (B_{\frac{\epsilon}{3}}(f(0)) \times [0,1]^\omega) \cap F"(X) \subseteq F"(X)$, for if $y \in X$ is such that $F(y) \in B_\epsilon(f(0)) \times [0,1]^\omega$, then:
$$ \begin{align*}
d(x,y) \leq d(x,x_0) + d(x_0,y) \leq 2d(x,x_0) + \frac{\epsilon}{3} < \epsilon
\end{align*}$$ Thus $y \in B_\epsilon(x) \subseteq V$, so $F"(V)$ is open.
It remains to show that $F"(X)$ is a $G_\delta$ subspace of $[0,1]^\omega$. We shall show that:
$$ \begin{align*}
F"(X) = \bigcap_{n=1}^\infty \bigcup_{x \in X} B_{\frac{1}{n}}(f(x)) =: Y
\end{align*}$$ Clearly $F"(X) \subseteq Y$. Conversely, we see that $y \in Y$ precisely if $(\forall n > 0)(\exists x \in X) \, d(f(x),y) < \frac{1}{n}$. In other words, $y$ is in the closure of $F"(X)$. But since $F$ is a homeomorphism and $X$ is complete, $F"(X)$ is closed and therefore $y \in F"(X)$.
$\square$