Let $\kappa$ be the cardinal of $\{f\in\mathbb{R}^{\mathbb{R}}:f\text{ is continuous}\}$. For all $k\in\mathbb{R}$, $f:x\rightarrow k$ is continuous so

$\kappa\geq|\mathbb{R}|=2^{{\aleph_{0}}}$ |

Let $f,g$ be two continuous functions that have the same restriction on $\mathbb{Q}$. For all $x\in\mathbb{R}$, find a sequence ${(a_{n})}_{{n\in\mathbb{N}}}\in\mathbb{Q}^{\mathbb{N}}$ such that $x=\lim_{{n\to\infty}}a_{n}$. Then the limit of the sequence $f(a_{n})=g(a_{n})$ is $f(x)=g(x)$ by continuity. This means that:

$\kappa\leq|\mathbb{R}^{\mathbb{Q}}|={(2^{{\aleph_{0}}})}^{{\aleph_{0}}}=2^{{% \aleph_{0}}}$ |

Let $\zeta$ be the order-type of $\mathbb{R}$. For every sequence $a\in\mathbb{N}^{\mathbb{N}}$ we define the sum of order types $\tau_{a}=a_{0}+\zeta+a_{1}+\zeta+...$. For each $x\in\tau_{a}$, there is some $n\in\mathbb{N}$ such that $x$ belongs to the $n$-th term. In this $n$-th order-type, $x$ can be viewed as an element of $\mathbb{R}$. Hence we have a one-to-one map from $\tau_{a}$ to $\mathbb{N}\times\mathbb{R}$ which is countable by exercise 3.3. Hence $\tau_{a}$ is a countable order-type and obviously linear.

Suppose that we have an isomorphism $f$ between two sets $A=A_{0}\sqcup X_{0}\sqcup A_{1}\sqcup X_{1}\sqcup A_{2}...$ and $B=B_{0}\sqcup Y_{0}\sqcup B_{1}\sqcup Y_{1}\sqcup B_{2}...$ of respective order-types $\tau_{a}$ and $\tau_{b}$. The elements of $A_{i}$ can be written $0_{{i,a}},1_{{i,a}},...{(a_{0}-1)}_{{i,a}}$ those of $X_{i}$ can be written $\{p_{{i,a}}^{*}:p\in\mathbb{R}\}$ and similarly for $B_{i},Y_{i}$. Suppose that $a\neq b$ and let $n$ be the least index such that $a_{n}\neq b_{n}$ or without loss of generality $a_{n}<b_{n}$.

First note that for any $m\in\mathbb{N}$ and any $p\in\mathbb{R}$, we can not find $y\in B$ such that $f(p_{{m,a}}^{*})<y<f({(p+1)}_{{m,a}}^{*})$ for otherwise $p_{{m,a}}^{*}<f^{{-1}}(y)<{(p+1)}_{{m,a}}^{*}$ which is inconsistent with the order-type of $X_{m}$. In particular, for any $q\in\mathbb{R}$, $q_{{m,a}}^{*}$ is not sent on an element $i_{{m^{{\prime}},b}}$ for otherwise ${(q-1)}_{{m,a}}^{*}$ is necessarily sent to ${(i-1)}_{{m^{{\prime}},b}}$, ${(q-2)}_{{m,a}}^{*}$ to ${(i-2)}_{{m^{{\prime}},b}}$… and so forth until ${(q-i)}_{{m,a}}$ which is sent to $0_{{m^{{\prime}},b}}$. Then for $p=q-i-1$ we can find a $y$ leading to a contradiction (we could also have considered $p=q+(b_{{m^{{\prime}}}}-1-i_{{m^{{\prime}},b}}))$. This means that $f(q_{{m,a}}^{*})={q^{{\prime}}}_{{m^{{\prime}},b}}^{*}$. From this and using the same method as above we deduce that $f(X_{m})=Y_{{m^{{\prime}}}}$.

We shall prove that for all $k<n$ we have $f(A_{k})=B_{k}$ and $f(X_{k})=Y_{k}$. If $k<n$ and these equalities hold for all $k^{{\prime}}<k$, then necessarily $f(0_{{k,a}})\geq 0_{{k,b}}$ and $f^{{-1}}(0_{{k,b}})\geq 0_{{k,a}}$ because $f$ is one-to-one. But because $f$ is increasing, the later equality becomes $0_{{k,b}}\geq f(0_{{k,a}})$ and hence $f(0_{{k,a}})=0_{{k,b}}$. By induction, we show that $f(i_{{k,a}})=i_{{k,b}}$ for all $i<a_{k}=b_{k}$ and so $f(A_{k})=B_{k}$ as expected. By the discussion above, $f(X_{k})=Y_{l}$. By the induction hypothesis and because $f$ is one-to-one, $l\geq k$. If $l>k$ then because $f$ is increasing, $f$ does not reach any element of $Y_{l}$. This contradicts the fact that it is onto. Hence the only possibility is $f(X_{k})=Y_{k}$.

As above, $f(A_{n})=\{i_{{n,b}}:i<a_{0}\}$ and $f(X_{n})=Y_{l}$ for some $l\geq n$. Thus since $f$ is increasing, the element ${(a_{0})}_{{n,b}}$ (for example) is not reached and this contradicts the fact that is onto. Finally, the map $a\rightarrow\tau_{a}$ is one-to-one and so there are at least $|\mathbb{N}^{\mathbb{N}}|=2^{{\aleph_{0}}}$ such order-types.

Let $A$ be the set of algebraic reals. $\mathbb{Q}\subseteq A$ so $|A|\geq\aleph_{0}$. The set of all integer polynomials can be viewed as a subsets of integer sequences of finite length $\bigcup_{{1\leq n<\omega}}\mathbb{R}^{n}$. This set is countable by exercises 3.3 and 3.4, so is well-ordonable. For any $x\in A$, we consider the smallest polynomial $P$ of which $x$ is a root. We have $P\in\mathbb{R}^{n}$ for some $n\geq 1$ and by the fundamental theorem, $P$ has roots $m\leq n$ roots. $a_{1},a_{2},...,a_{m}$. If $k$ is such that $x=a_{k}$ then we define $f(x)=(P,k)$. Hence we have constructed (without the axiom of choice) a one-to-one map from $A$ to $\bigcup_{{1\leq n<\omega}}\mathbb{R}^{n}\times\mathbb{N}$. Again, by exercises 3.3 and 3.4 this set is countable and hence $|A|\leq\aleph_{0}$.

Let $S$ be countable. We have $\sup\{|\mathbb{R}\setminus S|,|S|\}=|\mathbb{R}\setminus S|+|S|=|\mathbb{R}|=2% ^{{\aleph_{0}}}$. But because $|S|=\aleph_{0}<2^{{\aleph_{0}}}$ the only possibility is $|\mathbb{R}\setminus S|=2^{{\aleph_{0}}}$.

((i)) $|\mathbb{R}\setminus\mathbb{Q}|=2^{{\aleph_{0}}}$ because $|\mathbb{Q}|=\aleph_{0}$.

((ii)) $|\mathbb{R}\setminus A|=2^{{\aleph_{0}}}$ because $|A|=\aleph_{0}$ by exercise 4.3.

Let $O$ be an open set of $\mathbb{R}$. We shall define a sequence of open interval $I_{\alpha}=(a_{\alpha},b_{\alpha})$ with rational endpoints and included in $O$. The $I_{\alpha}$’s will be pairwise disjoint and we define open sets $O_{\alpha}=\bigcup_{{\beta<\alpha}}I_{\beta}$. Because every collection of disjoint open nonempty intervals is at most countable (see p. 39) we get a sequence ${(I_{\alpha})}_{{\alpha<\lambda}}$ for some ordinal $\lambda$ at most countable.

Hence let $\alpha$ be an ordinal and suppose that we have contructed $I_{\beta}$ for all $\beta<\alpha$. If $O\setminus O_{\alpha}$ is empty then $O=\bigcup_{{\beta<\alpha}}I_{\beta}$ and we stop the construction. Otherwise, we start by reducing each $I_{\beta}$ and replacing its endpoints by $a^{{\prime}}_{\beta}=\frac{3}{4}a_{\beta}+\frac{1}{4}b_{\beta}$ and $b^{{\prime}}_{\beta}=\frac{1}{4}a_{\beta}+\frac{3}{4}b_{\beta}$ so that the open set $U=O\setminus\overline{O_{\alpha}}$ is still nonempty. Hence there is a point $x\in U$ and a positive number $r>0$ such that the ball of center $x$ and radius $r$ is included in $U$. By density of $\mathbb{Q}$ in $\mathbb{R}$, there are rationals $a,b$ such that $x-r\leq a<x<b\leq x+r$ and so a nonempty interval $I=(a,b)$ with rational endpoints such that $I\subseteq U$. Because there are at most $|\mathbb{Q}^{2}|=\aleph_{0}$ such intervals, we can pick one such interval $I_{\alpha}$, without using the axiom of choice.

Finally, reordering the $I_{\alpha}$ and adding empty ones (i.e. $a_{\alpha}=b_{\alpha}$) we have shown that every open interval can be written

$O=\bigcup_{{n\in\mathbb{N}}}I_{n}$ |

for $I_{n}$ an open interval with rational endpoints. Hence there are at most

$\left|{(\mathbb{Q}^{2})}^{\mathbb{N}}\right|={(\aleph_{0}^{2})}^{{\aleph_{0}}}% =\aleph_{0}^{{\aleph_{0}}}=2^{{\aleph_{0}}}$ |

open sets in $\mathbb{R}$. The collection of open intervals $(0,r)$ for $r\in\mathbb{R}$ clearly shows that there are exactly $2^{{\aleph_{0}}}$ many open intervals.

Alternative construction of the decomposition of $O$: let ${(I_{n})}_{{n\in\mathbb{N}}}$ be an enumeration of nonempty open interval with rational endpoints which are included in $O$. Let $X$ the element of $n\in\mathbb{N}$ for which $I_{n}$ is minimal i.e. not included in any other $I_{m}$. The ${(I_{n})}_{{n\in\mathbb{N}}}$ are pairwise disjoint: if $I_{n}$ and $I_{m}$ ($n,m\in X$) have nonempty intersection then $I_{n},I_{m}\supseteq I_{n}\cap I_{m}=I_{p}\supseteq I_{q}$ for some $p\in\mathbb{N}$ and $q\in X$. Hence $I_{n}=I_{q}=I_{m}$. Moreover, we have

$O=\bigcup_{{n\in X}}I_{n}$ |

We define for all the closed set (union of $2^{m}<\aleph_{0}$ closed intervals):

$F_{m}=\left\{x=\sum_{{n=1}}^{{+\infty}}\frac{a_{n}}{3^{n}}:a_{n}\in\{0,1,2\}% \wedge(\forall n<m,a_{n}\neq 1)\right\}$ |

Clearly, the Cantor set is defined by $C=\bigcap_{{m\in\mathbb{N}}}F_{m}$ and hence is closed as an intersection of closed sets. Moreover, it is nonempty (for example $0\in C$). We shall see that it has no isolated point and so is perfect.

Let $x=\sum_{{n=1}}^{{+\infty}}\frac{a_{n}}{3^{n}}$ be a point in $C$. If there is $n_{0}$ such that $a_{n}=0$ for all $n>n_{0}$ then we define for all $m\in\mathbb{N},x_{m}=x+\frac{2}{3^{{n_{0}+1+m}}}$. Otherwise we define $x_{m}=\sum_{{n=1}}^{{m}}\frac{a_{n}}{3^{n}}$. In both case, $x_{m}\neq x$ and $x_{m}\in C$ for all $m\in\mathbb{N}$ and $\lim_{{m\rightarrow+\infty}}x_{m}=x$, thus $x$ is not isolated.

Let $P$ be a perfect set and $P\cap(a,b)=\emptyset$. Define $I_{{\emptyset}}=[a,b]$. Then $I_{{\emptyset}}\cap P$ is perfect and we can use the proof of theorem 4.5 to construct a one-to-one function from ${\{0,1\}}^{\omega}$ to $P\cap(a,b)$. Hence $|P\cap(a,b)|=\mathfrak{c}=2^{{\aleph_{0}}}$.

Suppose $P_{2}\nsubseteq P_{1}$ are perfect sets. Let $p\in P_{2}\backslash P_{1}$. Since $P_{1}$ is closed, there is some open interval $(a,b)\subseteq{\mathbb{R}\backslash P_{1}}$ such that $p\in(a,b)$. Hence $P_{2}\cap(a,b)\neq\emptyset$ and by exercise 4.8, $|P_{2}\cap(a,b)|=\mathfrak{c}$. But we have

$\displaystyle{P_{2}\backslash P_{1}}\cap(a,b)$ | $\displaystyle={P_{2}\cap{\mathbb{R}\backslash P_{1}}}\cap(a,b)$ | ||

$\displaystyle=P_{2}\cap(a,b)$ |

Finally $\mathfrak{c}\geq|P_{2}\backslash P_{1}|\geq|P_{2}\cap(a,b)|\geq\mathfrak{c}$

For any set $P$ we have $P^{*}\subseteq\bar{P}$ and if $P$ is closed, $\bar{P}=P$. If moreover $P$ is perfect and $p\in P$, then a neighborhood of $p$ contains an open neighborhood $(a,b)\cap P$ of $p$ which is uncountable by 4.8 and thus $p\in P^{*}$. Finally, $P=P^{*}$.

For any sets $A\subseteq B$ it is easy to see that $A^{*}\subseteq B^{*}$. Let $P$ be a perfect set and $F$ a closed set, such that $P\subseteq F$. Then $P^{*}\subseteq F^{*}$ and by 4.10 $P^{*}=P$. So $P\subseteq F^{*}$.

Let $F$ be an uncountable closed set and $P$ the perfect set contructed in theorem 4.6. By exercise 4.11, we have $P\subseteq F^{*}$. Let $a\in F^{*}$. For any $r>0$, we have $|B(a,r)\cap F|$ is uncountable by definition of $F^{*}$. But we also have $F\backslash P$ countable so $B(a,r)\cap P$ is uncountable and a fortiori nonempty. As a consequence, $a\in\bar{P}=P$. Finally $P=F^{*}$.

By 4.12, we get $F=F^{*}\cup(F\backslash F^{*})$. Let’s see it is the only such decomposition. Suppose $F=P_{1}\cup(F\backslash P_{1})=P_{2}\cup(F\backslash P_{2})$ where $P_{1},P_{2}$ are perfect. We have $F\backslash P_{1}=((F\backslash P_{2})\backslash P_{1})\cup(P_{2}\backslash P_% {1})$. But $F\backslash P_{1}$ is countable so $P_{2}\backslash P_{1}$ is at most countable. By 4.9, this means $P_{2}\subseteq P_{1}$. Similarly, we get $P_{1}\subseteq P_{2}$ and so $P_{1}=P_{2}$.

Suppose there are open sets $U_{i}$ such that

$\mathbb{Q}=\bigcap_{{i\in\mathbb{N}}}U_{i}$ |

In particular, for each $i\in\mathbb{N}$, $\mathbb{Q}\subseteq U_{i}$ and because $\mathbb{Q}$ is dense, so is $U_{i}$. Taking the complement, we get

$\mathbb{Q}^{c}=\bigcup_{{i\in\mathbb{N}}}U_{i}^{c}$ |

Because the $U_{i}$’s are open dense, the $U_{i}^{c}$’s are nowhere dense. Hence the Baire space (which is homeomorphic to $\mathbb{Q}^{c}$) is the union of countably many nowhere dense sets. This contradicts the Baire Category Theorem.

Denote by $\mathcal{B}$ the set of borel sets. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function and define $S$ to be the set of $X\subseteq\mathbb{R}$ such that $f_{{-1}}(X)\in\mathcal{B}$. If $U$ is open, then because $f$ is continuous $f_{{-1}}(X)$ is open and so belongs to $\mathcal{B}$. If we have a countable family ${(X_{i})}_{{i\in\mathbb{N}}}$ of elements of $S$ then we have

$f_{{-1}}\left({\bigcup_{{i\in\mathbb{N}}}X_{i}}\right)=\bigcup_{{i\in\mathbb{N% }}}f_{{-1}}(X_{i})\in\mathcal{B}$ |

and so $\bigcup_{{i\in\mathbb{N}}}X_{i}\in S$. Similarly, if $X\in S$ then $f_{{-1}}(X^{c})=f_{{-1}}(\mathbb{R})\cap{f_{{-1}}(X)}^{c}\in\mathcal{B}$ and thus $X^{c}\in S$. Finally, $S$ is a $\sigma$-algebra which contains the open sets and so $S\supseteq\mathcal{B}$. In particular, the preimage of a borel set is a borel set.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function and $S\subseteq\mathbb{R}$ the set of points where it is continuous. For all $x\in\mathbb{R}$ and $n\in\mathbb{N}$ define the open set

$U_{n}=\bigcup_{{x\in S}}B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}$ |

where for each $x\in S$, $N_{{x,n}}$ is the least positive integer such that $f\left(B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}\right)\subseteq B{\left(f(x% ),\frac{1}{2^{n}}\right)}$. Such an integer exists because $f$ is continuous at $x$. We shall prove that

$S=\bigcap_{{n\in\mathbb{N}}}U_{n}$ |

i.e. $S$ is of type $G_{{\delta}}$. The inclusion from left to right is trivial so let’s consider the other direction. Consider a point $y\in\bigcap_{{n\in\mathbb{N}}}U_{n}$ and a neighborhood $V$ of $f(y)$. Let $n$ be an integer such that $B{\left(f(y),\frac{1}{2^{{n-1}}}\right)}\subseteq V$. Because $y\in U_{n}$, there exists $x\in S$ such that $y\in B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}$. In particular, $f(y)\in B{\left(f(x),\frac{1}{2^{n}}\right)}$. Moreover, if $z\in B{\left(f(x),\frac{1}{2^{n}}\right)}$, the triangular inequality gives

$\displaystyle|z-f(y)|\leq|z-f(x)|+|f(x)-f(y)|<\frac{1}{2^{n}}+\frac{1}{2^{n}}=% \frac{1}{2^{{n-1}}}$ |

Hence if $U=B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}$, we have $y\in U$ and $f(U)\subseteq B{\left(f(x),\frac{1}{2^{n}}\right)}\subseteq B{\left(f(y),\frac% {1}{2^{{n-1}}}\right)}\subseteq V$. So $f$ is continuous at $y$ i.e. $y\in S$.

((i)) Consider the function $f:\mathcal{N}\times\mathcal{N}\rightarrow\mathcal{N}$ given by ${(a_{i})}_{{i<\omega}}\times{(b_{i})}_{{i<\omega}}\overset{f}{\mapsto}{(c_{i})% }_{{i<\omega}}$ where for all $i<\omega$, $c_{{2i}}=a_{i}$ and $c_{{2i+1}}=b_{i}$. $f$ is clearly a bijection. We shall prove it is actually a homeomorphism.

Now pick $(a,b)\in\mathcal{N}\times\mathcal{N}$ and define $c=f(a,b)$. For any neighborhood $V$ of $c$ there exists some $n$ such that $O(c_{{|2n}})\subseteq V$. Define the open neighborhood $U=O(a_{{|n}})\times O(b_{{|n}}$ of $(a,b)$. For any $(x,y)\in U$ and $0\leq i<n$, we have ${f(x,y)}_{{2i}}=x_{i}=a_{i}=c_{{2i}}$ and ${f(x,y)}_{{2i+1}}=y_{i}=b_{i}=c_{{2i+1}}$. Hence $f(x,y)\in O(c_{{|2n}})$. Finally $f(U)\subseteq V$ and $f$ is continuous.

Conversely, choose $c\in\mathcal{N}$ define $(a,b)=f^{{-1}}(c)$. Let $V$ be a neighborhood of $(a,b)$. There exists some $n$ such that $O(a_{{|n}})\times O(b_{{|n}})\subseteq V$. Define $U=O(c_{{|2n}})$. Let $z\in U$ and $(x,y)=f^{{-1}}(z)$. For all $0\leq i<n$, we have $x_{i}=z_{{2i}}=c_{{2i}}=a_{i}$ and $y_{i}=z_{{2i+1}}=c_{{2i+1}}=b_{i}$. So $(x,y)\in O(a_{{|n}})\times O(b_{{|n}})$. Finally $f^{{-1}}(U)\subseteq V$ and $f^{{-1}}$ is continuous.

((ii)) We proceed the same way. Let $g:\omega\cong\omega^{2}$ be a bijection and define the bijection $f:\mathcal{N}^{\omega}\rightarrow\mathcal{N}$ by ${\left({(a_{{ji}})}_{{i<\omega}}\right)}_{{j<\omega}}\overset{f}{\mapsto}{(a_{% {g(k)}})}_{{k<\omega}}$. Again, this is a homeomorphism. Let $a\in\mathcal{N}^{\omega}$ and $b=f(a)$. If $V$ is a neighborhood of $b$ then there exists $N$ such that $O(b_{{|N}})\subseteq V$. Now let $n=1+\max_{{0\leq k<N}}{\{\pi_{1}{(g^{{-1}}(k))},\pi_{2}{(g^{{-1}}(k))}\}}$ where $\pi_{1}:\mathcal{N}\times\mathcal{N}\rightarrow\mathcal{N}$ and $\pi_{2}:\mathcal{N}\times\mathcal{N}\rightarrow\mathcal{N}$ are the canonical projection. Define the open neighborhood of $a$ $U=\prod_{{{0\leq j<n}}}{O({(a_{{ji}})}_{{0\leq i<n}})}\times\mathcal{N}^{\omega}$. Then it’s easy to prove $f(U)\subseteq V$ and so $f$ is continuous. Conversely, with $a=f^{{-1}}(b)$ as above and $V$ a neighborhood of $a$ there exists $n$ such that $U=\prod_{{{0\leq j<n}}}{O({(a_{{ji}})}_{{0\leq i<n}})}\times\mathcal{N}^{% \omega}\subseteq V$. We define $N=1+\max_{{0\leq i,j<n}}{\{g(i,j)\}}$ and $U=O(b_{{|n}})$ and open neighborhood of $b$. Then $f^{{-1}}(U)\subseteq V$. Thus $f^{{-1}}$ is continuous.

Let $F$ be a closed subset of $\mathcal{N}$. We have $T_{F}=\{s\in\omega^{{<\omega}}:\exists f\in F,s\subseteq f\}$. $T_{F}$ is a tree and for $s\in T_{F}$, there exists $f\in F$ and $n<\omega$ such that $s=f_{{|n}}$. If we define $t=s\cup{(n,f(n))}$ then $t\in T_{F}$ (because $t\subseteq f$) and $s\subsetneq t$. So $T_{F}$ does not have any maximal node.

Let’s consider the map from the set of closed subsets of $\mathcal{N}$ to the set of sequential trees without a maximal node given by $F\mapsto T_{F}$. By a result of the book, it is obvious that the function is one-to-one: If $T_{F}=T_{G}$, $F=[T_{F}]=[T_{G}]=G$. Another (similar) proof: for all $\epsilon>0$, pick $n$ such that $\frac{1}{2^{{n+1}}}<\epsilon$. We have $f_{{|n}}\in T_{F}=T_{G}$. So there exists some $g\in G$ such that $f_{{|n}}=g_{{|n}}$. In particular, $d(f,g)\leq\frac{1}{2^{{n+1}}}<\epsilon$. So $f$ is in $\bar{G}=G$. We do the same the other direction and get $F=G$.

Let’s prove that the map $F\mapsto T_{F}$ is surjective. We consider $T$ a sequential tree without maximal node. We define $F=[T]=\{f\in\mathcal{N},\forall n<\omega,f_{{|n}}\in T\}$. Let $g\in\bar{F}$. For all $n<\omega$, $O(g_{{|n}})\cap F\neq\emptyset$. So there is $f\in F$ such that $g_{{|n}}=f{|n}\in T$. Thus $g\in F$ and $\bar{F}=F$ i.e. $F$ is closed. By definition $T_{F}=\{s\in\omega^{{<\omega}}:\exists f\in F,s\subseteq f\}=\{s\in\omega^{{<% \omega}}:\exists f\in\mathcal{N},\forall n<\omega,f_{{|n}}\in T\wedge s% \subseteq f\}$. In particular it is clear that $T_{F}\subseteq T$. Suppose that the previous inequality is strict. Let $s\in T$ such that $\forall f\in\mathcal{N},\forall n<\omega,f_{{|n}}\notin T\vee s\nsubseteq f$. Let $t$ be another element of $T$, of length $n_{0}$. Because $T$ is a tree, we can define for all $n\leq n_{0}$, $t_{n}=t_{{|n}}\in T$. Because $T$ has no maximal element and is countable, we can construct by induction for all $n\geq n_{0}$, an element $t_{{n+1}}$ that extends $t_{n}$. Moreover, because $T$ is a tree, we can suppose $t_{{n+1}}$ of length $n+1$. $t_{n}$ is a Cauchy sequence and $t_{n}\rightarrow f\in\mathcal{N}$. By construction, $f_{{|n}}=t_{n}\in T$. Hence $s\nsubseteq f$ and a fortiori $s\nsubseteq f_{{n_{0}}}=t$. So $s$ is a maximal element. Contradiction.

Let $P$ be a perfect Polish space. Consider a metric such that $P$ is complete and $Q\subseteq P$ a countable dense subset. We denote by $\tilde{B}(x,r)$ the closed ball of center $x$ and radius $r$. For any sequence $s\in 2^{{<\omega}}$, we shall construct by induction such a ball $\tilde{B}(x_{s},r_{s})$. We shall take $x_{s}\in Q$ and $r_{s}\in\mathbb{Q}$, so there are at most countably many possibilities and so we won’t need the axiom of choice.

We start by choosing $x_{0}\in Q$ and setting $r_{0}=1$. If $B_{s}$ is contructed, then because $P$ is perfect, the open ball $B(x_{s},r_{s})$ contains at least another point $p\neq x_{s}$. For $\epsilon>0$ small enough, the ball $B(p,\epsilon)$ is included in $B(x_{s},r_{s})$ and does not contain $x_{s}$. Then because $Q$ is dense, there exists some $q\in B(p,\epsilon)\cap Q$. For a nonzero $\mu\in\mathbb{Q}$ small enough, $\tilde{B}(x_{s},\mu)$ and $\tilde{B}(q,\mu)$ are two disjoint balls included in $B_{s}$ of diameters $\leq 2^{{-(length(s)+1)}}$. Thus we can choose two such balls $\tilde{B}(x_{{(s,0)}},r_{{(s,0)}})$ and $\tilde{B}(x_{{(s,1)}},r_{{(s,1)}})$.

Now for each $f\in 2^{{\omega}}$, the set $\bigcap_{{n\in\mathbb{N}}}B_{{f_{{|n}}}}$ contains exactly one element because $P$ is complete. We let $F(f)$ be this element of $P$. $F$ is one-to-one: if $f,g\in 2^{{\omega}}$ are distinct and $n$ is the least index such that $f(n)\neq g(n)$, then $F(f)\in B_{{f_{{|{(n+1)}}}}}$ and $F(g)\in B_{{g_{{|{(n+1)}}}}}$ but these two balls are disjoint by construction. We note $F^{{-1}}$ the inverse function defined on the image $F(2^{\omega})$.

Given $f\in 2^{{\omega}}$ and $\epsilon>0$ we take $n$ such that $2^{{-n}}<\epsilon$ then if $g\in O(f_{{|n}})$ i.e. $f_{{|n}}=g_{{|n}}$, we have $F(f),F(g)\in\tilde{B}_{{f_{{|n}}}}$ but this ball is of diameter $\leq 2^{{-n}}<\epsilon$. Conversely, given $F(f)\in F(2^{{\omega}})$ and $V$ a neighborhood of $F(f)$, we choose $n$ large enough so that $O(f_{{|n}})\subseteq V$. Then if $F(g)\in F(2^{{\omega}})$ is such that $F(g)$ is in the open neighborhood $B(x_{{f_{{|n}}}},r_{{f_{{|n}}}})$ of $F(f)$, we necessarily have $g_{{|n}}=f_{{|n}}$ and so $g\in V$. Hence both $F$ and $F^{{-1}}$ are continuous and $F$ is an homeomophism.

As in exercise 4.7, we consider for each $m<\aleph_{0}$ the set $F_{m}$, union of the of $2^{m}$ closed balls $B_{s}$ where $s$ is of length $m$. By construction, $F(2^{\omega})=\bigcap_{{m\in\mathbb{N}}}F_{m}$ and is closed as an intersection of closed sets. Finally, $F(2^{\omega})\subset P$ is a closed subset homeomorphic to the Cantor space $2^{\omega}$.

First recall that $\frac{2}{\pi}\operatorname{atan}:[0,+\infty)\rightarrow[0,1]$ is a homeomorphism and from that we obtain a homeomorphism between ${[0,+\infty)}^{{\omega}}$ and the Hilbert cube ${[0,1]}^{{\omega}}$.

Let $P$ be a Polish space and $\{x_{n},n\in\mathbb{N}\}$ be a countable dense subset of $P$. We define the map $f:P\rightarrow{[0,+\infty)}^{{\omega}}$ by $f(x)=\left\langle d(x,x_{n}):n\in\mathbb{N}\right\rangle$. This map is one-to-one: if $x\neq y$ then $d(x,y)\neq 0$ and because $Q$ is dense we can find some $x_{n}\in B\left(x,\frac{d(x,y)}{2}\right)$. Then $d(x,x_{n})<\frac{d(x,y)}{2}$ while $d(y,x_{n})\geq\left|d(y,x)-d(x,x_{n})\right|>\frac{d(x,y)}{2}$ and so $f(x)\neq f(y)$. We will note $f^{{-1}}$ its inverse function, defined on its image $f(P)$.

Let $x\in P$ and $V$ be a neighborhood of $f(x)$. There are some $n\in\mathbb{N}$ and $\epsilon>0$ such that $U={\prod_{{0\leq i<n}}{B(x_{i},\epsilon)}}\times{[0,+\infty)}^{{\omega}}$ is an open subset of $V$. If $y\in P$ is such that $d(x,y)<\epsilon$ then $\forall 0\leq i<n,\left|d(x,x_{i})-d(y,x_{i})\right|\leq d(x,y)<\epsilon$ so $f(y)\in U\subseteq V$. Conversely, let $f(x)\in f(P)$ and $\epsilon>0$. Because $Q$ is dense, there is some $n$ such that $x_{n}\in B\left(x,\frac{\epsilon}{3}\right)$. Then if $f(y)$ is in the open subset $U={[0,+\infty)}^{{n-1}}\times B\left(x,\frac{\epsilon}{3}\right)\times{[0,+% \infty)}^{{\omega}}$ we have in particular $d(y,x_{n})\leq d(y,x_{n})+\frac{\epsilon}{3}$ and so $d(x,y)\leq d(x,x_{n})+d(y,x_{n})\leq 2d(x,x_{n})+\frac{\epsilon}{3}<\epsilon$. Hence both $f$ and $f^{{-1}}$ are continuous and $f$ is an homeomophism.

Finally, we have $f(P)=\bigcap_{{n\in\mathbb{N}}}U_{n}$ where $U_{n}$ are open sets defined by

$U_{n}=\bigcup_{{x\in P}}\left(\left({\prod_{{0\leq i<n}}{B\left(d(x,x_{i}),2^{% {-n}}\right)}}\right)\times{[0,+\infty)}^{{\omega}}\right)$ |

Hence, any Polish space is homeomorphic to a $G_{\delta}$ subspace of the Hilbert Cube.