Infinite simple continued fractions parametrise the irrational real numbers

Write $\mathcal{A}$ for the set of sequences $a=(a_0,a_1,\dots )$ with $a_0\in \mathbb{Z}$ and $a_n\ge 1$ for $n\ge 1$, and for a finite prefix put $C(a_0,\dots ,a_n):=\{b\in \mathcal{A}:b_k=a_k\text{ for }k\le n\}$; by [F1] the assignment $x\mapsto a$ with $a_0=z(x_0)$ and $a_n=x_n+1$ for $n\ge 1$ is a bijection of $\mathcal{N}={\mathbb{N}}^{\mathbb{N}}$ onto $\mathcal{A}$, since $z$ is a bijection of $\mathbb{N}$ onto $\mathbb{Z}$ and $m\mapsto m+1$ is a bijection of $\mathbb{N}$ onto the integers that are at least $1$, and it acts coordinatewise, so it carries the cylinder $N_s$ of [F6] onto the cylinder $C(t_0,\dots ,t_{k-1})$ of the decoded prefix $t_0=z(s_0)$ and $t_i=s_i+1$ for $1\le i<k$, and every $C(a_0,\dots ,a_n)$ arises from exactly one $N_s$ in this way; give $\mathcal{A}$ the topology transported along the bijection, so that the bijection is a homeomorphism and, the $N_s$ being a basis of $\mathcal{N}$ by [F6] and their images being exactly the sets $C(a_0,\dots ,a_n)$, those sets are a basis of $\mathcal{A}$, which is the cylinder topology of the Statement, while $\mathbb{R}\setminus \mathbb{Q}$ carries the subspace topology of [F7] inherited from $\mathbb{R}$.

For $a\in \mathcal{A}$ let $p_n$ and $q_n$ be as in [F2] and, for $n\ge -1$, put $M_n(t):=(p_{n}t+p_{n-1})\mathrm{/}(q_{n}t+q_{n-1})$; the initial values of [F2] give $M_{-1}(t)=(1\cdot t+0)\mathrm{/}(0\cdot t+1)=t$ for every real $t$, and for $n\ge 0$ and every real $t\ge 1$ the denominator satisfies $q_{n}t+q_{n-1}>0$, since $q_{0}t+q_{-1}=t>0$ and, for $n\ge 1$, $q_n>0$ and $q_{n-1}>0$ by [F2], so $M_n$ is defined at every real $t\ge 1$ when $n\ge 0$ and at every real $t$ when $n=-1$.

For $a\in \mathcal{A}$ the intervals $J(a_0,\dots ,a_n)$ are closed and bounded, are nested as $n$ increases, and satisfy $\mathrm{diam}J(a_0,\dots ,a_n)=1\mathrm{/}(q_n(q_n+q_{n-1}))\to 0$, all by [F2], so by claims 1 and 2 of [F5] their intersection is a single real, written $V(a)$, and $V(a)\in J(a_0,\dots ,a_n)$ for every $n\in \mathbb{N}$; moreover $V(a)$ is irrational, for suppose $V(a)=u\mathrm{/}v$ with $u,v$ integers and $v\ge 1$, and note first that the $q_n$ are positive for $n\ge 0$ and strictly increasing for $n\ge 1$ by [F2] and are integers, so $q_1\ge 1$ and $q_{n+1}\ge q_n+1$ give $q_n\ge n$ for $n\ge 1$; for $n\ge 1$ both $V(a)$ and the endpoint $p_n\mathrm{/}{q}_n$ lie in $J(a_0,\dots ,a_n)$, so $\mathrm{\mid }V(a)-p_n\mathrm{/}{q}_n\mathrm{\mid }\le 1\mathrm{/}(q_n(q_n+q_{n-1}))<1\mathrm{/}{q}_n^2$ because $q_{n-1}>0$, and hence $\mathrm{\mid }u{q}_n-v{p}_n\mathrm{\mid }=v{q}_n\mathrm{\mid }V(a)-p_n\mathrm{/}{q}_n\mathrm{\mid }<v\mathrm{/}{q}_n$; taking $n>v$ makes $\mathrm{\mid }u{q}_n-v{p}_n\mathrm{\mid }<1$, and it is a nonnegative integer, hence $0$, so $V(a)=p_n\mathrm{/}{q}_n$ for every $n>v$, which is impossible because the determinant identity of [F2] gives $p_{n+1}\mathrm{/}{q}_{n+1}-p_n\mathrm{/}{q}_n=(p_{n+1}{q}_n-p_n{q}_{n+1})\mathrm{/}(q_n{q}_{n+1})=(-1{)}^n\mathrm{/}(q_n{q}_{n+1})\mathrm{\ne }0$.

For $n\ge -1$ and reals $s,t$ at which $M_n$ is defined, clearing denominators gives $M_n(s)-M_n(t)=(s-t)(p_n{q}_{n-1}-p_{n-1}{q}_n)\mathrm{/}((q_{n}s+q_{n-1})(q_{n}t+q_{n-1}))$, and by [F2] the determinant $p_n{q}_{n-1}-p_{n-1}{q}_n$ equals $(-1{)}^{n-1}$ for $n\ge 0$ while for $n=-1$ the initial values give $p_{-1}{q}_{-2}-p_{-2}{q}_{-1}=1\cdot 1-0\cdot 0=1$; in every case it is $1$ or $-1$, and the denominators are positive by step 1.2, so $M_n$ is strictly monotone on its domain, strictly increasing when that determinant is $1$ and strictly decreasing when it is $-1$.

Fix $a\in \mathcal{A}$ and $n\ge 0$; the recurrences of [F2] give $M_{n-1}(a_n)=(p_{n-1}{a}_n+p_{n-2})\mathrm{/}(q_{n-1}{a}_n+q_{n-2})=p_n\mathrm{/}{q}_n$ and $M_{n-1}(a_n+1)=(p_{n-1}(a_n+1)+p_{n-2})\mathrm{/}(q_{n-1}(a_n+1)+q_{n-2})=(p_n+p_{n-1})\mathrm{/}(q_n+q_{n-1})$, both arguments lying in the domain found in step 1.2 because $a_n\ge 1$ when $n\ge 1$ while $M_{-1}$ is defined on all of $\mathbb{R}$, so by [F2] the interval $J(a_0,\dots ,a_n)$ is the closed interval with endpoints $M_{n-1}(a_n)$ and $M_{n-1}(a_n+1)$, both of which are rational by [F2]; the interval is nondegenerate because $\mathrm{diam}J(a_0,\dots ,a_n)=1\mathrm{/}(q_n(q_n+q_{n-1}))>0$, and $M_{n-1}$ depends only on $a_0,\dots ,a_{n-1}$, since $p_{n-1},p_{n-2},q_{n-1},q_{n-2}$ do.

Let $x\in \mathbb{R}\setminus \mathbb{Q}$ and define $x_0:=x$, $a_n:=\lfloor x_n\rfloor$ and $x_{n+1}:=1\mathrm{/}(x_n-a_n)$; by induction on $n$ every $x_n$ is irrational and $x_n>1$ for $n\ge 1$, since given $x_n$ irrational [F3] supplies the unique integer $a_n$ with $a_n\le x_n<a_n+1$, so that $0\le x_n-a_n<1$ with $x_n-a_n\mathrm{\ne }0$ because $x_n$ is irrational while $a_n$ is an integer, whence $0<x_n-a_n<1$ and $x_{n+1}=1\mathrm{/}(x_n-a_n)>1$, and $x_{n+1}$ is irrational because a rational nonzero $x_{n+1}$ would make $x_n=a_n+1\mathrm{/}{x}_{n+1}$ rational; consequently $a_n=\lfloor x_n\rfloor \ge 1$ for $n\ge 1$, the recursion never divides by zero, and $\mathrm{\Phi }(x):=(a_0,a_1,\dots )$ is a member of the set $\mathcal{A}$ of step 1.1.

Let $S$ be open in $\mathbb{R}\setminus \mathbb{Q}$, so $S=T\cap (\mathbb{R}\setminus \mathbb{Q})$ for some $T$ open in $\mathbb{R}$ by [F7], and let $a\in \mathcal{A}$ satisfy $V(a)\in S$; claim 3 of [F8] gives a real $r>0$ with $(V(a)-r,V(a)+r)\subseteq T$, and the diameters $\mathrm{diam}J(a_0,\dots ,a_n)$ tend to $0$ by [F2], so some $n$ has $\mathrm{diam}J(a_0,\dots ,a_n)<r$; for every $b\in C(a_0,\dots ,a_n)$ the interval $J(b_0,\dots ,b_n)$ equals $J(a_0,\dots ,a_n)$, because [F2] computes it from the prefix alone, and both $V(b)$ and $V(a)$ lie in it by step 1.3, so $\mathrm{\mid }V(b)-V(a)\mathrm{\mid }<r$ and $V(b)\in T\cap (\mathbb{R}\setminus \mathbb{Q})=S$; hence every point of the preimage lies in a cylinder contained in it, so that preimage is the union of the cylinders it contains and is open by step 1.1, and the coding map is continuous.

Fix $a\in \mathcal{A}$ and $n\ge 0$; subtracting and using the determinant identity of [F2] gives $M_n(t)-p_n\mathrm{/}{q}_n=(q_n(p_{n}t+p_{n-1})-p_n(q_{n}t+q_{n-1}))\mathrm{/}(q_n(q_{n}t+q_{n-1}))=-(p_n{q}_{n-1}-p_{n-1}{q}_n)\mathrm{/}(q_n(q_{n}t+q_{n-1}))=(-1{)}^n\mathrm{/}(q_n(q_{n}t+q_{n-1}))$ for every real $t\ge 1$, and at $t=1$ the value $M_n(1)=(p_n+p_{n-1})\mathrm{/}(q_n+q_{n-1})$ is the endpoint of $J(a_0,\dots ,a_n)$ other than $p_n\mathrm{/}{q}_n$ identified in step 2.2; for $t>1$ the two differences $M_n(t)-p_n\mathrm{/}{q}_n$ and $M_n(1)-p_n\mathrm{/}{q}_n$ therefore have the same sign $(-1{)}^n$ and satisfy $\mathrm{\mid }{M}_n(t)-p_n\mathrm{/}{q}_n\mathrm{\mid }<\mathrm{\mid }{M}_n(1)-p_n\mathrm{/}{q}_n\mathrm{\mid }$, because $q_{n}t+q_{n-1}>q_n+q_{n-1}>0$, so $M_n(t)$ lies strictly between the two endpoints of $J(a_0,\dots ,a_n)$ and is neither of them.

Let $x\in \mathbb{R}\setminus \mathbb{Q}$, let $(a_n)$ and $(x_n)$ be as in step 2.3, and let $p_n$ and $q_n$ be the convergent numerators and denominators of the code $\mathrm{\Phi }(x)$; then $x=M_n(x_{n+1})$ for every $n\in \mathbb{N}$, by induction on $n$: at $n=0$ the values $p_0=a_0$, $q_0=1$, $p_{-1}=1$ and $q_{-1}=0$ of [F2] give $M_0(x_1)=(a_0{x}_1+1)\mathrm{/}{x}_1=a_0+1\mathrm{/}{x}_1=a_0+(x_0-a_0)=x$ because $x_1=1\mathrm{/}(x_0-a_0)$, and for the inductive step $x_{n+1}=a_{n+1}+1\mathrm{/}{x}_{n+2}$ with $x_{n+2}>1$, so multiplying numerator and denominator by $x_{n+2}$ gives $M_n(x_{n+1})=((a_{n+1}{p}_n+p_{n-1})x_{n+2}+p_n)\mathrm{/}((a_{n+1}{q}_n+q_{n-1})x_{n+2}+q_n)=(p_{n+1}{x}_{n+2}+p_n)\mathrm{/}(q_{n+1}{x}_{n+2}+q_n)=M_{n+1}(x_{n+2})$ by the recurrences of [F2]; every denominator here is nonzero, since $x_{n+1}>1$ and $x_{n+2}>1$ lie in the domains found in step 1.2.

Let $a,b\in \mathcal{A}$ with $a\mathrm{\ne }b$ and let $m$ be the least index with $a_m\mathrm{\ne }{b}_m$; the quantities $p_{m-1},p_{m-2},q_{m-1},q_{m-2}$ are computed from the common initial segment $a_0,\dots ,a_{m-1}$ alone, which is empty when $m=0$, the four quantities then being the initial values of [F2], so by step 2.2 the codes $a$ and $b$ determine the same function $M:=M_{m-1}$, and $J(a_0,\dots ,a_m)$ is the closed interval with endpoints $M(a_m)$ and $M(a_m+1)$ while $J(b_0,\dots ,b_m)$ is the closed interval with endpoints $M(b_m)$ and $M(b_m+1)$; assume $a_m<b_m$, which costs nothing by symmetry, so that $a_m+1\le b_m$ as these are integers, and all four arguments lie in the domain of $M$; if $M$ is strictly increasing there, which is one of the two cases of step 2.1, then $J(a_0,\dots ,a_m)=[M(a_m),M(a_m+1)]$ and $J(b_0,\dots ,b_m)=[M(b_m),M(b_m+1)]$ with $M(a_m+1)\le M(b_m)$, so the two meet in $\mathrm{\varnothing }$ when $M(a_m+1)<M(b_m)$ and in $\{M(b_m)\}$ when $M(a_m+1)=M(b_m)$, while if $M$ is strictly decreasing the same computation with the endpoints exchanged gives $J(a_0,\dots ,a_m)=[M(a_m+1),M(a_m)]$ and $J(b_0,\dots ,b_m)=[M(b_m+1),M(b_m)]$ with $M(b_m)\le M(a_m+1)$, so the two meet in at most the single point $M(a_m+1)$; in both cases $J(a_0,\dots ,a_m)\cap J(b_0,\dots ,b_m)$ has at most one point, and any such point is an endpoint of both intervals and so is rational by step 2.2.

Let $x\in \mathbb{R}\setminus \mathbb{Q}$ and let $a:=\mathrm{\Phi }(x)$ be the code produced in step 2.3; for every $n\in \mathbb{N}$ the tail identity of step 3.2 gives $x=M_n(x_{n+1})$ with $x_{n+1}>1$, so step 3.1 places $x$ strictly between the two endpoints of $J(a_0,\dots ,a_n)$ and in particular inside it, whence $x\in {\bigcap }_{n\in \mathbb{N}}J(a_0,\dots ,a_n)$, which by step 1.3 is the single point $V(a)$; therefore $V(\mathrm{\Phi }(x))=x$, and $V$ maps $\mathcal{A}$ onto $\mathbb{R}\setminus \mathbb{Q}$.

Let $a,b\in \mathcal{A}$ with $a\mathrm{\ne }b$ and let $m$ be least with $a_m\mathrm{\ne }{b}_m$; by step 1.3 the values $V(a)$ and $V(b)$ are irrational with $V(a)\in J(a_0,\dots ,a_m)$ and $V(b)\in J(b_0,\dots ,b_m)$, so if $V(a)=V(b)$ then that common value would lie in $J(a_0,\dots ,a_m)\cap J(b_0,\dots ,b_m)$ and hence be rational by step 3.3, a contradiction; therefore $V(a)\mathrm{\ne }V(b)$ and $V$ is injective.

By step 1.3 the map $V$ sends every member of $\mathcal{A}$ to an irrational real, it is surjective onto $\mathbb{R}\setminus \mathbb{Q}$ by step 4.1 and injective by step 4.2, so $V:\mathcal{A}\to \mathbb{R}\setminus \mathbb{Q}$ is a bijection, its inverse sends an irrational $x$ to the code $\mathrm{\Phi }(x)$ of step 2.3, and composing with the coordinatewise bijection of step 1.1 presents it as a bijection defined on ${\mathbb{N}}^{\mathbb{N}}$.

Let $C(a_0,\dots ,a_n)$ be a cylinder and let $x\in V[C(a_0,\dots ,a_n)]$, say $x=V(c)$ with $c\in C(a_0,\dots ,a_n)$, so that $J(c_0,\dots ,c_n)=J(a_0,\dots ,a_n)$ is a closed interval $[\alpha ,\beta ]$ with $\alpha <\beta$ rational by step 2.2 and $x\in [\alpha ,\beta ]$ irrational by step 1.3, giving $\alpha <x<\beta$, and by [F4] there are rationals $u,v$ with $\alpha <u<x<v<\beta$; if $x^{\mathrm{\prime }}$ is irrational with $u<x^{\mathrm{\prime }}<v$ then $x^{\mathrm{\prime }}=V(b)$ for exactly one $b\in \mathcal{A}$ by step 5.1 and $x^{\mathrm{\prime }}\in J(a_0,\dots ,a_n)$, and were $b_k\mathrm{\ne }{a}_k$ for some $k\le n$ then, taking $m$ to be the least such index, so that $m\le n$, the nesting of [F2] would put $x^{\mathrm{\prime }}$ in $J(a_0,\dots ,a_m)\cap J(b_0,\dots ,b_m)$, which by step 3.3 has at most one point and that point is rational, contradicting irrationality of $x^{\mathrm{\prime }}$, so $b\in C(a_0,\dots ,a_n)$ and $x^{\mathrm{\prime }}\in V[C(a_0,\dots ,a_n)]$; consequently the set $T:=\bigcup \{(s,t):s<t\text{ rational and }(s,t)\cap (\mathbb{R}\setminus \mathbb{Q})\subseteq V[C(a_0,\dots ,a_n)]\}$ is a union of open intervals and so is open in $\mathbb{R}$ by [F8], and $T\cap (\mathbb{R}\setminus \mathbb{Q})=V[C(a_0,\dots ,a_n)]$, the inclusion from left to right being the defining condition of the union and the reverse holding because the interval $(u,v)$ just produced for the arbitrary point $x$ of the image is one of the united intervals; hence that image is open in $\mathbb{R}\setminus \mathbb{Q}$ by [F7], and since every open subset of $\mathcal{A}$ is a union of cylinders by step 1.1 and the image of a union is the union of the images, $V$ maps open sets to open sets, which for the bijection of step 5.1 says exactly that its inverse is continuous.

The coding map is a bijection onto $\mathbb{R}\setminus \mathbb{Q}$ by step 5.1, it is continuous by step 2.4, and its inverse is continuous by step 6.1, which is the assertion.