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Infinite simple continued fractions parametrise the irrational real numbers

Statement

The continued-fraction coding determined by Simple continued fractions, convergents, and the integer-coordinate coding of $N^{N}$ gives a bijection from the sequences $(a_{0}, a_{1}, \dots)$ with $a_{0} \in Z$ and $a_{n} \geq 1$ for $n \geq 1$ onto $R ∖ Q$ . Both the coding map and its inverse are continuous for the cylinder and subspace topologies.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

Define a bijection $z : N \to Z$ by $z (2 k) = k$ and $z (2 k + 1) = - (k + 1)$ ; the division algorithm makes these two cases exhaustive (thm-division-algorithm-in-z). For $x \in N^{N}$ put $a_{0} = z (x_{0})$ and $a_{n} = x_{n} + 1$ for $n \geq 1$ . Its finite simple continued fractions $[a_{0}; \dots, a_{n}]$ are defined by recursion on the length, evaluated in $Q$ (def-rationals, def-rat-operations): $[a_{n}] : = a_{n}$ and $[a_{k}; a_{k + 1}, \dots, a_{n}] : = a_{k} + 1 / [a_{k + 1}; \dots, a_{n}]$ for $k < n$ . The recursion never divides by zero, because $a_{j} \geq 1$ for $j \geq 1$ makes every tail value $[a_{k + 1}; \dots, a_{n}]$ at least $1$ . A finite prefix determines the cylinder of all codes extending it. Infinite continued-fraction values are established, rather than assumed, in this item. (Simple continued fractions, convergents, and the integer-coordinate coding of $N^{N}$ ).

[F2]

Let $a_{0} \in Z$ and $a_{n} \in Z$ with $a_{n} \geq 1$ for $n \geq 1$ . With the initial values $p_{- 2} = 0$ , $p_{- 1} = 1$ , $q_{- 2} = 1$ , $q_{- 1} = 0$ and the recurrences $p_{n} = a_{n} p_{n - 1} + p_{n - 2}$ , $q_{n} = a_{n} q_{n - 1} + q_{n - 2}$ for $n \geq 0$ : $p_{0} = a_{0}$ and $q_{0} = 1$ ; the $q_{n}$ are positive for $n \geq 0$ and strictly increasing for $n \geq 1$ ; and $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1}$ for $n \geq 0$ . For a finite prefix $(a_{0}, \dots, a_{n})$ , $C (a_{0}, \dots, a_{n}) \subseteq N^{N}$ is the code cylinder of all codes extending that prefix, and $J (a_{0}, \dots, a_{n})$ is the closed real interval with endpoints $p_{n} / q_{n}$ and $(p_{n} + p_{n - 1}) / (q_{n} + q_{n - 1})$ . The intervals $J$ are nested as the prefix is extended, and $diam J (a_{0}, \dots, a_{n}) = 1 / (q_{n} (q_{n} + q_{n - 1}))$ , which tends to $0$ . A code cylinder and a real interval are different objects and the two are not identified. Both endpoints of $J (a_{0}, \dots, a_{n})$ are rational, being ratios of integers; whether an infinite code's value can equal such an endpoint is not settled there. (Continued-fraction convergents, determinant identities, and nested irrational cylinders).

[F3]

Identify $Z$ with its canonical copy inside $R$ . Then for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1,$ written $⌊ x ⌋$ and called the integer part, or floor, of $x$ . Existence is the Archimedean property (thm-of-archimedean) together with the well-ordering of $N$ (thm-well-ordering-principle); uniqueness is the discreteness of $Z$ , no integer lying strictly between $m$ and $m + 1$ . (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[F4]

The map $q \mapsto \hat{q}$ (def-real-numbers) is an embedding of ordered fields. Every real is approximated by rationals: for $x \in R$ and rational $ε > 0$ there is $q \in Q$ with $∣ x - \hat{q} ∣ < \hat{ε}$ . Consequently, strictly between any two reals lies a rational. (The rationals embed densely in the reals).

[F5]

For each $k \in N$ let $I_{k} = [c_{k}, d_{k}]$ be a closed bounded interval with $c_{k} \leq d_{k}$ (def-interval), and suppose the family is nested: $I_{k + 1} \subseteq I_{k}$ for every $k \in N$ . Write $ℓ_{k} = d_{k} - c_{k} \geq 0$ for the length of $I_{k}$ . Then: 1. $⋂_{k \in N} I_{k}$ is nonempty; more precisely, with $c = \sup {c_{k} : k \in N}$ and $d = \inf {d_{k} : k \in N}$ , both of which exist, one has $c \leq d$ and $⋂_{k \in N} I_{k} = [c, d]$ . 2. $⋂_{k \in N} I_{k}$ is a single point if and only if $ℓ_{k} \to 0$ . (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ).

[F6]

The Baire sequence space is $N : = N^{N}$ , the set of functions from $N$ to itself, with the product topology obtained by giving each copy of $N$ the discrete topology (def-product-topology, def-standard-topologies). For a finite sequence $s = (s_{0}, \dots, s_{k - 1})$ , its cylinder is $N_{s} : = {x \in N : x_{i} = s_{i} for i < k}$ . The empty sequence has cylinder $N$ , and these cylinders form a basis. (Baire sequence space $N^{N}$ and its cylinder topology).

[F7]

For $S \subseteq X$ the subspace topology on $S$ is $T_{S} : = {U \cap S : U \in T}$ , the family of traces on $S$ of the open sets of $X$ ; a subset of $S$ lying in $T_{S}$ is said to be open in $S$ . (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[F8]

For $d_{R} (x, y) : = ∣ x - y ∣$ : $d_{R}$ is a metric on $R$ , the open ball is the bounded open interval $B (x, r) = (x - r, x + r)$ , and consequently $U \subseteq R$ is open in the metric topology of $d_{R}$ exactly when for every $x \in U$ there is $r > 0$ with $(x - r, x + r) \subseteq U$ ; this topology is called the usual topology of $R$ . (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, claims 1 to 3).

Proof

technique · direct

1.1

Write $A$ for the set of sequences $a = (a_{0}, a_{1}, \dots)$ with $a_{0} \in Z$ and $a_{n} \geq 1$ for $n \geq 1$ , and for a finite prefix put $C (a_{0}, \dots, a_{n}) : = {b \in A : b_{k} = a_{k} for k \leq n}$ ; by [F1] the assignment $x \mapsto a$ with $a_{0} = z (x_{0})$ and $a_{n} = x_{n} + 1$ for $n \geq 1$ is a bijection of $N = N^{N}$ onto $A$ , since $z$ is a bijection of $N$ onto $Z$ and $m \mapsto m + 1$ is a bijection of $N$ 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 $1 \leq i < k$ , and every $C (a_{0}, \dots, a_{n})$ arises from exactly one $N_{s}$ in this way; give $A$ the topology transported along the bijection, so that the bijection is a homeomorphism and, the $N_{s}$ being a basis of $N$ by [F6] and their images being exactly the sets $C (a_{0}, \dots, a_{n})$ , those sets are a basis of $A$ , which is the cylinder topology of the Statement, while $R ∖ Q$ carries the subspace topology of [F7] inherited from $R$ .

givenF1F6F7

1.2

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

givenF2algebra

1.3

For $a \in A$ the intervals $J (a_{0}, \dots, a_{n})$ are closed and bounded, are nested as $n$ increases, and satisfy $diam J (a_{0}, \dots, a_{n}) = 1 / (q_{n} (q_{n} + q_{n - 1})) \to 0$ , 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 $n \in N$ ; moreover $V (a)$ is irrational, for suppose $V (a) = u / v$ with $u, v$ integers and $v \geq 1$ , and note first that the $q_{n}$ are positive for $n \geq 0$ and strictly increasing for $n \geq 1$ by [F2] and are integers, so $q_{1} \geq 1$ and $q_{n + 1} \geq q_{n} + 1$ give $q_{n} \geq n$ for $n \geq 1$ ; for $n \geq 1$ both $V (a)$ and the endpoint $p_{n} / q_{n}$ lie in $J (a_{0}, \dots, a_{n})$ , so $∣ V (a) - p_{n} / q_{n} ∣ \leq 1 / (q_{n} (q_{n} + q_{n - 1})) < 1 / q_{n}^{2}$ because $q_{n - 1} > 0$ , and hence $∣ u q_{n} - v p_{n} ∣ = v q_{n} ∣ V (a) - p_{n} / q_{n} ∣ < v / q_{n}$ ; taking $n > v$ makes $∣ u q_{n} - v p_{n} ∣ < 1$ , and it is a nonnegative integer, hence $0$ , so $V (a) = p_{n} / q_{n}$ for every $n > v$ , which is impossible because the determinant identity of [F2] gives $p_{n + 1} / q_{n + 1} - p_{n} / q_{n} = (p_{n + 1} q_{n} - p_{n} q_{n + 1}) / (q_{n} q_{n + 1}) = (- 1)^{n} / (q_{n} q_{n + 1}) \neq 0$ .

givenF2F5

2.1

For $n \geq - 1$ and reals $s, t$ at which $M_{n}$ is defined, clearing denominators gives $M_{n} (s) - M_{n} (t) = (s - t) (p_{n} q_{n - 1} - p_{n - 1} q_{n}) / ((q_{n} s + q_{n - 1}) (q_{n} t + q_{n - 1}))$ , and by [F2] the determinant $p_{n} q_{n - 1} - p_{n - 1} q_{n}$ equals $(- 1)^{n - 1}$ for $n \geq 0$ 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 $- 1$ .

step 1.2F2algebra

2.2

Fix $a \in A$ and $n \geq 0$ ; the recurrences of [F2] give $M_{n - 1} (a_{n}) = (p_{n - 1} a_{n} + p_{n - 2}) / (q_{n - 1} a_{n} + q_{n - 2}) = p_{n} / q_{n}$ and $M_{n - 1} (a_{n} + 1) = (p_{n - 1} (a_{n} + 1) + p_{n - 2}) / (q_{n - 1} (a_{n} + 1) + q_{n - 2}) = (p_{n} + p_{n - 1}) / (q_{n} + q_{n - 1})$ , both arguments lying in the domain found in step 1.2 because $a_{n} \geq 1$ when $n \geq 1$ while $M_{- 1}$ is defined on all of $R$ , 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 $diam J (a_{0}, \dots, a_{n}) = 1 / (q_{n} (q_{n} + q_{n - 1})) > 0$ , 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.

step 1.2F2algebra

2.3

Let $x \in R ∖ Q$ and define $x_{0} : = x$ , $a_{n} : = ⌊ x_{n} ⌋$ and $x_{n + 1} : = 1 / (x_{n} - a_{n})$ ; by induction on $n$ every $x_{n}$ is irrational and $x_{n} > 1$ for $n \geq 1$ , since given $x_{n}$ irrational [F3] supplies the unique integer $a_{n}$ with $a_{n} \leq x_{n} < a_{n} + 1$ , so that $0 \leq x_{n} - a_{n} < 1$ with $x_{n} - a_{n} \neq 0$ because $x_{n}$ is irrational while $a_{n}$ is an integer, whence $0 < x_{n} - a_{n} < 1$ and $x_{n + 1} = 1 / (x_{n} - a_{n}) > 1$ , and $x_{n + 1}$ is irrational because a rational nonzero $x_{n + 1}$ would make $x_{n} = a_{n} + 1 / x_{n + 1}$ rational; consequently $a_{n} = ⌊ x_{n} ⌋ \geq 1$ for $n \geq 1$ , the recursion never divides by zero, and $Φ (x) : = (a_{0}, a_{1}, \dots)$ is a member of the set $A$ of step 1.1.

step 1.1F3algebra

2.4

Let $S$ be open in $R ∖ Q$ , so $S = T \cap (R ∖ Q)$ for some $T$ open in $R$ by [F7], and let $a \in A$ 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 $diam J (a_{0}, \dots, a_{n})$ tend to $0$ by [F2], so some $n$ has $diam J (a_{0}, \dots, a_{n}) < r$ ; 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 $∣ V (b) - V (a) ∣ < r$ and $V (b) \in T \cap (R ∖ Q) = S$ ; 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.

step 1.1step 1.3F2F7F8

3.1

Fix $a \in A$ and $n \geq 0$ ; subtracting and using the determinant identity of [F2] gives $M_{n} (t) - p_{n} / q_{n} = (q_{n} (p_{n} t + p_{n - 1}) - p_{n} (q_{n} t + q_{n - 1})) / (q_{n} (q_{n} t + q_{n - 1})) = - (p_{n} q_{n - 1} - p_{n - 1} q_{n}) / (q_{n} (q_{n} t + q_{n - 1})) = (- 1)^{n} / (q_{n} (q_{n} t + q_{n - 1}))$ for every real $t \geq 1$ , and at $t = 1$ the value $M_{n} (1) = (p_{n} + p_{n - 1}) / (q_{n} + q_{n - 1})$ is the endpoint of $J (a_{0}, \dots, a_{n})$ other than $p_{n} / q_{n}$ identified in step 2.2; for $t > 1$ the two differences $M_{n} (t) - p_{n} / q_{n}$ and $M_{n} (1) - p_{n} / q_{n}$ therefore have the same sign $(- 1)^{n}$ and satisfy $∣ M_{n} (t) - p_{n} / q_{n} ∣ < ∣ M_{n} (1) - p_{n} / q_{n} ∣$ , 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.

step 1.2step 2.2F2algebra

3.2

Let $x \in R ∖ Q$ , 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 $Φ (x)$ ; then $x = M_{n} (x_{n + 1})$ for every $n \in N$ , 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 $M_{0} (x_{1}) = (a_{0} x_{1} + 1) / x_{1} = a_{0} + 1 / x_{1} = a_{0} + (x_{0} - a_{0}) = x$ because $x_{1} = 1 / (x_{0} - a_{0})$ , and for the inductive step $x_{n + 1} = a_{n + 1} + 1 / x_{n + 2}$ with $x_{n + 2} > 1$ , so multiplying numerator and denominator by $x_{n + 2}$ gives $M_{n} (x_{n + 1}) = ((a_{n + 1} p_{n} + p_{n - 1}) x_{n + 2} + p_{n}) / ((a_{n + 1} q_{n} + q_{n - 1}) x_{n + 2} + q_{n}) = (p_{n + 1} x_{n + 2} + p_{n}) / (q_{n + 1} x_{n + 2} + q_{n}) = M_{n + 1} (x_{n + 2})$ 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.

step 1.2step 2.3F2algebra

3.3

Let $a, b \in A$ with $a \neq b$ and let $m$ be the least index with $a_{m} \neq b_{m}$ ; 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 $a_{m} + 1 \leq b_{m}$ 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 $M (a_{m} + 1) \leq M (b_{m})$ , so the two meet in $\emptyset$ 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 $M (b_{m}) \leq M (a_{m} + 1)$ , 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.

step 2.1step 2.2

4.1

Let $x \in R ∖ Q$ and let $a : = Φ (x)$ be the code produced in step 2.3; for every $n \in N$ 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 $x \in ⋂_{n \in N} J (a_{0}, \dots, a_{n})$ , which by step 1.3 is the single point $V (a)$ ; therefore $V (Φ (x)) = x$ , and $V$ maps $A$ onto $R ∖ Q$ .

step 1.3step 2.3step 3.1step 3.2

4.2

Let $a, b \in A$ with $a \neq b$ and let $m$ be least with $a_{m} \neq b_{m}$ ; 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 $V (a) \neq V (b)$ and $V$ is injective.

step 1.3step 3.3

5.1

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

step 1.1step 1.3step 4.1step 4.2

6.1

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 $[α, β]$ with $α < β$ rational by step 2.2 and $x \in [α, β]$ irrational by step 1.3, giving $α < x < β$ , and by [F4] there are rationals $u, v$ with $α < u < x < v < β$ ; if $x^{'}$ is irrational with $u < x^{'} < v$ then $x^{'} = V (b)$ for exactly one $b \in A$ by step 5.1 and $x^{'} \in J (a_{0}, \dots, a_{n})$ , and were $b_{k} \neq a_{k}$ for some $k \leq n$ then, taking $m$ to be the least such index, so that $m \leq n$ , the nesting of [F2] would put $x^{'}$ 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 $x^{'}$ , so $b \in C (a_{0}, \dots, a_{n})$ and $x^{'} \in V [C (a_{0}, \dots, a_{n})]$ ; consequently the set $T : = ⋃ {(s, t) : s < t rational and (s, t) \cap (R ∖ Q) \subseteq V [C (a_{0}, \dots, a_{n})]}$ is a union of open intervals and so is open in $R$ by [F8], and $T \cap (R ∖ Q) = V [C (a_{0}, \dots, a_{n})]$ , 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 $R ∖ Q$ by [F7], and since every open subset of $A$ 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.

step 1.1step 1.3step 2.2step 3.3step 5.1F2F4F7F8

7.1

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

step 2.4step 5.1step 6.1∎

Remarks

The tail identity is what makes the algorithm invert the coding. Running floors and reciprocals on an irrational $x$ produces a code, but nothing in that recipe by itself says the code's value is $x$ again. The identity $x = M_{n} (x_{n + 1})$ of step 3.2 says it: the whole of $x$ , not merely an approximation to it, is recovered from the first $n + 1$ partial quotients together with the exact remainder $x_{n + 1}$ , and since $x_{n + 1} > 1$ the value sits strictly inside the $n$ -th prefix interval. The intersection of those intervals is a single point, so it is $x$ .
Irrationality is used twice, and for different purposes. It is what makes the algorithm run forever, since a remainder equal to its own integer part would stop it; and it is what makes prefixes separate, since two prefix intervals of the same length belonging to different codes can share only a rational endpoint. The second use is what gives injectivity and the continuity of the inverse at once.
Where the parametrisation fails for rationals. Nothing above extends to a rational target: the algorithm terminates, and the coding map is onto the irrationals only. That is the reason the companion identification is with $R ∖ Q$ and not with $R$ , and the reason $Q$ cannot be homeomorphic to $N^{N}$ by this route.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 127 results over 31 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

David Marker, Descriptive Set Theory, §§1–2 (standard reference, not scraped)
Michael Kunzinger, General Topology, §§11.3–11.4 (standard reference, not scraped)
MFF General Topology course summary, §4.3 (standard reference, not scraped)
Jesse Peterson, Real Analysis, §§3.6–3.7 (standard reference, not scraped)