Read-back: unique normal form for a non-identity element of $F$

This is a single statement. Everything it rests on is imported, so the account below unfolds all of it; nothing needs to be looked up elsewhere.

1. The ambient group

Let $I = [0,1] \subseteq \mathbb{R}$, carrying the order inherited from $\mathbb{R}$.

By an order automorphism of $I$ we mean a bijection $g : I \to I$ such that for all $x, y \in I$,

$$x \le y \iff g(x) \le g(y).$$

Such a $g$ is strictly increasing, and since $0$ and $1$ are the least and greatest elements of $I$, necessarily $g(0) = 0$ and $g(1) = 1$.

Write $\mathrm{Aut}(I)$ for the set of order automorphisms of $I$. It is a group under the following operations, and this is the group structure used throughout:

• the product $g \cdot h$ is the composite $x \mapsto g(h(x))$ — the left factor is applied last;
• the identity element $1$ is the identity map of $I$;
• $g^{-1}$ is the inverse bijection;
• for $n \in \mathbb{N}$, $g^{n}$ is the $n$-fold composite of $g$ with itself, with $g^{0} = 1$.

The product is associative, so unparenthesised products below are unambiguous; but the composition order above matters and is fixed as stated.

2. The subgroup $F$

Call a real number dyadic if it can be written as $m / 2^{k}$ with $m \in \mathbb{Z}$ and $k \in \mathbb{N}$.

Call $g \in \mathrm{Aut}(I)$ admissible if there exists a finite set $S \subseteq \mathbb{R}$ such that

1. every element of $S$ is dyadic, and
2. for all $x, y \in I$ with $x < y$ and $(x,y) \cap S = \emptyset$ (the open interval $(x,y)$ meets $S$ nowhere), there exist $n \in \mathbb{Z}$ and $c \in \mathbb{R}$ with

$$g(z) = 2^{n} z + c \quad \text{for every } z \in I \text{ with } x \le z \le y$$

(the closed interval, so the endpoints $x$ and $y$ are included even though they are themselves allowed to lie in $S$).

Note the exact shape of this: $S$ is a finite set of real numbers, not required to be nonempty and not required to lie inside $[0,1]$; the exponent $n$ ranges over all of $\mathbb{Z}$, so the slope $2^{n}$ is an arbitrary positive integer power of $2$ (positive or negative exponent); and $n$ and $c$ may depend on the pair $(x,y)$.

$F$ is defined to be the smallest subgroup of $\mathrm{Aut}(I)$ containing every admissible map: equivalently, $g \in F$ if and only if $g$ lies in every subgroup of $\mathrm{Aut}(I)$ that contains all admissible maps; equivalently again, $g$ is a finite product of admissible maps and inverses of admissible maps.

So the hypothesis "$f \in F$" is exactly membership in that generated subgroup. It is not the hypothesis that $f$ is itself admissible, although every admissible map does lie in $F$.

3. Two specific elements

Two order automorphisms of $I$ are singled out. Each is the restriction to $I$ of an explicitly given increasing bijection of $\mathbb{R}$ which is the identity outside a bounded region; the restriction is well defined because each of those bijections carries $[0,1]$ onto $[0,1]$.

$$A(x) = \begin{cases} x/2, & 0 \le x \le \tfrac12, \\[2pt] x - \tfrac14, & \tfrac12 \le x \le \tfrac34, \\[2pt] 2x - 1, & \tfrac34 \le x \le 1. \end{cases}$$ $$B(x) = \begin{cases} x, & 0 \le x \le \tfrac12, \\[2pt] x/2 + \tfrac14, & \tfrac12 \le x \le \tfrac34, \\[2pt] x - \tfrac18, & \tfrac34 \le x \le \tfrac78, \\[2pt] 2x - 1, & \tfrac78 \le x \le 1. \end{cases}$$

(The clauses agree at the shared endpoints, so both formulas are consistent; both maps fix $0$ and $1$, and $B$ fixes all of $[0,\tfrac12]$ pointwise.)

4. The generators $X_i$

An infinite family $X_0, X_1, X_2, \dots$ of elements of $\mathrm{Aut}(I)$ is defined by recursion on the index:

$$X_0 = A, \qquad X_{n+1} = (A^{n})^{-1} \, B \, A^{n} \quad (n \in \mathbb{N}).$$

With the composition convention of §1, $X_{n+1}$ is the map $x \mapsto A^{-n}\bigl(B(A^{n}(x))\bigr)$, i.e. the conjugate of $B$ by $A^{n}$ in that order. In particular $X_1 = (A^{0})^{-1} B A^{0} = B$.

5. Words

For a finite list $c = (c_0, c_1, \dots, c_{m-1})$ of natural numbers (so each $c_i \ge 0$; the list may be empty and its length $m$ is arbitrary), define

$$w(c) = X_0^{\,c_0} \cdot X_1^{\,c_1} \cdots X_{m-1}^{\,c_{m-1}} \in \mathrm{Aut}(I),$$

with $w(\text{empty list}) = 1$.

Precisely: the defining recursion walks the list from its head, starting at generator index $0$ and advancing the index by exactly one at each step, so the $i$-th entry of the list (counting from $0$) is the exponent of $X_i$, and the recursion terminates with the identity element once the list is exhausted. The factors appear in the displayed order, and by §1 the leftmost factor $X_0^{c_0}$ is the one applied last. Exponents are natural numbers, so no factor $X_i^{-1}$ can occur inside $w(c)$; an entry $c_i = 0$ contributes the identity factor.

6. The normal-form condition on a pair of lists

Let $a = (a_0,\dots,a_{m-1})$ and $b = (b_0,\dots,b_{m'-1})$ be finite lists of natural numbers. Throughout, an out-of-range lookup returns $0$: the value at index $k$ of a list is its $k$-th entry when $k$ is less than the length, and is $0$ for every $k$ at or beyond the length.

Say the pair $(a,b)$ is normal-form data when all four of the following hold.

1. $a$ is not the empty list.
2. $a$ and $b$ have the same length. (With (1), both lists are nonempty; write $m \ge 1$ for this common length.)
3. Looking at the final index $m-1$ of both lists, exactly one of the two entries there is zero and the other is strictly positive; formally, either

$$a_{m-1} = 0 \ \text{ and } \ b_{m-1} > 0, \qquad \text{or} \qquad a_{m-1} > 0 \ \text{ and } \ b_{m-1} = 0.$$

(Because of (1) the index $m-1$ is a genuine last position, and by (2) it is the last position of $b$ as well, so no out-of-range lookup occurs here.) 4. For every index $k$ with $k + 1 < m$ — that is, every position other than the last — if $a_k > 0$ and $b_k > 0$, then $a_{k+1} > 0$ or $b_{k+1} > 0$.

Condition (4) imposes nothing at the last position and nothing beyond the end of the lists. Condition (3) is a disjunction of two conjunctions and, as written, is exclusive in effect: it cannot hold with both last entries zero, nor with both strictly positive.

These conditions are satisfiable. For instance $a = (0)$, $b = (1)$ is normal-form data, with (4) vacuous; for this pair $w(b) \cdot w(a)^{-1} = X_0^{1} \cdot (X_0^{0})^{-1} = A$.

7. The statement

Let $f$ be an order automorphism of $I = [0,1]$. Assume $f \in F$ and $f \ne 1$ (that is, $f$ is not the identity map of $[0,1]$). Then there is exactly one ordered pair $(a,b)$ of finite lists of natural numbers such that

$$(a,b) \text{ is normal-form data, and } \quad f = w(b) \cdot w(a)^{-1}.$$

The variable $f$ is quantified universally over all order automorphisms of $[0,1]$ (it is an implicit argument of the statement, but a genuine universally quantified variable). The two hypotheses are $f \in F$ and $f \ne 1$. There are no further hypotheses and no side conditions of any other kind.

Some points of precision about the conclusion.

• The quantifier is $\exists!$, not $\exists$. Unfolded, it asserts: there exists a pair $(a,b)$ satisfying both displayed conditions, and every pair $(a',b')$ satisfying both displayed conditions equals $(a,b)$.
• Uniqueness is uniqueness of the pair. Since two pairs are equal exactly when their corresponding components are equal, this does determine both lists individually — both $a$ and $b$ are uniquely determined by $f$. But it says so only jointly, among pairs satisfying both conditions at once.
• The pair ranges over all ordered pairs of finite lists of natural numbers, with no a priori bound on length or on entries; the normal-form condition of §6 is what cuts this down, and it is part of the property being asserted-and-uniquely-determined, not a restriction on the range of the quantifier.
• The order of the two components is as displayed: the second component $b$ supplies the word on the left of the product, and the first component $a$ supplies the word that is inverted on the right. With the composition convention of §1, $f$ is the map $x \mapsto w(b)\bigl(w(a)^{-1}(x)\bigr)$: first undo $w(a)$, then apply $w(b)$.
• Nothing is asserted about $f = 1$; that case is excluded by hypothesis.

The hypotheses are jointly satisfiable, so the statement is not vacuous: $A$ from §3 is admissible (take $S = \{\tfrac12, \tfrac34\}$) and hence lies in $F$, and $A \ne 1$ because $A(\tfrac12) = \tfrac14$.