Read-back: what the statement asserts

0. The assertion in one sentence

Inside the group $G$ of all order isomorphisms of the closed unit interval $[0,1]$, the subgroup generated by a certain explicit two-element set $\{A,B\}$ of such isomorphisms is equal to the subgroup generated by the set of all order isomorphisms of $[0,1]$ satisfying an explicit "piecewise-affine with dyadic breakpoints and power-of-two slopes" condition.

It is a closed statement: it has no universally or existentially quantified variables of its own, no hypotheses, no explicit or implicit arguments, and no typeclass assumptions beyond those already baked into the fixed objects it names. Everything below is the unfolding of those fixed objects.

Sections 1–4 expand every ingredient; section 5 states the assertion; section 6 collects the degenerate and edge cases that the definitions silently admit.

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1. The ambient group $G$

Write

$$I \;=\; [0,1] \;=\; \{t \in \mathbb{R} \;:\; 0 \le t \le 1\},$$

regarded as a set of real numbers and ordered by the order it inherits from $\mathbb{R}$: for $a, b \in I$, $a \le b$ means exactly $a \le b$ as real numbers.

An order isomorphism of $I$ is a bijection $f : I \to I$ such that for all $a, b \in I$

$$a \le b \quad\Longleftrightarrow\quad f(a) \le f(b).$$

Equivalently: an increasing bijection of $[0,1]$ onto itself. Note carefully what is not required: no continuity, no differentiability, no measurability, no piecewise structure. The ambient object is a purely order-theoretic one; the order-isomorphism condition is an "if and only if", i.e. it says $f$ reflects as well as preserves $\le$, which for a bijection is the same as saying $f$ and $f^{-1}$ are both increasing.

$G$ denotes the set of all order isomorphisms of $I$, made into a group by

• multiplication $=$ composition, in the order $(fg)(x) = f\bigl(g(x)\bigr)$ — the right-hand factor is applied first;
• identity element $=$ the identity map of $I$;
• inverse $=$ the inverse function.

Two facts about $G$ that follow from the definitions alone and are used silently throughout: every $f \in G$ satisfies $f(0) = 0$ and $f(1) = 1$ (an increasing bijection of $[0,1]$ must fix the least and greatest elements), and $G$ is not required to consist of piecewise-linear maps — it is the full order-automorphism group of the interval.

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2. The two distinguished elements $A$ and $B$

Both are specific, fully explicit elements of $G$. Each is obtained by writing down an increasing bijection of the whole real line that is the identity outside $[0,1]$, and then restricting it to $[0,1]$; the restriction is legitimate because each of those maps fixes every point of $(-\infty,0]$ and every point of $[1,\infty)$, and in particular fixes $0$ and $1$. The resulting elements of $G$ are given on $[0,1]$ by the following formulas.

The map $A$.

$$A(x) \;=\; \begin{cases} \tfrac{1}{2}x, & 0 \le x \le \tfrac12,\\[2pt] x - \tfrac14, & \tfrac12 \le x \le \tfrac34,\\[2pt] 2x - 1, & \tfrac34 \le x \le 1. \end{cases}$$

The three formulas agree at the overlaps ($A(\tfrac12) = \tfrac14$, $A(\tfrac34) = \tfrac12$), and $A(0) = 0$, $A(1) = 1$. So $A$ is affine on each of $[0,\tfrac12]$, $[\tfrac12,\tfrac34]$, $[\tfrac34,1]$, with slopes $2^{-1}, 2^{0}, 2^{1}$ respectively and breakpoints $\tfrac12$ and $\tfrac34$.

The map $B$.

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

Again the formulas agree at the overlaps ($B(\tfrac12) = \tfrac12$, $B(\tfrac34) = \tfrac58$, $B(\tfrac78) = \tfrac34$), and $B(0) = 0$, $B(1) = 1$. So $B$ is the identity on $[0,\tfrac12]$ and is affine on each of $[\tfrac12,\tfrac34]$, $[\tfrac34,\tfrac78]$, $[\tfrac78,1]$, with slopes $2^{-1}, 2^{0}, 2^{1}$ and breakpoints $\tfrac12, \tfrac34, \tfrac78$.

$A$ and $B$ are distinct elements of $G$ (for instance $A(\tfrac14) = \tfrac18$ while $B(\tfrac14) = \tfrac14$), so $\{A,B\}$ is a genuine two-element subset of $G$.

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3. The dyadic-piecewise-affine condition $T$

Call a real number $t$ dyadic if

$$\exists\, m \in \mathbb{Z},\ \exists\, k \in \mathbb{N} \quad\text{with}\quad t = \frac{m}{2^{k}} .$$

Here $m$ ranges over all integers (positive, negative and zero) and $k$ over the natural numbers including $0$; so every integer is dyadic, as are $0$ and all negative dyadic rationals. The power $2^k$ is the ordinary $k$-fold product of the real number $2$.

For $f \in G$, say that $f$ satisfies condition $T$ when:

there exists a finite set $\mathcal{B} \subseteq \mathbb{R}$ such that

1. every element of $\mathcal{B}$ is dyadic, and
2. for all $x, y \in [0,1]$ with $x < y$ and with

$$(x,y) \cap \mathcal{B} = \varnothing$$

(the open interval $(x,y)$ containing no element of $\mathcal{B}$), there exist an integer $n \in \mathbb{Z}$ and a real number $c \in \mathbb{R}$ such that

$$f(z) \;=\; 2^{n} z + c \qquad\text{for every } z \in [x,y]$$

(the closed interval $[x,y]$).

Points of detail that the phrasing fixes, and which change the meaning if altered:

• The quantifier order is: $\mathcal{B}$ first (one finite set works for all pairs), then $x,y$, then $n$ and $c$. So $n$ and $c$ may depend on $x$ and $y$, but $\mathcal{B}$ may not.
• The hypothesis on $x,y$ concerns the open interval $(x,y)$, while the affine formula is demanded on the closed interval $[x,y]$. Consequently the formula is pinned down at the endpoints $x$ and $y$ as well, even when those endpoints themselves lie in $\mathcal{B}$.
• The exponent $n$ is an arbitrary integer, so the admissible slopes are exactly the numbers $\ldots, \tfrac14, \tfrac12, 1, 2, 4, \ldots$ — all integer powers of $2$, every one of them strictly positive. Negative $n$ is read as $2^{n} = 1/2^{-n}$.
• The additive constant $c$ is an arbitrary real number. Nothing in the condition requires $c$ to be dyadic, rational, or in any way constrained.
• $x$ and $y$ are constrained to lie in $[0,1]$; $z$ likewise ranges over the points of $[0,1]$ lying in $[x,y]$, which (since $x,y \in [0,1]$) is all of $[x,y]$.
• No relation between $\mathcal{B}$ and the actual set of breakpoints of $f$ is asserted: $\mathcal{B}$ is merely some finite dyadic set outside which $f$ is affine in the above sense. It need not be minimal, and it need not be contained in $[0,1]$.

Condition $T$ is a property of a single map, not of a set of maps; it is not asserted anywhere that the collection of maps satisfying $T$ is closed under composition or under inverses.

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4. The two subgroups

For a subset $S \subseteq G$, the subgroup generated by $S$ is the smallest subgroup of $G$ containing $S$ — literally, the intersection of all subgroups of $G$ that contain $S$. Equivalently, it is the set of all elements expressible as a finite product

$$g_1^{\varepsilon_1} g_2^{\varepsilon_2} \cdots g_r^{\varepsilon_r}, \qquad r \ge 0,\quad g_i \in S,\quad \varepsilon_i \in \{+1,-1\},$$

with the empty product ($r = 0$) equal to the identity. In particular the generated subgroup always contains the identity, whatever $S$ is, and is generated by $S$ as a set — no ordering, multiplicity or finiteness condition on $S$ is involved.

The two subgroups in play are:

• $\langle A, B\rangle$, the subgroup of $G$ generated by the two-element set $\{A,B\}$ described in section 2;
• $F \;=\; \bigl\langle\, \{\,f \in G : f \text{ satisfies condition } T \,\}\,\bigr\rangle$, the subgroup of $G$ generated by the (in general infinite) set of all elements of $G$ satisfying condition $T$ of section 3.

Note that $F$ is defined as the subgroup generated by the $T$-maps, not as the set of $T$-maps.

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5. What the declaration asserts

$$\boxed{\;\langle A, B\rangle \;=\; F\;}$$

as subgroups of $G$. Equality of subgroups here is equality of the underlying subsets of $G$: the assertion is the conjunction of the two inclusions, i.e. that for every order isomorphism $f : [0,1] \to [0,1]$,

$$f \text{ is a finite product of } A^{\pm 1}, B^{\pm 1} \qquad\Longleftrightarrow\qquad f \text{ is a finite product of maps satisfying } T \text{ and their inverses.}$$

This is an equality (iff), not an inclusion in either direction.

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6. Degenerate cases and things the quantifiers silently admit

• No hypotheses at all. The declaration has no antecedent, so there is no question of a hypothesis being vacuous or impossible to satisfy; the statement is an unconditional equation between two specific subgroups.

• The empty breakpoint set is allowed in condition $T$. If $\mathcal{B} = \varnothing$, clause (1) holds trivially and clause (2) applies to every pair $x < y$ in $[0,1]$; taking $x=0,y=1$ it then demands a single formula $f(z) = 2^n z + c$ valid on all of $[0,1]$. So the condition is satisfiable with $\mathcal{B} = \varnothing$ — the identity map does so, with $n = 0$, $c = 0$ — and the generating set of $F$ is in particular non-empty.

• $\mathcal{B}$ may contain points outside $[0,1]$. The condition on $\mathcal{B}$ is only that it be finite and consist of dyadic reals. Since $x,y$ are confined to $[0,1]$, only the points of $\mathcal{B}$ lying in $(0,1)$ can ever make the hypothesis $(x,y) \cap \mathcal{B} = \varnothing$ fail.

• $\mathcal{B}$ need not be a subset of the moved set, nor related to $f$ at all beyond clause (2); e.g. adding extra dyadic points to a working $\mathcal{B}$ never destroys condition $T$, since enlarging $\mathcal{B}$ only weakens the hypothesis in clause (2).

• Dyadicity of $\mathcal{B}$ is required; dyadicity of the intercepts is not. Nothing in condition $T$ says that the constants $c$, or the values $f(z)$, are dyadic.

• Slopes are integer powers of two, hence never zero or negative; but $n$ is unbounded in both directions, so arbitrarily large and arbitrarily small slopes are permitted, and a single $f$ may use different $n$ on different intervals.

• Degenerate pairs $x,y$ are excluded by $x < y$. The case $x = y$ is not covered by clause (2), so no constraint is imposed by single points on their own.

• Endpoint behaviour is automatic, not assumed. Every member of $G$ fixes $0$ and $1$ because it is an increasing bijection of $[0,1]$; this is not listed as part of condition $T$, and condition $T$ adds nothing about the endpoints beyond what clause (2) gives.

• No topology anywhere. Neither $G$, nor condition $T$, nor $A$ and $B$ involve continuity as a hypothesis. (Affinity on closed subintervals as in clause (2) is the only regularity condition present, and it is a consequence of the clause rather than an assumption about $f$.)

• The set $\{A,B\}$ is unordered and has exactly two elements ($A \neq B$), so the left-hand side is the subgroup generated by these two maps together with their inverses; no claim is made about either map individually.