3.1 The proof of Theorem 2.2

For any arbitrary positive integer m, define

$$ \begin{align*} \mathcal{L}_{m} & =\bigg[ 0, \frac{1}{3^{m}}\bigg], \quad \mathcal{M}_{m}=\bigg[ \frac{1}{3^{m}},1-\frac{1}{3^{m}}\bigg], \quad \mathcal{R}_{m}=\bigg[ 1-\frac{1}{3^{m}},1\bigg], \quad \mathcal{M}_{m}^{\circ } =\bigg( \frac{1}{3^{m}},1-\frac{1}{3^{m}}\bigg), \\ & \quad\mathcal{G}_{m} =\bigg\{ g_{m}\in \mathrm{C}( [ 0,1] ) :g_{m}( x) =\bigg\{ \begin{array}{@{}ll} 3^{m}x & \text{if } x\in \mathcal{L}_{m} \\ 3^{m}x-( 3^{m}-1) & \text{if } x\in \mathcal{R}_{m} \end{array} \bigg\}. \end{align*} $$

For any arbitrary positive integer n, partition $\mathcal {M}_{m}^{\circ }$ into $2^{n}$ mutually disjoint intervals of length $\delta _{m}^{n}={2^{-n}}\mu ( \mathcal {M}_{m}^{\circ }) $ with each interval defined as

$$ \begin{align*}M^n_{m,j}=\bigg( \frac{1}{3^{m}}+({\kern1.5pt}j- 1) \delta _{m}^{n},\frac{1}{3^{m}}+j \delta _{m}^{n}\bigg), \quad \mbox{for } 1\leq j\leq 2^{n}.\end{align*} $$

3.2 The construction

The construction is obtained through the following three steps. In step 1, divide the unit interval into three exhaustive segments $\mathcal {L}_{m}$ , $\mathcal {M}_{m}$ and $\mathcal {R}_{m}$ as defined above. We first construct $2^n$ mutually disjoint sets in $\mathcal {L}_{m}$ which are extremally scrambled sets for any function in $\mathcal {G}_{m}$ . Then, in step 2, for the middle interval $\mathcal {M}_{m}$ , we construct $2^n$ mutually disjoint Smith–Volterra–Cantor (SVC) sets each of which is contained in each of the $2^n$ intervals $M^n_{m,j}$ ( $1\leq j\leq 2^n$ ). Finally, in step 3, we construct a homeomorphism from the $2^n$ SVC sets to the $2^n$ extremally scrambled sets. Extending the homeomorphism to the unit interval gives the desired continuous map $f_{m,n}$ .

Step 1. Denote by $\Omega $ the space of infinite binary sequences, which is an ordered topological space when equipped with the lexicographic ordering and topologised with the topology of pointwise convergence. Consider the $2^{n+1}$ binary $( n+1) $ -tuples

$$ \begin{align*}\varepsilon _{j} =( e_{1}^{j},\ldots ,e_{n+1}^{j}) \quad\text{for } 1\leq j\leq 2^{n+1}, \text{ where } e_{k}^{j} \in \{ 0,1\}.\end{align*} $$

Define a binary relation $0\prec 1$ and order the $2^{n+1}$ binary $( n+1) $ -tuples lexicographically so that $\varepsilon _{1} \prec \varepsilon _{2}\prec \cdots \prec \varepsilon _{2^{n+1}}. $ Define

$$ \begin{align*} \Omega _{\varepsilon _{j}}=\{ ( \omega _{k}) _{k\geq 1}\in \Omega :( \omega _{k}) _{k\geq 1}=( \varepsilon _{j},\omega _{n+2},\omega _{n+3},\ldots ) \}. \end{align*} $$

Then $\Omega _{\varepsilon _{1}}\prec \cdots \prec \Omega _{\varepsilon _{2^{n+1}}}, \Omega _{\varepsilon _{i}}\cap \Omega _{\varepsilon _{j}}=\emptyset $ for $i\neq j$ and $\Omega =\Omega _{\varepsilon _{1}}\cup \cdots \cup \Omega _{\varepsilon _{2^{n+1}}}.$ Hence the subsets $ \{ \Omega _{\varepsilon _{1}},\ldots ,\Omega _{\varepsilon _{2^{n+1}}}\} $ form a partition of the space $\Omega $ into $2^{n+1}$ mutually disjoint subsets. Consider the shift mapping $\sigma :\Omega \rightarrow \Omega $ given by

$$ \begin{align*} \sigma ( \omega ) = ( \sigma ( \omega _{1}) ,\sigma ( \omega _{2}) ,\ldots ), \quad\text{where } \sigma ( \omega _{j}) =\omega _{j+1}\text{ for }j\geq 1 \text{ and } \omega \in \Omega. \end{align*} $$

Define $\phi _{m}:\Omega \rightarrow [ 0,1] $ by

$$ \begin{align*} \phi _{m}( \omega ) =\sum_{k=1}^{\infty }\frac{( 3^{m}-1) \omega _{k}}{3^{mk}}, \end{align*} $$

and observe the following routine facts:

1. (i) $\phi _{m}$ is an order-preserving homeomorphism;

2. (ii) $\phi _{m}( \Omega ) \cap \mathcal {M}_{m}^{\circ }=\emptyset $ ; and

3. (iii) $\phi _{m}( \Omega ) $ is homeomorphic to the classical ‘middle-thirds’ Cantor set [Reference Dovgoshey, Martio, Ryazanov and Vuorinen4].

Consider $( \omega _{k}) _{k\geq 1}\in \Omega _{\varepsilon _{j}}$ for $1\leq j\leq 2^{n}$ . For any positive integer t, define

$$ \begin{align*}\Lambda _{t} =\Big( \underbrace{\omega _{1}\cdots \omega _{1}}_{t\text{ times}}\underbrace{\omega _{2}\cdots \omega _{2}}_{t\text{ times}}\cdots \underbrace{\omega _{t}\cdots \omega _{t}}_{t\text{ times}}\Big) \quad\mbox{and}\quad B_{t} =\Big( \underbrace{0\cdots 0}_{t\text{ times}}\underbrace{1\cdots 1} _{t\text{ times}}\Big).\end{align*} $$

The mapping $Z_{\varepsilon _{j}} :\Omega _{\varepsilon _{j}}\rightarrow Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) \subset \Omega _{\varepsilon _{j}}$ defined by

$$ \begin{align*}Z_{\varepsilon _{j}}( \varepsilon _{j},\omega _{n+2},\ldots ) =( \varepsilon _{j},B_{1}\Lambda _{1}B_{2}\Lambda _{2}B_{3}\Lambda _{3}\cdots )\end{align*} $$

is an order-preserving homeomorphism. We define

$$ \begin{align*}K_{\varepsilon _{j}}=\phi _{m}\circ Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}), \quad\mbox{for } 1\leq j\leq 2^{n},\end{align*} $$

and observe that as $\Omega _{\varepsilon _{1}}\prec \cdots \prec \Omega _{\varepsilon _{2^{n}}}$ and $Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) \subset \Omega _{\varepsilon _{j}}$ for $1\leq j\leq 2^{n},$ we have $Z_{\varepsilon _{1}}( \Omega _{\varepsilon _{1}}) \prec \cdots \prec Z_{\varepsilon _{2^{n}}}( \Omega _{\varepsilon _{2^{n}}}) .$ Since the function $\phi _{m}$ is an order-preserving homeomorphism and since $\phi _{m}( \Omega _{\varepsilon _{j}}) \subset \mathcal {L}_{m}$ for $ 1\leq j\leq 2^{n},$ it follows that $\phi _{m}( Z_{\varepsilon _{1}}( \Omega _{\varepsilon _{1}}) ) =K_{\varepsilon _{1}}<\cdots <\phi _m ( Z_{\varepsilon _{2^{n}}}( \Omega _{\varepsilon _{2^{n}}}) ) =K_{\varepsilon _{2^{n}}}.$ Thus, the sets $K_{\varepsilon _{1}},\ldots ,K_{\varepsilon _{2^{n}}}$ are a collection of $2^{n}$ mutually disjoint subsets of the interval $ \mathcal {L}_{m}.$ Later we will establish in Proposition 3.6 that each of the $2^{n}$ mutually disjoint subsets $ K_{\varepsilon _{j}}$ is an extremally scrambled set for every function $g_{m}\in \mathcal {G}_{m}.$

Step 2. For ease of exposition, let $M_{\varepsilon _{j}}=( \alpha _{j},\beta _{j}) \equiv M^n_{m,j-2^n}$ for $2^{n}+1\leq j\leq 2^{n+1}$ . We construct positive measure SVC sets $\mathcal {C}_{\varepsilon _{j}}\subset M_{\varepsilon _{j}}$ such that $\mu ( \mathcal {C}_{\varepsilon _{j}}) =1/2^{n+2}$ . To this end, consider any one of the above intervals $M_{\varepsilon _{j}},$ and consider its midpoint $( \alpha _{j}+\beta _{j}) /2=m_{j}.$ Choose a positive real number $ \epsilon $ such that

$$ \begin{align*} \frac{1}{2^{n+2}}<\epsilon <\delta _{m}^{n} \quad\text{and let}\quad \delta =\bigg( \epsilon -\frac{1}{2^{n+2}}\bigg). \end{align*} $$

Define $E_{\varepsilon _{j}}^{0}=[ \ell _{\varepsilon _{j}},r_{\varepsilon _{j}} ] =[ m_{j}-\tfrac 12\epsilon ,m_{j}+\tfrac 12\epsilon ] $ and set $ \ell _{\varepsilon _{j}0} =\ell _{\varepsilon _{j}}+\tfrac 14\delta $ , $r_{\varepsilon _{j}0}=m_{j}, \ell _{\varepsilon _{j}1} =m_{j}+\tfrac 14\delta , r_{\varepsilon _{j}1}=r_{\varepsilon _{j}}$ and $E_{\varepsilon _{j}}^{1}=[ \ell _{\varepsilon _{j}0},r_{\varepsilon _{j}0}] \cup [ \ell _{\varepsilon _{j}1},r_{\varepsilon _{j}1}].$ Observe that $\mu ( E_{\varepsilon _{j}}^{1}) =\mu ( E_{\varepsilon _{j}}^{0}) -\tfrac 12\delta .$ Next, set $\ell _{\varepsilon _{j}00} =\ell _{\varepsilon _{j}0}+\tfrac {1}{16}\delta $ , $r_{\varepsilon _{j}00}=\tfrac 12(\ell _{\varepsilon _{j}0}+r_{\varepsilon _{j}0})$ , $\ell _{\varepsilon _{j}01} =r_{\varepsilon _{j}00}+\tfrac {1}{16}\delta $ , $r_{\varepsilon _{j}01}=r_{\varepsilon _{j}0},$ $\ell _{\varepsilon _{j}10} =\ell _{\varepsilon _{j}1}+\tfrac {1}{16}\delta $ , $r_{\varepsilon _{j}10}=\tfrac 12(\ell _{\varepsilon _{j}1}+r_{\varepsilon _{j}1}),$ $\ell _{\varepsilon _{j}11} =r_{\varepsilon _{j}10}+\tfrac {1}{16}\delta $ , $r_{\varepsilon _{j}11}=r_{\varepsilon _{j}1}$ and define $E_{\varepsilon _{j}}^{2}=[ \ell _{\varepsilon _{j}00},r_{\varepsilon _{j}00}] \cup [ \ell _{\varepsilon _{j}01},r_{\varepsilon _{j}01} ] \cup [ \ell _{\varepsilon _{j}10},r_{\varepsilon _{j}10}] \cup [ \ell _{\varepsilon _{j}11},r_{\varepsilon _{j}11}].$ Observe that $\mu ( E_{\varepsilon _{j}}^{2}) =\mu ( E_{\varepsilon _{j}}^{1}) -\tfrac 12\delta =\mu ( E_{\varepsilon _{j}}^{0}) -( \tfrac 12\delta +\tfrac 14\delta ).$ Repeating this construction yields a decreasing sequence of sets $ E_{\varepsilon _{j}}^{k}$ for $k=0,1,2,\ldots $ in which every $ E_{\varepsilon _{j}}^{k}$ is a finite union of disjoint closed and bounded intervals and is therefore compact. Observe that all the right end points of $E_{\varepsilon _{j}}^{k}$ are retained in $E_{\varepsilon _{j}}^{k+1},$ but that all the left end points of $E_{\varepsilon _{j}}^{k}$ have been replaced by new left end points which only survive one iteration of this construction of the SVC set. Thus, the decreasing sequence of nonempty compact sets $E_{\varepsilon _{j}}^{0}\supset E_{\varepsilon _{j}}^{1}\supset E_{\varepsilon _{j}}^{2}\supset \cdots $ must have a nonempty intersection $ \mathcal {C}_{\varepsilon _{j}}=\bigcap _{k=0}^{\infty }E_{\varepsilon _{j}}^{k}\neq \emptyset $ where

$$ \begin{align*} \mu ( \mathcal{C}_{\varepsilon _{j}}) =\mu ( E_{\varepsilon _{j}}^{0}) -\sum_{k=1}^{\infty }\frac{\delta }{2^{k}} =\mu ( E_{\varepsilon _{j}}^{0}) -\delta=\epsilon -\delta =\epsilon -\bigg( \epsilon -\frac{1}{2^{n+2}}\bigg) = \frac{1}{2^{n+2}}. \end{align*} $$

Since $\mathcal {C}_{\varepsilon _{j}}\subset M_{\varepsilon _{j}},$ the $2^{n}$ mutually disjoint SVC sets $\mathcal {C}_{\varepsilon _{_{2^n+1}}}, \ldots ,\mathcal {C}_{\varepsilon _{2^{n+1}}}$ in $\mathcal {M} _{m}^{\circ }$ all have positive Lebesgue measure $1/2^{n+2}.$

Step 3. Fix any particular j with $2^{n}+1\leq j\leq 2^{n+1}$ . We shall construct an order-preserving homeomorphism $\xi _{\varepsilon _{j}}:\Omega _{\varepsilon _{j}}\rightarrow \mathcal {C} _{\varepsilon _{j}},$ where $\Omega _{\varepsilon _{j}}$ is constructed in step 1 and $\mathcal {C} _{\varepsilon _{j}}$ is constructed in step 2. Towards this end, for every positive integer $k>1$ and for any $(k-1)$ -tuple $( \omega _{1},\ldots ,\omega _{k-1})$ , replace in $\Omega $

$$ \begin{align*}( \omega _{1},\ldots ,\omega _{k-1},1,0,0,\ldots ,0,0,\ldots ) \quad\mbox{by}\quad( \omega _{1},\ldots ,\omega _{k-1},0,1,1,\ldots ,1,1,\ldots ),\end{align*} $$

where the first terminates in an infinite sequence of 0s and the second terminates in an infinite sequence of 1s.

Consider the set $\Omega _{\varepsilon _{j}}^{\mathbf {1}}$ of all binary sequences in $\Omega _{\varepsilon _{j}}$ that terminate in $1$ after a finite number of coordinates starting at $\omega _{n+2}$ (after the first $n+1$ coordinates which are always equal to $\varepsilon _{j}$ for every element of $\Omega _{\varepsilon _{j}}$ ). It is clear that this is a countable dense subset of $ \Omega _{\varepsilon _{j}}.$ Let $\mathcal {C}_{\varepsilon _{j}}^{\mathbf {1}} $ consist of all the right end points that appear in the decreasing sequence of closed intervals $E_{\varepsilon _{j}}^{k}$ ( $k =0,1,2,\ldots $ ) whose intersection is the SVC set $\mathcal {C}_{\varepsilon _{j}}.$ Observe that $\mathcal {C} _{\varepsilon _{j}}^{\mathbf {1}}$ is also a countably infinite dense subset of $\mathcal {C}_{\varepsilon _{j}}.$

We contend that all the right end points that comprise $\mathcal {C} _{\varepsilon _{j}}^{\mathbf {1}}$ are indexed by subscripts which are sequences in $\Omega _{\varepsilon _{j}}^{\mathbf {1}}.$ To see that this is so, observe that the first time that the terminal binary digit $1$ appears as a coordinate in the subscript that indexes a given right end point in $\mathcal {C}_{\varepsilon _{j}}$ , in the Nth coordinate say, counted after $\varepsilon _{j}$ (which always appears as the first $n+1$ coordinates of every element in $\mathcal {C} _{\varepsilon _{j}}$ ) so that $\omega _{N}=1$ and $\omega _{N-1}=0$ for $N>n+2$ , is in the set $E_{\varepsilon _{j}}^{N-n-1}$ at which the right end point $r_{\scriptsize {\underbrace {\varepsilon _{j},\ldots ,\omega _{N-1},1}_{N \text { subscripts}}\text { }}}$ appears. Now observe that after the infinite intersection $\mathcal {C}_{\varepsilon _{j}}=\bigcap _{k=0}^{\infty }E_{\varepsilon _{j}}^{k}$ has been taken, this right end point $r_{\scriptsize {\underbrace {\varepsilon _{j},\ldots ,\omega _{N-1},1}_{N \text { subscripts}}\text { }}}$ in $\mathcal {C}_{\varepsilon _{j}}$ is indexed by the subscript

$$ \begin{align*} r_{\varepsilon _{j},\ldots,\omega_{N-1},1,1,1,\ldots}, \quad\text{where } ( \varepsilon _{j},\ldots ,\omega_{N-1},1,1,1,\ldots ) \in \Omega _{\varepsilon _{j}}^{\mathbf{1}}. \end{align*} $$

Thus, every right end point of each interval $E_{\varepsilon _{j}}^{n}$ is indexed by a subscript in $\Omega _{\varepsilon _{j}}^{\mathbf {1}}.$

We define an order-preserving bijection $\xi _{\varepsilon _{j}}^{\mathbf {1} }:\Omega _{\varepsilon _{j}}^{\mathbf {1}}\rightarrow \mathcal {C} _{\varepsilon _{j}}^{\mathbf {1}}$ by

$$ \begin{align*} \xi _{\varepsilon _{j}}^{\mathbf{1}}( \varepsilon _{j},\omega_{n+2},\ldots ,\omega _{N-1},1,1,1,1,\ldots ) =r_{\varepsilon _{j},\omega _{n+2},\ldots ,\omega _{N-1},111\ldots .} \end{align*} $$

This map is plainly an order-preserving bijection, and extends to an order-preserving homeomorphism $\xi _{\varepsilon _{j}}:\Omega _{\varepsilon _{j}}\rightarrow \mathcal {C}_{\varepsilon _{j}}$ as the set $\Omega _{\varepsilon _{j}}^{\mathbf {1}}$ is dense in $\Omega _{\varepsilon _{j}}$ and $\mathcal {C}_{\varepsilon _{j}}^{\mathbf {1}}$ is dense in $\mathcal {C} _{\varepsilon _{j}}.$

Now, consider the mapping $r:\Omega \rightarrow \Omega $ defined by $r( ( \omega _{k}) _{k\geq 1}) =( r( \omega _{k}) _{k\geq 1})$ where, for every $k\geq 1$ ,

$$ \begin{align*} r( \omega _{k}) = \begin{cases} 0 \quad& \mbox{if } \omega _{k}=1, \\ 1 \quad& \mbox{if } \omega _{k}=0. \end{cases} \end{align*} $$

It is immediate that r is an order-reversing homeomorphism and that

$$ \begin{align*} r( \Omega _{\varepsilon _{j}}) =\Omega _{\varepsilon _{( 2^{n+1}+1) -j}}, \quad\text{for }1\leq j\leq 2^{n}. \end{align*} $$

To sum up, for all j with $2^{n}+1\leq j\leq 2^{n+1},$

$$ \begin{align*} \mathcal{C}_{\varepsilon _{j}}\overset{\xi _{\varepsilon _{j}}^{-1}}{ \rightarrow }\Omega _{\varepsilon _{j}}\overset{r}{\rightarrow }\Omega _{\varepsilon _{( 2^{n+1}+1) -j}}\overset{Z_{\varepsilon _{( 2^{n+1}+1) -j}}}{\rightarrow }Z_{\varepsilon _{( 2^{n+1}+1) -j}}( \Omega _{\varepsilon _{( 2^{n+1}+1) -j}}) \overset{ \phi _{m}}{\rightarrow }K_{\varepsilon _{( 2^{n+1}+1) -j}}. \end{align*} $$

Since every mapping that appears in the sequence above is a homeomorphism, and all but one are order-preserving, the mapping $f_{\varepsilon _{j},m,n}:\mathcal {C}_{\varepsilon _{j}}\rightarrow K_{\varepsilon _{( 2^{n+1}+1) -j}}$ defined by

$$ \begin{align*} f_{\varepsilon _{j},m,n} =( \phi _{m}\circ Z_{\varepsilon _{( 2^{n+1}+1) -j}}\circ r\circ \xi _{\varepsilon _{j}}^{-1}), \quad\text{for } 2^{n}+1\leq j\leq 2^{n+1} \end{align*} $$

must be an order-reversing homeomorphism.

To finish the construction of a continuous mapping in $\mathcal {G}_{m},$ define $\mathcal {S} =\bigcup _{j=2^{n}+1}^{2^{n+1}}\mathcal {C}_{\varepsilon _{j}}$ , $\mathcal { K=}\bigcup _{j=1}^{2^{n}}K_{\varepsilon _{j}}$ and $f_{m,n}:\mathcal {S} \rightarrow \mathcal {K}$ by setting $f_{m,n}\mid _{\mathcal {C}_{\varepsilon _{j}}}=f_{\varepsilon _{j},m,n}.$ We define $f_{m,n}$ on $\mathcal {L}_{m}\cup \mathcal {R}_{m}$ to ensure that $f_{m,n}\in \mathcal {G}_{m}$ by setting

$$ \begin{align*} f_{m,n}( x) =\bigg\{ \begin{array}{@{}ll} 3^{m}x & \text{for } x\in \mathcal{L}_{m}, \\ 3^{m}x-( 3^{m}-1) & \text{for } x\in \mathcal{R}_{m}. \end{array} \end{align*} $$

We linearly interpolate to extend $f_{m,n}$ to $\mathcal {M}_m$ from S. So, for any $s\in \mathcal {M}_{m}\backslash \mathcal {S},$ we define $s_{\ell }=\sup ( \{ t\in \mathcal {S}:t<s\} \cup \{ 1/3^{m}\} ) , s_{r}=\inf ( \{ t\in S:t>s\} \cup \{ 1-{3^{-m}}\} )$ and

$$ \begin{align*} f_{m,n}( s) =f_{m,n}( s_{\ell }) +\frac{s-s_{\ell }}{s_{r}-s_{\ell }}( f_{m,n}( s_{r}) -f_{m,n}( s_{\ell }) ). \end{align*} $$

3.3 The proofs of the ancillary results

In this subsection, we furnish the proof of Proposition 3.1. To this end, we first state and prove two lemmas. These lemmas will then be used to establish Proposition 3.6, which itself will be used in the proof of Proposition 3.1.

Proof of Lemma 3.4

We argue by mathematical induction on $k.$ If $k=1,$ then

$$ \begin{align*} g_{m}\circ \phi _{m}( \omega ) &=g_{m}\bigg( \sum_{k=1}^{\infty }\frac{( 3^{m}-1) \omega _{k}}{3^{mk}}\bigg) =3^{m}\bigg( \sum_{k=1}^{\infty }\frac{( 3^{m}-1) \omega _{k}}{ 3^{mk}}\bigg) -( 3^{m}-1) \omega _{1} \\ &=\sum_{k=2}^{\infty }\frac{( 3^{m}-1) \omega _{k}}{3^{m( k-1) }} =\sum_{k=1}^{\infty }\frac{( 3^{m}-1) \omega _{k+1}}{3^{mk}} =\phi _{m}( \sigma ( \omega ) ). \end{align*} $$

Assume now, as our induction hypothesis, that for any positive integer $ k\geq 1,$

$$ \begin{align*} g_{m}^{k}\circ \phi _{m}( \omega ) =\phi _{m}( \sigma ^{k}( \omega ) ). \end{align*} $$

Then

$$ \begin{align*} g_{m}^{k+1}\circ \phi _{m}( \omega ) &=( g_{m}\circ g_{m}^{k}) \circ \phi _{m}( \omega ) =g_{m}\circ ( g_{m}^{k}\circ \phi _{m}) ( \omega ) \\ &=g_{m}\circ \phi _{m}( \sigma ^{k}( \omega ) ) =\phi _{m}( \sigma ( \sigma ^{k}( \omega ) ) ) =\phi _{m}( \sigma ^{k+1}( \omega ) ). \end{align*} $$

Hence, the lemma follows by mathematical induction.

Proof of Lemma 3.5

From Lemma 3.4,

$$ \begin{align*} g_{m}^{k}( \phi _{m}( \omega ) ) =\phi _{m}( \sigma ^{k}( \omega ) ) =\sum_{j=1}^{\infty }\frac{( 3^{m}-1) \omega _{k+j}}{3^{mj}} \end{align*} $$

and the three inequalities follow directly.

Proof of Proposition 3.6

In order to prove the proposition, we must show that each of the three defining conditions of an extremally scrambled set are satisfied by any particular $ K_{\varepsilon _{j}}$ for $1\leq j\leq 2^{n}$ and for every $g_{m}\in \mathcal {G}_{m}.$ To this end, choose any subset $\{ x,y\} \subset K_{\varepsilon _{j}}$ such that $x<y$ and any $g_{m}\in \mathcal {G} _{m}.$ Assume also that $x_{0}$ is a periodic point of $g_{m}$ having period $p.$ Let $( \omega _{x},\omega _{y}) \in Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) \times Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) $ be such that

$$ \begin{align*} ( \omega _{x},\omega _{y}) =( \phi _{m}^{-1}( x) ,\phi _{m}^{-1}( y) ). \end{align*} $$

Since $\phi _{m}$ is an order-preserving homeomorphism, $\omega _{x}$ and $ \omega _{y}$ are unique elements of $\Omega ,$ with $\omega _{x}\prec \omega _{y}.$ Now since $\{ \omega _{x},\omega _{y}\} \subset Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) $ and $ \omega _{x}\prec \omega _{y},$ to every positive integer u there corresponds a strictly increasing sequence $\{ t_{u_{r}}\} _{r\geq 1}$ of positive integers such that $ ( \omega _{x}) _{t_{u_{r}}+j} =0$ and $( \omega _{y}) _{t_{u_{r}}+j} =1 $ for $1\leq j\leq u$ . Thus, by Lemma 3.5,

$$ \begin{align*} \vert g_{m}^{t_{u_{r}}}( \phi _{m}( \omega _{x}) ) -g_{m}^{t_{u_{r}}}( \phi _{m}( \omega _{y}) ) \vert &\geq \vert g_{m}^{t_{u_{r}}}( \phi _{m}( \omega _{y}) ) \vert -\vert g_{m}^{t_{u_{r}}}( \phi _{m}( \omega _{x}) ) \vert \\ &\geq \bigg( 1-\frac{1}{3^{mu}}\bigg) -\frac{1}{3^{mu}}=1-\frac{2}{3^{mu}}. \end{align*} $$

Since

$$ \begin{align*} \vert g_{m}^{t_{u_{r}}}( \phi _{m}( \omega _{x}) ) -g_{m}^{t_{u_{r}}}( \phi _{m}( \omega _{y}) ) \vert \geq 1-\frac{2}{3^{mu}} \quad\text{for every }u\geq 1, \end{align*} $$

we conclude that

$$ \begin{align*} \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( y) \vert =1. \end{align*} $$

For the same pair $\{ \omega _{x},\omega _{y}\} ,$ since $\{ \omega _{x},\omega _{y}\} \subset Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) ,$ there is an increasing sequence $\{ t_{u_{s}}\} _{s\geq 1}$ of positive integers such that $( \omega _{x}) _{t_{u_{s}}+j}=( \omega _{y}) _{t_{u_{s}}+j}=0$ for $1\leq j\leq u$ . This implies, by Lemma 3.5, that

$$ \begin{align*} \vert g_{m}^{t_{u_{s}}}( \phi _{m}( \omega _{x}) ) -g_{m}^{t_{u_{s}}}( \phi _{m}( \omega _{y}) ) \vert &\leq \vert g_{m}^{t_{u_{s}}}( \phi _{m}( \omega _{x}) ) \vert +\vert g_{m}^{t_{u_{s}}}( \phi _{m}( \omega _{y}) ) \vert \\ &\leq \frac{1}{3^{mu}}+\frac{1}{3^{mu}}=\frac{2}{3^{mu}}. \end{align*} $$

As this estimate holds for every positive integer $u\geq 1,$ we conclude that

$$ \begin{align*} \lim \inf_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( y) \vert =0. \end{align*} $$

Finally, assume that $x_{0}$ is a periodic point of period p for the function $g_{m}.$ We must prove that for any $x\in K_{\varepsilon _{j}},$

$$ \begin{align*} \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( x_{0}) \vert \geq \tfrac{1}{2}. \end{align*} $$

Observe now that $\max _{k\geq 1}\vert g_{m}^{k}( x_{0}) \vert $ is a definite real number in $[ 0,1] $ since we are simply choosing the maximum value from a finite set consisting of p distinct real numbers in $[ 0,1] .$ Assume that $\omega _{x}=\phi _{m}^{-1}( x) \in Z_{\varepsilon _{j}}( \Omega _{\varepsilon _{j}}) $ where $x\in K_{\varepsilon _{j}}.$ Then to every positive integer u there correspond strictly increasing sequences $( u_{r_{t}}) _{t\geq 1}$ and $( u_{s_{t}}) _{t\geq 1}$ of positive integers such that $ ( \omega _{x}) _{u_{r_{t}}+j} =0$ and $( \omega _{x}) _{u_{s_{t}}+j} =1$ , for $1\leq j\leq (u+p)$ . Hence, by Lemma 3.5, $\vert g_{m}^{u_{r_{t}}}( x) \vert \leq {3^{-m( u+p) }}$ and

$$ \begin{align*}\vert g_{m}^{u_{r_{t}}+j}( x) -g_{m}^{u_{r_{t}}+j}( x_{0}) \vert \geq \vert g_{m}^{u_{r_{t}}+j}( x_{0}) \vert -\vert g_{m}^{u_{r_{t}}+j}( x) \vert \geq \vert g_{m}^{u_{r_{t}}+j}( x_{0}) \vert -\frac{1}{3^{mu}} \end{align*} $$

for $0\leq j\leq p-1$ . Since the above inequalities hold for every positive integer $u,$ we conclude that

$$ \begin{align*} \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( x_{0}) \vert \geq \max_{k\geq 1}\vert g_{m}^{k}( x_{0}) \vert. \end{align*} $$

Similarly, by Lemma 3.5,

$$ \begin{align*} \vert g_{m}^{u_{s_{t}}}( x) -g_{m}^{u_{s_{t}}}( x_{0}) \vert &\geq \vert g_{m}^{u_{s_{t}}}( x) \vert -\vert g_{m}^{u_{s_{t}}}( x_{0}) \vert \\ &\geq \bigg( 1-\frac{1}{3^{mu}}\bigg) -\vert g_{m}^{u_{s_{t}}}( x_{0}) \vert \geq \bigg( 1-\frac{1}{3^{mu}}\bigg) -\max_{k\geq 1}\vert g_{m}^{k}( x_{0}) \vert. \end{align*} $$

As this estimate holds for every positive integer $u,$ we conclude that

$$ \begin{align*} \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( x_{0}) \vert \geq 1-\max_{k\geq 1}\vert g_{m}^{k}( x_{0}) \vert. \end{align*} $$

Finally, since

$$ \begin{align*} \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( x_{0}) \vert &\geq \max_{k\geq 1}\vert g_{m}^{k}( x_{0}) \vert \quad\text{and} \\ \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( x_{0}) \vert &\geq 1-\max_{k\geq 1}\vert g_{m}^{k}( x_{0}) \vert , \end{align*} $$

we have

$$ \begin{align*} \lim \sup_{k\rightarrow \infty }\vert g_{m}^{k}( x) -g_{m}^{k}( x_{0}) \vert \geq \tfrac{1}{2}, \end{align*} $$

proving that $K_{\varepsilon _{j}}$ must be an extremally scrambled set for any j with $1\leq j\leq 2^{n}$ and for any particular $g_{m}\in \mathcal {G} _{m}. $

Proof of Proposition 3.1

For any positive integers n and k, define $n'=nk.$ From step 3 of the construction, we observe by definition that

$$ \begin{align*} f_{m,n'}( x) =\bigg\{ \begin{array}{@{}ll} 3^{m}x & \text{for } x\in \mathcal{L}_{m}, \\ 3^{m}x-( 3^{m}-1) & \text{for } x\in \mathcal{R}_{m}. \end{array} \end{align*} $$

Moreover, $f_{m,n'}$ is decreasing and continuous on both $\mathcal {S}$ and $\mathcal {M}_{m}^{\circ }\backslash \mathcal {S}$ since it has been extended from $\mathcal {S}$ to $\mathcal {M}_{m}$ by linear interpolation in such a way that $f_{m,n'}( 1/3^{m}) =1$ and $f_{m,n'}( 1-( 1/3^{m}) ) =0.$ Thus, $f_{m,n'}$ is in $\mathcal {G}_{m}.$ Since $f_{m,n'}\in \mathcal {G}_{m},$ by Proposition 3.6, for any j with $2^{n'}+1\leq j'\leq 2^{n'+1}, K_{\varepsilon _{( 2^{n'+1}+1) -j'}}$ must be an extremally scrambled set for $f_{m,n'},$ and since $f_{m,n'}( \mathcal {C}_{\varepsilon _{j'}}) =K_{\varepsilon _{( 2^{n'+1}+1) -j'}}\!,$ for every $ \mathcal {C}_{\varepsilon _{j'}}\!$ , it follows that every $ \mathcal {C}_{\varepsilon _{j'}}$ must also be an extremally scrambled set for $f_{m,n'}\!.$ By construction it is clear that $\mu ( \mathcal {C}_{\varepsilon _{j'}}) =1/2^{n'+2}$ and the $\mathcal {C}_{\varepsilon _{j'}}$ are homomorphic to each other. Since $n'=nk,$ for $1\leq j\leq 2^n$ , $M_{m,j}^n$ contains k mutually disjoint sets $M_{m,j'}^{n'}$ with $k({\kern1.5pt}j-1)+1\leq j'\leq kj,$ and thus k mutually disjoint extremally scrambled sets of identical positive Lebesgue measure, homeomorphic to each other. Thus, $f_{m,n'}$ is the desired map.