Proof. We have

$$ \begin{align*} \mathbf{1}_0(\chi(0)) = \theta(\mathbf{1}_0(0)) = \theta(\emptyset) = \emptyset. \end{align*} $$

Hence, $\chi (0) = 0$ . Similarly, $ \mathbf {1}_0(\chi (1)) = \theta (\mathbf {1}_0(1)) = \theta (e) = e, $ and so $\chi (1) = 1$ .

Now

$$ \begin{align*} \mathbf{1}_{2l}(\chi(P_1 P_2)) = \theta(\mathbf{1}_l(P_1 P_2)). \end{align*} $$

However, $\mathbf {1}_l(P_1 P_2) = \{a \mathbf {1}_{l-1}(P_1), b \mathbf {1}_{l-1}(P_2) \}$ and then

$$ \begin{align*} \theta(\mathbf{1}_l(P_1 P_2)) &= \theta( \{a \mathbf{1}_{l-1}(P_1), b \mathbf{1}_{l-1}(P_2) \} )\\ &= \{ ba \theta(\mathbf{1}_{l-1}(P_1)), aa \theta(\mathbf{1}_{l-1}(P_2)), ab \theta(\mathbf{1}_{l-1}(P_2)), bb \theta(\mathbf{1}_{l-1}(P_2)) \}\\ &= \{ aa \mathbf{1}_{2l-2}(\chi(P_2)), ab \mathbf{1}_{2l-2}(\chi(P_2)), ba \mathbf{1}_{2l-2}(\chi(P_1)), bb \mathbf{1}_{2l-2}(\chi(P_2)) \}\\ &=\mathbf{1}_{2l}(\chi(P_1 P_2)). \end{align*} $$

Hence, $\chi (P_1 P_2) = \chi (P_2) \chi (P_2) \chi (P_1) \chi (P_2) .$

Proof. The proof is done by induction on the number of lines of the finite colored tree ${\mathfrak A}$ (with full last line). If ${\mathfrak A}=0,1$ (that is only line 0), then with $\otimes ={\mathfrak A}$ . Moreover, $\text {bw}({\mathfrak A})=\otimes $ . Then,

$$ \begin{align*}\text{vbw}(H({\mathfrak A}))=\otimes=\chi(\otimes)=\chi(\text{bw}({\mathfrak A})).\end{align*} $$

Assume that the result holds for any finite colored tree with full last line with l lines. Let ${\mathfrak A}$ be a finite colored tree with full last line and with $l+1$ lines. Set and $\text {bw}({\mathfrak A})=w$ . Then $w=\text {bw}({\mathfrak L})\text {bw}({\mathfrak R})$ holds with ${\mathfrak L}$ and ${\mathfrak R}$ two colored trees with full last line and with l-lines.

Lemma 3.5 and the induction assumption yield

$$ \begin{align*} \text{vbw}(H({\mathfrak A}))&= \text{vbw}(H({\mathfrak R}))\text{vbw}(H({\mathfrak R}))\text{vbw}(H({\mathfrak L}))\text{vbw}(H({\mathfrak R}))\\ &= \chi(\text{bw}({\mathfrak R}))\chi(\text{bw}({\mathfrak R}))\chi(\text{bw}({\mathfrak L}))\chi(\text{bw}({\mathfrak R}))\\ &= \chi(\text{bw}({\mathfrak L})\text{bw}({\mathfrak R}))=\chi(\text{bw}({\mathfrak A})).\\[-37pt] \end{align*} $$

Proof. The proof is done by induction on k. The result is obvious if $k=1$ . Let us assume it holds for k, and let us prove it for $k+1$ .

Set $\omega '$ to be the concatenation of $2^{k}$ words $\omega $ and $\omega "=\omega '\omega '$ . Then,

$$ \begin{align*}\chi(\omega")=\chi(\mathbf{\omega'}\omega')=\chi(\omega')\chi(\omega')\chi(\mathbf{\omega'})\chi(\omega').\end{align*} $$

Hence, the property holds for $k+1$ just by using the induction hypothesis.

Proof. The proof is done by induction.

First, $\chi (10)=0010$ . The second half of the word is 10 which contains 1.

Let us assume that the result holds for $\chi ^{u}(10)$ and let us prove it for $\chi ^{u+1}(10)$ . Set $\chi ^{u}(10)=P_{1}P_{2}$ with $|P_{1}|=|P_{2}|$ .

By definition, $\chi ^{u+1}(10)=\chi (\chi ^{u}(10))=\chi (P_{1}P_{2})=\chi (P_{2})\chi (P_{2})\chi (P_{1})\chi (P_{2})$ . The second half of $\chi ^{u+1}(10)$ is the word $\chi (P_{1})\chi (P_{2})$ as the length of the image of a word under $\chi $ only depends on the length of the word.

By the induction hypothesis, $P_{2}$ contains some 1s. By the definition of $\chi $ , as $\chi (1)=1$ , $\chi (P_{2})$ also contains some 1s. This shows that both halves contain some 1s.

Proof. The proof is done by induction on u.

For $u=0$ , we look at an odd line that is only composed by blocks 10 by the definition of H.

We assume that this the property holds for any line of the form $2^{u}(2n+1)$ and we prove it for a line of the form $2^{u+1}(2n+1)$ .

At line $2^{u}(2n+1)$ , we consider all the consecutive and juxtaposed blocks of length $2^{2^{u}}$ . They form a sequence of family blocks with common $2^{u}$ -ancestors. More precisely, the line $2^{u}(2n+1)$ is the juxtaposition of the bottom-words of all the finite subtrees in ${\mathfrak J}$ with root at line $2^{u+1}n$ and with size $2^{u}$ .

The tree ${\mathfrak J}$ is invariant by H, and hence the line $2^{u+1}(2n+1)$ is obtained from the line $2^{u}(2n+1)$ . From Proposition 3.8 and the induction hypothesis, Lemma 3.9 shows that the line $2^{u+1}(2n+1)$ is the concatenation of blocks $\chi (\chi ^{u}(10))=\chi ^{u+1}(10)$ .

Proof. We remind that any odd line for ${\mathfrak J}$ is composed by a string of 10, and every even line of ${\mathfrak J}$ is composed by blocks 0000 or 0010. This yields that if $|\omega |$ is even and $|\omega '|$ is odd,

$$ \begin{align*}d(T_{\omega}({\mathfrak J}),T_{\omega'}({\mathfrak J}))\ge 2^{-3}.\end{align*} $$

A converging sequence is also a Cauchy sequence, and thus all lengths must have the same parity for sufficiently large index.

Proof. Let us set ${\mathfrak A}^{k}:=T_{\omega _{k}}({\mathfrak J})$ . The $u_{k}$ -source for $\omega _{k}$ is a site $\omega ^{\prime }_{k}$ that is at level $2n_{k}+1$ . The digit is either $0$ or $1$ (for every sufficiently large k, only one option is possible because H is marked). Hence, ${\mathfrak A}^{k}$ starts with $H^{u_{k}}(0)$ or $H^{u_{k}}(1)$ . To simplify the proof, we assume that the source is 0.

The assumption $u_{k}\to +\infty $ yields that for every integer m, and for every sufficiently large k, ${{\mathfrak A}^{k}}$ starts as $H^{m}(0)$ . This implies that ${\mathfrak A}$ starts as $H^{m}(0)$ , and this holds for every m. Hence, ${\mathfrak A}={\mathfrak J}$ .

The same argument holds if the source is 1. In that case, ${\mathfrak A}={\mathfrak J}'$ .

Proof. Let u and k be positive integers. Let $\omega $ and $\omega '$ in $\mathbb {F}_{2}^{+}$ such that $|\omega |=2^{u}(2n+1)$ and $|\omega '|=2^{u+k}(2m+1)$ . Set ${\mathfrak B}:=T_{\omega }({\mathfrak J})$ and ${\mathfrak B}':=T_{\omega '}({\mathfrak J})$ .

Line $2^{u}$ in ${\mathfrak B}$ is at level $2^{u+1}(n+1)=:2^{u+a}(2l+1)$ in ${\mathfrak J}$ , with $a\ge 1$ . By Proposition 3.11, this line is a concatenation of blocks $\chi ^{u+a}(10)$ . Now, line $2^{u}$ in ${\mathcal B}'$ is at level $2^{u}(2^{k}(2m+1)+1)$ . Hence, it is a concatenation of blocks $\chi ^{u}(10)$ . Therefore, Corollary 3.13 yields

(4) $$ \begin{align} d({\mathfrak B},{\mathfrak B}')\ge 2^{-u}. \end{align} $$

Now, consider $(\omega _{k})$ such that ${\mathfrak A}=\lim _{k\to \infty }T_{\omega _{k}}({\mathfrak J})$ with $|\omega _{k}|=2^{u_{k}}(2n_{k}+1)$ . Because we have assumed ${\mathfrak A}\neq {\mathfrak J},{\mathfrak J}'$ , Lemma 5.6 shows that $(u_{k})$ is bounded. Hence, it only takes finitely many values. We may consider the smallest one which is taken infinitely many times. We denote it by u.

The sequence $(T_{\omega _{k}}({\mathfrak J}))$ is converging and thus it is a Cauchy sequence. By equation (4), we must have $u_{k}\le u$ for sufficiently big k. This shows that the sequence $(u_{k})$ is eventually constant.

Furthermore, Corollary 3.13 yields that u is unique because trees in ${\mathfrak J}$ with root at lines $2^{u}(2n+1)$ and trees with root at lines $2^{u+t}(2m+1)$ are uniformly (with respect to n, m, and position of the sites) far away from each other.

Notation 2. In the following, we shall say that ${\mathfrak J}$ and ${\mathfrak J}'$ are of $2^{\infty }$ -type.

Proof. Assume, by contradiction, that this is not the case. Up to a sub-sequence, we can assume that $u_{k}=u\in \mathbb {N}$ for every k. We can also assume that for every k, ${\mathfrak A}^{k}_{e}=0$ because $\lim _{k\to +\infty }{\mathfrak A}^{k}={\mathfrak J}$ .

Set ${\mathfrak A}^{k}=:H^{u}({\mathfrak B}^{k})$ . Each ${\mathfrak B}^{k}$ is an odd-type tree. They all have root equal to 0 because H is marked and all ${\mathfrak A}^{k}$ have root equal to 0. Hence, taking the source, we see a configuration

Because H is marked, $H^{u}({\mathfrak B}^{k})$ differs with ${\mathfrak J}$ , either at line $2^{u}+1$ or at line $2^{u+1}+1$ . This means that $\lim _{k\to +\infty }{\mathfrak A}^{k}\neq {\mathfrak J}$ , which is a contradiction.

If now we assume ${\mathfrak J}'=\lim _{k\to +\infty }{\mathfrak A}^{k}$ , we re-employ the same notation, and mimic the proof of first point, in that case, we have

because a 1 at an odd level must have its two followers equal to 0. Hence, $H^{u}({\mathfrak B}^{k})$ differs with ${\mathfrak J}'$ at line $2^{u}+1$ .

Proof. This is an immediate consequence of equation (4) and the fact that a converging sequence is also a Cauchy sequence.

Proof. Let ${\mathfrak A}^{k}$ be a sequence of trees in ${\mathfrak J}$ which converges to ${\mathfrak A}$ . We may assume that all the roots appear at levels $2^{u}(2n_{k}+1)$ . Hence, any line numbered $2^{u}2n$ in ${\mathfrak A}^{k}$ corresponds to a piece of line $2^{u}(2(n_{k}+n)+1)$ in ${\mathfrak J}$ . By Proposition 3.11, such a line (in ${\mathfrak J}$ ) is a concatenation of $\chi ^{u}(10)$ . This cannot be detected at line 0 in ${\mathfrak A}^{k}$ but it is detectable from line $2^{u+1}$ where we already have a full block $\chi ^{u}(10)$ .

This yields that for any line in ${\mathfrak A}^{k}$ multiple of $2^{u+1}$ , the proportion of 1s with respect to the 0s is fixed and is equal to the proportion of 1s in $\chi ^{u}(10)$ . This passes to the limit, and thus this also holds in the tree ${\mathfrak A}$ .

This shows that the set of integers l such that any line $2^{l+1}n$ ( $n\in \mathbb {N}^{*}$ ) is a concatenation of blocks $\chi ^{l}(10)$ is non-empty and contains u. We now prove that u is the unique integer with this property.

• • Indeed, let us pick $v'<u$ . Choose any v such that $v'+1+v>u$ . Any line $2^{v'}2n$ in ${\mathfrak A}^{k}$ corresponds to a piece of line $2^{u}(2n_{k}+1)+2^{v'}2n=2^{v'+1}(n+2^{u-v'-1} (2n_{k}+1))$ in ${\mathfrak J}$ . Now, choose a positive n of the form $n=2^{v}(2m+1)-2^{u-v'-1}(2n_{k}+1)$ (with m sufficiently big). Then the line $2^{v'}2n$ in ${\mathfrak A}^{k}$ corresponds to a piece of line $2^{v'+1+v} (2m+1)$ in ${\mathfrak J}$ . It is a concatenation of blocks $\chi ^{v'+1+v}(10)$ , but it may be truncated. As m can increase as wanted, we can fix it such that the line $2^{v'}2n$ in ${\mathfrak A}^{k}$ is a non-truncated concatenation of blocks $\chi ^{v'+1+v}(10)$ . Now, we remind that Corollary 3.14 states that a concatenation of blocks $\chi ^{v'+1+v}(10)$ cannot be equal to a concatenation of blocks $\chi ^{u}(10)$ .

  Because v can be chosen arbitrarily, this shows that the proportion of 1s in lines $2^{v'}2n$ does not converge because it has several different accumulation points.

• • Now for $v'>u$ and for any n, a line $2^{v'}2n$ in ${\mathfrak A}$ is also a line $2^{u}2.2^{v'-u}n$ , and hence is a concatenation of blocks $\chi ^{u}(10)$ .

This finishes to prove that u is unique integer such that all lines $2^{l}2n$ in ${\mathfrak A}$ are concatenations of blocks $\chi ^{l}(10)$ .

Proof. The result is trivial if $u=0$ (that is, odd-type tree). We only do the proof for $u\ge 1$ (that is, even-type tree).

We consider a sequence $({\mathfrak A}^{k})$ in ${\mathcal O}({\mathfrak J})$ converging to ${\mathfrak A}$ . By Proposition 5.9, we can assume that all the ${\mathfrak A}^{k}$ are of the form

with $\otimes =0,1$ . The tree is of odd-type. With this notation, we prove that $({\mathfrak C}^{k})$ and $({\mathfrak D}^{k})$ do converge in X.

The tree is of the form $H^{u}(0)$ , and then at each slot at line $2^{u}+1$ , we connect either the tree $H^{u}({\mathfrak C}^{k})$ or the tree $H^{u}({\mathfrak D}^{k})$ . If we say that ${\mathfrak A}^{k}$ coincides with ${\mathfrak A}$ for the first $n_{k}$ -lines, the convergence ${\mathfrak A}^{k}\to {\mathfrak A}$ yields $n_{k}\to +\infty $ as $k\to +\infty $ . Hence, $H^{u}({\mathfrak C}^{k})$ and $H^{u}({\mathfrak D}^{k})$ coincide with the right $T_{\omega }({\mathfrak A})$ with $|\omega |=2^{u}$ for at least $n_{k}-2^{u}$ lines.

Let us fix N. If we assume that for any $k\ge k_{0}$ , $n_{k}>N+2^{u}$ , then for any k and $k'$ greater than $k_{0}$ , ${\mathfrak C}^{k}$ and ${\mathfrak C}^{k'}$ on the one hand, and ${\mathfrak D}^{k}$ and ${\mathfrak D}^{k'}$ on the other hand do coincide for at least $\lfloor N2^{-u}-1\rfloor $ digits. In other words, the sequences $({\mathfrak C}^{k})$ and $({\mathfrak D}^{k})$ are Cauchy sequences, and hence converge.

Proof. Due to Lemma 5.5 and Proposition 5.9, it suffices to prove the statement for configurations in ${\mathfrak J}$ . Hence, we consider that we see the configuration

in ${\mathfrak J}$ .

We denote by $\omega $ the site where the root of the configuration we study stands. At line $|\omega |+2$ , we see a block 0010, and hence line $|\omega |+2$ is even. Therefore, $|\omega |$ is even.

Because $\omega $ lays at an even level, we can consider its source. At the source, we see which yields at site $\omega $

This immediately yields $T_{aa \widetilde \omega }({\mathfrak J})=T_{ba \widetilde \omega }({\mathfrak J})=T_{bb \widetilde \omega }({\mathfrak J})=:{\mathfrak A}$ and ${\mathfrak B}:=H({\mathfrak A}')$ . By hypothesis, ${\mathfrak A}_{e}=0$ and ${\mathfrak B}_{e}=1$ .

Proof. Again, Lemma 5.5 and Proposition 5.9 show that it is sufficient to prove the statement for configurations in ${\mathfrak J}$ .

At the level of the roots for ${\mathfrak A}$ and ${\mathfrak B}$ , we see the block 00. This shows that this level is even. By Lemma 5.2, we only see blocks 0000 or 0010 at even levels. Hence, we must see either the configuration

or the reverse one

Then, Lemmas 5.3 and 5.14 show that ${\mathfrak A}={\mathfrak B}$ .

Proof. Again, by Proposition 5.9, we may only prove that the lemma holds for configurations in ${\mathfrak J}$ .

Then, the first three statements are a direct consequence of Lemma 5.3, plus the fact that a block 0000 can only appear at a level in $4\mathbb {N}$ . Note that the grandfather for ${\mathfrak A}_{e}$ must be at an even line which is not a multiple of $4$ .

Let us assume that the line where the ${\mathfrak A}$ terms have their root is $2^{u}(2n+1)$ with $u\ge 2$ . If we consider u-times the source, we are at level $2n+1$ . We must see

where . Actually, we have a better description because the root $\oplus $ is at an even level: we see

and hence we have ${\mathfrak E}={\mathfrak E}'={\mathfrak D}'$ .

Let us set and This means that we see the configuration

The image by H of the subtree at level is

Taking the image by H of the subtree in the dashed frame, we see

In the dotted framebox, we see the same kind of configuration as previously but at level $2^{2}(2n+1)$ . Iterating this process $u-2$ more times, we finally get the configuration:

Now, remember that $H^{u}({\mathfrak A}')={\mathfrak A}$ . Moreover, ${\mathfrak B}^{\prime }_{e}=1$ and H is marked, and thus ${H^{u}({\mathfrak B}')_{e}=1}$ .

Proof. Minimality comes from the fact that ${\mathfrak J}$ contains many finite subtrees $H^{u}(0)$ with u as big as wanted. Lemma 5.16 applied to ${\mathfrak J}$ shows that any level where ${\mathfrak J}$ contains $H^{u}(0)$ with $u\ge 2$ , also contains a subtree starting as $H^{u}(1)$ . Letting $u\to +\infty $ yields that ${{\mathfrak J}'=\lim _{n\to +\infty }H^{n}(1)}$ belongs to $X=\overline {{\mathcal O}({\mathfrak J})}$ .

Proof. Because ${\mathfrak A}$ is of even-0-type, this means ${\mathfrak A}_{e}$ stands at level $2^{u}(2n+1)$ with $u\ge 2$ . Considering the source, we see with $H({\mathfrak B}')={\mathfrak B}$ and $H({\mathfrak A}')={\mathfrak A}$ . Roots ${\mathfrak B}^{\prime }_{e}=1$ and ${\mathfrak A}^{\prime }_{e}=0$ are at the even level $2^{u-1}(2n+1)$ . By Lemma 5.14, we only see blocks of 0000 or 0010. Hence, we actually see

Taking the H-image of the dashed frame, we get

with the ${\mathfrak A}$ terms at the level $2^{u}(2n+1)$ .

Proof. We consider a sequence ${\mathfrak A}^{k}$ in ${\mathcal O}({\mathfrak J})$ which converges to ${\mathfrak A}$ . For simplicity, we set ${\mathfrak A}_{e}=\otimes $ . By Proposition 5.9, we may assume that all the ${\mathfrak A}^{k}$ are of the form

and is an odd-tree.

By Lemma 5.13, convergence for ${\mathfrak A}^{k}$ is equivalent to the convergence for ${\mathfrak E}^{k}$ to ${\mathfrak E}$ and for ${\mathfrak D}^{k}$ to ${\mathfrak D}$ .

Furthermore, the root of ${\mathfrak E}^{k}$ and ${\mathfrak D}^{k}$ stand at even level, say $2^{u_{k}}(2m_{k}+1)$ . Again by Proposition 5.9, we may assume that all are of fixed type $2^{u}$ with $u\ge 1$ .

At even level, we see blocks 0000 or 0010. Hence, the possibilities are ${\mathfrak E}_{e}=0={\mathfrak D}_{e}$ or ${\mathfrak E}_{e}=1$ and ${\mathfrak D}_{e}=0$ . Note that by Corollary 5.15, ${\mathfrak E}_{e}={\mathfrak D}_{e}$ yields ${\mathfrak E}={\mathfrak D}$ . Furthermore, ${\mathfrak E}_{e}=0$ is only possible if $u\ge 2$ .

Hence, the possibilities are:

1. (1) $u\ge 2$ , ;

2. (2) $u=1\ \otimes =1$ , ;

3. (3) $u\ge 1$ , $\otimes =0$ , and with ${\mathfrak E}_{e}=1$ and ${\mathfrak D}_{e}=0$ .

Now, we explain how we can determine the value of v. The case $v=+\infty $ is extremely simple because we must see either ${\mathfrak J}$ or ${\mathfrak J}'$ . Hence, we assume that ${\mathfrak A}$ is not ${\mathfrak J}$ nor ${\mathfrak J}'$ . Then, Lemma 5.11 shows that v is entirely determined by one of the branches of ${\mathfrak A}$ : for any u, and for any sufficiently large n, the line n in $T_{a}({\mathfrak A})$ or $T_{b}({\mathfrak A})$ only contains full blocks of length $|\chi ^{u}(10)|$ . Hence, v is the unique integer u where we see blocks $\chi ^{u}(10)$ at lines in $2^{u+1}\mathbb {N}$ (in ${\mathfrak A}$ ).

Proof. Still by Proposition 5.9, we only do the proof in the case that we see the configuration in ${\mathfrak J}$ .

Let $\omega $ be a site where we see

• • First, we consider the case $|\omega |$ is even. If we consider the source of $\omega $ , we see the configuration which yields at site $\omega $

  
  Hence, we get and

  The latter tree can be obtained from ${\mathfrak B}$ :

  1. (1) by exchanging the color of the root;

  2. (2) by replacing the left-hand side subtree by a copy of the one at the right-hand side.

  This corresponds to the result of the lemma with $u=0$ .

• • Assume that $|\omega |$ is odd. Then we set $|\omega |+1=2^{u}(2n+1)$ . We consider the u-source for ${\mathfrak A}$ and ${\mathfrak B}$ . By Lemma 5.1, they are either brothers or equal. As H is marked and ${\mathfrak A}_{e}=1\neq 0={\mathfrak B}_{e}$ , the u-sources are brothers. Let us denote them by ${\mathfrak A}^{u}$ and ${\mathfrak B}^{u}$ . They are odd-type trees at level $2n+1$ . Furthermore, ${\mathfrak A}^{u}_{e}=1$ and ${\mathfrak B}^{u}_{e}=0$ . By Lemma 5.2, their father is 0.

We use the part of Lemma 5.20 that is already proved (with father at even level). Hence, ${\mathfrak A}^{u}$ is obtained from ${\mathfrak B}^{u}$ by exchanging the color of the root and replacing $T_{a}({\mathfrak B}^{u})$ by $T_{b}({\mathfrak B}^{u})$ . Then, applying $H^{u}$ , we get for ${\mathfrak A}$ :

1. (1) ${\mathfrak A}$ starts as $H^{u}(1)$ whereas ${\mathfrak B}$ starts with $H^{u}(0)$ ;

2. (2) at any slot at line $2^{u}$ , we glue $H^{u}(T_{b}({\mathfrak B}^{u}))$ .

Employing notation from the statement of the lemma, we have ${\mathfrak B}=H^{u}({\mathfrak B}^{u})$ , ${\mathfrak A}=H^{u}({\mathfrak A}^{u})$ . If then