The field of the Reals and the Random Graph are not Finite-Word Ordinal-Automatic

A simple induction shows that all $\gamma\in U_{m}^{i}(\beta)$ satisfy $\gamma_{\sim m}=\beta_{\sim m}+\omega_{1}\cdot k$ for some $0\leq k\leq(i\cdot m)$.

One also proves inductively that the coefficient of $\omega^{j}$ of an element of $U_{m}^{i}(X,\alpha)$ is bounded by $(c_{m}(w)+im)$. ∎

Set $I=\Set{1,2,\dots,n}$. We define the function

such that $f(j)$ contains $(q,p,i)$ if and only if there is a run of $\mathcal{A}_{i}$ from state $q$ to state $p$ on $v\restriction{[\gamma,\gamma+\omega^{n}j)}\otimes\diamond^{\omega^{n}j}$. By choice of $K$ there are $n_{1}<n_{2}$ with $f(n_{1})=f(n_{2})$. Thus,

from which we immediately conclude that $v\sim^{\Phi}_{E}w$. ∎

The proof is by outer induction on $\lvert\mathsf{supp}(v)\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert$ and by inner (transfinite) induction on

Fix the presentation

with $b_{0},\dots,b_{K}\in\mathbb{N}$ and proceed as follows.

• •

  If there is some $n\leq K$ such that $b_{n}+1-K\geq 0$ and

  		$\displaystyle(\mathsf{supp}(E)\cup\{\alpha\})\cap[\epsilon_{1},\epsilon_{2})=\emptyset\text{\ where }$
  		$\displaystyle\epsilon_{1}=\beta_{\sim K}+\omega^{K}b_{K}+\omega^{K-1}b_{K-1}+\dots+\omega^{n}(b_{n}+1-K)\text{\ and}$
  		$\displaystyle\epsilon_{2}=\beta_{\sim K}+\omega^{K}b_{K}+\omega^{K-1}b_{K-1}+\dots+\omega^{n+1}(b_{n+1}+1),$

  then we can apply the previous lemma and obtain a word $v^{\prime}$ such that $v\sim^{\Phi}_{E}v^{\prime}$ and

  $$\lvert\mathsf{supp}(v^{\prime})\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert<\lvert\mathsf{supp}(v)\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert$$

  or

  $$\beta^{\prime}=\min(\mathsf{supp}(v^{\prime})\setminus U_{K}(\mathsf{supp}(E),\alpha)).$$

  has a presentation

  $$\beta^{\prime}=\beta_{\sim K}+\omega^{K}b_{K}+\omega^{K-1}b_{K-1}+\dots+\omega^{n+1}b_{n+1}+\omega^{n}b^{\prime}+\omega^{n-1}b_{n-1}+\dots+b_{0}$$

  with $b^{\prime}<b_{n}$.

• •

  Assume that the conditions for the previous case are not satisfied. Either $b_{i}\leq K-1$ for $0\leq i\leq K$ or there is a minimal $i\leq K$ such that $b_{i}\geq K$. We first show that the latter case cannot occur. Assuming $b_{i}\geq K$ we have

  		$\displaystyle(\mathsf{supp}(E)\cup\{\alpha\})\cap[\epsilon_{1},\epsilon_{2})\neq\emptyset\text{\ where}$
  		$\displaystyle\epsilon_{1}=\beta_{\sim K}+\omega^{K}b_{K}+\omega^{K-1}b_{K-1}+\dots+\omega^{i}(b_{i}+1-K)\text{\ and}$
  		$\displaystyle\epsilon_{2}=\beta_{\sim K}+\omega^{K}b_{K}+\omega^{K-1}b_{K-1}+\dots+\omega^{i+1}(b_{i+1}+1).$

  Thus, there is some $\gamma\in\mathsf{supp}(E)\cup\{\alpha\}$ with

  $$\gamma=\beta_{\sim K}+\omega^{K}b_{K}+\omega^{K-1}b_{K-1}+\dots+\omega^{i+1}b_{i+1}+\omega^{i}c_{i}+\omega^{i-1}c_{i-1}+\dots+c_{0}$$

  such that $c_{i}+K-1>b_{i}$ whence $\beta\in U_{K}(\gamma)\subseteq U_{K}(\mathsf{supp}(E),\alpha)$ contradicting the definition of $\beta$.

  Thus, we can assume that $b_{i}\leq K-1$ for all $0\leq i\leq K$. By definition of $\beta$ we conclude that $\beta_{\sim K}\neq\gamma_{\sim K}+\omega_{1}c$ for $c\in\{0,1,\dots,K\}$ for all $\gamma\in\mathsf{supp}(E)\cup\{0,\alpha\}$. We proceed with one of the following cases depending on the cofinality of $\beta_{\sim K}$.

  1. (1)

     If $\beta_{\sim K}$ has countable cofinality, let

     	$\displaystyle\gamma=\max((\mathsf{supp}(E\cup\{v\})\cap\beta_{\sim K})+1$
     	$\displaystyle\delta=\min((\mathsf{supp}(E)\cap[\beta,\alpha))\cup\{\alpha\})\text{\ and}$
     	$\displaystyle\delta^{\prime}=\max(\mathsf{supp}(v)\cap\delta_{\sim K})+1.$

     From the definition of $\beta$ it follows that $\gamma_{\sim K}<\beta_{\sim K}<\delta_{\sim K}$ and that $\sup(E)\cap[\gamma,\delta)=\emptyset$. Note that $[\gamma,\beta_{\sim K})$ is of shape $\omega^{K+1}\eta_{1}$ for some ordinal $\eta_{1}\geq 1$ of countable cofinality. By definition of $\delta^{\prime}$, $[\delta^{\prime},\delta_{\sim K})$ is of shape $\omega^{K+1}\eta_{2}$ for some ordinal $\eta_{2}\geq 1$. Choose an ordinal $\eta$ such that $[\beta_{\sim K},\delta^{\prime})+\eta$ is isomorphic to $[\gamma,\delta_{\sim K})$ and define

     $$w:=v\restriction{[0,\gamma)}+\diamond^{\omega^{K}}+v\restriction{[\beta_{\sim K},\delta^{\prime})}+\diamond^{\eta}+v\restriction{[\delta_{\sim K},\alpha)}.$$

     For all $\bar{e}\in E^{k}$ and $\mathcal{A}\in A$ we conclude that

     whence $w\sim^{\Phi}_{E}v$. Moreover

     $$\lvert\mathsf{supp}(w)\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert<\lvert\mathsf{supp}(v)\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert$$

     because the letter at position $\beta$ in $v$ has been shifted to position

     $$\gamma_{\sim K}+\omega^{K}(b_{K}+1)+\omega^{K-1}b_{k-1}+\omega^{K-2}b_{k-2}+\dots+b_{0}.$$

     Since all $b_{i}\leq K-1$ we conclude that this position belongs to $U_{K}(\mathsf{supp}(E),\alpha)$.

  2. (2)

     If $\beta_{\sim K}$ has uncountable cofinality, Let

     	$\displaystyle\gamma=\max(\mathsf{supp}(E)\cap\beta_{\sim K})+1$
     	$\displaystyle\delta=\min((\mathsf{supp}(E)\cap[\beta,\alpha))\cup\{\alpha\})\text{\ and}$
     	$\displaystyle\delta^{\prime}=\max(\mathsf{supp}(v)\cap\delta_{\sim K})+1.$

     From the definition of $\beta$ and the uncountable cofinality of $\beta_{\sim K}$ it follows that $\gamma_{\sim K}+\omega_{1}K<\beta_{\sim K}<\delta_{\sim K}$ and that $\sup(E)\cap[\gamma,\delta)=\emptyset$. Let

     $$\gamma^{\prime}=\max(\mathsf{supp}(E\cup\{v\})\cap\beta_{\sim K})+1$$

     and note that $\gamma^{\prime}\leq\gamma_{\sim K}+\omega_{1}(K+1)$ because $\gamma^{\prime}\in U_{K}(E,\alpha)$ since $\beta$ has been chosen minimal. If $\gamma^{\prime}\geq\gamma_{\sim K}+\omega_{1}K$, then we do the following preparatory step that locally changes $v$ to some $v^{\prime}$ such that $v\sim^{\Phi}_{E}v^{\prime}$ and shrinking the corresponding value of $\gamma^{\prime}$. For this purpose, note that $e\restriction{[\gamma,\beta)}=\diamond^{[\gamma,\beta)}$ for all $e\in E$. By choice of $K$ there are numbers $i<j\leq K$ such that for all $\mathcal{A}\in\Phi$, all $q,p\in Q$ and all $\bar{e}\in E^{k}$ we have

     $$q\xrightarrow[\mathcal{A}]{(v\otimes\bar{e})\restriction{[\gamma,\gamma+\omega_{1}i)}}p\Longleftrightarrow q\xrightarrow[\mathcal{A}]{(v\otimes\bar{e})\restriction{[\gamma,\gamma+\omega_{1}j)}}p$$

     (2)

     Choose an ordinal $\eta$ such that $\omega_{1}\cdot(j-i)+[\gamma+\omega_{1}j,\gamma^{\prime})=[\gamma+\omega_{1}j,\gamma^{\prime})+\eta$ and set

     $\displaystyle v^{\prime}=v\restriction{[0,\gamma_{\sim K}+\omega_{1}i)}+v\restriction{[\gamma_{\sim K}+\omega_{1}j,\gamma^{\prime})}+\diamond^{\eta}+v\restriction{[\gamma^{\prime},\alpha)}.$

     Now $v\sim^{\Phi}_{E}v^{\prime}$ because for all $\bar{e}\in E^{k}$

     Note that the definitions of $\beta$, $\gamma$, $\delta$ and $\delta^{\prime}$ with respect to $v^{\prime}$ agree with those for $v$. Thus, from now on we replace $v$ by $v^{\prime}$ whence we can assume that $\gamma^{\prime}<\gamma+\omega_{1}K$.

     Note that $[\gamma^{\prime},\beta_{\sim K})$ is of shape $\omega^{K+1}\eta_{1}$ for some ordinal $\eta_{1}\geq 1$ of uncountable cofinality. By definition of $\delta^{\prime}$, $[\delta^{\prime},\delta_{\sim K})$ is of shape $\omega^{K+1}\eta_{2}$ for some ordinal $\eta_{2}\geq 1$. Choose an ordinal $\eta$ such that $[\beta_{\sim K},\delta^{\prime})+\eta$ is isomorphic to $[\gamma^{\prime}+\omega_{1},\delta_{\sim K})$ and define

     $$w:=v\restriction{[0,\gamma^{\prime})}+\diamond^{\omega_{1}}+v\restriction{[\beta_{\sim K},\delta^{\prime})}+\diamond^{\eta}+v\restriction{[\delta_{\sim K},\alpha)}.$$

     For all $\bar{e}\in E^{k}$ and $\mathcal{A}\in A$ we conclude that

     whence $w\sim^{\Phi}_{E}v$. Moreover

     $$\lvert\mathsf{supp}(w)\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert<\lvert\mathsf{supp}(v)\setminus U_{K}(\mathsf{supp}(E),\alpha)\rvert$$

     because the letter at position $\beta$ in $v$ has been shifted to position

     $$\gamma^{\prime}_{\sim K}+\omega_{1}+\omega^{K}b_{K}+\omega^{K-1}b_{k-1}+\omega^{K-2}b_{k-2}+\dots+b_{0}$$

     and since all $b_{i}<K-1$ we conclude that this is position is contained in $U_{K}(\mathsf{supp}(E),\alpha)$ because $\gamma_{\sim K}=\eta_{\sim K}+\omega_{1}c$ for some $\eta\in\mathsf{supp}(E)\cup\{0\}$ and some $c\in\{0,1,\dots,K-1\}$. ∎

     The previous result allows us to directly deduce the following bound on the growth rates of ordinal-automatic relations.

     Theorem 17.

     Fix an infinite family $\mathcal{F}$ of sets of $(\alpha)$-words with $\emptyset\in\mathcal{F}$. For every $c>1$, $\nu^{\Phi}(n)\leq n^{c}$ for infinitely many $n\in\mathbb{N}$.

     Proof.

     Heading for a contradiction, assume that there is a natural number $N$ such that

     $$\forall E\in\mathcal{F}\text{\ with }\lvert E\rvert>N\,\forall G\text{\ maximal $E$-$\Phi$-free}\,\exists F\in\mathcal{F}\,(F\subseteq G\text{ and }\lvert F\rvert\geq\lvert E\rvert^{c})$$

     where $c>1$.

     Take a finite set $F_{0}$ of parameters with $\lvert F_{0}\rvert>N$. Having defined a finite set $F_{i-1}$ such that $\mathsf{supp}(F_{i-1})\in U_{K}^{i-1}(\mathsf{supp}(F_{0}))$ we can use the previous lemma to choose some maximal $F_{i-1}$-$\Phi$-free set $G_{i}$ with $\mathsf{supp}(G_{i})\in U_{K}^{i}(\mathsf{supp}(F_{0}))$. By assumption, there is some $F_{i}\in\mathcal{F}$ with $F_{i}\subseteq G_{i}$ and $\lvert F_{i}\rvert\geq\lvert F_{i-1}\rvert^{c}$. By induction we obtain $\lvert F_{i}\rvert\geq\lvert F_{0}\rvert^{c^{i}}$.

     On the other hand, all elements of $F_{i}$ have support in $U_{K}^{i}(\mathsf{supp}(F_{0}))$ which by Lemma 14 is at most of size

     $$(\lvert\Sigma\rvert+1)^{c_{0}(iK+1)(c_{1}+Ki)^{K}}$$

     for some constants $c_{0}$ and $c_{1}$. Since $c^{i}$ grows faster than any polynomial in $i$ we have $c^{i}>c_{0}(i+1)(c_{1}+Ki)^{K}$ for some large $i$ which leads to a contradiction. ∎

     5. Applications

     5.1. Random Graph

     The random graph (or Rado graph) $(V,E)$ is the unique countable graph that has the property that for any choice of finite subsets $V_{0},V_{1}\subseteq V$ there is a node $v^{\prime}$ which is adjacent to every element of $V_{0}$ but not adjacent to any element of $V_{1}$.

     Theorem 18.

     Given an ordinal $\alpha$, the random graph is not $(\alpha)$-automatic.

     Proof.

     Heading for a contradiction assume that the random graph was $(\alpha)$-automatic. As shown by Delhommé [3], the random graph satisfies $\nu_{\mathcal{F}}^{\Phi}(n)=2^{n}$ for all $n\in\mathbb{N}$ where $\Phi$ consists of only one automaton recognising the edge relation of the random graph and $\mathcal{F}$ contains all subsets of the domain of the random graph. This contradicts Proposition 15. ∎

     Remark 19.

     Similarly, taking $\mathcal{F}$ to be the family of all antichains, one proves that the random partial order is not $(\alpha)$-automatic (cf. [11] for an analogous result for automatic structures).

     5.2. Integral Domains

     In this part we show that there is no infinite $(\alpha)$-automatic integral domain for any $\alpha<{\omega_{1}+\omega^{\omega}}$. We cannot use the growth rate theorem directly but use a variant of its proof. The difference is that we do not use a fixed set of relations $\Phi$ when defining the sequence $(F_{i})_{i\in\mathbb{N}}$ but in each step we take a different relation but ensure that we can still apply Proposition 15 with a fixed constant $K$ in each step. This is ensured by using relations defined by a fixed automaton $\mathcal{A}$ which has an additional parameter which is chosen very carefully. In fact, we follow the proof of Khoussainov et al. [11] from the automatic case. Let us recall the basic definitions and some observations from their proof. An integral domain is a commutative ring with identity $(D,+,\cdot,0,1)$ such that $d\cdot e=0\Rightarrow d=0$ or $e=0$ for all $d,e\in D$.

     Lemma 20 (cf. Proof of Theorem 3.10 from [11]).

     Let $(D,+,\cdot,0,1)$ be an integral domain and $E\subseteq D$ a finite subset. There is some $d\in D$ such that for all $a_{1},a_{2},b_{1},b_{2}\in E$, if $a_{1}d+b_{1}=a_{2}d+b_{2}$, then $a_{1}=a_{2}$ and $b_{1}=b_{2}$, i.e., the function $f_{d}:E^{2}\to D$, $(e_{1},e_{2})\mapsto e_{1}d+e_{2}$ is injective.

     Proposition 21.

     Let $\mathfrak{A}=(D,+,\cdot,0,1)$ be an $(\alpha)$-automatic integral domain for some $\alpha<{\omega_{1}+\omega^{\omega}}$.¹¹1Without loss of generality, we can assume that $\mathfrak{A}$ has a injective representation by the automata $(\mathcal{A}_{D},\mathcal{A}_{+},\mathcal{A}_{\cdot},\mathcal{A}_{0},\mathcal{A}_{1})$ such that $L(\mathcal{A}_{D})=D$, i.e., $D$ is a set of $(\alpha)$-words (cf. [6]).

     There is a constant $m$ such that for every finite set $X\subseteq\alpha$ of ordinals we have

     $$\lvert\Set{d\in D\mid\mathsf{supp}(d)\subseteq U_{m}(X,\alpha)}\rvert\geq\lvert\Set{d\in D\mid\mathsf{supp}(d)\subseteq X}\rvert^{2}.$$

     Proof.

     As an abbreviation, we use the expression $x_{1},\dots,x_{n}\subseteq_{\mathsf{supp}}y$ for

     $$x_{1}\subseteq_{\mathsf{supp}}y\land\dots\land x_{n}\subseteq_{\mathsf{supp}}y.$$

     Let $\psi(x,p)$ denote the formula

     $$x\in D\land\forall a_{1},a_{2},b_{1},b_{2}\subseteq_{\mathsf{supp}}p\,\left((a_{1}x+b_{1}=a_{2}x+b_{2})\rightarrow(a_{1}=a_{2}\land b_{1}=b_{2})\right)$$

     and $\psi_{\min}(x,p)$ the formula

     $$\psi(x,p)\land\forall y\,(\psi(y,p)\rightarrow x\sqsubseteq y)$$

     where $\sqsubseteq$ denotes the $(\alpha)$-automatic well-order from Lemma  11. Due to the previous lemma for every $p\in{{\Sigma}^{(\alpha)}}$ there is a unique $x$ satisfying $\psi_{\min}(x,p)$. Moreover, the map $f:(a,b)\mapsto ax+b$ is injective when the domain is restricted to words with support contained in $\mathsf{supp}(p)$.

     Since $\sqsubseteq$ is $(\alpha)$-automatic and $(\alpha)$-automatic structures are close under first-order definitions, there is an automaton $\mathcal{A}_{\varphi}$ corresponding to the following formula

     $$\varphi(p,a,b,c)=a\subseteq_{\mathsf{supp}}p\land b\subseteq_{\mathsf{supp}}p\land\exists x\,\left(\psi_{\min}(x,P)\land c=ax+b\right).$$

     For each finite set $X$ of ordinals, choose an $(\alpha)$-word $p_{X}$ such that $\mathsf{supp}(p)=X$. Set

     	$\displaystyle D_{X}$	$\displaystyle=\Set{d\in D}{c\subseteq_{\mathsf{supp}}p}\text{\ and}$
     	$\displaystyle F_{X}$	$\displaystyle=\Set{c\in D}{\exists a,b\in{{\Sigma}^{(\alpha)}}\,\mathcal{A}_{\varphi}\text{\ accepts }(p_{X},a,b,c)}$

     Since we are dealing with an injective presentation, Proposition 15 implies that for every $a,b\in D_{X}$ there is some $c_{a,b}\in F_{X}$ such that $\mathcal{A}_{\varphi}$ accepts $(p_{X},a,b,c_{a,b})$ and $\mathsf{supp}(c_{a,b})\subseteq U_{K}(\mathsf{supp}(p_{X})\cup\mathsf{supp}(a)\cup\mathsf{supp}(c),\alpha)=U_{K}(X,\alpha)$. Moreover, $c_{a,b}=c_{a^{\prime},b^{\prime}}$ implies $a=a^{\prime}$ and $b=b^{\prime}$ whence we conclude that

     $$\lvert\Set{d\in D}{\mathsf{supp}(d)\subseteq U_{K}(X,\alpha)}\rvert\geq\lvert F_{X}\rvert\geq\lvert D_{X}\rvert^{2}.$$

     ∎

     Remark 22.

     We crucially rely on $\alpha<{\omega_{1}+\omega^{\omega}}$ because otherwise we cannot be sure that there is an automaton corresponding to $\psi(x,p)$.

     Corollary 23.

     Let $\alpha<{\omega_{1}+\omega^{\omega}}$. There is no $(\alpha)$-automatic infinite integral domain. In particular, there is no $(\alpha)$-automatic infinite field.

     Proof.

     Assume $D$ is the domain of an $(\alpha)$-automatic infinite integral domain. Choose two elements $d_{1}\neq d_{2}$ from $D$ and let $X=\mathsf{supp}(d_{1})\cup\mathsf{supp}(d_{2})$. Set

     $$F_{0}=\Set{d\in D}{\mathsf{supp}(d)\subseteq\mathsf{supp}(d_{1})\cup\mathsf{supp}(d_{2})}.$$

     Iterated application of the previous lemma yields that

     $$F_{i}:=\Set{d\in D}{\mathsf{supp}(d)\subseteq U^{i}_{K}(\mathsf{supp}(F_{0}),\alpha)}$$

     satisfies $\lvert F_{i+1}\rvert\geq\lvert F_{i}\rvert^{2}$. Since $\lvert F_{0}\rvert\geq 2$ we conclude that $\lvert F_{n}\rvert\geq 2^{2^{n}}$ and $\mathsf{supp}(F_{n})\subseteq U^{n}_{K}(X,\alpha)$. From Lemma 14 we conclude that there are only $2^{p(n)}$ many elements in $D$ with support in $U^{n}_{K}(X,\alpha)$ for some polynomial $p(n)$ which results in a contradiction for large $n$. ∎

     6. Optimality of the Bound on the Growth-Rate

     Recall that word-automatic structures satisfy that $\nu_{\mathcal{F}}^{\Phi}(n)<n\cdot k$ for some constant $k$ where $\mathcal{F}$ is a family as before and $\Phi$ is a finite set of word-automatic relations. In contrast, our bound for $(\alpha)$-automatic structures is only $n^{k}$ for every constant $k>1$. In this section, we give an example that, if $\alpha\geq\omega^{2}$, then there are $(\alpha)$-automatic structures violating any bound of the form $n\cdot k$ for every constant $k$.²²2It is easily shown that for $\alpha<\omega^{2}$ every $(\alpha)$-automatic structure is word-automatic and vice versa whence the stronger bound from the word-automatic case applies.

     Definition 24.

     For every $n\in\mathbb{N}$, let

     $$D_{n}=\Set{\omega n_{1}+n_{2}}{n_{1}+n_{2}\leq n}$$

     and let $T_{n}$ be the set $(\omega^{2})$-words over $\Sigma=\Set{a,b,\diamond}$ such that $w\in T_{n}$ if and only if $\mathsf{supp}(w)=D_{n}$. For $i\in\{a,b\}$, we also define functions

     	$\displaystyle f_{i}:{{\Sigma}^{(\omega^{2})}}\times{{\Sigma}^{(\omega^{2})}}\to{{\Sigma}^{(\omega^{2})}}\text{ by}$
     	$\displaystyle f_{i}(w,v)(\alpha)=\begin{cases}i&\text{if }\alpha=0,\\ w(\omega n_{1}+n_{2})&\text{if }\alpha=\omega n_{1}+n_{2}+1\\ v(\omega n_{1})&\text{if }\alpha=\omega(n_{1}+1).\end{cases}$

     It is not difficult to see that the graphs of $f_{a}$ and $f_{b}$ are $(\omega^{2})$-automatic relations. Let $\Phi$ consist of two automata, one corresponding to the graph of $f_{a}$ and one corresponding to the graph of $f_{b}$. It is straightforward to verify that $T_{n+1}=f_{a}(T_{n}\times T_{n})\cup f_{b}(T_{n}\times T_{n})$. Moreover, since $f_{a}$ and $f_{b}$ are functions, it follows that any maximal $T_{n}$-$\Phi$-free set is of the form $T_{n+1}\cup\{w\}$ for some $(\omega^{2})$-word $w\notin T_{n+1}$. A simple calculation shows that $\lvert D_{n}\rvert=\frac{(n+1)(n+2)}{2}$ whence $\lvert T_{n}\rvert=2^{\frac{(n+1)(n+2)}{2}}$.

     Corollary 25.

     Setting $\mathcal{F}$ to be the family of the sets $T_{n}$ for every $n\in\mathbb{N}$, we obtain that $\nu_{\mathcal{F}}^{\Phi}(m)=\begin{cases}m\cdot 2^{n+2}&\text{if }m=2^{\frac{(n+1)(n+2)}{2}},\\ \infty&\text{otherwise.}\end{cases}$

     Proof.

     Just note that $\lvert T_{n+1}\rvert=2^{\frac{(n+3)(n+2)}{2}}=2^{\frac{(n+1)(n+2)}{2}+(n+2)}=\lvert T_{n}\rvert\cdot 2^{n+2}$. ∎

     Since there for every constant $k$ there is some value $n_{0}\in\mathbb{N}$ such that $2^{n+2}\geq k$ this shows that $\nu_{\mathcal{F}}^{\Phi}(m)\leq m\cdot k$ only holds for finitely many $m\in\mathbb{N}$. This shows that there is no word-automatic presentation of the $(\omega^{2})$-automatic structure $({{\Sigma}^{(\omega^{2})}},f_{a},f_{b})$ and that the bound in Theorem 17 cannot be replaced by $n\cdot k$.

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