Transition Property for $\alpha$-Power Free Languages with $\alpha \ge 2$ and $k\ge 3$ Letters

Abstract

In 1985, Restivo and Salemi presented a list of five problems concerning power free languages. Problem 4 states: Given $\alpha$-power-free words u and v, decide whether there is a transition from u to v. Problem 5 states: Given $\alpha$-power-free words u and v, find a transition word w, if it exists.

Let $\varSigma _k$ denote an alphabet with k letters. Let $L_{k,\alpha }$ denote the $\alpha$-power free language over the alphabet $\varSigma _k$, where $\alpha$ is a rational number or a rational “number with $+$”. If $\alpha$ is a “number with $+$” then suppose $k\ge 3$ and $\alpha \ge 2$. If $\alpha$ is “only” a number then suppose $k=3$ and $\alpha >2$ or $k>3$ and $\alpha \ge 2$. We show that: If $u\in L_{k,\alpha }$ is a right extendable word in $L_{k,\alpha }$ and $v\in L_{k,\alpha }$ is a left extendable word in $L_{k,\alpha }$ then there is a (transition) word w such that $uwv\in L_{k,\alpha }$. We also show a construction of the word w.

1 Introduction

The power free words are one of the major themes in the area of combinatorics on words. An $\alpha$-power of a word r is the word $r^{\alpha }=rr\dots rt$ such that $\frac{\vert r^{\alpha }\vert }{\vert r\vert }=\alpha$ and t is a prefix of r, where $\alpha \ge 1$ is a rational number. For example $(1234)^{3}=123412341234$ and $(1234)^{\frac{7}{4}}=1234123$. We say that a finite or infinite word w is $\alpha$-power free if w has no factors that are $\beta$-powers for $\beta \ge \alpha$ and we say that a finite or infinite word w is $\alpha ^+$-power free if w has no factors that are $\beta$-powers for $\beta >\alpha$, where $\alpha , \beta \ge 1$ are rational numbers. In the following, when we write “$\alpha$-power free” then $\alpha$ denotes a number or a “number with $+$”. The power free words, also called repetitions free words, include well known square free (2-power free), overlap free ($2^+$-power free), and cube free words (3-power free). Two surveys on the topic of power free words can be found in [8] and [13].

One of the questions being researched is the construction of infinite power free words. We define the repetition threshold ${{\,\mathrm{RT}\,}}(k)$ to be the infimum of all rational numbers $\alpha$ such that there exists an infinite $\alpha$-power-free word over an alphabet with k letters. Dejean’s conjecture states that ${{\,\mathrm{RT}\,}}(2)=2$, ${{\,\mathrm{RT}\,}}(3)=\frac{7}{4}$, ${{\,\mathrm{RT}\,}}(4)=\frac{7}{5}$, and ${{\,\mathrm{RT}\,}}(k)=\frac{k}{k-1}$ for each $k>4$ [3]. Dejean’s conjecture has been proved with the aid of several articles [1,2,3, 5, 6, 9].

It is easy to see that $\alpha$-power free words form a factorial language [13]; it means that all factors of a $\alpha$-power free word are also $\alpha$-power free words. Then Dejean’s conjecture implies that there are infinitely many finite $\alpha$-power free words over $\varSigma _k$, where $\alpha >{{\,\mathrm{RT}\,}}(k)$.

In [10], Restivo and Salemi presented a list of five problems that deal with the question of extendability of power free words. In the current paper we investigate Problem 4 and Problem 5:

• Problem 4: Given $\alpha$-power-free words u and v, decide whether there is a transition word w, such that uwu is $\alpha$-power free.

• Problem 5: Given $\alpha$-power-free words u and v, find a transition word w, if it exists.

A recent survey on the progress of solving all the five problems can be found in [7]; in particular, the problems 4 and 5 are solved for some overlap free ($2^+$-power free) binary words. In addition, in [7] the authors prove that: For every pair (u, v) of cube free words (3-power free) over an alphabet with k letters, if u can be infinitely extended to the right and v can be infinitely extended to the left respecting the cube-freeness property, then there exists a “transition” word w over the same alphabet such that uwv is cube free.

In 2009, a conjecture related to Problems 4 and Problem 5 of Restivo and Salemi appeared in [12]:

Conjecture 1

[12, Conjecture 1] Let L be a power-free language and let $e(L)\subseteq L$ be the set of words of L that can be extended to a bi-infinite word respecting the given power-freeness. If $u, v \in e(L)$ then $uwv \in e(L)$ for some word w.

In 2018, Conjecture 1 was presented also in [11] in a slightly different form.

Let $\mathbb {N}$ denote the set of natural numbers and let $\mathbb {Q}$ denote the set of rational numbers.

Definition 1

Let

$$ \begin{aligned} \varUpsilon = \{(k,\alpha )\mid k\in \mathbb {N} \text{ and } \alpha \in \mathbb {Q} \text{ and } k=3 \text{ and } \alpha>2\}\\ \cup \{(k,\alpha )\mid k\in \mathbb {N} \text{ and } \alpha \in \mathbb {Q} \text{ and } k>3 \text{ and } \alpha \ge 2\}\\ \cup \{(k,\alpha ^+)\mid k\in \mathbb {N} \text{ and } \alpha \in \mathbb {Q} \text{ and } k\ge 3 \text{ and } \alpha \ge 2\}\text{. } \end{aligned} $$

Remark 1

The definition of $\varUpsilon$ says that: If $(k,\alpha )\in \varUpsilon$ and $\alpha$ is a “number with $+$” then $k\ge 3$ and $\alpha \ge 2$. If $(k,\alpha )\in \varUpsilon$ and $\alpha$ is “just” a number then $k=3$ and $\alpha >2$ or $k>3$ and $\alpha \ge 2$.

Let ${{\,\mathrm{L}\,}}$ be a language. A finite word $w\in {{\,\mathrm{L}\,}}$ is called left extendable (resp., right extendable) in ${{\,\mathrm{L}\,}}$ if for every $n\in \mathbb {N}$ there is a word $u\in {{\,\mathrm{L}\,}}$ with $\vert u\vert =n$ such that $uw\in {{\,\mathrm{L}\,}}$ (resp., $wu\in {{\,\mathrm{L}\,}}$).

In the current article we improve the results addressing Problems 4 and Problem 5 of Restivo and Salemi from [7] as follows. Let $\varSigma _k$ denote an alphabet with k letters. Let ${{\,\mathrm{L}\,}}_{k,\alpha }$ denote the $\alpha$-power free language over the alphabet $\varSigma _k$. We show that if $(k,\alpha )\in \varUpsilon$, $u\in {{\,\mathrm{L}\,}}_{k,\alpha }$ is a right extendable word in ${{\,\mathrm{L}\,}}_{k,\alpha }$, and $v\in {{\,\mathrm{L}\,}}_{k,\alpha }$ is a left extendable word in ${{\,\mathrm{L}\,}}_{k,\alpha }$ then there is a word w such that $uwv\in {{\,\mathrm{L}\,}}_{k,\alpha }$. We also show a construction of the word w.

We sketch briefly our construction of a “transition” word. Let u be a right extendable $\alpha$-power free word and let v be a left extendable $\alpha$-power free word over $\varSigma _k$ with $k>2$ letters. Let $\bar{u}$ be a right infinite $\alpha$-power free word having u as a prefix and let $\bar{v}$ be a left infinite $\alpha$-power free word having v as a suffix. Let x be a letter that is recurrent in both $\bar{u}$ and $\bar{v}$. We show that we may suppose that $\bar{u}$ and $\bar{v}$ have a common recurrent letter. Let t be a right infinite $\alpha$-power free word over $\varSigma _{k}\setminus \{x\}$. Let $\bar{t}$ be a left infinite $\alpha$-power free word such that the set of factors of $\bar{t}$ is a subset of the set of recurrent factors of t. We show that such $\bar{t}$ exists. We identify a prefix $\tilde{u}xg$ of $\bar{u}$ such that g is a prefix of t and $\tilde{u}xt$ is a right infinite $\alpha$-power free word. Analogously we identify a suffix $\bar{g}x\tilde{v}$ of $\bar{v}$ such that $\bar{g}$ is a suffix of $\bar{t}$ and $\bar{t}x\tilde{v}$ is a left infinite $\alpha$-power free word. Moreover our construction guarantees that u is a prefix of $\tilde{u}xt$ and v is a suffix of $\bar{t}x\tilde{v}$. Then we find a prefix hp of t such that $px\tilde{v}$ is a suffix of $\bar{t}x\tilde{v}$ and such that both h and p are “sufficiently long”. Then we show that $\tilde{u}xhpx\tilde{v}$ is an $\alpha$-power free word having u as a prefix and v as a suffix.

The very basic idea of our proof is that if u, v are $\alpha$-power free words and x is a letter such that x is not a factor of both u and v, then clearly uxv is $\alpha$-power free on condition that $\alpha \ge 2$. Just note that there cannot be a factor in uxv which is an $\alpha$-power and contains x, because x has only one occurrence in uxv. Our constructed words $\tilde{u}xt$, $\bar{t}x\tilde{v}$, and $\tilde{u}xhpx\tilde{v}$ have “long” factors which does not contain a letter x. This will allow us to apply a similar approach to show that the constructed words do not contain square factor rr such that r contains the letter x.

Another key observation is that if $k\ge 3$ and $\alpha >{{\,\mathrm{RT}\,}}(k-1)$ then there is an infinite $\alpha$-power free word $\bar{w}$ over $\varSigma _k\setminus \{x\}$, where $x\in \varSigma _k$. This is an implication of Dejean’s conjecture. Less formally said, if u, v are $\alpha$-power free words over an alphabet with k letters, then we construct a “transition” word w over an alphabet with $k-1$ letters such that uwv is $\alpha$-power free.

Dejean’s conjecture imposes also the limit to possible improvement of our construction. The construction cannot be used for ${{\,\mathrm{RT}\,}}(k)\le \alpha <{{\,\mathrm{RT}\,}}(k-1)$, where $k\ge 3$, because every infinite (or “sufficiently long”) word w over an alphabet with $k-1$ letters contains a factor which is an $\alpha$-power. Also for $k=2$ and $\alpha \ge 1$ our technique fails. On the other hand, based on our research, it seems that our technique, with some adjustments, could be applied also for ${{\,\mathrm{RT}\,}}(k-1)\le \alpha \le 2$ and $k\ge 3$. Moreover it seems to be possible to generalize our technique to bi-infinite words and consequently to prove Conjecture 1 for $k\ge 3$ and $\alpha \ge {{\,\mathrm{RT}\,}}(k-1)$.

2 Preliminaries

Recall that $\varSigma _k$ denotes an alphabet with k letters. Let $\epsilon$ denote the empty word. Let $\varSigma _k^*$ denote the set of all finite words over $\varSigma _k$ including the empty word $\epsilon$, let $\varSigma _k^{\mathbb {N},R}$ denote the set of all right infinite words over $\varSigma _k$, and let $\varSigma _k^{\mathbb {N},L}$ denote the set of all left infinite words over $\varSigma _k$. Let $\varSigma _k^{\mathbb {N}}=\varSigma _k^{\mathbb {N},L}\cup \varSigma _k^{\mathbb {N},R}$. We call $w\in \varSigma _k^{\mathbb {N}}$ an infinite word.

Let ${{\,\mathrm{occur}\,}}(w,t)$ denote the number of occurrences of the nonempty factor $t\in \varSigma _k^*\setminus \{\epsilon \}$ in the word $w\in \varSigma _k^*\cup \varSigma _k^{\mathbb {N}}$. If $w\in \varSigma _k^{\mathbb {N}}$ and ${{\,\mathrm{occur}\,}}(w,t)=\infty$, then we call t a recurrent factor in w.

Let ${{\,\mathrm{F}\,}}(w)$ denote the set of all finite factors of a finite or infinite word $w\in \varSigma _k^*\cup \varSigma _k^{\mathbb {N}}$. The set ${{\,\mathrm{F}\,}}(w)$ contains the empty word and if w is finite then also $w\in {{\,\mathrm{F}\,}}(w)$. Let ${{\,\mathrm{F}\,}}_r(w)\subseteq {{\,\mathrm{F}\,}}(w)$ denote the set of all recurrent nonempty factors of $w\in \varSigma _k^{\mathbb {N}}$.

Let ${{\,\mathrm{Prf}\,}}(w)\subseteq {{\,\mathrm{F}\,}}(w)$ denote the set of all prefixes of $w\in \varSigma _k^*\cup \varSigma _k^{\mathbb {N},R}$ and let ${{\,\mathrm{Suf}\,}}(w)\subseteq {{\,\mathrm{F}\,}}(w)$ denote the set of all suffixes of $w\in \varSigma _k^*\cup \varSigma _k^{\mathbb {N},L}$. We define that $\epsilon \in {{\,\mathrm{Prf}\,}}(w)\,\cap \,{{\,\mathrm{Suf}\,}}(w)$ and if w is finite then also $w\in {{\,\mathrm{Prf}\,}}(w)\,\cap \,{{\,\mathrm{Suf}\,}}(w)$.

We have that ${{\,\mathrm{L}\,}}_{k,\alpha }\subseteq \varSigma _k^*$. Let ${{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N}}\subseteq \varSigma _k^{\mathbb {N}}$ denote the set of all infinite $\alpha$-power free words over $\varSigma _k$. Obviously ${{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N}}=\{w\in \varSigma _k^{\mathbb {N}}\mid {{\,\mathrm{F}\,}}(w)\subseteq {{\,\mathrm{L}\,}}_{k,\alpha }\}$. In addition we define ${{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}={{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N}}\,\cap \,\varSigma _k^{\mathbb {N},R}$ and ${{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}={{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N}}\,\cap \,\varSigma _k^{\mathbb {N},L}$; it means the sets of right infinite and left infinite $\alpha$-power free words.

3 Power Free Languages

Let $(k,\alpha )\in \varUpsilon$ and let u, v be $\alpha$-power free words. The first lemma says that uv is $\alpha$-power free if there are no word r and no nonempty prefix $\bar{v}$ of v such that rr is a suffix of $u\bar{v}$ and rr is longer than $\bar{v}$.

Lemma 1

Suppose $(k,\alpha )\in \varUpsilon$, $u\in {{\,\mathrm{L}\,}}_{k,\alpha }$, and $v\in {{\,\mathrm{L}\,}}_{k,\alpha }\cup {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$. Let

$$ \begin{aligned}\varPi =\{(r,\bar{v})\mid r\in \varSigma _k^*\setminus \{\epsilon \} \text{ and } \bar{v}\in {{\,\mathrm{Prf}\,}}(v)\setminus \{\epsilon \} \text{ and } \\ rr\in {{\,\mathrm{Suf}\,}}(u\bar{v}) \text{ and } \vert rr\vert >\vert \bar{v}\vert \}\text{. }\end{aligned}$$

If $\varPi =\emptyset$ then $uv\in {{\,\mathrm{L}\,}}_{k,\alpha }\cup {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$.

Proof

Suppose that uv is not $\alpha$-power free. Since u is $\alpha$-power free, then there are $t\in \varSigma _k^*$ and $x\in \varSigma _k$ such that $tx\in {{\,\mathrm{Prf}\,}}(v)$, $ut\in {{\,\mathrm{L}\,}}_{k,\alpha }$ and $utx\not \in {{\,\mathrm{L}\,}}_{k,\alpha }$. It means that there is $r\in {{\,\mathrm{Suf}\,}}(utx)$ such that $r^{\beta }\in {{\,\mathrm{Suf}\,}}(utx)$ for some $\beta \ge \alpha$ or $\beta >\alpha$ if $\alpha$ is a “number with $+$”; recall Definition 1 of $\varUpsilon$. Because $\alpha \ge 2$, this implies that $rr\in {{\,\mathrm{Suf}\,}}(r^{\beta })$. If follows that $(tx,r)\in \varPi$. We proved that $uv\not \in {{\,\mathrm{L}\,}}_{k,\alpha }\cup {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$ implies that $\varPi \not =\emptyset$. The lemma follows.    $\square$

The following technical set $\varGamma (k,\alpha )$ of 5-tuples $(w_1,w_2,x,g,t)$ will simplify our propositions.

Definition 2

Given $(k,\alpha )\in \varUpsilon$, we define that $(w_1,w_2,x,g,t)\in \varGamma (k,\alpha )$ if

• 1. $w_1,w_2,g\in \varSigma _k^*$,

• 2. $x\in \varSigma _k$,

• 3. $w_1w_2xg\in {{\,\mathrm{L}\,}}_{k,\alpha }$,

• 4. $t\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$,

• 5. ${{\,\mathrm{occur}\,}}(t,x)=0$,

• 6. $g\in {{\,\mathrm{Prf}\,}}(t)$,

• 7. ${{\,\mathrm{occur}\,}}(w_2xgy,xgy)=1$, where $y\in \varSigma _k$ is such that $gy\in {{\,\mathrm{Prf}\,}}(t)$, and

• 8. ${{\,\mathrm{occur}\,}}(w_2,x)\ge {{\,\mathrm{occur}\,}}(w_1,x)$.

Remark 2

Less formally said, the 5-tuple $(w_1,w_2,x,g,t)$ is in $\varGamma (k,\alpha )$ if $w_1w_2xg$ is $\alpha$-power free word over $\varSigma _k$, t is a right infinite $\alpha$-power free word over $\varSigma _k$, t has no occurrence of x (thus t is a word over $\varSigma _k\setminus \{x\}$), g is a prefix of t, xgy has only one occurrence in $w_2xgy$, where y is a letter such that gy is a prefix of t, and the number of occurrences of x in $w_2$ is bigger than the number of occurrences of x in $w_1$, where $w_1,w_2, g$ are finite words and x is a letter.

The next proposition shows that if $(w_1,w_2,x,g,t)$ is from the set $\varGamma (k,\alpha )$ then $w_1w_2xt$ is a right infinite $\alpha$-power free word, where $(k,\alpha )$ is from the set$\varUpsilon$.

Proposition 1

If $(k,\alpha )\in \varUpsilon$ and $(w_1,w_2,x,g,t)\in \varGamma (k,\alpha )$ then $w_1w_2xt\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$.

Proof

Lemma 1 implies that it suffices to show that there are no $u\in {{\,\mathrm{Prf}\,}}(t)$ with $\vert u\vert >\vert g\vert$ and no $r\in \varSigma _k^*\setminus \{\epsilon \}$ such that $rr\in {{\,\mathrm{Suf}\,}}(w_1w_2xu)$ and $\vert rr\vert >\vert u\vert$. Recall that $w_1w_2xg$ is an $\alpha$-power free word, hence we consider $\vert u\vert >\vert g\vert$. To get a contradiction, suppose that such r, u exist. We distinguish the following distinct cases.

• If $\vert r\vert \le \vert u\vert$ then: Since $u\in {{\,\mathrm{Prf}\,}}(t)\subseteq {{\,\mathrm{L}\,}}_{k,\alpha }$ it follows that $xu\in {{\,\mathrm{Suf}\,}}(r^2)$ and hence $x\in {{\,\mathrm{F}\,}}(r^2)$. It is clear that ${{\,\mathrm{occur}\,}}(r^2,x)\ge 1$ if and only if ${{\,\mathrm{occur}\,}}(r,x)\ge 1$. Since $x\not \in {{\,\mathrm{F}\,}}(u)$ and thus $x\not \in {{\,\mathrm{F}\,}}(r)$, this is a contradiction.

• If $\vert r\vert >\vert u\vert$ and $rr\in {{\,\mathrm{Suf}\,}}(w_2xu)$ then: Let $y\in \varSigma _k$ be such that $gy\in {{\,\mathrm{Prf}\,}}(t)$. Since $\vert u\vert >\vert g\vert$ we have that $gy\in {{\,\mathrm{Prf}\,}}(u)$ and $xgy\in {{\,\mathrm{Prf}\,}}(xu)$. Since $\vert r\vert >\vert u\vert$ we have that $xgy\in {{\,\mathrm{F}\,}}(r)$. In consequence ${{\,\mathrm{occur}\,}}(rr,xgy)\ge 2$. But Property 7 of Definition 2 states that ${{\,\mathrm{occur}\,}}(w_2xgy, xgy)=1$. Since $rr\in {{\,\mathrm{Suf}\,}}(w_2xu)$, this is a contradiction.

• If $\vert r\vert >\vert u\vert$ and $rr\not \in {{\,\mathrm{Suf}\,}}(w_2xu)$ and $r\in {{\,\mathrm{Suf}\,}}(w_2xu)$ then: Let $w_{11}, w_{12}, w_{13}, w_{21}, w_{22}\in \varSigma _k^*$ be such that $w_1=w_{11}w_{12}w_{13}$, $w_2=w_{21}w_{22}$, $w_{12}w_{13}w_{21}=r$, $w_{12}w_{13}w_{2}xu=rr$, and $w_{13}w_{21}=xu$; see Figure below.

  		xu
  $w_{11}$	$w_{12}$	$w_{13}$	$w_{21}$	$w_{22}$	x	u
  	r			r

  It follows that $w_{22}xu=r$ and $w_{22}=w_{12}$. It is easy to see that $w_{13}w_{21}=xu$. From ${{\,\mathrm{occur}\,}}(u,x)=0$ we have that ${{\,\mathrm{occur}\,}}(w_2,x)={{\,\mathrm{occur}\,}}(w_{22},x)$ and ${{\,\mathrm{occur}\,}}(w_{13},x)=1$. From $w_{22}=w_{12}$ it follows that ${{\,\mathrm{occur}\,}}(w_1,x)>{{\,\mathrm{occur}\,}}(w_2,x)$. This is a contradiction to Property 8 of Definition 2.

• If $\vert r\vert >\vert u\vert$ and $rr\not \in {{\,\mathrm{Suf}\,}}(w_2xu)$ and $r\not \in {{\,\mathrm{Suf}\,}}(w_2xu)$ then: Let $w_{11}, w_{12}, w_{13}\in \varSigma _k^*$ be such that $w_1=w_{11}w_{12}w_{13}$, $w_{12}=r$ and $w_{13}w_2xu=r$; see Figure below.

  $w_{11}$	$w_{12}$	$w_{13}$	$w_{2}$	x	u
  	r	r

  It follows that

  $$\begin{aligned} {{\,\mathrm{occur}\,}}(w_{12},x)={{\,\mathrm{occur}\,}}(w_{13},x)+{{\,\mathrm{occur}\,}}(w_2,x)+{{\,\mathrm{occur}\,}}(xu,x)\text{. } \end{aligned}$$

  This is a contradiction to Property 8 of Definition 2.

We proved that the assumption of existence of r, u leads to a contradiction. Thus we proved that for each prefix $u\in {{\,\mathrm{Prf}\,}}(t)$ we have that $w_1w_2xu\in {{\,\mathrm{L}\,}}_{k,\alpha }$. The proposition follows.    $\square$

We prove that if $(k,\alpha )\in \varUpsilon$ then there is a right infinite $\alpha$-power free word over $\varSigma _{k-1}$. In the introduction we showed that this observation could be deduced from Dejean’s conjecture. Here additionally, to be able to address Problem 5 from the list of Restivo and Salemi, we present in the proof also examples of such words.

Lemma 2

If $(k,\alpha )\in \varUpsilon$ then the set ${{\,\mathrm{L}\,}}_{k-1,\alpha }^{\mathbb {N},R}$ is not empty.

Proof

If $k=3$ then $\vert \varSigma _{k-1}\vert = 2$. It is well known that the Thue Morse word is a right infinite $2^+$-power free word over an alphabet with 2 letters [11]. It follows that the Thue Morse word is $\alpha$-power free for each $\alpha >2$.

If $k>3$ then $\vert \varSigma _{k-1}\vert \ge 3$. It is well known that there are infinite 2-power free words over an alphabet with 3 letters [11]. Suppose $0,1,2\in \varSigma _k$. An example is the fixed point of the morphism $\theta$ defined by $\theta (0)=012$, $\theta (1)=02$, and $\theta (2)=1$ [11]. If an infinite word t is 2-power free then obviously t is $\alpha$-power free and $\alpha ^+$-power free for each $\alpha \ge 2$.

This completes the proof.    $\square$

We define the sets of extendable words.

Definition 3

Let ${{\,\mathrm{L}}}\subseteq \varSigma _k^*$. We define

$$\begin{aligned} {{\,\mathrm{lext}}}({{\mathrm{L}}})=\{w\in {{\,\mathrm{L}\,}}\mid w \text{ is } \text{ left } \text{ extendable } \text{ in } {{\,\mathrm{L}}}\} \end{aligned}$$

and

$$\begin{aligned} {{\,\mathrm{rext}}}({{\mathrm{L}}})=\{w\in {{\,\mathrm{L}\,}}\mid w \text{ is } \text{ right } \text{ extendable } \text{ in } {{\,\mathrm{L}}}\}\text{. } \end{aligned}$$

If $u\in {{\,\mathrm{lext}\,}}({{\mathrm{L}}})$ then let ${{\,\mathrm{lext}\,}}(u,{{\mathrm{L}}})$ be the set of all left infinite words $\bar{u}$ such that ${{\,\mathrm{Suf}\,}}(\bar{u})\subseteq {{\mathrm{L}}}$ and $u\in {{\,\mathrm{Suf}\,}}(\bar{u})$. Analogously if $u\in {{\,\mathrm{rext}\,}}({{\mathrm{L}}})$ then let ${{\,\mathrm{rext}\,}}(u,{{\mathrm{L}}})$ be the set of all right infinite words $\bar{u}$ such that ${{\,\mathrm{Prf}\,}}(\bar{u})\subseteq {{\,\mathrm{L}}}$ and $u\in {{\,\mathrm{Prf}\,}}(\bar{u})$.

We show the sets ${{\,\mathrm{lext}\,}}(u,{{\,\mathrm{L}}})$ and ${{\,\mathrm{rext}\,}}(v,{{\,\mathrm{L}}})$ are nonempty for left extendable and right extendable words.

Lemma 3

If ${{\,\mathrm{L}\,}}\subseteq \varSigma _k^*$ and $u\in {{\,\mathrm{lext}\,}}({{\mathrm{L}}})$ (resp., $v\in {{\,\mathrm{rext}\,}}({{\mathrm{L}}})$) then ${{\,\mathrm{lext}\,}}(u,{{\,\mathrm{L}}})\not =\emptyset$ (resp., ${{\,\mathrm{rext}\,}}(v,{{\,\mathrm{L}}})\not =\emptyset$).

Proof

Realize that $u\in {{\,\mathrm{lext}\,}}({{\mathrm{L}}})$ (resp., $v\in {{\,\mathrm{rext}\,}}({{\mathrm{L}}})$) implies that there are infinitely many finite words in ${{\,\mathrm{L}}}$ having u as a suffix (resp., v as a prefix). Then the lemma follows from König’s Infinity Lemma [4, 8].    $\square$

The next proposition proves that if $(k,\alpha )\in \varUpsilon$, w is a right extendable $\alpha$-power free word, $\bar{w}$ is a right infinite $\alpha$-power free word having the letter x as a recurrent factor and having w as a prefix, and t is a right infinite $\alpha$-power free word over $\varSigma _k\setminus \{x\}$, then there are finite words $w_1,w_2,g$ such that the 5-tuple $(w_1,w_2,x,g,t)$ is in the set $\varGamma (k,\alpha )$ and w is a prefix of $w_1w_2xg$.

Proposition 2

If $(k,\alpha )\in \varUpsilon$, $w\in {{\,\mathrm{rext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$, $\bar{w}\in {{\,\mathrm{rext}\,}}(w,{{\,\mathrm{L}\,}}_{k,\alpha })$, $x\in {{\,\mathrm{F}\,}}_r(\bar{w})\,\cap \,\varSigma _k$, $t\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$, and ${{\,\mathrm{occur}\,}}(t,x)=0$ then there are finite words $w_1,w_2,g$ such that $(w_1,w_2,x,g,t)\in \varGamma (k,\alpha )$ and $w\in {{\,\mathrm{Prf}\,}}(w_1w_2xg)$.

Proof

Let $\omega ={{\,\mathrm{F}\,}}(\bar{w})\,\cap \,{{\,\mathrm{Prf}\,}}(xt)$ be the set of factors of $\bar{w}$ that are also prefixes of the word xt. Based on the size of the set $\omega$ we construct the words $w_1,w_2,g$ and we show that $(w_1,w_2,x,g,t)\in \varGamma (k,\alpha )$ and $w_1w_2xg\in {{\,\mathrm{Prf}\,}}(\bar{w})\subseteq {{\,\mathrm{L}\,}}_{k,\alpha }$. The Properties 1, 2, 3, 4, 5, and 6 of Definition 2 are easy to verify. Hence we explicitly prove only properties 7 and 8 and that $w\in {{\,\mathrm{Prf}\,}}(w_1w_2xg)$.

• If $\omega$ is an infinite set. It follows that ${{\,\mathrm{Prf}\,}}(xt)=\omega$. Let $g\in {{\,\mathrm{Prf}\,}}(t)$ be such that $\vert g\vert =\vert w\vert$; recall that t is infinite and hence such g exists. Let $w_2\in {{\,\mathrm{Prf}\,}}(\bar{w})$ be such that $w_2xg \in {{\,\mathrm{Prf}\,}}(\bar{w})$ and ${{\,\mathrm{occur}\,}}(w_2xg, xg)=1$. Let $w_1=\epsilon$. Property 7 of Definition 2 follows from ${{\,\mathrm{occur}\,}}(w_2xg, xg)=1$. Property 8 of Definition 2 is obvious, because $w_1$ is the empty word. Since $\vert g\vert =\vert w\vert$ and $w\in {{\,\mathrm{Prf}\,}}(\bar{w})$ we have that $w\in {{\,\mathrm{Prf}\,}}(w_1w_2xg)$.

• If $\omega$ is a finite set. Let $\bar{\omega }= \omega \,\cap \, {{\,\mathrm{F}\,}}_r(\bar{w})$ be the set of prefixes of xt that are recurrent in $\bar{w}$. Since x is recurrent in $\bar{w}$ we have that $x\in \bar{\omega }$ and thus $\bar{\omega }$ is not empty. Let $g\in {{\,\mathrm{Prf}\,}}(t)$ be such that xg is the longest element in $\bar{\omega }$. Let $w_1\in {{\,\mathrm{Prf}\,}}(w)$ be the shortest prefix of $\bar{w}$ such that if $u\in \omega \setminus \bar{\omega }$ is a non-recurrent prefix of xt in $\bar{w}$ then ${{\,\mathrm{occur}\,}}(w_1,u)={{\,\mathrm{occur}\,}}(\bar{w},u)$. Such $w_1$ obviously exists, because $\omega$ is a finite set and non-recurrent factors have only a finite number of occurrences. Let $w_2$ be the shortest factor of $\bar{w}$ such that $w_1w_2xg\in {{\,\mathrm{Prf}\,}}(\bar{w})$, ${{\,\mathrm{occur}\,}}(w_1,x)<{{\,\mathrm{occur}\,}}(w_2,x)$, and $w\in {{\,\mathrm{Prf}\,}}(w_1w_2xg)$. Since xg is recurrent in $\bar{w}$ and $w\in {{\,\mathrm{Prf}\,}}(\bar{w})$ it is clear such $w_2$ exists. We show that Property 7 of Definition 2 holds. Let $y\in \varSigma _k$ be such that $gy\in {{\,\mathrm{Prf}\,}}(t)$. Suppose that ${{\,\mathrm{occur}\,}}(w_2xg,xgy)>0$. It would imply that xgy is recurrent in $\bar{w}$, since all occurrences of non-recurrent words from $\omega$ are in $w_1$. But we defined xg to be the longest recurrent word $\omega$. Hence it is contradiction to our assumption that ${{\,\mathrm{occur}\,}}(w_2xg,xgy)>0$. Property 8 of Definition 2 and $w\in {{\,\mathrm{Prf}\,}}(w_1w_2xg)$ are obvious from the construction of $w_2$.

This completes the proof.    $\square$

We define the reversal $w^R$ of a finite or infinite word $w=\varSigma _k^*\cup \varSigma _k^{\mathbb {N}}$ as follows: If $w\in \varSigma _k^*$ and $w=w_1w_2\dots w_m$, where $w_i\in \varSigma _k$ and $1\le i\le m$, then $w^R=w_mw_{m-1}\dots w_2w_1$. If $w\in \varSigma _k^{\mathbb {N},L}$ and $w=\dots w_2w_1$, where $w_i\in \varSigma _k$ and $i\in \mathbb {N}$, then $w^R=w_1w_2\dots \in \varSigma _k^{\mathbb {N},R}$. Analogously if $w\in \varSigma _k^{\mathbb {N},R}$ and $w=w_1w_2\dots$, where $w_i\in \varSigma _k$ and $i\in \mathbb {N}$, then $w^R=\dots w_2w_1\in \varSigma _k^{\mathbb {N},L}$.

Proposition 1 allows one to construct a right infinite $\alpha$-power free word with a given prefix. The next simple corollary shows that in the same way we can construct a left infinite $\alpha$-power free word with a given suffix.

Corollary 1

If $(k,\alpha )\in \varUpsilon$, $w\in {{\,\mathrm{lext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$, $\bar{w}\in {{\,\mathrm{lext}\,}}(w,{{\,\mathrm{L}\,}}_{k,\alpha })$, $x\in {{\,\mathrm{F}\,}}_r(\bar{w})\,\cap \,\varSigma _k$, $t\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}$, and ${{\,\mathrm{occur}\,}}(t,x)=0$ then there are finite words $w_1,w_2,g$ such that $(w_1^R,w_2^R,x,g^R,t^R)\in \varGamma (k,\alpha )$, $w\in {{\,\mathrm{Suf}\,}}(gxw_2w_1)$, and $txw_2w_1\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}$.

Proof

Let $u\in \varSigma _k^*\cup \varSigma _k^{\mathbb {N}}$. Realize that $u\in {{\,\mathrm{L}\,}}_{k,\alpha }\cup {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N}}$ if and only if $u^R\in {{\,\mathrm{L}\,}}_{k,\alpha }\cup {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N}}$. Then the corollary follows from Proposition 1 and Proposition 2.    $\square$

Given $k\in \mathbb {N}$ and a right infinite word $t\in \varSigma _k^{\mathbb {N},R}$, let $\varPhi (t)$ be the set of all left infinite words $\bar{t}\in \varSigma _k^{\mathbb {N},L}$ such that ${{\,\mathrm{F}\,}}(\bar{t})\subseteq {{\,\mathrm{F}\,}}_r(t)$. It means that all factors of $\bar{t}\in \varPhi (t)$ are recurrent factors of t. We show that the set $\varPhi (t)$ is not empty.

Lemma 4

If $k\in \mathbb {N}$ and $t\in \varSigma _k^{\mathbb {N},R}$ then $\varPhi (t)\not =\emptyset$.

Proof

Since t is an infinite word, the set of recurrent factors of t is not empty. Let g be a recurrent nonempty factor of t; g may be a letter. Obviously there is $x\in \varSigma _k$ such that xg is also recurrent in t. This implies that the set $\{h\mid hg\in {{\,\mathrm{F}\,}}_r(t)\}$ is infinite. The lemma follows from König’s Infinity Lemma [4, 8].    $\square$

The next lemma shows that if u is a right extendable $\alpha$-power free word then for each letter x there is a right infinite $\alpha$-power free word $\bar{u}$ such that x is recurrent in $\bar{u}$ and u is a prefix of $\bar{u}$.

Lemma 5

If $(k,\alpha )\in \varUpsilon$, $u\in {{\,\mathrm{rext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$, and $x\in \varSigma _{k}$ then there is $\bar{u}\in {{\,\mathrm{rext}\,}}(u,{{\,\mathrm{L}\,}}_{k,\alpha })$ such that $x\in {{\,\mathrm{F}\,}}_r(\bar{u})$.

Proof

Let $w\in {{\,\mathrm{rext}\,}}(u,{{\,\mathrm{L}\,}}_{k,\alpha })$; Lemma 3 implies that ${{\,\mathrm{rext}\,}}(u,{{\,\mathrm{L}\,}}_{k,\alpha })$ is not empty. If $x\in {{\,\mathrm{F}\,}}_r(w)$ then we are done. Suppose that $x\not \in {{\,\mathrm{F}\,}}_r(w)$. Let $y\in {{\,\mathrm{F}\,}}_r(w)\,\cap \,\varSigma _k$. Clearly $x\not =y$. Proposition 2 implies that there is $(w_1,w_2,y,g,t)\in \varGamma (k,\alpha )$ such that $u\in {{\,\mathrm{Prf}\,}}(w_1w_2yg)$. The proof of Lemma 2 implies that we can choose t in such a way that x is recurrent in t. Then $w_1w_2yt\in {{\,\mathrm{rext}\,}}(u,{{\,\mathrm{L}\,}}_{k,\alpha })$ and $x\in {{\,\mathrm{F}\,}}_r(w_1w_2yt)$. This completes the proof.    $\square$

The next proposition shows that if u is left extendable and v is right extendable then there are finite words $\tilde{u}, \tilde{v}$, a letter x, a right infinite word t, and a left infinite word $\bar{t}$ such that $\tilde{u}xt, \bar{t}x\tilde{v}$ are infinite $\alpha$-power free words, t has no occurrence of x, every factor of $\bar{t}$ is a recurrent factor in t, u is a prefix of $\tilde{u}xt$, and v is a suffix of $\bar{t}x\tilde{v}$.

Proposition 3

If $(k,\alpha )\in \varUpsilon$, $u\in {{\,\mathrm{rext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$, and $v\in {{\,\mathrm{lext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$ then there are $\tilde{u}, \tilde{v}\in \varSigma _k^*$, $x\in \varSigma _k$, $t\in \varSigma _k^{\mathbb {N},R}$, and $\bar{t}\in \varSigma _k^{\mathbb {N},L}$ such that $\tilde{u}xt\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$, $\bar{t}x\tilde{v}\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}$, ${{\,\mathrm{occur}\,}}(t,x)=0$, ${{\,\mathrm{F}\,}}(\bar{t})\subseteq {{\,\mathrm{F}\,}}_r(t)$, $u\in {{\,\mathrm{Prf}\,}}(\tilde{u}xt)$, and $v\in {{\,\mathrm{Suf}\,}}(\bar{t}x\tilde{v})$.

Proof

Let $\bar{u}\in {{\,\mathrm{rext}\,}}(u,{{\,\mathrm{L}\,}}_{k,\alpha })$ and $\bar{v}\in {{\,\mathrm{lext}\,}}(v,{{\,\mathrm{L}\,}}_{k,\alpha })$ be such that ${{\,\mathrm{F}\,}}_r(\bar{u})\,\cap \,{{\,\mathrm{F}\,}}_r(\bar{v})\,\cap \, \varSigma _k\not =\emptyset$. Lemma 5 implies that such $\bar{u}, \bar{v}$ exist. Let $x\in {{\,\mathrm{F}\,}}_r(\bar{u})\,\cap \,{{\,\mathrm{F}\,}}_r(\bar{v})\,\cap \, \varSigma _k$. It means that the letter x is recurrent in both $\bar{u}$ and $\bar{v}$.

Let t be a right infinite $\alpha$-power free word over $\varSigma _k\setminus \{x\}$. Lemma 2 asserts that such t exists. Let $\bar{t}\in \varPhi (t)$; Lemma 4 shows that $\varPhi (t)\not =\emptyset$. It is easy to see that $\bar{t}\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}$, because ${{\,\mathrm{F}\,}}(\bar{t})\subseteq {{\,\mathrm{F}\,}}_r(t)$ and $t\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$.

Proposition 2 and Corollary 1 imply that there are $u_1,u_2,g,v_1,v_2,\bar{g}\in {{\,\mathrm{L}\,}}_{k,\alpha }$ such that

• $(u_1,u_2,x,g,t)\in \varGamma (k,\alpha )$,

• $(v_1^R,v_2^R,x,\bar{g}^R,\bar{t}^R)\in \varGamma (k,\alpha )$,

• $u\in {{\,\mathrm{Prf}\,}}(u_1u_2xg)$, and

• $v^R\in {{\,\mathrm{Prf}\,}}(v_1^Rv_2^Rx\bar{g}^R)$; it follows that $v\in {{\,\mathrm{Suf}\,}}(\bar{g}xv_2v_1)$.

Proposition 1 implies that $u_1u_2xt, v_1^Rv_2^Rx\bar{t}^R\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$. It follows that $\bar{t}xv_2v_1\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}$. Let $\tilde{u}=u_1u_2$ and $\tilde{v}=v_2v_1$. This completes the proof.    $\square$

The main theorem of the article shows that if u is a right extendable $\alpha$-power free word and v is a left extendable $\alpha$-power free word then there is a word w such that uwv is $\alpha$-power free. The proof of the theorem shows also a construction of the word w.

Theorem 1

If $(k,\alpha )\in \varUpsilon$, $u\in {{\,\mathrm{rext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$, and $v\in {{\,\mathrm{lext}\,}}({{\,\mathrm{L}\,}}_{k,\alpha })$ then there is $w\in {{\,\mathrm{L}\,}}_{k,\alpha }$ such that $uwv\in {{\,\mathrm{L}\,}}_{k,\alpha }$.

Proof

Let $\tilde{u},\tilde{v},x,t,\bar{t}$ be as in Proposition 3. Let $p\in {{\,\mathrm{Suf}\,}}(\bar{t})$ be the shortest suffix such that $\vert p\vert >\max \{\vert \tilde{u}x\vert ,\vert x\tilde{v}\vert , \vert u\vert , \vert v\vert \}$. Let $h\in {{\,\mathrm{Prf}\,}}(t)$ be the shortest prefix such that $hp\in {{\,\mathrm{Prf}\,}}(t)$ and $\vert h\vert >\vert p\vert$; such h exists, because p is a recurrent factor of t; see Proposition 3. We show that $\tilde{u}xhpx\tilde{v}\in {{\,\mathrm{L}\,}}_{k,\alpha }$.

We have that $\tilde{u}xhp\in {{\,\mathrm{L}\,}}_{k,\alpha }$, since $hp\in {{\,\mathrm{Prf}\,}}(t)$ and Proposition 3 states that $\tilde{u}xt\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},R}$. Lemma 1 implies that it suffices to show that there are no $g\in {{\,\mathrm{Prf}\,}}(\tilde{v})$ and no $r\in \varSigma _k^*\setminus \{\epsilon \}$ such that $rr\in {{\,\mathrm{Suf}\,}}(\tilde{u}xhpxg)$ and $\vert rr\vert >\vert xg\vert$. To get a contradiction, suppose there are such r, g. We distinguish the following cases.

• If $\vert r\vert \le \vert xg\vert$ then $rr\in {{\,\mathrm{Suf}\,}}(pxg)$, because $\vert p\vert >\vert x\tilde{v}\vert$ and $xg\in {{\,\mathrm{Prf}\,}}(x\tilde{v})$. This is a contradiction, since $px\tilde{v}\in {{\,\mathrm{Suf}\,}}(\bar{t}x\tilde{v})$ and $\bar{t}x\tilde{v}\in {{\,\mathrm{L}\,}}_{k,\alpha }^{\mathbb {N},L}$; see Proposition 3.

• If $\vert r\vert >\vert xg\vert$ then $\vert r\vert \le \frac{1}{2}\vert \tilde{u}xhpxg\vert$, otherwise rr cannot be a suffix of $\tilde{u}xhpxg$. Because $\vert h\vert>\vert p\vert > \max \{\vert \tilde{u}x\vert ,\vert x\tilde{v}\vert \}$ we have that $r\in {{\,\mathrm{Suf}\,}}(hpxg)$. Since ${{\,\mathrm{occur}\,}}(hp,x)=0$, $\vert h\vert>\vert p\vert >\vert x\tilde{v}\vert$, and $xg\in {{\,\mathrm{Suf}\,}}(r)$ it follows that there are words $h_1,h_2$ such that $\tilde{u}xhpxg=\tilde{u}xh_1h_2pxg$, $r=h_2pxg$ and $r\in {{\,\mathrm{Suf}\,}}(\tilde{u}xh_1)$. It follows that $xg\in {{\,\mathrm{Suf}\,}}(\tilde{u}xh_1)$ and because ${{\,\mathrm{occur}\,}}(h_1,x)=0$ we have that $\vert h_1\vert \le \vert g\vert$. Since $\vert p\vert >\vert \tilde{u}x\vert$ we get that $\vert h_2pxg\vert >\vert \tilde{u}xg\vert \ge \vert \tilde{u}xh_1\vert$; hence $\vert r\vert >\vert \tilde{u}xh_1\vert$. This is a contradiction.

We conclude that there is no word r and no prefix $g\in {{\,\mathrm{Prf}\,}}(\tilde{v})$ such that $rr\in {{\,\mathrm{Suf}\,}}(\tilde{u}xhpxg)$. Hence $\tilde{u}xhpx\tilde{v}\in {{\,\mathrm{L}\,}}_{k,\alpha }$. Due to the construction of p and h we have that $u\in {{\,\mathrm{Prf}\,}}(\tilde{u}xhpx\tilde{v})$ and $v\in {{\,\mathrm{Suf}\,}}(\tilde{u}xhpx\tilde{v})$. This completes the proof.    $\square$