Do any of these sequences have infinitely-many distinct iterates under run-length substitution?

Question

Let $$S = \{x \in \{1,2\}^\mathbb{N}: \ \text{every run in }x\text{ has finite length}\}$$ and define $$T: S\to \mathbb{N}^\mathbb{N} $$ such that for any $x\in S$, ${T}x$ is the sequence of run-lengths in $x$; that is, $Tx$ is the result of replacing each (maximal) run by its length. The $T$-iterates of $x$ (when they exist) are then $x, Tx, T^2x, T^3x,...$

Terminology: Given an infinite sequence $x$, a run in $x$ is any subsequence of $x$ comprising one or more contiguous equal elements. A run is called maximal if it is not adjacent to an element equal to those in the run. If a run has finitely many elements, then their number is called the length of the run; otherwise, the length of the run is said to be infinite.

Question: Is there any $x \in S$ having infinitely-many distinct $T$-iterates? If "no", how to prove? If "yes", how to construct an example?

I suspect that no $x\in S$ has infinitely-many distinct iterates, and that every $T$-trajectory either terminates at some iterate not in $S$, or eventually enters a cycle (as in the following examples).

Each point in $S$ has exactly two immediate $T$-predecessors, and these are mutual "complements"; i.e., for any $x\in S$, there exist exactly two points in $S$, say $w$ and $\overline{w}$, such that $Tw = T \ \overline{w} = x$, where $\overline{w}$ is the result of replacing (in $w$) each $1$ by $2$ and vice versa. (Consequently, in the above picture, it must be the case that $y=\overline{z}$ and $q=\overline{t}$. This applies to the numerical examples given below in $1b$.)

---

The question can be formally restated by partitioning $S$ as follows:

$$S = (A_{1a} \cup A_{1b}) \cup A_2$$ where (with $i,j,k$ restricted to nonnegative integers) $$\begin{align} A_{1a} & = \{x \in S : \exists i (T^i x \notin S) \}\\ A_{1b} & = \{x \in S : \forall i (T^i x \in S) \ \mathrm{and} \ \exists\ j\ne k \ (T^jx = T^kx) \}\\ A_2 & = \{x \in S : \forall i (T^i x \in S) \ \mathrm{and} \ \forall\ j\ne k \ (T^jx \ne T^kx) \}. \end{align} $$

In other words, for any infinite sequence $x\in S$, iterating $T$ repeatedly must result in exactly one of the following cases:

1. $x$ has only finitely-many distinct iterates in $S$.

   1a. The iterations terminate due to an iterate that's not in $S$ (i.e., it has some element that's neither $1$ nor $2$, and/or it has a run of infinite length). E.g., $1112...\to 3...$, or $121212...\to 1^\infty$.

   1b. All iterates remain in $S$, but there are only finitely many of them (i.e., they eventually enter a finite cycle). E.g., the two Kolakoski sequences are the only 1-cycles (fixed points) of $T$: $$12211212212211211221211212211... \to 12211212212211211221211212211...\\ 2211212212211211221211212211... \to 2211212212211211221211212211... $$ Using the first of these two, here's an example corresponding to the above picture: $$\begin{align} &.\\ &.\\ \to\ v =\ &22122112112212112112212212112...\\ \to\ w =\ &21221221121221211211221221211...\\ \to\ x =\ &11212211211212212112212211212...\\ \to\ y =\ &21122121121122122112122121122... \ (= \overline{z}) \\ \to\ z =\ &\mathbf{12211212212211211221211212211}...\\ \to\ z \to\ &z\ \to \ ... \end{align} $$ By starting with the first element of each sequence of a cycle, then working "backwards", it's straightforward to construct cycles of arbitrary finite size; e.g., here's a 3-cycle construction (with labels corresponding to the above picture): $$\begin{align} &.\\ &.\\ \to\ p =\ &211221221211211221211212211211212212112212211...\\ \to\ q =\ &122121121221121122121121122122121121221121121...\ (= \overline{t})\\ \to\ r =\ &\mathbf{121121122122112122121121122121121221121121221...}\\ \to\ s =\ &\mathbf{112122122112112122112112212112...}\\ \to\ t =\ &\mathbf{2112122121122122112\dots}\\ \to\ r =\ &\underline{121121122122}112122121121122121121221121121221...\\ \to\ s\ \to &\ t\ \to\ r\ \to\ s\ \to t\ \to\ r\ \to\ ... \end{align}$$

2. $x$ has infinitely-many distinct iterates in $S$.

Question: How to prove whether $A_2$ is empty? If $A_2$ is not empty, how to construct an example element?

---

Since, for any $x \in \{1,2\}^\mathbb{N}$, an infinite run can occur only as a suffix, $S$ seems to have the same cardinality as the set of irrationals in the real interval $[0,1]$ (i.e. uncountably infinite); cf. all the reals in that interval except those with "terminating" binary expansions. On the other hand, although it seems that $A_{1b}$ is countable, I'm unsure of the cardinality of $A_{1a}$ (uncountable?). So I'm unable to deduce even the cardinality of $A_2$.

Answer 1

The answer is yes.

Let $\mathbb N$ denote the set $\{1,2,...\}$ of positive integers and $P=\{1,2\}^{\mathbb N}$. Given any $\rho\in P$ we will define $\varphi_\rho\in S$ such that all $T$-iterates of $\varphi_\rho$ remain in $S$ and $(T^{k-1}(\varphi_\rho))(1)=\rho(k)$ for all $k\in\mathbb N$, where $\rho(k)$ denotes the $k$-th element of $\rho$, and $(T^{k-1}(\varphi_\rho))(1)$ denotes the first element of the $(k-1)$-st iterate, $(T^{k-1}(\varphi_\rho))$, of $\varphi_\rho$.

If $T^k(\varphi_\rho)$, $k\in\mathbb N$, is eventually cyclic, that is, if there are $n,m$ such that $T^{k+m}(\varphi_\rho)=T^k(\varphi_\rho)$ for all $k\ge n-1$, then clearly the sequence of the corresponding first coordinates, namely $(T^k(\varphi_\rho))(1)$, $k\in\mathbb N$, would also be eventually cyclic. Hence $\rho$ turns out to be eventually cyclic, with $\rho(k+m)=\rho(k)$ for all $k\ge n$.

Since $P$ is uncountable (with $|P|=\mathfrak c =2^{\aleph_0}$) and since there are only countably many eventually cyclic sequences $\rho\in P$, it follows that for all the uncountably many remaining choices of $\rho$ the iterates under $T$ of $\varphi_\rho$ would never repeat. Thus, to complete the proof, it only remains to define $\varphi_\rho$ with the above properties, for each $\rho\in P$.

I will just illustrate the construction of $\varphi_\rho$, given $\rho\in P$, with a couple of examples. First, say $\rho$ starts as $\langle 2,1,2,2,1,... \rangle$, or, abbreviated, as $21221..$. List the elements of $\rho$ in a column, from top to bottom, as illustrated:
$\\ 2\\ 1\\ 2\\ 2\\ 1\\ ... $

Then, starting with lower and lower rows, work backwards, to define larger and larger initial segments of the first row, which will be our $\varphi_\rho$. That is, start with row $2$, work backwards, then start with row $3$, work backwards, etc. Here is how it works when we start with row $2$ (and work backwards, that is, partially define row $1$ so that an application of $T$ to row $1$ results in row $2$):
$\\ 21\\ 1\\ 2\\ 2\\ 1\\ ... $

Then start with row $3$ and work backwards (so that an application of $T$ to row $2$ results in row $3$, and an application of $T$ to row $1$ results in row $2$:
$\\ 21221\\ 112\\ 2\\ 2\\ 1\\ ... $
We do all this subject to the requirement that each row remains in $S$ and has runs of length at most $2$.

From the given initial segment of $\rho$ following this procedure we obtain:
$\\ 2122112112212\\ 11221221\\ 2212\\ 21\\ 1\\ ... $
where the first row represents an initial segment of the $\varphi_\rho$ that we construct.

As a different example say $\rho$ starts at $\rho=\langle 2,1,1,1,2,1,,... \rangle=211121..$. Listing the elements of $\rho$ in a column, from top to bottom, we have:
$\\ 2\\ 1\\ 1\\ 1\\ 2\\ 1\\ ... $

Working from row $3$ backwards we have:
$\\ 2112\\ 12\\ 1\\ 1\\ 2\\ 1\\ ... $
and working from row $6$ backwards we have:
$\\ 21122122121121\\ 122121121\\ 121121\\ 1121\\ 21\\ 1\\ ... $
where, again, the first row represents an initial segment of the corresponding $\varphi_\rho$ that we construct.

Clearly, by construction, each $T^k(\varphi_\rho)$ remains in $S$, and $(T^{k-1}(\varphi_\rho))(1)=\rho(k)$, which completes our answer.

What follows below an an older and more complicated recursive construction of $\varphi_\rho$. You need not read it, though I will leave at available.

I post a second answer (compared not to the above, but to versions posted even earlier), that uses the same ideas as my previous answer (that is, earlier answer posted separately), but is written better (I think (well, at hindsight, clearly not as good as the simple description given above)). I would keep the previous answer too (... someone voted it up (no longer, perhaps the vote was retracted or there was also another vote down, but anyway) so I hate to erase it now, besides it might be of some interest, but please do not read my older answer unless you would like to see how I came up with this example, and unless you are ready to put up with the inconsistencies there, which I do not intend to clean).

So here is what I hope the better example.

The answer is ``yes''. I describe a procedure that produces many examples (it only fails to produce an example for a countable subset of a certain product space of uncountable cardinality).

Let $\mathbb N$ denote the set of positive integers and $P=\{1,2\}^{\mathbb N}$. (There are some fine notational adjustments that could be made in what follows, depending on whether one takes $0\in\mathbb N$ or $0\not\in\mathbb N$; my notation is consistent with the latter choice, that is, $1=\min\mathbb N$.)

We will define a map $\varphi$ from $P$ to $S$. That is, $$\varphi: P \to S, \mathrm{\ if\ } \rho\in P \mathrm{\ then\ } \varphi_\rho\in S$$

The $n$-th element of a sequence will be denoted like $\rho(n)$ or $\varphi_\rho(n)$. The map $\varphi$ will have the following two properties:

1. $T^k(\varphi_\rho)\in S$ for all $k\in \mathbb N\cup\{0\}$ and all $\rho\in P$,

2. For all $k\in\mathbb N$, $(T^{k-1}(\varphi_\rho))(1)=\rho(k)$.

Condition 2 implies that if the sequence $\langle \varphi_\rho,T(\varphi_\rho),T^2(\varphi_\rho),..., T^k(\varphi_\rho),... \rangle$ is eventually cyclic, that is, if there are $n,m\in\mathbb N$ with $T^k(\varphi_\rho)=T^{k+m}(\varphi_\rho)$ for all $k\ge n-1$, then the sequence $\rho=\langle\rho(1),\rho(2),...\rangle$ is eventually cyclic, with $\rho(k)= \rho(k+m)$ for all $k\ge n$.

Let $P_0$ be the subset of $P$ consisting of all eventually cyclic sequences, and let $P_1=P\setminus P_0$. Since $P_0$ is countable and $P$ is uncountable (with $|P|=\mathfrak c = 2^{\aleph_0}$, the cardinality of the continuum), we have that $P_1$ is uncountable (with $|P_1|=\mathfrak c$). Thus, all $T$-iterates of $\varphi_\rho$ remain in $S$ and are never repeating, whenever $\rho\in P_1$. This shows that not only examples do exist, but there are uncountably (continuum) many of them.

In order to construct $\varphi$ we first construct another map $$\beta: P \to P, \mathrm{\ if\ } \rho\in P \mathrm{\ then\ } \beta_\rho\in P$$

To define $\beta$, we take all initial segments of $\rho$, reverse each, and put them one after the other. To illustrate, say $\rho$ starts as $\langle2,1,1,1,2,1,... \rangle$, or abbreviated $211121..$. Then the initial segments of $\rho$ are $\langle2 \rangle$, $\langle2,1 \rangle$, $\langle2,1,1 \rangle$, $\langle2,1,1,1 \rangle$, $\langle2,1,1,1,2, \rangle$, $\langle2,1,1,1,2,1 \rangle$, ... , or abbreviated,

$$2,21,211,2111,21112,211121,...$$

Reversing each initial segment we obtain

$$2,12,112,1112,21112,121112,...$$

and concatenating these we obtain that $\beta_\rho$ starts as

$$212112111221112121112...= \langle2,1,2,1,1,2,1,1,1,2,2,1,1,1,2,1,2,1,1,1,2,... \rangle $$

Formally let $t_n=\frac{n(n+1)}2$ denote the $n$-th triangular number. Then $\beta_\rho(t_n-m)=\rho(m+1)$ for all $n$, and all $m\in\{0,1,...,n-1\}$.

We now start the construction of $\varphi_\rho$. To illustrate, let again $\rho$ start as $211121..$., so $\beta_\rho$ starts as $212112111221112121112..$. List the elements of $\beta_\rho$ in a vertical column starting at the bottom and going up like this:

$\\ ... \\ 2\\ 1\\ 1\\ 2\\ 1\\ 2 $

Working "backwards" define partial rows consistent with the leftmost column, as described below. Let $p^j$ denote partial row $j$ (counted from bottom) and $p^j(n)$ denote the $n$-th element of $p^j$. The requirement is that $T(p^{j+1})=p^j$. For example, $p^2$ must be a partial sequence that starts with $\beta_\rho(2)$, in our example $\beta_\rho(2)=1$, and such that $T(p^2)$ starts as $p^1$. Thus, $p^2$ must start as $11$, and for convenience (and to make sure that the lengths of the $p^j$ strictly increase even in cases when $\beta_\rho(1)=1$) we also add a terminating element consistent with $p^2$ being an initial segment of a sequence in $S$ and having runs of length at most $2$, so $p^2$ starts as $112$. (Here we only apply the restriction of $T$ to just a finite sequence, so the rest of the sequence (which cannot be determined by the given information) could be denoted by a $*$, e.g. $p^2=112*$.) Then, $p^3$ must be a partial sequence that starts with $\beta_\rho(3)$, in our example $\beta_\rho(3)=2$, and such that $T(p^3)$ starts as $p^2$. Thus, $p^3$ starts as $21221$. Since the first element of $p^j$ is given, namely $p^j(1)=\beta_\rho(j)$ the construction as described so far is unambiguous, and produces the following partial rows for our example.

$\\ ... \\ 211212212211212212112\\ 12112122112112\\ 112112212\\ 21221\\ 112\\ 2 $

Let $|p^j|$ denote the length of (the defined part of) $p^j$, and let $|p^j|_2$ count how many times the number $2$ occurs in $p^j$. For example $|21221|=5$ and $|21221|_2=3$. It is easily shown that $|p^{j+1}|=|p^j|+|p^j|_2 +1$ for all $j$, and (using that the last two elements of $p^j$ are always different) that $|p^j|_2\le |p^j|-1$ for all $j\ge2$. It follows that $|p^j|<|p^{j+1}|\le 2 |p^j|$ for $j\ge2$.

Let $J_1=\{\frac{n(n+1)}2: n\in\mathbb N\}$ be the set of all triangular numbers. We will construct a decreasing sequence of infinite sets $J_1\supset J_2 \supset J_3 \supset ...$, and simultaneously define $\varphi_\rho(n)$ as follows. Set $\varphi_\rho(1)=\rho(1)$. Note that $p^j(1)=\beta_\rho(j)=\beta_\rho(t_n)=\rho(1)$ for all $j\in J_1$ (where $j=t_n$ is a triangular number for some $n$).

For each $j\ge2$, we have that $p^j(2)$ is defined and is in $\{1,2\}$. Split $J_1\setminus\{1\}$ into the disjoint union of two sets, $J_1(1)=\{j\in J_1\setminus\{1\}:p^j(2)=1\}$, and $J_1(2)=\{j\in J_1\setminus\{1\}:p^j(2)=2\}$. At least one of these two sets must be infinite. If $J_1(1)$ is infinite then let $J_2=J_1(1)$ and $\varphi_\rho(2)=1$. Else let $J_2=J_1(2)$ and $\varphi_\rho(2)=2$. Proceed by recursion as follows.

Split $J_n\setminus\{1,...,n\}$ into two sets, $J_n(1)=\{j\in J_n\setminus\{1,...,n\}:p^j(n+1)=1\}$, and $J_n(2)=\{j\in J_n\setminus\{1,...,n\}:p^j(n+1)=2\}$. If $J_n(1)$ is infinite then let $J_{n+1}=J_n(1)$ and $\varphi_\rho(n+1)=1$. Else let $J_{n+1}=J_n(2)$ and $\varphi_\rho(n+1)=2$.

This defines $\varphi_\rho(n)$ for all $n$. Note that if $j\in J_n$ with $|p^j|\ge n$ then $p^j(k)=\varphi_\rho(k)$ for all $k\in\{1,2,...,n\}$. (Use that $J_n\subset J_{n-1}\subset ... \subset J_1$ .)

Next we show that for all $k\in\mathbb N$ we have $(T^{k-1}(\varphi_\rho))(1)=\rho(k)$. When $k=1$, $(T^{0}(\varphi_\rho))(1)=\varphi_\rho(1)=\rho(1)$. Now fix $k>1$. Note that if $|p^j|$ is large enough, e.g. if $|p^j|\ge3^k$, then $(T^{k-1}(p^j))(1)$ is defined (use that $|T(p^{m+1})|=|p^m|\ge\frac{|p^{m+1}|}2$ for all $m\ge2$). Pick $j$ large enough subject to the following conditions: $j\in J_{3^k}$, $|p^j|\ge3^k$, and $j=t_n$ with $n>k$. Then $(T^{k-1}(\varphi_\rho))(1)=(T^{k-1}(p^j))(1) =\beta_\rho(t_n-(k-1))=\rho((k-1)+1)=\rho(k)$.

The proof that $T^k(\varphi_\rho)\in S$ for all $k$, and all $\rho\in P$ is similar. Indeed if $T^k(\varphi_\rho)\not\in S$ then either $(T^k(\varphi_\rho))(m)\ge3$ for some $m$, or $(T^k(\varphi_\rho))$ has an infinite run. In the latter case $(T^{k+1}(\varphi_\rho))(m)\ge3$ for some $m$, so it reduces to the former. But if $(T^k(\varphi_\rho))(m)\ge3$ then we could pick large enough $n$ and $j$ with $j\in J_n$ and $|p^j|\ge n$ such that $3\le(T^k(\varphi_\rho))(m)= (T^k(p^j))(m)\in\{1,2\}$, a contradiction.

Answer 2

This is my older answer please do not read it, and read my newer answer instead. If you read this one, you may see how I arrived to my answer, but it might be difficult to read (and I plan to leave it as is, without attempting to edit it anymore to clean it). The other one is written much better (I hope).

This is the third version I post today, and I am running out of steam. I think I have enough to post (and perhaps edit later if necessary). Here it is.

I am constructing an example again. I think compactness plays a role (both as in topology and as in logic, and it has something to do with inverse limits too), but I don't quite get it yet myself, so let me try to describe my current idea, to the point when someone would be able to verify it. So, trying to construct a sequence $\sigma$ such that all iterates under $T$ remain in $S$ but never repeat.

First fix the sequence $\beta=\langle\beta(1),\beta(2),\beta(3),...\rangle= \langle 1, 2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,...\rangle$.
That is, we have groups of $2$'s separated by single $1$'s, and the number of $2$'s in each group increase. (This will eventually "push the length of the cycle to infinity".)

Now, $\beta$ is for $\beta$ackwards. That is, we start by listing the elements of $\beta$ backwards, in a single column, from bottom to top, like this:
.
.
1
2
2
2
1
2
2
1
2
1

Then we start at the bottom and proceed backwards (in a manner described by the OP, in the question), and preserving the first column (which tells us whether to start a sequence with $1$ or with $2$). We get:
.
.
1
2
2
2 etc, going backwards
112212211212212112
22122112112
212212
1121
21
1

To illustrate, if we pick any (partial) row and apply $T$ we obtain the next (partial) row (going down this time), e.g.
$T(112212211212212112*)=22122112112*$
$T(22122112112*)=212212*$
$T(212212*)=1121*$
$T(1121*)=21*$
$T(21*)=1*$

So we have
.
.
1121
21
1
and we consider these rows to be labeled with the natural numbers, starting at the bottom, so the first raw starts with 1, the second row starts 21, the third row starts 1121, etc.

Let me introduce more notation. The above procedure describes how to get partial sequences, or partial rows, starting from the bottom and going backwards. Let me call these partial rows $p^j$, so $p^1=1$, $p^2=21$, $p^3=1121$, etc. Also, $p^j(k)$ will denote the $k$-th element of row $j$, e.g. $p^3(1)=1$, $p^3(3)=2$, $p^3(5)$ is undefined by our construction.

Let $\sigma(1)=1$, and let $J_1=\{j:$ row $j$ starts with $\sigma(1)\}= \{j: p^j(1)=\sigma(1)\}$.
By construction, $J_1$ is infinite.

The set $J_1\setminus\{1\}$ naturally splits into two sets:
$J_1(1)=\{j\in J_1\setminus\{1\}: p^j(2)=1\}$ and
$J_1(2)=\{j\in J_1\setminus\{1\}: p^j(2)=2\}$.
As least one of these two sets is infinite.
If $J_1(1)$ is infinite then let $\sigma(2)=1$ and $J_2=J_1(1)$.
Else, let $\sigma(2)=2$ and $J_2=J_1(2)$.

Recursively define infinite sets $J_{n+1}\subseteq J_{n}\setminus\{1,2,...,n\}$ and $\sigma(n+1)\in\{1,2\}$ such that $p^j(n+1)=\sigma(n+1)$ whenever $j\in J_{n+1}$.

This defines the sequence $\sigma$, that is, the above construction eventually defines $\sigma(n)$ for all natural numbers $n$.

Let us first show that $T^k(\sigma)\in S$ for all $k$. Suppose the contrary. The idea here resembles compactness (as in logic), or if I don't get my terminology right, the idea is that, if there is some $k$ such that $T^k(\sigma)\not\in S$, then there is some coordinate $m$ (depending on $k$) such that $(T^k(\sigma))(m)\ge3$. There is a long enough initial segment of $\sigma$, call it $\tau$, such that the first $m$-many coordinates of $T^k(\sigma)$ coincide with the first $m$-many coordinates of $T^k(\tau)$. (Infinitely many coordinates of $T^k(\tau)$ remain undefined but we need not worry about them. Also I keep thinking of $S$ as consisting of $x\in \{1,2\}^\mathbb{N}$ such that each run of $x$ has length at most $2$, whereas the OP says finite length: But it doesn't make much difference. If $T^k(\sigma)\not\in S$ then either $(T^k(\sigma))(m)\ge3$ for some $m$, or else $(T^{k+1}(\sigma))(m)\ge3$ for some $m$, so if necessary we may replace $k$ with $k+1$.) There is a big enough $n$ such that $\tau$ is an initial segment of the partial row $p^n$. But then $T^k(p^n)\ge3$, a contradiction.

The idea to show that $T^k(\sigma)$ never repeats is similar. Taking long enough initial segments of $\sigma$ we could show that:
1. For each $m$ there is $n$ such that $(T^k(\sigma))(1)=2$ for all $k\in\{n,n+1,...,n+m\}$, yet
2. there are infinitely many $k$ such that $(T^k(\sigma))(1)=1$.

I may combine the above two conditions into one (which is not exactly equivalent to the conjunction of the above two, but does the same job):
(*) For each $m$ there is $n$ such that $(T^n(\sigma))(1)=1$ but $(T^k(\sigma))(1)=2$ for all $k\in\{n+1,...,n+m\}$.

I think one may vary $\beta$ to obtain uncountably many examples (but I have not verified all details of this part). At least, I believe the above $\sigma$ works. (I thought of it a little more, and see no gaps, and I am convinced that it is correct. It may benefit from a better exposition and perhaps a bit more details, but please let me know if you have any specific questions.)

I will stick with the above answer (of course perhaps adding more details later). What follows below are my previous answers, which should be ignored, unless you are interested to see some of my initial steps and ideas.

I changed my mind (from what was posted at the bottom) so let me post an opposite answer (here) to what I had posted earlier (at the bottom). (Warning, I have a gap in this proof too.)

There is no counterexample. I mean there is no sequence $\sigma=\langle\sigma(1),\sigma(2),\sigma(3),...\rangle$ such that all $T$ iterates are different. For a contradiction suppose $\sigma$ was such a sequence. Let $K_1=\{n:T^n\sigma$ starts with $1\}$, and $K_2=\{n:T^n\sigma$ starts with $2\}$. At least one of $K_1$ and $K_2$ must be infinite (perhaps both are). Consider the case when $K_1$ is infinite. Then we could pick $n$ as big as we wish, in $K_1$, and starting with the first element of $T^n\sigma$, which is $1$, work backwards, and as we do so, we will get, at the top, bigger and bigger segments of the Kolakoski sequence that starts with $1$. So at the top we must have the Kolakowski sequence that starts with $1$, a contradiction. A similar contradiction is obtained if $K_2$ is infinite.

There are some details to be filled in, I will do it now (unless I get bogged again, but I think I have it right this time:) Well, I don't know, the gap this time is that as we go backwards, say from $T^n\sigma$ to $T^{n-1}\sigma$, we do not know if $T^{n-1}\sigma$ starts with $1$ or with $2$. (If it always started with $1$ then we do indeed get longer and longer initial segments of $\sigma$ to coincide with the initial segments of the Kolakowski sequence that starts with $1$, but of course one needs to take handle the other possibilities, and I do not know how.)

What follows is my old answer (from the time when I though there was an example). I guess you could pick the answer that you wish ... now that I have gaps in each version of my answer ...

I believe there is an example. I will edit my answer to provide more formal details the next few hours, but here is an informal description that might be good enough to follow.

The main idea is to work backwards, and at the same time diagonalize, so as to push the length of the cycle to infinity.

Just to be definite, the example will start with 1. When we work backwards we could start each preceding sequence with either 1 or 2. We could work backwards starting the sequence with 2 for as many steps as we wish (this in the long run will account for pushing the length of the cycle to infinity), then we could work backwards for as many steps as necessary, this time starting the sequence with 1, till we match the initial segment that is currently under consideration. (There is an argument that this has to happen, for otherwise we get a cycle where all sequences start with 1, on the other hand we already have sequences down the road that start with 2.) Thus we obtain a new initial segment, work backward again, this time with the new initial segment, using a larger number of sequences that start with 2 (before we revert to sequences that start with 1 to match the new initial segment), forcing the length of the cycle to be even longer, etc. Let me post this and then try to write something more formal.

Given I don't have all details yet, perhaps this example should not be taken too seriously, but I am working on it. The "for otherwise we get a cycle where all sequences start with 1" part of my arguments needs a more careful examination. So, for now this is just an idea, will see if I can make it work. (If it does, then it would also show that there are continuum many examples, i.e. uncountably many with the same cardinality as the reals.) I think it works, but it is a bit messy. The idea of inverse limit does play a role. But I have yet to put all the details together.