Is the set of allowed sequences for the golden mean shift uncountable? Consider infinite sequences of two symbols, $L$ and $R$. The set of infinite sequences of these two symbols that do not contain consecutive $R$'s, i.e. $LRRL...$, is called the golden mean shift space.
Is the set of all such sequences uncountable?
COMMENT: Standard diagonalization arguments don't work, since you can produce forbidden sequences through the diagonalization process. However, there are certain properties that lead me to believe that this set is uncountable—for example, that there are an infinite number of sequences that contain $n$ $R$'s for every integer $n\geq 1$.
 A: Actually a standard diagonalization argument will work here - we just have to be careful how to set it up.

Before presenting a diagonal argument which works, let me give a simpler proof. To show that the set in question - which I'll call "$F$" - is uncountable, it will be enough to find an injection $2^\omega\rightarrow F$ (here $2^\omega$ is the set of infinite binary sequences, which the standard diagonal argument shows is uncountable). The idea is to use a coding process which transforms bits in the input sequence to blocks of bits, rather than individual bits, in the output sequence. Specifically, given finite sequences $\sigma_0,\sigma_1$ of $R$s and $L$s, define a function from infinite binary sequences to infinite $R/L$-sequences as follows: $$f_{\sigma_0,\sigma_1}(\alpha)=\sigma_{\alpha(0)}{}^\smallfrown \sigma_{\alpha(1)}{}^\smallfrown \sigma_{\alpha(2)}{}^\smallfrown ...$$ where "${}^\smallfrown$" denotes concatenation. For example, we have $$f_{\color{red}{RLR}, \color{green}{LRL}}(01010101...)=\color{red}{RLR}\color{green}{LRL}\color{red}{RLR}\color{green}{LRL}\color{red}{RLR}\color{green}{LRL}...$$
Now not every function of this form is nice: some $f_{\sigma_0,\sigma_1}$s are not injective, and some do not always output elements of $F$. Can you find a choice of $\sigma_0,\sigma_1$ such that $f_{\sigma_0,\sigma_1}$ is an injection from $2^\omega$ to $F$?

 Take for example $\sigma_0=LLL$ and $\sigma_1=LRL$ - the $L$s on each end of each string prevent double $R$s, and $f_{LLL,LRL}$ is clearly injective.


Now let me package that as a diagonal argument:
Suppose we have a sequence $(\alpha_i)_{i\in\mathbb{N}}$ of elements of $F$; we want to build an element $\beta$ of $F$ not in this sequence. To do this we'll block diagonalize: rather than have the $i$th bit of $\beta$ "attack" the $i$th bit of $\alpha_i$, we'll have the $i$th block of bits of $\beta$ "attack" the $i$th block of bits of $\alpha_i$. The possible behaviors of $\beta$ will be chosen analogously to the argument above, to make sure that we don't produce a bad sequence at the end.
For example, the choice of $\sigma_0$ and $\sigma_1$ above suggests the following block diagonalization:

 Let $\tau_i=LLL$ if $\langle\alpha_i(3i),\alpha_i(3i+1), \alpha_i(3i+2)\rangle= LRL$ and let $\tau_i=LRL$ otherwise. Then set $$\beta=\tau_0{}^\smallfrown\tau_1{}^\smallfrown\tau_2{}^\smallfrown...$$

As in the usual argument, we can now show that $\beta\not=\alpha_i$ for each $i\in\mathbb{N}$.
