Hausdorff dimension and critical exponent of words What is the Hausdorff dimension of the subset $S_c \subset [0,1]$ of points such that the critical exponent of their binary expansion is $c$? It's clear that $\dim_H S_{\infty}=1$, but what can be said for $c<\infty$?. Also, what is $\dim_H \cup_{c\in [2,\infty)} S_c$? Are there techniques for the evaluation of the Hausdorff dimension which are applicable to these cases?
The text by Falconer [1] doesn't seem to cover this kind of questions.
Update: it occurred to me that even the estimate of the (upper) box dimension of $S_c$ is not trivial. Already for integer $c$, it leads to a tricky combinatorial problem, that is, how many distinct binary words of length $n$ are there having no factor of the form $w^c$ ($w \in \{0,1\}^*$)?
It's also not easy for me to guess whether the Hausdorff and box dimensions could disagree.
[1]: Falconer, K. (2004). Fractal geometry: mathematical foundations and applications. John Wiley & Sons.
 A: Here are a few observations: (the last one is maybe the main point of interest)
$\bullet$ Your $S_c$ is a closed subset of $[0,1]$ invariant under $x \mapsto 2x \pmod 1$ (usually there's a subtlety about points with multiple expansions, but none of those are in $S_c$ for $c < \infty$), and so I think a result of Furstenberg ([1]) shows that it has equal Hausdorff and box dimension. They should both be equal to the topological entropy $h(S_c)$ of $S_c$ viewed as a subshift on $\{0,1\}$ (using $\log_2$ in definition of entropy).
$\bullet$ In [1], Kolpakov proves that $dim(S_3) = h(S_3) \geq 0.5435$ (I've converted to $\log_2$, he used $\ln$)
$\bullet$ for all $c < \infty$, $dim(S_c) = h(S_c) < 1$, since at the least you forbid the word $0^{\lceil c \rceil}$, and all proper subshifts of $\{0,1\}^{\mathbb{N}}$ have entropy less than $1$.
$\bullet$ (this is quite long and probably hard to read, sorry) There are probably better methods, but you can use the techniques of my paper [3] to bound the entropy (and therefore dimension) of $S_c$ from below. Let's restrict to integer $c$ for now. Then $S_c$ is a subshift defined by forbidden list $\{w^c \ : \ w \in \{0,1\}^*\}$, where $w^c$ is the $c$th power of $w$.
The technique from that paper to bound entropy from below is to rescale at some length $j$, and then view your subshift $X$ (here $S_c$) as a new subshift whose alphabet is $A^{(j)} = L_j(X)$ (the set of $j$-letter words in $X$) and whose forbidden list is $F^{(j)}$, a set of forbidden concatenations of `letters' $A^{(j)}$ which induces the same forbidden words as those for $X$. More formally,
$a_1 \ldots a_i$ is in $F^{(j)}$ if the $ij$-letter concatenation $a_1 \ldots a_i$ contains a forbidden word for $X$, but neither $a_1 \ldots a_i-1$ nor $a_2 \ldots a_i$ contains a forbidden word for $X$. This forbidden list $F^{(j)}$ on $A^{(j)}$ induces a subshift $X^{(j)}$ where each point is a sequence of $c$-letter blocks whose concatenation is in $X$. (Formally, $X^{(j)}$ is isomorphic to $X$ under the $c$th power of the shift.)
Here, if we use $X = S_c$ and set $j = c$, then it's easy to bound from above the number of $i$-'letter' words in $F^{(c)}$. There are none for $i = 1$ by definition. For larger $i$, if $a_1 \ldots a_i \in F^{(c)}$, then either $a_1 \ldots a_i$ is a $c$th power of an $i$-letter word (less than $2^i$ possibilities) or contains a $c$th power of an $(i-1)$-letter word ($c$ choices for where in $a_1$ the power starts, less than $2^{i-1}$ possibilities for the base of the power, less than $2^c$ choices for the remaining letters). If only shorter $c$th powers were in $a_1 \ldots a_i$, then one of $a_1 \ldots a_{i-1}$ or $a_2 \ldots a_i$ would contain a $c$th power, disallowed by definition. So, the number of $i$-letter forbidden words in $F^{(c)}$ is less than $2^i + c2^c 2^{i-1} \leq c 2^{c+i}$.
Now, Theorem 4.1 from [3] gives a lower bound for entropy of a subshift. It states that if $X$ is a subshift with alphabet $A$ and an infinite forbidden list $F$ with $F_n$ words of length $n$ for every $n$, and if
$\sum F_i x^i < x(|A| - k + 1) - 1$ for some $x > 1/|A|$ and $k < |A|$, then
$h(X) \geq \log k$. This is hideous, but we can make it easier.
We'll apply to the $c$-letter rescaling $S_c^{(c)}$ of $S_c$ defined above, and assume that $c \leq 10$. Then $|A^{(c)}| = 2^c - 2$ (the only illegal $c$-letter words in $S_c$ are $0^c$ and $1^c$). Let's take $k = 2^{c-1} - 1$ and $x = 4/2^c$. Then $x(|A^{(c)}| - k + 1) = 1$, so we need only verify that
$\sum (F^{(c)})_i x^i < 1$; recall that $(F^{(c)})_i \leq c 2^{c+i}$. Finally,
$\sum_{i = 2}^\infty (F^{(c)})_i x^i \leq \sum_{i=2}^\infty c 2^{c+i} (4/2^c)^i = 
c 2^c \sum_{i=2}^\infty (8/2^c)^i = c 2^c \frac{64/2^{2c}}{1 - (8/2^c)}
= \frac{64c}{2^c-8}$.
This is less than $1$ for $c \leq 10$, so Theorem 4.1 shows that in that case,
$h(S_c^{(c)}) \geq \log k = \log(2^{c-1} - 1)$. But $h(S_c^{(c)}) = c h(S_c)$ (since the former is isomorphic to the latter under the $c$th power of the shift). So, at long last, we get $h(S_c) \geq \frac{\log(2^{c-1} - 1)}{c}$.
In particular, this approaches $1$ as $c \rightarrow \infty$, so this should answer one of your questions: $dim(\bigcup S_c) = 1$.
I should say that I found several papers estimating entropies of $S_c$ from below, and would not at all be surprised if one gives a much easier proof that $dim(\bigcup S_c) = 1$. But all I found were for specific values of $c$.
[1] H. Furstenberg, Disjointness in ergodic theory, minimal sets, and a problem in Diophantine approximation. Math. Systems Theory, 1 (1967)
[2] R. Kolpakov, Efficient lower bounds on the number of repetition-free words. J. Integer Seq. 10 (2007), no. 3
[3] R. Pavlov, Ronnie, On subshifts with slow forbidden word growth.
Ergodic Theory Dynam. Systems 42 (2022), no. 4
