Denote $A=\{0\}, B=\{0,1\}$. Then any subset of $\Omega:=\{A,B\}^{\mathbb N}$ is a continuum provided the number of $B$'s is infinite. We treat these as binary expansions of numbers in $[0,1]$.

For instance, $(AB)^\omega$ is the set $\left\{\sum_{n=1}^\infty a_n4^{-n}\mid a_n\in\{0,1\}\right\}$, which is a Cantor set of Hausdorff dimension $\frac12$. Similarly, it is easy to show that $(AB^N)^\infty$ has Hausdorff dimension $\frac{N}{N+1}$.

Now, suppose the sequence of $A$'s and $B$'s is random: i.i.d. with distribution $\bigl(\frac1{N+1}, \frac{N}{N+1}\bigr)$. It is obvious that a generic sequence will be a Cantor set.

Question. Is it true that a subset of $\Omega$ has Hausdorff dimension $\frac{N}{N+1}$ almost surely? A weaker version: is this a null set almost surely?

  • 1
    $\begingroup$ Weaker version: The finite version of the set to level $k$ is a collection of intervals with Lebesgue measure that falls by a factor of two for each $A$. The sequence has infinitely many $A$s a.s., so the measure of the limiting set is bounded by a sequence tending to zero. Or am I missing something? $\endgroup$
    – user25199
    Commented Jul 20, 2016 at 13:24
  • $\begingroup$ Fat Cantor sets? $\endgroup$ Commented Jul 20, 2016 at 13:27
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    $\begingroup$ Just to clarify. For any infinite word $w$ in the alphabet $\{A,B\}$ the associated subset of $\{0,1\}^{\mathbb N}$ is obtained from $w$ by replacing $A$ with $0$ and $B$ with either $0$ or $1$. Is this what you mean? $\endgroup$
    – R W
    Commented Jul 20, 2016 at 14:51
  • $\begingroup$ This is what I mean, yes. $\endgroup$ Commented Jul 20, 2016 at 15:08

1 Answer 1


Like the classical Cantor set, your set has a natural representation $C=\bigcap C_n$, where $C_1\supseteq C_2\supseteq C_3\supseteq\ldots$, and each $C_n$ is a disjoint union of $2^{n-X_n}$ intervals of length $2^{-n}$, with $X_n$ denoting the number of $A$'s in your sequence up to position $n$.

So, covering by intervals of that length, we have that $H_{2^{-n}}^d(C_n)=2^{n(1-d)-X_n}$. If we write $p=P(A)$, then $X_n/n\to p$ almost surely, so $\dim C\le 1-p$ almost surely.

For the lower bound, we observe that $C$ will contain at least one point from each of the $2^{n-X_n}$ intervals of length $2^{-n}$ whose union is $C_n$. This means that we cannot find a $2^{-n}$ covering of $C$ that decreases the number of intervals needed by more than a fixed factor, compared to the one of $C_n$ from above.

So $\dim C=1-p$ almost surely, as claimed.


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