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Given a real number $r$, and an integer $b$>0, we can define $B_b(r)$ as the set of numbers which are obtained from $r$ by writing $r$ in base $b$ and then altering a density zero subset of its digits. For example, if $r=0.0000...$ then $B_2(r)$ would include $0.10100100001000001 \cdots$ . There is a slight ambiguity here in the definition if r has a finite expansion in base $b$; for example whether for 1 we use $r$ written as 1.0000... or we use r written as 0.9999... For our purposes, it is probably ok to include both, and won't alter the answers to any of the things we care about. Given S a subset of the real numbers we can then define $B_b(S)$ as the union of $B_b(s)$ for every s in S. We'll define $T(S)$ as the union of $B_b(S)$ for every base $b$.

Set $U(r)$ to be $\bigcup_{i=0}^{\infty} T^i(r)$.

Question 1) Is it true that $U(0)= \mathbb{R}$?. My guess is that the answer is probably "No" but it isn't obvious. One might try a measure theoretic argument, but since the union of an uncountable number of measure zero sets is not necessarily of measure zero, it doesn't seem to get a result. Note that for any $r$, $T(r)$ contains every number of the form $r+q$ for $q$ a non-negative rational, since adding $\frac{a}{b}$ changes only a finite number of digits in base $b$.

We can also define versions of $B_b$ for complex numbers, where we do the same thing changing a zero density set of the real part and the complex part, and then define $T_{\mathrm{com}}(s)$ accordingly. Given a set of complex numbers Let $\overline{S}$ be the algebraic closure of the smallest field containing a given set S of complex numbers. We can then define $U_{\mathrm{com}}(r)$ as the union of $T(r)$, $\overline{(T(r))}$, $T(\overline{(T(r))})$, $\overline{T(\overline{(T(r))})} \cdots$

Question 2: Is $U_{\mathrm{com}}(0)$ all complex numbers? This question seems to be possibly substantially much harder, and I'm less certain what the answer is.

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    $\begingroup$ These are nice problems, but putting both of them in the same question can be discouraging. It is better to have each one posed separately. $\endgroup$ Mar 16 at 15:10
  • $\begingroup$ @YuvalPeres They seem like they are closely connected. A yes answer for the first one would immediately imply yes for the second. $\endgroup$
    – JoshuaZ
    Mar 16 at 16:11
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    $\begingroup$ But the answer for the first one is negative, $U(0)$ has zero packing dimension. Writing down the proof takes effort but in this case will only answer half a question....en.wikipedia.org/wiki/…. $\endgroup$ Mar 16 at 16:23
  • $\begingroup$ @YuvalPeres Hmm, interesting. I don't see how to get that argument to work, but I'm only weakly familiar with the packing dimension. If you can write up an argument that sketches it out as an answer, I'll accept that answer and then ask question 2 as a separate question. $\endgroup$
    – JoshuaZ
    Mar 17 at 1:22
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We will use the fact

$(*) \quad$ The number of subsets of $\{1,\ldots,n\}$ of cardinality less than $\alpha n$ is at most $e^n H(\alpha)$, where $H(\alpha)=-\alpha \log \alpha -(1-\alpha) \log (1-\alpha)$; see Theorem 3.1 in [1] for a nice proof. Write $H_b(\alpha)=\frac{H(\alpha)}{\log b}$.

Recall that the upper Minkowski dimension of a set $S \subset {\mathbb R}$ is $$\dim_M(S)=\limsup_{h \to 0} \frac {\log N(S;h)}{\log 1/h}$$ where $N(S;h)$ is the minimal cardinality of a cover of $S$ by intervals of length $h$. The limit is unchanged if we restrict $h$ to negative powers of $b>1$. Further, define the packing dimension $\dim_P(S)$ as the infimum of all $\beta$ such that there exists a countable cover $S \subset \cup_{j} S_j$ with $\dim_M(S_j) \le \beta$ for all $j$. This definition implies stability under countable unions: $$ (**) \quad \dim_P (\cup_{k \ge 1} A_k) = \sup_{k \ge 1} \dim_P(A_k) \, .$$

For background on dimension notions, see, e.g., [2] or [3]. We will not use any theorems about these notions, just $(**)$ and the fact that sets of zero packing dimension also have zero Hausdorff dimension and zero Lebesgue measure.

Without loss of generality, we will work in $[0,1]$ rather than all of $\mathbb R$. Given $s,r \in [0,1]$ let $D_b(s,r,n)$ be the number of $i \in [1,n]$ such that the $i'$th digit of $s$ in base $b$ differs from the $i'$th digit of $r$. For $\alpha>0$ and $k \ge 1$ let
$$B_b(s,k,\alpha)= \{r \in [0,1] \,: \, \forall n>k \quad \text{we have} \quad D_b(s,r,n) < n\alpha\} \,,$$ and
$$B_b(s,\alpha)= \cup_{k \ge 1} B_b(s,k,\alpha) \,.$$

Lemma 1 Let $s \in S \subset [0,1]$ and $\alpha>0$. For any $ k>1$ and $h=b^{-m} < b^{-k}$, we have $$N(B_b(s,k, \alpha) ; h) \le 2 e^{m H(\alpha)}b^{m\alpha} \,,$$ and $$N(B_b(S,k, \alpha) ; h) \le 2N(S; h) e^{m H(\alpha)}b^{m\alpha} \,.$$ Proof: Since any interval of length $b^{-m}$ can be covered by at most two $b$-adic intervals that are determined by specifying the first $m$ digits in base $b$, the lemma follows readily from (*).

Lemma 2 Let $S \subset [0,1]$ and $\alpha>0$. For any $ k>1$, we have $$\dim_M(B_b(S,k, \alpha)) \le \dim_M(S)+ H_b(\alpha)+m\alpha \,.$$ Proof: Take logarithms in Lemma 1, divide both sides by $\log(b^m)$, and take limits as $m \to \infty$.

Lemma 3 Let $S \subset [0,1]$ and $\alpha>0$. Then $$\dim_P(B_b(S,\alpha)) \le \dim_P(S)+ H_b(\alpha)+m\alpha \,.$$ Proof: Given $\epsilon>0$ and a cover of $S$ by sets $S_j$ with $\dim_M(S_j) \le \dim_P(S)+\epsilon$, apply lemma 2 to each $S_j$ and take a countable union over $j$ and $k$.

Lemma 4 Let $S \subset [0,1]$ and recall the definition of $B_b(S)$ in the question. Then for all $b>1$ and $\ell \ge 1$, $$(i) \quad \dim_P(B_b(S)) =\dim_P(S)$$ $$ (ii) \quad \dim_P(T(S)) =\dim_P(S)$$ $$ (iii) \quad \dim_P(T^\ell(S)) =\dim_P(S)$$ $$ (iv) \quad \dim_P(U(S)) =\dim_P(S)$$

Proof: (i) follows from the inclusion (valid for all $\alpha>0$) $$B_b(s) \subset B_b(s,\alpha) \,.$$

(ii) holds by $(*)$. Then (iii) for all $\ell \ge 1$ is obtained by induction, and (iv) follows by another application of $(**)$.

In particular, $\dim_P(U(0))=0$.

[1] https://arxiv.org/pdf/1406.7872.pdf

[2] Bishop, Christopher J.; Peres, Yuval (2017). Fractals in Probability and Analysis. Cambridge University Press. Available at https://www.math.stonybrook.edu/~bishop/classes/math324.F15/book1Dec15.pdf

[3] Mattila, Pertti. Geometry of sets and measures in Euclidean spaces: fractals and rectifiability. No. 44. Cambridge university press, 1999.

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  • $\begingroup$ Thanks for the writeup. This seems to work. Before I make the second question as a separate question, can you point out why this same sort of argument wouldn't also work there? $\endgroup$
    – JoshuaZ
    Apr 6 at 18:48
  • $\begingroup$ In the real case, although the idea seemed clear a couple of weeks ago, it took me half a day to produce a precise writeup (I am slow these days.) In the complex case you consider another operation of algebraic closure that I have not thought about in this context. This operation can certainly increase packing dimension in some cases, e.g. from 1 to 2. $\endgroup$ Apr 6 at 22:25

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