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.



Reaching real numbers from other real numbers by changing a small number of digits in the base b expansion
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](https://www.math.stonybrook.edu/%7Ebishop/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.


