Is this additive subgroup of $\Bbb R$ a Bernstein set? This Q Construct an additive group by transfinite induction asked for proof-verification and also asked for other ways to construct a Bernstein subset of $\Bbb R$ that is also an additive group.
Consider $\Bbb R$ as a vector space over the field $\Bbb Q$ and let $B$ be a Hamel basis for $\Bbb R$ with $1\in B.$ For each $x\in \Bbb R$ there is a unique $\{x_b:b\in B\}\subset \Bbb Q$ such that $x=\sum_{b\in B}bx_b.$ ( Only finitely many $x_b$ are non-$0$.).
Q: Is the additive group  $S=\{x\in \Bbb R: x_1\in \Bbb Z\}$ a Bernstein set?
The set $S\cap [0,1)$ is an old textbook example of a non-Lebesgue-measurable set and it has inner Lebesgue measure $0.$ But does $S$ have an uncountable closed Lebesgue-null subset? I have not got anywhere on this.
 A: It is at least possible for $S$ to have a perfect subset (which is by definition uncountable and closed, and must be null since $S$ has inner measure $0$). I suspect some $B$s will yield Bernstein $S$s, though I haven't verified that.

Consider the following result (where $\{q_w: w\in\mathbb{N}\}$ is some fixed enumeration of the rationals):

Fix a natural number $k$ and finitely many disjoint nondegenerate closed intervals $A_1,...,A_n$ with $1\in A_1$. There are disjoint nondegenerate closed intervals $B_1,...,B_{2n}$ such that $1\in B_1$ and for each $i\in\{1,...,2n\}$ we have:

*

*$B_{2i}\cup B_{2i+1}\subseteq A_i$, and


*for each $f\in B_i$ and each $g_1,...,g_k\in \bigcup_{j\in\{1,...,2n\}\setminus\{i\}}B_j$, there are no $c_1,...,c_k\in \{q_1,...,q_k\}\cup\{0\}$ such that $$f=\sum_{u=1}^kc_ug_u.$$

The proof is straightforward but tedious. The point is that by iterating this - increasing $k$ as we go along - we get a Cantor-set-like construction of a perfect $\mathbb{Q}$-linearly-independent set $Y$ with $1\in Y$; by cutting out a small open neighborhood of $1$, we get a $\mathbb{Q}$-linearly-independent perfect set $X$ whose span doesn't contain $1$. But now if we take a Hamel basis $B\supseteq Y$ (by Zorn), we get that $X$ is a perfect subset of the corresponding $S$.
